Řešení:

Nejprve si zapíšeme matici koeficientů a vektor pravých stran:

\( A = \begin{pmatrix} 2 & 3 \\ 4 & -1 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 5 \\ 1 \end{pmatrix} \)

Determinant matice \( A \) je:

\( \det A = 2 \cdot (-1) – 3 \cdot 4 = -2 – 12 = -14 \)

Pro výpočet \( x \) vytvoříme matici \( A_x \), kde první sloupec nahradíme vektorem \( \mathbf{b} \):

\( A_x = \begin{pmatrix} 5 & 3 \\ 1 & -1 \end{pmatrix} \)

Její determinant je:

\( \det A_x = 5 \cdot (-1) – 3 \cdot 1 = -5 – 3 = -8 \)

Pro výpočet \( y \) vytvoříme matici \( A_y \), kde druhý sloupec nahradíme vektorem \( \mathbf{b} \):

\( A_y = \begin{pmatrix} 2 & 5 \\ 4 & 1 \end{pmatrix} \)

Její determinant je:

\( \det A_y = 2 \cdot 1 – 5 \cdot 4 = 2 – 20 = -18 \)

Podle Cramerova pravidla platí:

\( x = \frac{\det A_x}{\det A} = \frac{-8}{-14} = \frac{4}{7}, \quad y = \frac{\det A_y}{\det A} = \frac{-18}{-14} = \frac{9}{7} \)

Řešením soustavy je tedy:

\( x = \frac{4}{7}, \quad y = \frac{9}{7} \)

Řešení:

Zápis matice koeficientů a vektoru pravých stran:

\( A = \begin{pmatrix} 1 & -2 & 1 \\ 3 & 1 & -1 \\ 2 & -1 & 3 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 4 \\ 1 \\ 7 \end{pmatrix} \)

Determinant \( \det A \) spočítáme pomocí rozvoje podle prvního řádku:

\( \det A = 1 \cdot \begin{vmatrix} 1 & -1 \\ -1 & 3 \end{vmatrix} – (-2) \cdot \begin{vmatrix} 3 & -1 \\ 2 & 3 \end{vmatrix} + 1 \cdot \begin{vmatrix} 3 & 1 \\ 2 & -1 \end{vmatrix} \)

\( = 1 \cdot (1 \cdot 3 – (-1) \cdot (-1)) + 2 \cdot (3 \cdot 3 – (-1) \cdot 2) + 1 \cdot (3 \cdot (-1) – 1 \cdot 2) \)

\( = 1 \cdot (3 – 1) + 2 \cdot (9 + 2) + 1 \cdot (-3 – 2) = 1 \cdot 2 + 2 \cdot 11 – 5 = 2 + 22 – 5 = 19 \)

Vypočítáme determinanty \( \det A_x, \det A_y, \det A_z \):

\( A_x = \begin{pmatrix} 4 & -2 & 1 \\ 1 & 1 & -1 \\ 7 & -1 & 3 \end{pmatrix} \)

\( \det A_x = 4 \cdot \begin{vmatrix} 1 & -1 \\ -1 & 3 \end{vmatrix} – (-2) \cdot \begin{vmatrix} 1 & -1 \\ 7 & 3 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 1 \\ 7 & -1 \end{vmatrix} \)

\( = 4 \cdot (1 \cdot 3 – (-1) \cdot (-1)) + 2 \cdot (1 \cdot 3 – (-1) \cdot 7) + 1 \cdot (1 \cdot (-1) – 1 \cdot 7) \)

\( = 4 \cdot (3 – 1) + 2 \cdot (3 + 7) + 1 \cdot (-1 – 7) = 4 \cdot 2 + 2 \cdot 10 – 8 = 8 + 20 – 8 = 20 \)

\( A_y = \begin{pmatrix} 1 & 4 & 1 \\ 3 & 1 & -1 \\ 2 & 7 & 3 \end{pmatrix} \)

\( \det A_y = 1 \cdot \begin{vmatrix} 1 & -1 \\ 7 & 3 \end{vmatrix} – 4 \cdot \begin{vmatrix} 3 & -1 \\ 2 & 3 \end{vmatrix} + 1 \cdot \begin{vmatrix} 3 & 1 \\ 2 & 7 \end{vmatrix} \)

\( = 1 \cdot (1 \cdot 3 – (-1) \cdot 7) – 4 \cdot (3 \cdot 3 – (-1) \cdot 2) + 1 \cdot (3 \cdot 7 – 1 \cdot 2) \)

\( = 1 \cdot (3 + 7) – 4 \cdot (9 + 2) + 1 \cdot (21 – 2) = 10 – 4 \cdot 11 + 19 = 10 – 44 + 19 = -15 \)

\( A_z = \begin{pmatrix} 1 & -2 & 4 \\ 3 & 1 & 1 \\ 2 & -1 & 7 \end{pmatrix} \)

\( \det A_z = 1 \cdot \begin{vmatrix} 1 & 1 \\ -1 & 7 \end{vmatrix} – (-2) \cdot \begin{vmatrix} 3 & 1 \\ 2 & 7 \end{vmatrix} + 4 \cdot \begin{vmatrix} 3 & 1 \\ 2 & -1 \end{vmatrix} \)

\( = 1 \cdot (1 \cdot 7 – 1 \cdot (-1)) + 2 \cdot (3 \cdot 7 – 1 \cdot 2) + 4 \cdot (3 \cdot (-1) – 1 \cdot 2) \)

\( = 1 \cdot (7 + 1) + 2 \cdot (21 – 2) + 4 \cdot (-3 – 2) = 8 + 2 \cdot 19 – 20 = 8 + 38 – 20 = 26 \)

Podle Cramerova pravidla platí:

\( x = \frac{\det A_x}{\det A} = \frac{20}{19}, \quad y = \frac{\det A_y}{\det A} = \frac{-15}{19}, \quad z = \frac{\det A_z}{\det A} = \frac{26}{19} \)

Řešením soustavy je tedy:

\( x = \frac{20}{19}, \quad y = -\frac{15}{19}, \quad z = \frac{26}{19} \)

Řešení:

Zápis matice koeficientů a vektoru pravých stran:

\( A = \begin{pmatrix} 5 & 1 & -2 \\ -3 & 4 & 1 \\ 2 & -1 & 3 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 9 \\ -2 \\ 7 \end{pmatrix} \)

Determinant matice \( A \) spočítáme rozvojem podle prvního řádku:

\( \det A = 5 \cdot \begin{vmatrix} 4 & 1 \\ -1 & 3 \end{vmatrix} – 1 \cdot \begin{vmatrix} -3 & 1 \\ 2 & 3 \end{vmatrix} – 2 \cdot \begin{vmatrix} -3 & 4 \\ 2 & -1 \end{vmatrix} \)

\( = 5 \cdot (4 \cdot 3 – 1 \cdot (-1)) – 1 \cdot (-3 \cdot 3 – 1 \cdot 2) – 2 \cdot (-3 \cdot (-1) – 4 \cdot 2) \)

\( = 5 \cdot (12 + 1) – 1 \cdot (-9 – 2) – 2 \cdot (3 – 8) = 5 \cdot 13 + 11 – 2 \cdot (-5) = 65 + 11 + 10 = 86 \)

Vypočítáme determinanty \( \det A_x, \det A_y, \det A_z \):

\( A_x = \begin{pmatrix} 9 & 1 & -2 \\ -2 & 4 & 1 \\ 7 & -1 & 3 \end{pmatrix} \)

\( \det A_x = 9 \cdot \begin{vmatrix} 4 & 1 \\ -1 & 3 \end{vmatrix} – 1 \cdot \begin{vmatrix} -2 & 1 \\ 7 & 3 \end{vmatrix} – 2 \cdot \begin{vmatrix} -2 & 4 \\ 7 & -1 \end{vmatrix} \)

\( = 9 \cdot (12 + 1) – 1 \cdot (-2 \cdot 3 – 1 \cdot 7) – 2 \cdot (-2 \cdot (-1) – 4 \cdot 7) \)

\( = 9 \cdot 13 – 1 \cdot (-6 – 7) – 2 \cdot (2 – 28) = 117 + 13 – 2 \cdot (-26) = 117 + 13 + 52 = 182 \)

\( A_y = \begin{pmatrix} 5 & 9 & -2 \\ -3 & -2 & 1 \\ 2 & 7 & 3 \end{pmatrix} \)

\( \det A_y = 5 \cdot \begin{vmatrix} -2 & 1 \\ 7 & 3 \end{vmatrix} – 9 \cdot \begin{vmatrix} -3 & 1 \\ 2 & 3 \end{vmatrix} – 2 \cdot \begin{vmatrix} -3 & -2 \\ 2 & 7 \end{vmatrix} \)

\( = 5 \cdot (-2 \cdot 3 – 1 \cdot 7) – 9 \cdot (-3 \cdot 3 – 1 \cdot 2) – 2 \cdot (-3 \cdot 7 – (-2) \cdot 2) \)

\( = 5 \cdot (-6 – 7) – 9 \cdot (-9 – 2) – 2 \cdot (-21 + 4) = 5 \cdot (-13) + 9 \cdot 11 – 2 \cdot (-17) = -65 + 99 + 34 = 68 \)

\( A_z = \begin{pmatrix} 5 & 1 & 9 \\ -3 & 4 & -2 \\ 2 & -1 & 7 \end{pmatrix} \)

\( \det A_z = 5 \cdot \begin{vmatrix} 4 & -2 \\ -1 & 7 \end{vmatrix} – 1 \cdot \begin{vmatrix} -3 & -2 \\ 2 & 7 \end{vmatrix} + 9 \cdot \begin{vmatrix} -3 & 4 \\ 2 & -1 \end{vmatrix} \)

\( = 5 \cdot (4 \cdot 7 – (-2) \cdot (-1)) – 1 \cdot (-3 \cdot 7 – (-2) \cdot 2) + 9 \cdot (-3 \cdot (-1) – 4 \cdot 2) \)

\( = 5 \cdot (28 – 2) – 1 \cdot (-21 + 4) + 9 \cdot (3 – 8) = 5 \cdot 26 + 17 + 9 \cdot (-5) = 130 + 17 – 45 = 102 \)

Podle Cramerova pravidla:

\( x = \frac{182}{86} = \frac{91}{43}, \quad y = \frac{68}{86} = \frac{34}{43}, \quad z = \frac{102}{86} = \frac{51}{43} \)

Řešení soustavy je tedy:

\( x = \frac{91}{43}, \quad y = \frac{34}{43}, \quad z = \frac{51}{43} \)

Řešení příkladu 4:

Nejprve sestavíme matici koeficientů: \[ A = \begin{vmatrix} 2 & 3 & -1 \\ -1 & 4 & 2 \\ 3 & -1 & 1 \end{vmatrix} \] Spočítáme determinant: \[ \det(A) = 2 \cdot \begin{vmatrix}4 & 2 \\ -1 & 1\end{vmatrix} – 3 \cdot \begin{vmatrix}-1 & 2 \\ 3 & 1\end{vmatrix} + (-1) \cdot \begin{vmatrix}-1 & 4 \\ 3 & -1\end{vmatrix} \] Poddeterminanty: \[ \begin{vmatrix}4 & 2 \\ -1 & 1\end{vmatrix} = 4 \cdot 1 – 2 \cdot (-1) = 4 + 2 = 6 \] \[ \begin{vmatrix}-1 & 2 \\ 3 & 1\end{vmatrix} = -1 \cdot 1 – 2 \cdot 3 = -1 – 6 = -7 \] \[ \begin{vmatrix}-1 & 4 \\ 3 & -1\end{vmatrix} = -1 \cdot (-1) – 4 \cdot 3 = 1 – 12 = -11 \] Dosadíme zpět: \[ \det(A) = 2 \cdot 6 – 3 \cdot (-7) + (-1) \cdot (-11) = 12 + 21 + 11 = 44 \] Protože \(\det(A) \neq 0\), soustava má jednoznačné řešení. Matice pro \(x\): \[ A_x = \begin{vmatrix} 7 & 3 & -1 \\ 4 & 4 & 2 \\ 10 & -1 & 1 \end{vmatrix} \] Spočítáme determinant: \[ \det(A_x) = 7 \cdot \begin{vmatrix}4 & 2 \\ -1 & 1\end{vmatrix} – 3 \cdot \begin{vmatrix}4 & 2 \\ 10 & 1\end{vmatrix} + (-1) \cdot \begin{vmatrix}4 & 4 \\ 10 & -1\end{vmatrix} \] Poddeterminanty: \[ \begin{vmatrix}4 & 2 \\ -1 & 1\end{vmatrix} = 6 \] \[ \begin{vmatrix}4 & 2 \\ 10 & 1\end{vmatrix} = 4 \cdot 1 – 2 \cdot 10 = 4 – 20 = -16 \] \[ \begin{vmatrix}4 & 4 \\ 10 & -1\end{vmatrix} = 4 \cdot (-1) – 4 \cdot 10 = -4 – 40 = -44 \] Dosadíme: \[ \det(A_x) = 7 \cdot 6 – 3 \cdot (-16) + (-1) \cdot (-44) = 42 + 48 + 44 = 134 \] Matice pro \(y\): \[ A_y = \begin{vmatrix} 2 & 7 & -1 \\ -1 & 4 & 2 \\ 3 & 10 & 1 \end{vmatrix} \] Spočítáme determinant: \[ \det(A_y) = 2 \cdot \begin{vmatrix}4 & 2 \\ 10 & 1\end{vmatrix} – 7 \cdot \begin{vmatrix}-1 & 2 \\ 3 & 1\end{vmatrix} + (-1) \cdot \begin{vmatrix}-1 & 4 \\ 3 & 10\end{vmatrix} \] Poddeterminanty: \[ \begin{vmatrix}4 & 2 \\ 10 & 1\end{vmatrix} = -16 \] \[ \begin{vmatrix}-1 & 2 \\ 3 & 1\end{vmatrix} = -7 \] \[ \begin{vmatrix}-1 & 4 \\ 3 & 10\end{vmatrix} = -1 \cdot 10 – 4 \cdot 3 = -10 – 12 = -22 \] Dosadíme: \[ \det(A_y) = 2 \cdot (-16) – 7 \cdot (-7) + (-1) \cdot (-22) = -32 + 49 + 22 = 39 \] Matice pro \(z\): \[ A_z = \begin{vmatrix} 2 & 3 & 7 \\ -1 & 4 & 4 \\ 3 & -1 & 10 \end{vmatrix} \] Spočítáme determinant: \[ \det(A_z) = 2 \cdot \begin{vmatrix}4 & 4 \\ -1 & 10\end{vmatrix} – 3 \cdot \begin{vmatrix}-1 & 4 \\ 3 & 10\end{vmatrix} + 7 \cdot \begin{vmatrix}-1 & 4 \\ 3 & -1\end{vmatrix} \] Poddeterminanty: \[ \begin{vmatrix}4 & 4 \\ -1 & 10\end{vmatrix} = 4 \cdot 10 – 4 \cdot (-1) = 40 + 4 = 44 \] \[ \begin{vmatrix}-1 & 4 \\ 3 & 10\end{vmatrix} = -1 \cdot 10 – 4 \cdot 3 = -10 – 12 = -22 \] \[ \begin{vmatrix}-1 & 4 \\ 3 & -1\end{vmatrix} = -1 \cdot (-1) – 4 \cdot 3 = 1 – 12 = -11 \] Dosadíme: \[ \det(A_z) = 2 \cdot 44 – 3 \cdot (-22) + 7 \cdot (-11) = 88 + 66 – 77 = 77 \] Výsledky: \[ x = \frac{134}{44} = \frac{67}{22} \approx 3{,}05, \quad y = \frac{39}{44} \approx 0{,}89, \quad z = \frac{77}{44} = \frac{7}{4} = 1{,}75 \] Soustava má řešení \( (x, y, z) = \left(\frac{67}{22}, \frac{39}{44}, \frac{7}{4}\right) \).

Řešení příkladu 5:

Matice koeficientů je \[ A = \begin{vmatrix} 1 & 2 & 3 \\ 2 & -1 & 1 \\ 4 & 1 & -2 \end{vmatrix} \] Spočítáme determinant: \[ \det(A) = 1 \cdot \begin{vmatrix}-1 & 1 \\ 1 & -2\end{vmatrix} – 2 \cdot \begin{vmatrix}2 & 1 \\ 4 & -2\end{vmatrix} + 3 \cdot \begin{vmatrix}2 & -1 \\ 4 & 1\end{vmatrix} \] Poddeterminanty: \[ \begin{vmatrix}-1 & 1 \\ 1 & -2\end{vmatrix} = (-1)(-2) – 1 \cdot 1 = 2 – 1 = 1 \] \[ \begin{vmatrix}2 & 1 \\ 4 & -2\end{vmatrix} = 2 \cdot (-2) – 1 \cdot 4 = -4 – 4 = -8 \] \[ \begin{vmatrix}2 & -1 \\ 4 & 1\end{vmatrix} = 2 \cdot 1 – (-1) \cdot 4 = 2 + 4 = 6 \] Dosadíme: \[ \det(A) = 1 \cdot 1 – 2 \cdot (-8) + 3 \cdot 6 = 1 + 16 + 18 = 35 \] Protože \(\det(A) \neq 0\), soustava má jednoznačné řešení. Matice pro \(x\): \[ A_x = \begin{vmatrix} 14 & 2 & 3 \\ 3 & -1 & 1 \\ 2 & 1 & -2 \end{vmatrix} \] Spočítáme determinant: \[ \det(A_x) = 14 \cdot \begin{vmatrix}-1 & 1 \\ 1 & -2\end{vmatrix} – 2 \cdot \begin{vmatrix}3 & 1 \\ 2 & -2\end{vmatrix} + 3 \cdot \begin{vmatrix}3 & -1 \\ 2 & 1\end{vmatrix} \] Poddeterminanty: \[ \begin{vmatrix}-1 & 1 \\ 1 & -2\end{vmatrix} = 1 \] \[ \begin{vmatrix}3 & 1 \\ 2 & -2\end{vmatrix} = 3 \cdot (-2) – 1 \cdot 2 = -6 – 2 = -8 \] \[ \begin{vmatrix}3 & -1 \\ 2 & 1\end{vmatrix} = 3 \cdot 1 – (-1) \cdot 2 = 3 + 2 = 5 \] Dosadíme: \[ \det(A_x) = 14 \cdot 1 – 2 \cdot (-8) + 3 \cdot 5 = 14 + 16 + 15 = 45 \] Matice pro \(y\): \[ A_y = \begin{vmatrix} 1 & 14 & 3 \\ 2 & 3 & 1 \\ 4 & 2 & -2 \end{vmatrix} \] Spočítáme determinant: \[ \det(A_y) = 1 \cdot \begin{vmatrix}3 & 1 \\ 2 & -2\end{vmatrix} – 14 \cdot \begin{vmatrix}2 & 1 \\ 4 & -2\end{vmatrix} + 3 \cdot \begin{vmatrix}2 & 3 \\ 4 & 2\end{vmatrix} \] Poddeterminanty: \[ \begin{vmatrix}3 & 1 \\ 2 & -2\end{vmatrix} = -8 \] \[ \begin{vmatrix}2 & 1 \\ 4 & -2\end{vmatrix} = -8 \] \[ \begin{vmatrix}2 & 3 \\ 4 & 2\end{vmatrix} = 2 \cdot 2 – 3 \cdot 4 = 4 – 12 = -8 \] Dosadíme: \[ \det(A_y) = 1 \cdot (-8) – 14 \cdot (-8) + 3 \cdot (-8) = -8 + 112 – 24 = 80 \] Matice pro \(z\): \[ A_z = \begin{vmatrix} 1 & 2 & 14 \\ 2 & -1 & 3 \\ 4 & 1 & 2 \end{vmatrix} \] Spočítáme determinant: \[ \det(A_z) = 1 \cdot \begin{vmatrix}-1 & 3 \\ 1 & 2\end{vmatrix} – 2 \cdot \begin{vmatrix}2 & 3 \\ 4 & 2\end{vmatrix} + 14 \cdot \begin{vmatrix}2 & -1 \\ 4 & 1\end{vmatrix} \] Poddeterminanty: \[ \begin{vmatrix}-1 & 3 \\ 1 & 2\end{vmatrix} = (-1) \cdot 2 – 3 \cdot 1 = -2 – 3 = -5 \] \[ \begin{vmatrix}2 & 3 \\ 4 & 2\end{vmatrix} = 2 \cdot 2 – 3 \cdot 4 = 4 – 12 = -8 \] \[ \begin{vmatrix}2 & -1 \\ 4 & 1\end{vmatrix} = 2 \cdot 1 – (-1) \cdot 4 = 2 + 4 = 6 \] Dosadíme: \[ \det(A_z) = 1 \cdot (-5) – 2 \cdot (-8) + 14 \cdot 6 = -5 + 16 + 84 = 95 \] Výsledky: \[ x = \frac{45}{35} = \frac{9}{7} \approx 1{,}29, \quad y = \frac{80}{35} = \frac{16}{7} \approx 2{,}29, \quad z = \frac{95}{35} = \frac{19}{7} \approx 2{,}71 \] Soustava má řešení \( (x, y, z) = \left(\frac{9}{7}, \frac{16}{7}, \frac{19}{7}\right) \).

Řešení příkladu 6:

Nejprve zapíšeme matici koeficientů: \[ A = \begin{vmatrix} 4 & -1 & 2 \\ -2 & 3 & -1 \\ 1 & 5 & 3 \end{vmatrix} \] Spočítáme determinant matice \( A \): \[ \det(A) = 4 \cdot (3 \cdot 3 – (-1) \cdot 5) – (-1) \cdot (-2 \cdot 3 – (-1) \cdot 1) + 2 \cdot (-2 \cdot 5 – 3 \cdot 1) \] \[ = 4 \cdot (9 + 5) – (-1) \cdot (-6 + 1) + 2 \cdot (-10 – 3) = 4 \cdot 14 – 1 \cdot 5 + 2 \cdot (-13) \] \[ = 56 – 5 – 26 = 25 \] Protože \(\det(A) \neq 0\), soustava má jednoznačné řešení. Vypočítáme determinanty pro \(x\), \(y\) a \(z\). Matice pro \(x\) (první sloupec nahradíme vektorem pravých stran): \[ A_x = \begin{vmatrix} 9 & -1 & 2 \\ -4 & 3 & -1 \\ 17 & 5 & 3 \end{vmatrix} \] \[ \det(A_x) = 9 \cdot (3 \cdot 3 – (-1) \cdot 5) – (-1) \cdot (-4 \cdot 3 – (-1) \cdot 17) + 2 \cdot (-4 \cdot 5 – 3 \cdot 17) \] \[ = 9 \cdot (9 + 5) – (-1) \cdot (-12 + 17) + 2 \cdot (-20 – 51) = 9 \cdot 14 – 1 \cdot 5 + 2 \cdot (-71) \] \[ = 126 – 5 – 142 = -21 \] Matice pro \(y\) (druhý sloupec nahradíme vektorem pravých stran): \[ A_y = \begin{vmatrix} 4 & 9 & 2 \\ -2 & -4 & -1 \\ 1 & 17 & 3 \end{vmatrix} \] \[ \det(A_y) = 4 \cdot (-4 \cdot 3 – (-1) \cdot 17) – 9 \cdot (-2 \cdot 3 – (-1) \cdot 1) + 2 \cdot (-2 \cdot 17 – (-4) \cdot 1) \] \[ = 4 \cdot (-12 + 17) – 9 \cdot (-6 + 1) + 2 \cdot (-34 + 4) = 4 \cdot 5 – 9 \cdot (-5) + 2 \cdot (-30) \] \[ = 20 + 45 – 60 = 5 \] Matice pro \(z\) (třetí sloupec nahradíme vektorem pravých stran): \[ A_z = \begin{vmatrix} 4 & -1 & 9 \\ -2 & 3 & -4 \\ 1 & 5 & 17 \end{vmatrix} \] \[ \det(A_z) = 4 \cdot (3 \cdot 17 – (-4) \cdot 5) – (-1) \cdot (-2 \cdot 17 – (-4) \cdot 1) + 9 \cdot (-2 \cdot 5 – 3 \cdot 1) \] \[ = 4 \cdot (51 + 20) – (-1) \cdot (-34 + 4) + 9 \cdot (-10 – 3) = 4 \cdot 71 – 1 \cdot 30 + 9 \cdot (-13) \] \[ = 284 – 30 – 117 = 137 \] Nyní spočítáme jednotlivé neznámé: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-21}{25} = -0{,}84 \] \[ y = \frac{\det(A_y)}{\det(A)} = \frac{5}{25} = 0{,}2 \] \[ z = \frac{\det(A_z)}{\det(A)} = \frac{137}{25} = 5{,}48 \] Výsledné řešení soustavy je \( (x, y, z) = (-0{,}84; 0{,}2; 5{,}48) \).

Řešení příkladu 7:

Matice koeficientů je \[ A = \begin{vmatrix} m & 1 & -1 \\ 1 & m+1 & 2 \\ 2 & -1 & m-1 \end{vmatrix} \] Spočítáme determinant \( \det(A) \): \[ \det(A) = m \cdot \begin{vmatrix} m+1 & 2 \\ -1 & m-1 \end{vmatrix} – 1 \cdot \begin{vmatrix} 1 & 2 \\ 2 & m-1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 1 & m+1 \\ 2 & -1 \end{vmatrix} \] Nejprve spočítáme poddeterminanty: \[ \begin{vmatrix} m+1 & 2 \\ -1 & m-1 \end{vmatrix} = (m+1)(m-1) – (-1)\cdot 2 = m^2 – 1 + 2 = m^2 +1 \] \[ \begin{vmatrix} 1 & 2 \\ 2 & m-1 \end{vmatrix} = 1 \cdot (m-1) – 2 \cdot 2 = m-1 – 4 = m – 5 \] \[ \begin{vmatrix} 1 & m+1 \\ 2 & -1 \end{vmatrix} = 1 \cdot (-1) – 2 \cdot (m+1) = -1 – 2m – 2 = -2m – 3 \] Dosadíme zpět: \[ \det(A) = m (m^2 + 1) – 1 (m – 5) + (-1)(-2m – 3) = m^3 + m – m + 5 + 2m + 3 = m^3 + 2m + 8 \] Pro řešení pomocí Cramerova pravidla platí, že pokud \(\det(A) \neq 0\), soustava má jednoznačné řešení. Vypočítáme determinanty pro \(x\), \(y\), \(z\): \[ A_x = \begin{vmatrix} 2 & 1 & -1 \\ 3 & m+1 & 2 \\ 1 & -1 & m-1 \end{vmatrix} \] \[ \det(A_x) = 2 \cdot \begin{vmatrix} m+1 & 2 \\ -1 & m-1 \end{vmatrix} – 1 \cdot \begin{vmatrix} 3 & 2 \\ 1 & m-1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 3 & m+1 \\ 1 & -1 \end{vmatrix} \] Poddeterminanty: \[ \begin{vmatrix} m+1 & 2 \\ -1 & m-1 \end{vmatrix} = m^2 + 1 \] \[ \begin{vmatrix} 3 & 2 \\ 1 & m-1 \end{vmatrix} = 3(m-1) – 2 \cdot 1 = 3m – 3 – 2 = 3m – 5 \] \[ \begin{vmatrix} 3 & m+1 \\ 1 & -1 \end{vmatrix} = 3 \cdot (-1) – 1 \cdot (m+1) = -3 – m – 1 = -m – 4 \] Dosadíme: \[ \det(A_x) = 2 (m^2 +1) – 1 (3m -5) + (-1)(-m -4) = 2m^2 + 2 – 3m + 5 + m + 4 = 2m^2 – 2m + 11 \] Matice pro \(y\): \[ A_y = \begin{vmatrix} m & 2 & -1 \\ 1 & 3 & 2 \\ 2 & 1 & m-1 \end{vmatrix} \] \[ \det(A_y) = m \cdot \begin{vmatrix} 3 & 2 \\ 1 & m-1 \end{vmatrix} – 2 \cdot \begin{vmatrix} 1 & 2 \\ 2 & m-1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} \] Poddeterminanty: \[ \begin{vmatrix} 3 & 2 \\ 1 & m-1 \end{vmatrix} = 3(m-1) – 2 = 3m – 3 – 2 = 3m – 5 \] \[ \begin{vmatrix} 1 & 2 \\ 2 & m-1 \end{vmatrix} = 1(m-1) – 2 \cdot 2 = m – 1 – 4 = m – 5 \] \[ \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = 1 \cdot 1 – 3 \cdot 2 = 1 – 6 = -5 \] Dosadíme: \[ \det(A_y) = m (3m – 5) – 2 (m – 5) – 5 = 3m^2 – 5m – 2m + 10 – 5 = 3m^2 – 7m + 5 \] Matice pro \(z\): \[ A_z = \begin{vmatrix} m & 1 & 2 \\ 1 & m+1 & 3 \\ 2 & -1 & 1 \end{vmatrix} \] \[ \det(A_z) = m \cdot \begin{vmatrix} m+1 & 3 \\ -1 & 1 \end{vmatrix} – 1 \cdot \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} + 2 \cdot \begin{vmatrix} 1 & m+1 \\ 2 & -1 \end{vmatrix} \] Poddeterminanty: \[ \begin{vmatrix} m+1 & 3 \\ -1 & 1 \end{vmatrix} = (m+1) \cdot 1 – (-1) \cdot 3 = m + 1 + 3 = m + 4 \] \[ \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = 1 \cdot 1 – 3 \cdot 2 = 1 – 6 = -5 \] \[ \begin{vmatrix} 1 & m+1 \\ 2 & -1 \end{vmatrix} = 1 \cdot (-1) – 2 \cdot (m+1) = -1 – 2m – 2 = -2m – 3 \] Dosadíme: \[ \det(A_z) = m (m + 4) – 1 (-5) + 2 (-2m – 3) = m^2 + 4m + 5 – 4m – 6 = m^2 – 1 \] Nakonec řešíme: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{2m^2 – 2m + 11}{m^3 + 2m + 8} \] \[ y = \frac{\det(A_y)}{\det(A)} = \frac{3m^2 – 7m + 5}{m^3 + 2m + 8} \] \[ z = \frac{\det(A_z)}{\det(A)} = \frac{m^2 – 1}{m^3 + 2m + 8} \] Pro hodnoty \(m\), kdy \(\det(A) = 0\), soustava nemusí mít řešení nebo má nekonečně mnoho řešení. Pro ostatní hodnoty \(m\) platí výše uvedené řešení.

Řešení příkladu 8:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} \] Determinant \( \det(A) = 1 \cdot (-1) – 2 \cdot 3 = -1 – 6 = -7 \neq 0 \). Determinant pro \(x\): \[ A_x = \begin{vmatrix} \sin t & 2 \\ e^t & -1 \end{vmatrix} = \sin t \cdot (-1) – 2 \cdot e^t = -\sin t – 2 e^t \] Determinant pro \(y\): \[ A_y = \begin{vmatrix} 1 & \sin t \\ 3 & e^t \end{vmatrix} = 1 \cdot e^t – 3 \sin t = e^t – 3 \sin t \] Řešení je: \[ x = \frac{A_x}{\det(A)} = \frac{-\sin t – 2 e^t}{-7} = \frac{\sin t + 2 e^t}{7} \] \[ y = \frac{A_y}{\det(A)} = \frac{e^t – 3 \sin t}{-7} = \frac{3 \sin t – e^t}{7} \] Tím jsme vyjádřili \(x(t)\) a \(y(t)\) pomocí funkcí \(\sin t\) a \(e^t\).

Řešení příkladu 9:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 1 & 4 & -1 \end{vmatrix} \] Vypočítáme determinant \(\det(A)\): \[ \det(A) = 1 \cdot \begin{vmatrix} -1 & 3 \\ 4 & -1 \end{vmatrix} – 1 \cdot \begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & -1 \\ 1 & 4 \end{vmatrix} \] \[ = 1((-1)(-1) – 3 \cdot 4) – 1(2 \cdot (-1) – 3 \cdot 1) + 1(2 \cdot 4 – (-1) \cdot 1) \] \[ = 1(1 – 12) – 1(-2 – 3) + 1(8 + 1) = 1(-11) – 1(-5) + 1(9) = -11 + 5 + 9 = 3 \] Determinanty pro neznámé: \[ A_x = \begin{vmatrix} t & 1 & 1 \\ e^t & -1 & 3 \\ \cos t & 4 & -1 \end{vmatrix}, \quad A_y = \begin{vmatrix} 1 & t & 1 \\ 2 & e^t & 3 \\ 1 & \cos t & -1 \end{vmatrix}, \quad A_z = \begin{vmatrix} 1 & 1 & t \\ 2 & -1 & e^t \\ 1 & 4 & \cos t \end{vmatrix} \] Vypočítáme \(A_x\): \[ A_x = t \cdot \begin{vmatrix} -1 & 3 \\ 4 & -1 \end{vmatrix} – 1 \cdot \begin{vmatrix} e^t & 3 \\ \cos t & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} e^t & -1 \\ \cos t & 4 \end{vmatrix} \] \[ = t((-1)(-1) – 3 \cdot 4) – 1(e^t \cdot (-1) – 3 \cos t) + 1(e^t \cdot 4 – (-1) \cos t) \] \[ = t(1 – 12) – ( -e^t – 3 \cos t ) + (4 e^t + \cos t) \] \[ = -11 t + e^t + 3 \cos t + 4 e^t + \cos t = -11 t + 5 e^t + 4 \cos t \] Vypočítáme \(A_y\): \[ A_y = 1 \cdot \begin{vmatrix} e^t & 3 \\ \cos t & -1 \end{vmatrix} – t \cdot \begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & e^t \\ 1 & \cos t \end{vmatrix} \] \[ = 1(e^t \cdot (-1) – 3 \cos t) – t(2 \cdot (-1) – 3 \cdot 1) + 1(2 \cos t – e^t) \] \[ = -e^t – 3 \cos t – t(-2 -3) + 2 \cos t – e^t \] \[ = -e^t – 3 \cos t + 5 t + 2 \cos t – e^t = 5 t – 2 e^t – \cos t \] Vypočítáme \(A_z\): \[ A_z = 1 \cdot \begin{vmatrix} -1 & e^t \\ 4 & \cos t \end{vmatrix} – 1 \cdot \begin{vmatrix} 2 & e^t \\ 1 & \cos t \end{vmatrix} + t \cdot \begin{vmatrix} 2 & -1 \\ 1 & 4 \end{vmatrix} \] \[ = 1((-1) \cdot \cos t – 4 e^t) – 1(2 \cos t – e^t) + t(2 \cdot 4 – (-1) \cdot 1) \] \[ = -\cos t – 4 e^t – 2 \cos t + e^t + t(8 + 1) \] \[ = -3 \cos t – 3 e^t + 9 t \] Řešení je: \[ x = \frac{A_x}{\det(A)} = \frac{-11 t + 5 e^t + 4 \cos t}{3} \] \[ y = \frac{A_y}{\det(A)} = \frac{5 t – 2 e^t – \cos t}{3} \] \[ z = \frac{A_z}{\det(A)} = \frac{-3 \cos t – 3 e^t + 9 t}{3} = -\cos t – e^t + 3 t \]

Řešení příkladu 10:

Koeficientová matice: \[ A = \begin{vmatrix} 2 & -1 & 1 \\ -1 & 3 & -2 \\ 1 & 1 & 4 \end{vmatrix} \] Vypočítáme determinant \(\det(A)\): \[ \det(A) = 2 \cdot \begin{vmatrix}3 & -2 \\ 1 & 4 \end{vmatrix} – (-1) \cdot \begin{vmatrix} -1 & -2 \\ 1 & 4 \end{vmatrix} + 1 \cdot \begin{vmatrix} -1 & 3 \\ 1 & 1 \end{vmatrix} \] \[ = 2(3 \cdot 4 – (-2) \cdot 1) + 1(-1 \cdot 4 – (-2) \cdot 1) + 1(-1 \cdot 1 – 3 \cdot 1) \] \[ = 2(12 + 2) + 1(-4 + 2) + 1(-1 – 3) = 2 \cdot 14 – 2 – 4 = 28 – 2 – 4 = 22 \] Determinanty pro neznámé: \[ A_x = \begin{vmatrix} \sin t & -1 & 1 \\ e^t & 3 & -2 \\ t^2 & 1 & 4 \end{vmatrix}, \quad A_y = \begin{vmatrix} 2 & \sin t & 1 \\ -1 & e^t & -2 \\ 1 & t^2 & 4 \end{vmatrix}, \quad A_z = \begin{vmatrix} 2 & -1 & \sin t \\ -1 & 3 & e^t \\ 1 & 1 & t^2 \end{vmatrix} \] Vypočítáme \(A_x\): \[ A_x = \sin t \cdot \begin{vmatrix}3 & -2 \\ 1 & 4 \end{vmatrix} – (-1) \cdot \begin{vmatrix} e^t & -2 \\ t^2 & 4 \end{vmatrix} + 1 \cdot \begin{vmatrix} e^t & 3 \\ t^2 & 1 \end{vmatrix} \] \[ = \sin t (3 \cdot 4 – (-2) \cdot 1) + 1(e^t \cdot 4 – (-2) t^2) + 1(e^t \cdot 1 – 3 t^2) \] \[ = \sin t (12 + 2) + 4 e^t + 2 t^2 + e^t – 3 t^2 = 14 \sin t + 5 e^t – t^2 \] Vypočítáme \(A_y\): \[ A_y = 2 \cdot \begin{vmatrix} e^t & -2 \\ t^2 & 4 \end{vmatrix} – \sin t \cdot \begin{vmatrix} -1 & -2 \\ 1 & 4 \end{vmatrix} + 1 \cdot \begin{vmatrix} -1 & e^t \\ 1 & t^2 \end{vmatrix} \] \[ = 2(e^t \cdot 4 – (-2) t^2) – \sin t (-1 \cdot 4 – (-2) \cdot 1) + (-1 \cdot t^2 – e^t \cdot 1) \] \[ = 2(4 e^t + 2 t^2) – \sin t (-4 + 2) – t^2 – e^t = 8 e^t + 4 t^2 + 2 \sin t – t^2 – e^t \] \[ = 7 e^t + 3 t^2 + 2 \sin t \] Vypočítáme \(A_z\): \[ A_z = 2 \cdot \begin{vmatrix} 3 & e^t \\ 1 & t^2 \end{vmatrix} – (-1) \cdot \begin{vmatrix} -1 & e^t \\ 1 & t^2 \end{vmatrix} + \sin t \cdot \begin{vmatrix} -1 & 3 \\ 1 & 1 \end{vmatrix} \] \[ = 2(3 t^2 – e^t) + 1(-1 \cdot t^2 – e^t \cdot 1) + \sin t(-1 \cdot 1 – 3 \cdot 1) \] \[ = 2(3 t^2 – e^t) – t^2 – e^t + \sin t (-1 – 3) = 6 t^2 – 2 e^t – t^2 – e^t – 4 \sin t \] \[ = 5 t^2 – 3 e^t – 4 \sin t \] Řešení je: \[ x = \frac{14 \sin t + 5 e^t – t^2}{22}, \quad y = \frac{7 e^t + 3 t^2 + 2 \sin t}{22}, \quad z = \frac{5 t^2 – 3 e^t – 4 \sin t}{22} \]

Řešení příkladu 11:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & 2 & 3 \\ 4 & -1 & 1 \\ 2 & 1 & -1 \end{vmatrix} \] Vypočítáme determinant: \[ \det(A) = 1 \cdot \begin{vmatrix} -1 & 1 \\ 1 & -1 \end{vmatrix} – 2 \cdot \begin{vmatrix} 4 & 1 \\ 2 & -1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 4 & -1 \\ 2 & 1 \end{vmatrix} \] \[ = 1((-1)(-1) – 1 \cdot 1) – 2(4 \cdot (-1) – 1 \cdot 2) + 3(4 \cdot 1 – (-1) \cdot 2) \] \[ = 1(1 – 1) – 2(-4 – 2) + 3(4 + 2) = 0 + 12 + 18 = 30 \] Determinanty pro neznámé: \[ A_x = \begin{vmatrix} e^{2t} & 2 & 3 \\ \sin 2t & -1 & 1 \\ \cos 2t & 1 & -1 \end{vmatrix}, \quad A_y = \begin{vmatrix} 1 & e^{2t} & 3 \\ 4 & \sin 2t & 1 \\ 2 & \cos 2t & -1 \end{vmatrix}, \quad A_z = \begin{vmatrix} 1 & 2 & e^{2t} \\ 4 & -1 & \sin 2t \\ 2 & 1 & \cos 2t \end{vmatrix} \] Výpočet \(A_x\): \[ A_x = e^{2t} \cdot \begin{vmatrix} -1 & 1 \\ 1 & -1 \end{vmatrix} – 2 \cdot \begin{vmatrix} \sin 2t & 1 \\ \cos 2t & -1 \end{vmatrix} + 3 \cdot \begin{vmatrix} \sin 2t & -1 \\ \cos 2t & 1 \end{vmatrix} \] \[ = e^{2t}((-1)(-1) – 1 \cdot 1) – 2(\sin 2t \cdot (-1) – 1 \cdot \cos 2t) + 3(\sin 2t \cdot 1 – (-1) \cdot \cos 2t) \] \[ = e^{2t}(1 – 1) – 2(-\sin 2t – \cos 2t) + 3(\sin 2t + \cos 2t) = 0 + 2(\sin 2t + \cos 2t) + 3(\sin 2t + \cos 2t) \] \[ = 5(\sin 2t + \cos 2t) \] Výpočet \(A_y\): \[ A_y = 1 \cdot \begin{vmatrix} \sin 2t & 1 \\ \cos 2t & -1 \end{vmatrix} – e^{2t} \cdot \begin{vmatrix} 4 & 1 \\ 2 & -1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 4 & \sin 2t \\ 2 & \cos 2t \end{vmatrix} \] \[ = 1(\sin 2t \cdot (-1) – 1 \cdot \cos 2t) – e^{2t}(4 \cdot (-1) – 1 \cdot 2) + 3(4 \cdot \cos 2t – \sin 2t \cdot 2) \] \[ = -\sin 2t – \cos 2t – e^{2t}(-4 – 2) + 3(4 \cos 2t – 2 \sin 2t) \] \[ = -\sin 2t – \cos 2t + 6 e^{2t} + 12 \cos 2t – 6 \sin 2t = 6 e^{2t} + 11 \cos 2t – 7 \sin 2t \] Výpočet \(A_z\): \[ A_z = 1 \cdot \begin{vmatrix} -1 & \sin 2t \\ 1 & \cos 2t \end{vmatrix} – 2 \cdot \begin{vmatrix} 4 & \sin 2t \\ 2 & \cos 2t \end{vmatrix} + e^{2t} \cdot \begin{vmatrix} 4 & -1 \\ 2 & 1 \end{vmatrix} \] \[ = 1((-1) \cdot \cos 2t – 1 \cdot \sin 2t) – 2(4 \cdot \cos 2t – 2 \cdot \sin 2t) + e^{2t}(4 \cdot 1 – (-1) \cdot 2) \] \[ = -\cos 2t – \sin 2t – 2(4 \cos 2t – 2 \sin 2t) + e^{2t}(4 + 2) \] \[ = -\cos 2t – \sin 2t – 8 \cos 2t + 4 \sin 2t + 6 e^{2t} = 6 e^{2t} – 9 \cos 2t + 3 \sin 2t \] Řešení je: \[ x = \frac{5(\sin 2t + \cos 2t)}{30} = \frac{\sin 2t + \cos 2t}{6} \] \[ y = \frac{6 e^{2t} + 11 \cos 2t – 7 \sin 2t}{30} \] \[ z = \frac{6 e^{2t} – 9 \cos 2t + 3 \sin 2t}{30} = \frac{6 e^{2t}}{30} – \frac{9 \cos 2t}{30} + \frac{3 \sin 2t}{30} \]

Řešení příkladu 12:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & 1 & -1 \\ 2 & -1 & 3 \\ -3 & 4 & 2 \end{vmatrix} \] Determinant: \[ \det(A) = 1 \cdot \begin{vmatrix} -1 & 3 \\ 4 & 2 \end{vmatrix} – 1 \cdot \begin{vmatrix} 2 & 3 \\ -3 & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 2 & -1 \\ -3 & 4 \end{vmatrix} \] \[ = 1((-1) \cdot 2 – 3 \cdot 4) – 1(2 \cdot 2 – 3 \cdot (-3)) – 1(2 \cdot 4 – (-1) \cdot (-3)) \] \[ = 1(-2 – 12) – 1(4 + 9) – 1(8 – 3) = -14 – 13 – 5 = -32 \] Determinanty: \[ A_x = \begin{vmatrix} \ln(t+1) & 1 & -1 \\ t^3 & -1 & 3 \\ t & 4 & 2 \end{vmatrix}, \quad A_y = \begin{vmatrix} 1 & \ln(t+1) & -1 \\ 2 & t^3 & 3 \\ -3 & t & 2 \end{vmatrix}, \quad A_z = \begin{vmatrix} 1 & 1 & \ln(t+1) \\ 2 & -1 & t^3 \\ -3 & 4 & t \end{vmatrix} \] Výpočet \(A_x\): \[ A_x = \ln(t+1) \cdot \begin{vmatrix} -1 & 3 \\ 4 & 2 \end{vmatrix} – 1 \cdot \begin{vmatrix} t^3 & 3 \\ t & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} t^3 & -1 \\ t & 4 \end{vmatrix} \] \[ = \ln(t+1)((-1) \cdot 2 – 3 \cdot 4) – (t^3 \cdot 2 – 3 \cdot t) – (t^3 \cdot 4 – (-1) \cdot t) \] \[ = \ln(t+1)(-2 – 12) – (2 t^3 – 3 t) – (4 t^3 + t) = -14 \ln(t+1) – 2 t^3 + 3 t – 4 t^3 – t \] \[ = -14 \ln(t+1) – 6 t^3 + 2 t \] Výpočet \(A_y\): \[ A_y = 1 \cdot \begin{vmatrix} t^3 & 3 \\ t & 2 \end{vmatrix} – \ln(t+1) \cdot \begin{vmatrix} 2 & 3 \\ -3 & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 2 & t^3 \\ -3 & t \end{vmatrix} \] \[ = 1 (t^3 \cdot 2 – 3 \cdot t) – \ln(t+1)(2 \cdot 2 – 3 \cdot (-3)) – (2 \cdot t – t^3 \cdot (-3)) \] \[ = 2 t^3 – 3 t – \ln(t+1)(4 + 9) – (2 t + 3 t^3) = 2 t^3 – 3 t – 13 \ln(t+1) – 2 t – 3 t^3 \] \[ = -t^3 – 5 t – 13 \ln(t+1) \] Výpočet \(A_z\): \[ A_z = 1 \cdot \begin{vmatrix} -1 & \ln(t+1) \\ 4 & t^3 \end{vmatrix} – 1 \cdot \begin{vmatrix} 2 & \ln(t+1) \\ -3 & t^3 \end{vmatrix} + \ln(t+1) \cdot \begin{vmatrix} 2 & -1 \\ -3 & 4 \end{vmatrix} \] \[ = 1((-1) \cdot t^3 – 4 \cdot \ln(t+1)) – (2 \cdot t^3 – (-3) \cdot \ln(t+1)) + \ln(t+1)(2 \cdot 4 – (-1) \cdot (-3)) \] \[ = -t^3 – 4 \ln(t+1) – 2 t^3 + 3 \ln(t+1) + \ln(t+1)(8 – 3) = -3 t^3 – \ln(t+1) + 5 \ln(t+1) \] \[ = -3 t^3 + 4 \ln(t+1) \] Výsledné řešení: \[ x = \frac{A_x}{\det(A)} = \frac{-14 \ln(t+1) – 6 t^3 + 2 t}{-32} = \frac{14 \ln(t+1) + 6 t^3 – 2 t}{32} \] \[ y = \frac{A_y}{\det(A)} = \frac{-t^3 – 5 t – 13 \ln(t+1)}{-32} = \frac{t^3 + 5 t + 13 \ln(t+1)}{32} \] \[ z = \frac{A_z}{\det(A)} = \frac{-3 t^3 + 4 \ln(t+1)}{-32} = \frac{3 t^3 – 4 \ln(t+1)}{32} \]

Řešení příkladu 13:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 4 \\ 1 & -3 & 5 \end{vmatrix} \] Výpočet determinantu: \[ \det(A) = 1 \cdot \begin{vmatrix} -1 & 4 \\ -3 & 5 \end{vmatrix} – 1 \cdot \begin{vmatrix} 2 & 4 \\ 1 & 5 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & -1 \\ 1 & -3 \end{vmatrix} \] \[ = 1((-1) \cdot 5 – 4 \cdot (-3)) – 1(2 \cdot 5 – 4 \cdot 1) + 1(2 \cdot (-3) – (-1) \cdot 1) \] \[ = 1(-5 + 12) – 1(10 – 4) + 1(-6 + 1) = 7 – 6 – 5 = -4 \] Determinanty pro \(x, y, z\): \[ A_x = \begin{vmatrix} \cos t & 1 & 1 \\ t^2 & -1 & 4 \\ e^t & -3 & 5 \end{vmatrix}, \quad A_y = \begin{vmatrix} 1 & \cos t & 1 \\ 2 & t^2 & 4 \\ 1 & e^t & 5 \end{vmatrix}, \quad A_z = \begin{vmatrix} 1 & 1 & \cos t \\ 2 & -1 & t^2 \\ 1 & -3 & e^t \end{vmatrix} \] Výpočet \(A_x\): \[ A_x = \cos t \cdot \begin{vmatrix} -1 & 4 \\ -3 & 5 \end{vmatrix} – 1 \cdot \begin{vmatrix} t^2 & 4 \\ e^t & 5 \end{vmatrix} + 1 \cdot \begin{vmatrix} t^2 & -1 \\ e^t & -3 \end{vmatrix} \] \[ = \cos t((-1) \cdot 5 – 4 \cdot (-3)) – (t^2 \cdot 5 – 4 e^t) + (t^2 \cdot (-3) – (-1) e^t) \] \[ = \cos t(-5 + 12) – (5 t^2 – 4 e^t) + (-3 t^2 + e^t) = 7 \cos t – 5 t^2 + 4 e^t – 3 t^2 + e^t \] \[ = 7 \cos t – 8 t^2 + 5 e^t \] Výpočet \(A_y\): \[ A_y = 1 \cdot \begin{vmatrix} t^2 & 4 \\ e^t & 5 \end{vmatrix} – \cos t \cdot \begin{vmatrix} 2 & 4 \\ 1 & 5 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & t^2 \\ 1 & e^t \end{vmatrix} \] \[ = 1(t^2 \cdot 5 – 4 e^t) – \cos t (2 \cdot 5 – 4 \cdot 1) + (2 e^t – t^2) \] \[ = 5 t^2 – 4 e^t – \cos t (10 – 4) + 2 e^t – t^2 = 5 t^2 – 4 e^t – 6 \cos t + 2 e^t – t^2 \] \[ = 4 t^2 – 2 e^t – 6 \cos t \] Výpočet \(A_z\): \[ A_z = 1 \cdot \begin{vmatrix} -1 & t^2 \\ -3 & e^t \end{vmatrix} – 1 \cdot \begin{vmatrix} 2 & t^2 \\ 1 & e^t \end{vmatrix} + \cos t \cdot \begin{vmatrix} 2 & -1 \\ 1 & -3 \end{vmatrix} \] \[ = 1((-1) e^t – (-3) t^2) – (2 e^t – t^2) + \cos t (2 \cdot (-3) – (-1) \cdot 1) \] \[ = (- e^t + 3 t^2) – 2 e^t + t^2 + \cos t (-6 + 1) = -3 e^t + 4 t^2 – 5 \cos t \] Výsledné řešení: \[ x = \frac{7 \cos t – 8 t^2 + 5 e^t}{-4} = \frac{8 t^2 – 7 \cos t – 5 e^t}{4} \] \[ y = \frac{4 t^2 – 2 e^t – 6 \cos t}{-4} = \frac{-4 t^2 + 2 e^t + 6 \cos t}{4} \] \[ z = \frac{-3 e^t + 4 t^2 – 5 \cos t}{-4} = \frac{3 e^t – 4 t^2 + 5 \cos t}{4} \]

Řešení příkladu 14:

Koeficientová matice: \[ A = \begin{vmatrix} t & 1 & 1 \\ 1 & t & 1 \\ 1 & 1 & t \end{vmatrix} \] Výpočet determinantu: \[ \det(A) = t \begin{vmatrix} t & 1 \\ 1 & t \end{vmatrix} – 1 \begin{vmatrix} 1 & 1 \\ 1 & t \end{vmatrix} + 1 \begin{vmatrix} 1 & t \\ 1 & 1 \end{vmatrix} \] \[ = t(t \cdot t – 1 \cdot 1) – (1 \cdot t – 1 \cdot 1) + (1 \cdot 1 – t \cdot 1) = t(t^2 – 1) – (t – 1) + (1 – t) \] \[ = t^3 – t – t + 1 + 1 – t = t^3 – 3 t + 2 \] Determinanty pro \(x, y, z\): \[ A_x = \begin{vmatrix} e^{2t} & 1 & 1 \\ \sin t & t & 1 \\ \cos t & 1 & t \end{vmatrix}, \quad A_y = \begin{vmatrix} t & e^{2t} & 1 \\ 1 & \sin t & 1 \\ 1 & \cos t & t \end{vmatrix}, \quad A_z = \begin{vmatrix} t & 1 & e^{2t} \\ 1 & t & \sin t \\ 1 & 1 & \cos t \end{vmatrix} \] Výpočet \(A_x\): \[ A_x = e^{2t} \begin{vmatrix} t & 1 \\ 1 & t \end{vmatrix} – 1 \begin{vmatrix} \sin t & 1 \\ \cos t & t \end{vmatrix} + 1 \begin{vmatrix} \sin t & t \\ \cos t & 1 \end{vmatrix} \] \[ = e^{2t}(t \cdot t – 1 \cdot 1) – (\sin t \cdot t – 1 \cdot \cos t) + (\sin t \cdot 1 – t \cdot \cos t) \] \[ = e^{2t}(t^2 – 1) – (t \sin t – \cos t) + (\sin t – t \cos t) \] \[ = e^{2t}(t^2 – 1) – t \sin t + \cos t + \sin t – t \cos t \] Výpočet \(A_y\): \[ A_y = t \begin{vmatrix} \sin t & 1 \\ \cos t & t \end{vmatrix} – e^{2t} \begin{vmatrix} 1 & 1 \\ 1 & t \end{vmatrix} + 1 \begin{vmatrix} 1 & \sin t \\ 1 & \cos t \end{vmatrix} \] \[ = t(\sin t \cdot t – 1 \cdot \cos t) – e^{2t}(1 \cdot t – 1 \cdot 1) + (1 \cdot \cos t – \sin t \cdot 1) \] \[ = t^2 \sin t – t \cos t – e^{2t}(t – 1) + (\cos t – \sin t) \] Výpočet \(A_z\): \[ A_z = t \begin{vmatrix} t & 1 \\ 1 & t \end{vmatrix} – 1 \begin{vmatrix} 1 & \sin t \\ 1 & \cos t \end{vmatrix} + e^{2t} \begin{vmatrix} 1 & t \\ 1 & 1 \end{vmatrix} \] \[ = t(t \cdot t – 1 \cdot 1) – (1 \cdot \cos t – \sin t \cdot 1) + e^{2t}(1 \cdot 1 – t \cdot 1) \] \[ = t(t^2 – 1) – (\cos t – \sin t) + e^{2t}(1 – t) \] \[ = t^3 – t – \cos t + \sin t + e^{2t}(1 – t) \] Výsledné řešení: \[ x = \frac{A_x}{\det(A)} = \frac{e^{2t}(t^2 – 1) – t \sin t + \cos t + \sin t – t \cos t}{t^3 – 3 t + 2} \] \[ y = \frac{A_y}{\det(A)} = \frac{t^2 \sin t – t \cos t – e^{2t}(t – 1) + \cos t – \sin t}{t^3 – 3 t + 2} \] \[ z = \frac{A_z}{\det(A)} = \frac{t^3 – t – \cos t + \sin t + e^{2t}(1 – t)}{t^3 – 3 t + 2} \]

Řešení příkladu 15:

Koeficientová matice: \[ A = \begin{vmatrix} 2 & 3 & -1 \\ -1 & 4 & 2 \\ 3 & -1 & 1 \end{vmatrix} \] Vypočítáme determinant \( \det(A) \): \[ \det(A) = 2 \begin{vmatrix}4 & 2 \\ -1 & 1\end{vmatrix} – 3 \begin{vmatrix}-1 & 2 \\ 3 & 1\end{vmatrix} – 1 \begin{vmatrix}-1 & 4 \\ 3 & -1\end{vmatrix} \] \[ = 2(4 \cdot 1 – 2 \cdot (-1)) – 3(-1 \cdot 1 – 2 \cdot 3) – 1(-1 \cdot (-1) – 4 \cdot 3) \] \[ = 2(4 + 2) – 3(-1 – 6) – 1(1 – 12) = 2 \cdot 6 – 3(-7) – 1(-11) = 12 + 21 + 11 = 44 \] Determinant pro \(x\): \[ A_x = \begin{vmatrix} 5 & 3 & -1 \\ 6 & 4 & 2 \\ 4 & -1 & 1 \end{vmatrix} \] \[ A_x = 5 \begin{vmatrix}4 & 2 \\ -1 & 1\end{vmatrix} – 3 \begin{vmatrix}6 & 2 \\ 4 & 1\end{vmatrix} – 1 \begin{vmatrix}6 & 4 \\ 4 & -1\end{vmatrix} \] \[ = 5(4 \cdot 1 – 2 \cdot (-1)) – 3(6 \cdot 1 – 2 \cdot 4) – 1(6 \cdot (-1) – 4 \cdot 4) \] \[ = 5(4 + 2) – 3(6 – 8) – 1(-6 – 16) = 5 \cdot 6 – 3(-2) – 1(-22) = 30 + 6 + 22 = 58 \] Determinant pro \(y\): \[ A_y = \begin{vmatrix} 2 & 5 & -1 \\ -1 & 6 & 2 \\ 3 & 4 & 1 \end{vmatrix} \] \[ A_y = 2 \begin{vmatrix}6 & 2 \\ 4 & 1\end{vmatrix} – 5 \begin{vmatrix}-1 & 2 \\ 3 & 1\end{vmatrix} – 1 \begin{vmatrix}-1 & 6 \\ 3 & 4\end{vmatrix} \] \[ = 2(6 \cdot 1 – 2 \cdot 4) – 5(-1 \cdot 1 – 2 \cdot 3) – 1(-1 \cdot 4 – 6 \cdot 3) \] \[ = 2(6 – 8) – 5(-1 – 6) – 1(-4 – 18) = 2(-2) – 5(-7) – 1(-22) = -4 + 35 + 22 = 53 \] Determinant pro \(z\): \[ A_z = \begin{vmatrix} 2 & 3 & 5 \\ -1 & 4 & 6 \\ 3 & -1 & 4 \end{vmatrix} \] \[ A_z = 2 \begin{vmatrix}4 & 6 \\ -1 & 4\end{vmatrix} – 3 \begin{vmatrix}-1 & 6 \\ 3 & 4\end{vmatrix} + 5 \begin{vmatrix}-1 & 4 \\ 3 & -1\end{vmatrix} \] \[ = 2(4 \cdot 4 – 6 \cdot (-1)) – 3(-1 \cdot 4 – 6 \cdot 3) + 5(-1 \cdot (-1) – 4 \cdot 3) \] \[ = 2(16 + 6) – 3(-4 – 18) + 5(1 – 12) = 2 \cdot 22 – 3(-22) + 5(-11) = 44 + 66 – 55 = 55 \] Řešení soustavy: \[ x = \frac{A_x}{\det(A)} = \frac{58}{44} = \frac{29}{22} \] \[ y = \frac{A_y}{\det(A)} = \frac{53}{44} \] \[ z = \frac{A_z}{\det(A)} = \frac{55}{44} = \frac{5}{4} \]

Řešení příkladu 16:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 1 & 4 & 2 \end{vmatrix} \] Výpočet determinantu: \[ \det(A) = 1 \begin{vmatrix} -1 & 3 \\ 4 & 2 \end{vmatrix} – 1 \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} + 1 \begin{vmatrix} 2 & -1 \\ 1 & 4 \end{vmatrix} \] \[ = 1(-1 \cdot 2 – 3 \cdot 4) – 1(2 \cdot 2 – 3 \cdot 1) + 1(2 \cdot 4 – (-1) \cdot 1) \] \[ = 1(-2 – 12) – 1(4 – 3) + 1(8 + 1) = -14 – 1 + 9 = -6 \] Determinant pro \(x\): \[ A_x = \begin{vmatrix} 3 & 1 & 1 \\ 9 & -1 & 3 \\ 8 & 4 & 2 \end{vmatrix} \] \[ A_x = 3 \begin{vmatrix} -1 & 3 \\ 4 & 2 \end{vmatrix} – 1 \begin{vmatrix} 9 & 3 \\ 8 & 2 \end{vmatrix} + 1 \begin{vmatrix} 9 & -1 \\ 8 & 4 \end{vmatrix} \] \[ = 3(-1 \cdot 2 – 3 \cdot 4) – 1(9 \cdot 2 – 3 \cdot 8) + 1(9 \cdot 4 – (-1) \cdot 8) \] \[ = 3(-2 – 12) – 1(18 – 24) + 1(36 + 8) = 3(-14) – 1(-6) + 44 = -42 + 6 + 44 = 8 \] Determinant pro \(y\): \[ A_y = \begin{vmatrix} 1 & 3 & 1 \\ 2 & 9 & 3 \\ 1 & 8 & 2 \end{vmatrix} \] \[ A_y = 1 \begin{vmatrix} 9 & 3 \\ 8 & 2 \end{vmatrix} – 3 \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} + 1 \begin{vmatrix} 2 & 9 \\ 1 & 8 \end{vmatrix} \] \[ = 1(9 \cdot 2 – 3 \cdot 8) – 3(2 \cdot 2 – 3 \cdot 1) + 1(2 \cdot 8 – 9 \cdot 1) \] \[ = 1(18 – 24) – 3(4 – 3) + 1(16 – 9) = -6 – 3 + 7 = -2 \] Determinant pro \(z\): \[ A_z = \begin{vmatrix} 1 & 1 & 3 \\ 2 & -1 & 9 \\ 1 & 4 & 8 \end{vmatrix} \] \[ A_z = 1 \begin{vmatrix} -1 & 9 \\ 4 & 8 \end{vmatrix} – 1 \begin{vmatrix} 2 & 9 \\ 1 & 8 \end{vmatrix} + 3 \begin{vmatrix} 2 & -1 \\ 1 & 4 \end{vmatrix} \] \[ = 1(-1 \cdot 8 – 9 \cdot 4) – 1(2 \cdot 8 – 9 \cdot 1) + 3(2 \cdot 4 – (-1) \cdot 1) \] \[ = 1(-8 – 36) – 1(16 – 9) + 3(8 + 1) = -44 – 7 + 27 = -24 \] Řešení soustavy: \[ x = \frac{A_x}{\det(A)} = \frac{8}{-6} = -\frac{4}{3} \] \[ y = \frac{A_y}{\det(A)} = \frac{-2}{-6} = \frac{1}{3} \] \[ z = \frac{A_z}{\det(A)} = \frac{-24}{-6} = 4 \]

Řešení příkladu 17:

Koeficientová matice: \[ A = \begin{vmatrix} 4 & -1 & 2 \\ 1 & 2 & -1 \\ 3 & -1 & 3 \end{vmatrix} \] Výpočet determinantu: \[ \det(A) = 4 \begin{vmatrix} 2 & -1 \\ -1 & 3 \end{vmatrix} – (-1) \begin{vmatrix} 1 & -1 \\ 3 & 3 \end{vmatrix} + 2 \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} \] \[ = 4(2 \cdot 3 – (-1) \cdot (-1)) + 1(1 \cdot 3 – (-1) \cdot 3) + 2(1 \cdot (-1) – 2 \cdot 3) \] \[ = 4(6 – 1) + 1(3 + 3) + 2(-1 – 6) = 4 \cdot 5 + 6 + 2 \cdot (-7) = 20 + 6 – 14 = 12 \] Determinant pro \(x\): \[ A_x = \begin{vmatrix} 7 & -1 & 2 \\ 4 & 2 & -1 \\ 10 & -1 & 3 \end{vmatrix} \] \[ A_x = 7 \begin{vmatrix} 2 & -1 \\ -1 & 3 \end{vmatrix} – (-1) \begin{vmatrix} 4 & -1 \\ 10 & 3 \end{vmatrix} + 2 \begin{vmatrix} 4 & 2 \\ 10 & -1 \end{vmatrix} \] \[ = 7(6 – 1) + 1(4 \cdot 3 – (-1) \cdot 10) + 2(4 \cdot (-1) – 2 \cdot 10) \] \[ = 7 \cdot 5 + 1(12 + 10) + 2(-4 – 20) = 35 + 22 + 2 \cdot (-24) = 35 + 22 – 48 = 9 \] Determinant pro \(y\): \[ A_y = \begin{vmatrix} 4 & 7 & 2 \\ 1 & 4 & -1 \\ 3 & 10 & 3 \end{vmatrix} \] \[ A_y = 4 \begin{vmatrix} 4 & -1 \\ 10 & 3 \end{vmatrix} – 7 \begin{vmatrix} 1 & -1 \\ 3 & 3 \end{vmatrix} + 2 \begin{vmatrix} 1 & 4 \\ 3 & 10 \end{vmatrix} \] \[ = 4(4 \cdot 3 – (-1) \cdot 10) – 7(1 \cdot 3 – (-1) \cdot 3) + 2(1 \cdot 10 – 4 \cdot 3) \] \[ = 4(12 + 10) – 7(3 + 3) + 2(10 – 12) = 4 \cdot 22 – 7 \cdot 6 + 2 \cdot (-2) = 88 – 42 – 4 = 42 \] Determinant pro \(z\): \[ A_z = \begin{vmatrix} 4 & -1 & 7 \\ 1 & 2 & 4 \\ 3 & -1 & 10 \end{vmatrix} \] \[ A_z = 4 \begin{vmatrix} 2 & 4 \\ -1 & 10 \end{vmatrix} – (-1) \begin{vmatrix} 1 & 4 \\ 3 & 10 \end{vmatrix} + 7 \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} \] \[ = 4(2 \cdot 10 – 4 \cdot (-1)) + 1(1 \cdot 10 – 4 \cdot 3) + 7(1 \cdot (-1) – 2 \cdot 3) \] \[ = 4(20 + 4) + 1(10 – 12) + 7(-1 – 6) = 4 \cdot 24 – 2 + 7 \cdot (-7) = 96 – 2 – 49 = 45 \] Řešení soustavy: \[ x = \frac{A_x}{\det(A)} = \frac{9}{12} = \frac{3}{4} \] \[ y = \frac{A_y}{\det(A)} = \frac{42}{12} = \frac{7}{2} \] \[ z = \frac{A_z}{\det(A)} = \frac{45}{12} = \frac{15}{4} \]

Řešení příkladu 18:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & 2 & 3 \\ 2 & -1 & 1 \\ 4 & 1 & -2 \end{vmatrix} \] Výpočet determinantu: \[ \det(A) = 1 \begin{vmatrix} -1 & 1 \\ 1 & -2 \end{vmatrix} – 2 \begin{vmatrix} 2 & 1 \\ 4 & -2 \end{vmatrix} + 3 \begin{vmatrix} 2 & -1 \\ 4 & 1 \end{vmatrix} \] \[ = 1(-1 \cdot (-2) – 1 \cdot 1) – 2(2 \cdot (-2) – 1 \cdot 4) + 3(2 \cdot 1 – (-1) \cdot 4) \] \[ = 1(2 – 1) – 2(-4 – 4) + 3(2 + 4) = 1 – 2(-8) + 3 \cdot 6 = 1 + 16 + 18 = 35 \] Determinant pro \(x\): \[ A_x = \begin{vmatrix} 14 & 2 & 3 \\ 3 & -1 & 1 \\ 1 & 1 & -2 \end{vmatrix} \] \[ A_x = 14 \begin{vmatrix} -1 & 1 \\ 1 & -2 \end{vmatrix} – 2 \begin{vmatrix} 3 & 1 \\ 1 & -2 \end{vmatrix} + 3 \begin{vmatrix} 3 & -1 \\ 1 & 1 \end{vmatrix} \] \[ = 14(-1 \cdot (-2) – 1 \cdot 1) – 2(3 \cdot (-2) – 1 \cdot 1) + 3(3 \cdot 1 – (-1) \cdot 1) \] \[ = 14(2 – 1) – 2(-6 – 1) + 3(3 + 1) = 14 – 2(-7) + 3 \cdot 4 = 14 + 14 + 12 = 40 \] Determinant pro \(y\): \[ A_y = \begin{vmatrix} 1 & 14 & 3 \\ 2 & 3 & 1 \\ 4 & 1 & -2 \end{vmatrix} \] \[ A_y = 1 \begin{vmatrix} 3 & 1 \\ 1 & -2 \end{vmatrix} – 14 \begin{vmatrix} 2 & 1 \\ 4 & -2 \end{vmatrix} + 3 \begin{vmatrix} 2 & 3 \\ 4 & 1 \end{vmatrix} \] \[ = 1(3 \cdot (-2) – 1 \cdot 1) – 14(2 \cdot (-2) – 1 \cdot 4) + 3(2 \cdot 1 – 3 \cdot 4) \] \[ = 1(-6 – 1) – 14(-4 – 4) + 3(2 – 12) = -7 + 14 \cdot 8 + 3(-10) = -7 + 112 – 30 = 75 \] Determinant pro \(z\): \[ A_z = \begin{vmatrix} 1 & 2 & 14 \\ 2 & -1 & 3 \\ 4 & 1 & 1 \end{vmatrix} \] \[ A_z = 1 \begin{vmatrix} -1 & 3 \\ 1 & 1 \end{vmatrix} – 2 \begin{vmatrix} 2 & 3 \\ 4 & 1 \end{vmatrix} + 14 \begin{vmatrix} 2 & -1 \\ 4 & 1 \end{vmatrix} \] \[ = 1(-1 \cdot 1 – 3 \cdot 1) – 2(2 \cdot 1 – 3 \cdot 4) + 14(2 \cdot 1 – (-1) \cdot 4) \] \[ = 1(-1 – 3) – 2(2 – 12) + 14(2 + 4) = -4 – 2(-10) + 14 \cdot 6 = -4 + 20 + 84 = 100 \] Řešení soustavy: \[ x = \frac{A_x}{\det(A)} = \frac{40}{35} = \frac{8}{7} \] \[ y = \frac{A_y}{\det(A)} = \frac{75}{35} = \frac{15}{7} \] \[ z = \frac{A_z}{\det(A)} = \frac{100}{35} = \frac{20}{7} \]

Řešení příkladu 19:

Koeficientová matice: \[ A = \begin{vmatrix} 3 & 1 & -2 \\ 1 & -4 & 1 \\ 2 & 3 & 5 \end{vmatrix} \] Výpočet determinantu: \[ \det(A) = 3 \begin{vmatrix} -4 & 1 \\ 3 & 5 \end{vmatrix} – 1 \begin{vmatrix} 1 & 1 \\ 2 & 5 \end{vmatrix} + (-2) \begin{vmatrix} 1 & -4 \\ 2 & 3 \end{vmatrix} \] \[ = 3(-4 \cdot 5 – 1 \cdot 3) – 1(1 \cdot 5 – 1 \cdot 2) – 2(1 \cdot 3 – (-4) \cdot 2) \] \[ = 3(-20 – 3) – (5 – 2) – 2(3 + 8) = 3(-23) – 3 – 2 \cdot 11 = -69 – 3 – 22 = -94 \] Determinant pro \(x\): \[ A_x = \begin{vmatrix} 5 & 1 & -2 \\ -2 & -4 & 1 \\ 12 & 3 & 5 \end{vmatrix} \] \[ A_x = 5 \begin{vmatrix} -4 & 1 \\ 3 & 5 \end{vmatrix} – 1 \begin{vmatrix} -2 & 1 \\ 12 & 5 \end{vmatrix} + (-2) \begin{vmatrix} -2 & -4 \\ 12 & 3 \end{vmatrix} \] \[ = 5(-4 \cdot 5 – 1 \cdot 3) – 1(-2 \cdot 5 – 1 \cdot 12) – 2(-2 \cdot 3 – (-4) \cdot 12) \] \[ = 5(-20 – 3) – (-10 – 12) – 2(-6 + 48) = 5(-23) + 22 – 2 \cdot 42 = -115 + 22 – 84 = -177 \] Determinant pro \(y\): \[ A_y = \begin{vmatrix} 3 & 5 & -2 \\ 1 & -2 & 1 \\ 2 & 12 & 5 \end{vmatrix} \] \[ A_y = 3 \begin{vmatrix} -2 & 1 \\ 12 & 5 \end{vmatrix} – 5 \begin{vmatrix} 1 & 1 \\ 2 & 5 \end{vmatrix} + (-2) \begin{vmatrix} 1 & -2 \\ 2 & 12 \end{vmatrix} \] \[ = 3(-2 \cdot 5 – 1 \cdot 12) – 5(1 \cdot 5 – 1 \cdot 2) – 2(1 \cdot 12 – (-2) \cdot 2) \] \[ = 3(-10 – 12) – 5(5 – 2) – 2(12 + 4) = 3(-22) – 5 \cdot 3 – 2 \cdot 16 = -66 – 15 – 32 = -113 \] Determinant pro \(z\): \[ A_z = \begin{vmatrix} 3 & 1 & 5 \\ 1 & -4 & -2 \\ 2 & 3 & 12 \end{vmatrix} \] \[ A_z = 3 \begin{vmatrix} -4 & -2 \\ 3 & 12 \end{vmatrix} – 1 \begin{vmatrix} 1 & -2 \\ 2 & 12 \end{vmatrix} + 5 \begin{vmatrix} 1 & -4 \\ 2 & 3 \end{vmatrix} \] \[ = 3(-4 \cdot 12 – (-2) \cdot 3) – 1(1 \cdot 12 – (-2) \cdot 2) + 5(1 \cdot 3 – (-4) \cdot 2) \] \[ = 3(-48 + 6) – (12 + 4) + 5(3 + 8) = 3(-42) – 16 + 5 \cdot 11 = -126 – 16 + 55 = -87 \] Řešení soustavy: \[ x = \frac{A_x}{\det(A)} = \frac{-177}{-94} = \frac{177}{94} \] \[ y = \frac{A_y}{\det(A)} = \frac{-113}{-94} = \frac{113}{94} \] \[ z = \frac{A_z}{\det(A)} = \frac{-87}{-94} = \frac{87}{94} \]

Řešení příkladu 20:

Koeficientová matice: \[ A = \begin{vmatrix} 4 & -1 & 2 \\ -2 & 3 & -1 \\ 1 & 1 & 1 \end{vmatrix} \] Výpočet determinantu: \[ \det(A) = 4 \begin{vmatrix} 3 & -1 \\ 1 & 1 \end{vmatrix} – (-1) \begin{vmatrix} -2 & -1 \\ 1 & 1 \end{vmatrix} + 2 \begin{vmatrix} -2 & 3 \\ 1 & 1 \end{vmatrix} \] \[ = 4(3 \cdot 1 – (-1) \cdot 1) + 1(-2 \cdot 1 – (-1) \cdot 1) + 2(-2 \cdot 1 – 3 \cdot 1) \] \[ = 4(3 + 1) + 1(-2 + 1) + 2(-2 – 3) = 4 \cdot 4 – 1 + 2 \cdot (-5) = 16 – 1 – 10 = 5 \] Determinant pro \(x\): \[ A_x = \begin{vmatrix} 7 & -1 & 2 \\ -1 & 3 & -1 \\ 4 & 1 & 1 \end{vmatrix} \] \[ A_x = 7 \begin{vmatrix} 3 & -1 \\ 1 & 1 \end{vmatrix} – (-1) \begin{vmatrix} -1 & -1 \\ 4 & 1 \end{vmatrix} + 2 \begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix} \] \[ = 7(3 \cdot 1 – (-1) \cdot 1) + 1(-1 \cdot 1 – (-1) \cdot 4) + 2(-1 \cdot 1 – 3 \cdot 4) \] \[ = 7(3 + 1) + 1(-1 + 4) + 2(-1 – 12) = 7 \cdot 4 + 3 + 2 \cdot (-13) = 28 + 3 – 26 = 5 \] Determinant pro \(y\): \[ A_y = \begin{vmatrix} 4 & 7 & 2 \\ -2 & -1 & -1 \\ 1 & 4 & 1 \end{vmatrix} \] \[ A_y = 4 \begin{vmatrix} -1 & -1 \\ 4 & 1 \end{vmatrix} – 7 \begin{vmatrix} -2 & -1 \\ 1 & 1 \end{vmatrix} + 2 \begin{vmatrix} -2 & -1 \\ 1 & 4 \end{vmatrix} \] \[ = 4(-1 \cdot 1 – (-1) \cdot 4) – 7(-2 \cdot 1 – (-1) \cdot 1) + 2(-2 \cdot 4 – (-1) \cdot 1) \] \[ = 4(-1 + 4) – 7(-2 + 1) + 2(-8 + 1) = 4 \cdot 3 – 7 \cdot (-1) + 2 \cdot (-7) = 12 + 7 – 14 = 5 \] Determinant pro \(z\): \[ A_z = \begin{vmatrix} 4 & -1 & 7 \\ -2 & 3 & -1 \\ 1 & 1 & 4 \end{vmatrix} \] \[ A_z = 4 \begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} – (-1) \begin{vmatrix} -2 & -1 \\ 1 & 4 \end{vmatrix} + 7 \begin{vmatrix} -2 & 3 \\ 1 & 1 \end{vmatrix} \] \[ = 4(3 \cdot 4 – (-1) \cdot 1) + 1(-2 \cdot 4 – (-1) \cdot 1) + 7(-2 \cdot 1 – 3 \cdot 1) \] \[ = 4(12 + 1) + 1(-8 + 1) + 7(-2 – 3) = 4 \cdot 13 – 7 – 35 = 52 – 7 – 35 = 10 \] Řešení soustavy: \[ x = \frac{A_x}{\det(A)} = \frac{5}{5} = 1 \] \[ y = \frac{A_y}{\det(A)} = \frac{5}{5} = 1 \] \[ z = \frac{A_z}{\det(A)} = \frac{10}{5} = 2 \]

Řešení příkladu 21:

Koeficientová matice: \[ A = \begin{vmatrix} 2 & 3 & -1 \\ -1 & 4 & 2 \\ 3 & -1 & 1 \end{vmatrix} \] Výpočet determinantu: \[ \det(A) = 2 \begin{vmatrix} 4 & 2 \\ -1 & 1 \end{vmatrix} – 3 \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} + (-1) \begin{vmatrix} -1 & 4 \\ 3 & -1 \end{vmatrix} \] \[ = 2(4 \cdot 1 – 2 \cdot (-1)) – 3(-1 \cdot 1 – 2 \cdot 3) – 1(-1 \cdot (-1) – 4 \cdot 3) \] \[ = 2(4 + 2) – 3(-1 – 6) – 1(1 – 12) = 2 \cdot 6 + 3 \cdot 7 – ( -11 ) = 12 + 21 + 11 = 44 \] Determinant pro \(x\): \[ A_x = \begin{vmatrix} 4 & 3 & -1 \\ 1 & 4 & 2 \\ 5 & -1 & 1 \end{vmatrix} \] \[ A_x = 4 \begin{vmatrix} 4 & 2 \\ -1 & 1 \end{vmatrix} – 3 \begin{vmatrix} 1 & 2 \\ 5 & 1 \end{vmatrix} + (-1) \begin{vmatrix} 1 & 4 \\ 5 & -1 \end{vmatrix} \] \[ = 4(4 \cdot 1 – 2 \cdot (-1)) – 3(1 \cdot 1 – 2 \cdot 5) – 1(1 \cdot (-1) – 4 \cdot 5) \] \[ = 4(4 + 2) – 3(1 – 10) – 1(-1 – 20) = 4 \cdot 6 – 3(-9) – (-21) = 24 + 27 + 21 = 72 \] Determinant pro \(y\): \[ A_y = \begin{vmatrix} 2 & 4 & -1 \\ -1 & 1 & 2 \\ 3 & 5 & 1 \end{vmatrix} \] \[ A_y = 2 \begin{vmatrix} 1 & 2 \\ 5 & 1 \end{vmatrix} – 4 \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} + (-1) \begin{vmatrix} -1 & 1 \\ 3 & 5 \end{vmatrix} \] \[ = 2(1 \cdot 1 – 2 \cdot 5) – 4(-1 \cdot 1 – 2 \cdot 3) – 1(-1 \cdot 5 – 1 \cdot 3) \] \[ = 2(1 – 10) – 4(-1 – 6) – 1(-5 – 3) = 2(-9) – 4(-7) – (-8) = -18 + 28 + 8 = 18 \] Determinant pro \(z\): \[ A_z = \begin{vmatrix} 2 & 3 & 4 \\ -1 & 4 & 1 \\ 3 & -1 & 5 \end{vmatrix} \] \[ A_z = 2 \begin{vmatrix} 4 & 1 \\ -1 & 5 \end{vmatrix} – 3 \begin{vmatrix} -1 & 1 \\ 3 & 5 \end{vmatrix} + 4 \begin{vmatrix} -1 & 4 \\ 3 & -1 \end{vmatrix} \] \[ = 2(4 \cdot 5 – 1 \cdot (-1)) – 3(-1 \cdot 5 – 1 \cdot 3) + 4(-1 \cdot (-1) – 4 \cdot 3) \] \[ = 2(20 + 1) – 3(-5 – 3) + 4(1 – 12) = 2 \cdot 21 + 3 \cdot 8 + 4(-11) = 42 + 24 – 44 = 22 \] Řešení soustavy: \[ x = \frac{A_x}{\det(A)} = \frac{72}{44} = \frac{18}{11} \] \[ y = \frac{A_y}{\det(A)} = \frac{18}{44} = \frac{9}{22} \] \[ z = \frac{A_z}{\det(A)} = \frac{22}{44} = \frac{1}{2} \]

Řešení příkladu 22:

Koeficientová matice soustavy je \[ A = \begin{vmatrix} 2 & 3 & -1 \\ -1 & 4 & 2 \\ 3 & -1 & 1 \end{vmatrix}. \] Spočítáme determinant matice \(A\): \[ \det(A) = 2 \begin{vmatrix}4 & 2 \\ -1 & 1 \end{vmatrix} – 3 \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} + (-1) \begin{vmatrix} -1 & 4 \\ 3 & -1 \end{vmatrix}. \] Vypočteme jednotlivé determinanty menších matic: \[ \begin{vmatrix}4 & 2 \\ -1 & 1 \end{vmatrix} = 4 \cdot 1 – 2 \cdot (-1) = 4 + 2 = 6, \] \[ \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} = -1 \cdot 1 – 2 \cdot 3 = -1 – 6 = -7, \] \[ \begin{vmatrix} -1 & 4 \\ 3 & -1 \end{vmatrix} = -1 \cdot (-1) – 4 \cdot 3 = 1 – 12 = -11. \] Dosadíme zpět: \[ \det(A) = 2 \cdot 6 – 3 \cdot (-7) – 1 \cdot (-11) = 12 + 21 + 11 = 44 \neq 0. \] Determinant není nulový, soustava má tedy jednoznačné řešení. Pro \(x\) vytvoříme matici \(A_x\), kde první sloupec nahradíme pravými stranami: \[ A_x = \begin{vmatrix} 5 & 3 & -1 \\ 6 & 4 & 2 \\ 4 & -1 & 1 \end{vmatrix}. \] Spočítáme determinant: \[ \det(A_x) = 5 \begin{vmatrix}4 & 2 \\ -1 & 1 \end{vmatrix} – 3 \begin{vmatrix}6 & 2 \\ 4 & 1 \end{vmatrix} + (-1) \begin{vmatrix}6 & 4 \\ 4 & -1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix}4 & 2 \\ -1 & 1 \end{vmatrix} = 4 \cdot 1 – 2 \cdot (-1) = 6, \] \[ \begin{vmatrix}6 & 2 \\ 4 & 1 \end{vmatrix} = 6 \cdot 1 – 2 \cdot 4 = 6 – 8 = -2, \] \[ \begin{vmatrix}6 & 4 \\ 4 & -1 \end{vmatrix} = 6 \cdot (-1) – 4 \cdot 4 = -6 – 16 = -22. \] Dosadíme: \[ \det(A_x) = 5 \cdot 6 – 3 \cdot (-2) – 1 \cdot (-22) = 30 + 6 + 22 = 58. \] Podobně pro \(y\) nahradíme druhý sloupec: \[ A_y = \begin{vmatrix} 2 & 5 & -1 \\ -1 & 6 & 2 \\ 3 & 4 & 1 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 2 \begin{vmatrix}6 & 2 \\ 4 & 1 \end{vmatrix} – 5 \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} + (-1) \begin{vmatrix} -1 & 6 \\ 3 & 4 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix}6 & 2 \\ 4 & 1 \end{vmatrix} = 6 \cdot 1 – 2 \cdot 4 = 6 – 8 = -2, \] \[ \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} = -1 – 6 = -7, \] \[ \begin{vmatrix} -1 & 6 \\ 3 & 4 \end{vmatrix} = -1 \cdot 4 – 6 \cdot 3 = -4 – 18 = -22. \] Dosadíme: \[ \det(A_y) = 2 \cdot (-2) – 5 \cdot (-7) – 1 \cdot (-22) = -4 + 35 + 22 = 53. \] Nakonec pro \(z\) nahradíme třetí sloupec: \[ A_z = \begin{vmatrix} 2 & 3 & 5 \\ -1 & 4 & 6 \\ 3 & -1 & 4 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 2 \begin{vmatrix}4 & 6 \\ -1 & 4 \end{vmatrix} – 3 \begin{vmatrix} -1 & 6 \\ 3 & 4 \end{vmatrix} + 5 \begin{vmatrix} -1 & 4 \\ 3 & -1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix}4 & 6 \\ -1 & 4 \end{vmatrix} = 4 \cdot 4 – 6 \cdot (-1) = 16 + 6 = 22, \] \[ \begin{vmatrix} -1 & 6 \\ 3 & 4 \end{vmatrix} = -1 \cdot 4 – 6 \cdot 3 = -4 – 18 = -22, \] \[ \begin{vmatrix} -1 & 4 \\ 3 & -1 \end{vmatrix} = -1 \cdot (-1) – 4 \cdot 3 = 1 – 12 = -11. \] Dosadíme: \[ \det(A_z) = 2 \cdot 22 – 3 \cdot (-22) + 5 \cdot (-11) = 44 + 66 – 55 = 55. \] Výsledky dosadíme do vzorců Cramerova pravidla: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{58}{44} = \frac{29}{22}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{53}{44}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{55}{44} = \frac{5}{4}. \]

Řešení příkladu 22:

Koeficientová matice soustavy je \[ A = \begin{vmatrix} 2 & 3 & -1 \\ -1 & 4 & 2 \\ 3 & -1 & 1 \end{vmatrix}. \] Spočítáme determinant matice \(A\): \[ \det(A) = 2 \begin{vmatrix}4 & 2 \\ -1 & 1 \end{vmatrix} – 3 \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} + (-1) \begin{vmatrix} -1 & 4 \\ 3 & -1 \end{vmatrix}. \] Vypočteme jednotlivé determinanty menších matic: \[ \begin{vmatrix}4 & 2 \\ -1 & 1 \end{vmatrix} = 4 \cdot 1 – 2 \cdot (-1) = 4 + 2 = 6, \] \[ \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} = -1 \cdot 1 – 2 \cdot 3 = -1 – 6 = -7, \] \[ \begin{vmatrix} -1 & 4 \\ 3 & -1 \end{vmatrix} = -1 \cdot (-1) – 4 \cdot 3 = 1 – 12 = -11. \] Dosadíme zpět: \[ \det(A) = 2 \cdot 6 – 3 \cdot (-7) – 1 \cdot (-11) = 12 + 21 + 11 = 44 \neq 0. \] Determinant není nulový, soustava má tedy jednoznačné řešení. Pro \(x\) vytvoříme matici \(A_x\), kde první sloupec nahradíme pravými stranami: \[ A_x = \begin{vmatrix} 5 & 3 & -1 \\ 6 & 4 & 2 \\ 4 & -1 & 1 \end{vmatrix}. \] Spočítáme determinant: \[ \det(A_x) = 5 \begin{vmatrix}4 & 2 \\ -1 & 1 \end{vmatrix} – 3 \begin{vmatrix}6 & 2 \\ 4 & 1 \end{vmatrix} + (-1) \begin{vmatrix}6 & 4 \\ 4 & -1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix}4 & 2 \\ -1 & 1 \end{vmatrix} = 4 \cdot 1 – 2 \cdot (-1) = 6, \] \[ \begin{vmatrix}6 & 2 \\ 4 & 1 \end{vmatrix} = 6 \cdot 1 – 2 \cdot 4 = 6 – 8 = -2, \] \[ \begin{vmatrix}6 & 4 \\ 4 & -1 \end{vmatrix} = 6 \cdot (-1) – 4 \cdot 4 = -6 – 16 = -22. \] Dosadíme: \[ \det(A_x) = 5 \cdot 6 – 3 \cdot (-2) – 1 \cdot (-22) = 30 + 6 + 22 = 58. \] Podobně pro \(y\) nahradíme druhý sloupec: \[ A_y = \begin{vmatrix} 2 & 5 & -1 \\ -1 & 6 & 2 \\ 3 & 4 & 1 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 2 \begin{vmatrix}6 & 2 \\ 4 & 1 \end{vmatrix} – 5 \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} + (-1) \begin{vmatrix} -1 & 6 \\ 3 & 4 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix}6 & 2 \\ 4 & 1 \end{vmatrix} = 6 \cdot 1 – 2 \cdot 4 = 6 – 8 = -2, \] \[ \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} = -1 – 6 = -7, \] \[ \begin{vmatrix} -1 & 6 \\ 3 & 4 \end{vmatrix} = -1 \cdot 4 – 6 \cdot 3 = -4 – 18 = -22. \] Dosadíme: \[ \det(A_y) = 2 \cdot (-2) – 5 \cdot (-7) – 1 \cdot (-22) = -4 + 35 + 22 = 53. \] Nakonec pro \(z\) nahradíme třetí sloupec: \[ A_z = \begin{vmatrix} 2 & 3 & 5 \\ -1 & 4 & 6 \\ 3 & -1 & 4 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 2 \begin{vmatrix}4 & 6 \\ -1 & 4 \end{vmatrix} – 3 \begin{vmatrix} -1 & 6 \\ 3 & 4 \end{vmatrix} + 5 \begin{vmatrix} -1 & 4 \\ 3 & -1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix}4 & 6 \\ -1 & 4 \end{vmatrix} = 4 \cdot 4 – 6 \cdot (-1) = 16 + 6 = 22, \] \[ \begin{vmatrix} -1 & 6 \\ 3 & 4 \end{vmatrix} = -1 \cdot 4 – 6 \cdot 3 = -4 – 18 = -22, \] \[ \begin{vmatrix} -1 & 4 \\ 3 & -1 \end{vmatrix} = -1 \cdot (-1) – 4 \cdot 3 = 1 – 12 = -11. \] Dosadíme: \[ \det(A_z) = 2 \cdot 22 – 3 \cdot (-22) + 5 \cdot (-11) = 44 + 66 – 55 = 55. \] Výsledky dosadíme do vzorců Cramerova pravidla: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{58}{44} = \frac{29}{22}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{53}{44}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{55}{44} = \frac{5}{4}. \]

Řešení příkladu 23:

Koeficientová matice soustavy je \[ A = \begin{vmatrix} 1 & 2 & -1 \\ 3 & -1 & 4 \\ 2 & 1 & 3 \end{vmatrix}. \] Spočítáme determinant matice \(A\): \[ \det(A) = 1 \cdot \begin{vmatrix} -1 & 4 \\ 1 & 3 \end{vmatrix} – 2 \cdot \begin{vmatrix} 3 & 4 \\ 2 & 3 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix}. \] Vypočteme determinanty menších matic: \[ \begin{vmatrix} -1 & 4 \\ 1 & 3 \end{vmatrix} = (-1)(3) – 4(1) = -3 – 4 = -7, \] \[ \begin{vmatrix} 3 & 4 \\ 2 & 3 \end{vmatrix} = 3 \cdot 3 – 4 \cdot 2 = 9 – 8 = 1, \] \[ \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} = 3 \cdot 1 – (-1) \cdot 2 = 3 + 2 = 5. \] Dosadíme do výrazu pro determinant: \[ \det(A) = 1 \cdot (-7) – 2 \cdot 1 – 1 \cdot 5 = -7 – 2 – 5 = -14 \neq 0. \] Determinant není nulový, soustava má jednoznačné řešení. Vytvoříme matice pro výpočet neznámých: \[ A_x = \begin{vmatrix} 4 & 2 & -1 \\ 10 & -1 & 4 \\ 9 & 1 & 3 \end{vmatrix}, \quad A_y = \begin{vmatrix} 1 & 4 & -1 \\ 3 & 10 & 4 \\ 2 & 9 & 3 \end{vmatrix}, \quad A_z = \begin{vmatrix} 1 & 2 & 4 \\ 3 & -1 & 10 \\ 2 & 1 & 9 \end{vmatrix}. \] Spočítáme determinant \(A_x\): \[ \det(A_x) = 4 \cdot \begin{vmatrix} -1 & 4 \\ 1 & 3 \end{vmatrix} – 2 \cdot \begin{vmatrix} 10 & 4 \\ 9 & 3 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 10 & -1 \\ 9 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 4 \\ 1 & 3 \end{vmatrix} = -1 \cdot 3 – 4 \cdot 1 = -3 – 4 = -7, \] \[ \begin{vmatrix} 10 & 4 \\ 9 & 3 \end{vmatrix} = 10 \cdot 3 – 4 \cdot 9 = 30 – 36 = -6, \] \[ \begin{vmatrix} 10 & -1 \\ 9 & 1 \end{vmatrix} = 10 \cdot 1 – (-1) \cdot 9 = 10 + 9 = 19. \] Dosadíme: \[ \det(A_x) = 4 \cdot (-7) – 2 \cdot (-6) – 1 \cdot 19 = -28 + 12 – 19 = -35. \] Spočítáme determinant \(A_y\): \[ \det(A_y) = 1 \cdot \begin{vmatrix} 10 & 4 \\ 9 & 3 \end{vmatrix} – 4 \cdot \begin{vmatrix} 3 & 4 \\ 2 & 3 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 3 & 10 \\ 2 & 9 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 10 & 4 \\ 9 & 3 \end{vmatrix} = 10 \cdot 3 – 4 \cdot 9 = 30 – 36 = -6, \] \[ \begin{vmatrix} 3 & 4 \\ 2 & 3 \end{vmatrix} = 3 \cdot 3 – 4 \cdot 2 = 9 – 8 = 1, \] \[ \begin{vmatrix} 3 & 10 \\ 2 & 9 \end{vmatrix} = 3 \cdot 9 – 10 \cdot 2 = 27 – 20 = 7. \] Dosadíme: \[ \det(A_y) = 1 \cdot (-6) – 4 \cdot 1 – 1 \cdot 7 = -6 – 4 – 7 = -17. \] Spočítáme determinant \(A_z\): \[ \det(A_z) = 1 \cdot \begin{vmatrix} -1 & 10 \\ 1 & 9 \end{vmatrix} – 2 \cdot \begin{vmatrix} 3 & 10 \\ 2 & 9 \end{vmatrix} + 4 \cdot \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 10 \\ 1 & 9 \end{vmatrix} = (-1) \cdot 9 – 10 \cdot 1 = -9 – 10 = -19, \] \[ \begin{vmatrix} 3 & 10 \\ 2 & 9 \end{vmatrix} = 3 \cdot 9 – 10 \cdot 2 = 27 – 20 = 7, \] \[ \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} = 3 \cdot 1 – (-1) \cdot 2 = 3 + 2 = 5. \] Dosadíme: \[ \det(A_z) = 1 \cdot (-19) – 2 \cdot 7 + 4 \cdot 5 = -19 – 14 + 20 = -13. \] Nakonec: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-35}{-14} = \frac{35}{14} = \frac{5}{2}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-17}{-14} = \frac{17}{14}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{-13}{-14} = \frac{13}{14}. \]

Řešení příkladu 24:

Koeficientová matice: \[ A = \begin{vmatrix} 4 & -1 & 1 \\ -1 & 5 & -2 \\ 2 & 3 & 4 \end{vmatrix}. \] Determinant matice \(A\): \[ \det(A) = 4 \begin{vmatrix}5 & -2 \\ 3 & 4 \end{vmatrix} + 1 \begin{vmatrix} -1 & -2 \\ 2 & 4 \end{vmatrix} + 1 \begin{vmatrix} -1 & 5 \\ 2 & 3 \end{vmatrix}. \] Výpočty determinantů: \[ \begin{vmatrix}5 & -2 \\ 3 & 4 \end{vmatrix} = 5 \cdot 4 – (-2) \cdot 3 = 20 + 6 = 26, \] \[ \begin{vmatrix} -1 & -2 \\ 2 & 4 \end{vmatrix} = (-1) \cdot 4 – (-2) \cdot 2 = -4 + 4 = 0, \] \[ \begin{vmatrix} -1 & 5 \\ 2 & 3 \end{vmatrix} = (-1) \cdot 3 – 5 \cdot 2 = -3 – 10 = -13. \] Dosadíme: \[ \det(A) = 4 \cdot 26 + 1 \cdot 0 + 1 \cdot (-13) = 104 + 0 – 13 = 91 \neq 0. \] Matice pro \(x\): \[ A_x = \begin{vmatrix} 7 & -1 & 1 \\ -3 & 5 & -2 \\ 20 & 3 & 4 \end{vmatrix}, \] determinant: \[ \det(A_x) = 7 \begin{vmatrix}5 & -2 \\ 3 & 4 \end{vmatrix} + 1 \begin{vmatrix} -3 & -2 \\ 20 & 4 \end{vmatrix} + 1 \begin{vmatrix} -3 & 5 \\ 20 & 3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix}5 & -2 \\ 3 & 4 \end{vmatrix} = 26, \] \[ \begin{vmatrix} -3 & -2 \\ 20 & 4 \end{vmatrix} = (-3) \cdot 4 – (-2) \cdot 20 = -12 + 40 = 28, \] \[ \begin{vmatrix} -3 & 5 \\ 20 & 3 \end{vmatrix} = (-3) \cdot 3 – 5 \cdot 20 = -9 – 100 = -109. \] Dosadíme: \[ \det(A_x) = 7 \cdot 26 + 1 \cdot 28 + 1 \cdot (-109) = 182 + 28 – 109 = 101. \] Matice pro \(y\): \[ A_y = \begin{vmatrix} 4 & 7 & 1 \\ -1 & -3 & -2 \\ 2 & 20 & 4 \end{vmatrix}, \] determinant: \[ \det(A_y) = 4 \begin{vmatrix} -3 & -2 \\ 20 & 4 \end{vmatrix} – 7 \begin{vmatrix} -1 & -2 \\ 2 & 4 \end{vmatrix} + 1 \begin{vmatrix} -1 & -3 \\ 2 & 20 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -3 & -2 \\ 20 & 4 \end{vmatrix} = 28, \] \[ \begin{vmatrix} -1 & -2 \\ 2 & 4 \end{vmatrix} = 0, \] \[ \begin{vmatrix} -1 & -3 \\ 2 & 20 \end{vmatrix} = (-1) \cdot 20 – (-3) \cdot 2 = -20 + 6 = -14. \] Dosadíme: \[ \det(A_y) = 4 \cdot 28 – 7 \cdot 0 + 1 \cdot (-14) = 112 + 0 – 14 = 98. \] Matice pro \(z\): \[ A_z = \begin{vmatrix} 4 & -1 & 7 \\ -1 & 5 & -3 \\ 2 & 3 & 20 \end{vmatrix}, \] determinant: \[ \det(A_z) = 4 \begin{vmatrix} 5 & -3 \\ 3 & 20 \end{vmatrix} – (-1) \begin{vmatrix} -1 & -3 \\ 2 & 20 \end{vmatrix} + 7 \begin{vmatrix} -1 & 5 \\ 2 & 3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 5 & -3 \\ 3 & 20 \end{vmatrix} = 5 \cdot 20 – (-3) \cdot 3 = 100 + 9 = 109, \] \[ \begin{vmatrix} -1 & -3 \\ 2 & 20 \end{vmatrix} = -14, \] \[ \begin{vmatrix} -1 & 5 \\ 2 & 3 \end{vmatrix} = -1 \cdot 3 – 5 \cdot 2 = -3 – 10 = -13. \] Dosadíme: \[ \det(A_z) = 4 \cdot 109 + 1 \cdot (-14) + 7 \cdot (-13) = 436 – 14 – 91 = 331. \] Výsledky: \[ x = \frac{101}{91}, \quad y = \frac{98}{91}, \quad z = \frac{331}{91}. \]

Řešení příkladu 25:

Koeficientová matice: \[ A = \begin{vmatrix} 2 & 1 & 3 \\ -1 & 4 & -1 \\ 5 & -2 & 2 \end{vmatrix}. \] Determinant: \[ \det(A) = 2 \begin{vmatrix}4 & -1 \\ -2 & 2 \end{vmatrix} – 1 \begin{vmatrix} -1 & -1 \\ 5 & 2 \end{vmatrix} + 3 \begin{vmatrix} -1 & 4 \\ 5 & -2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix}4 & -1 \\ -2 & 2 \end{vmatrix} = 4 \cdot 2 – (-1) \cdot (-2) = 8 – 2 = 6, \] \[ \begin{vmatrix} -1 & -1 \\ 5 & 2 \end{vmatrix} = (-1) \cdot 2 – (-1) \cdot 5 = -2 + 5 = 3, \] \[ \begin{vmatrix} -1 & 4 \\ 5 & -2 \end{vmatrix} = (-1) \cdot (-2) – 4 \cdot 5 = 2 – 20 = -18. \] Dosadíme: \[ \det(A) = 2 \cdot 6 – 1 \cdot 3 + 3 \cdot (-18) = 12 – 3 – 54 = -45. \] Matice pro \(x\): \[ A_x = \begin{vmatrix} 15 & 1 & 3 \\ 1 & 4 & -1 \\ 10 & -2 & 2 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 15 \begin{vmatrix}4 & -1 \\ -2 & 2 \end{vmatrix} – 1 \begin{vmatrix} 1 & -1 \\ 10 & 2 \end{vmatrix} + 3 \begin{vmatrix} 1 & 4 \\ 10 & -2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix}4 & -1 \\ -2 & 2 \end{vmatrix} = 6, \] \[ \begin{vmatrix} 1 & -1 \\ 10 & 2 \end{vmatrix} = 1 \cdot 2 – (-1) \cdot 10 = 2 + 10 = 12, \] \[ \begin{vmatrix} 1 & 4 \\ 10 & -2 \end{vmatrix} = 1 \cdot (-2) – 4 \cdot 10 = -2 – 40 = -42. \] Dosadíme: \[ \det(A_x) = 15 \cdot 6 – 1 \cdot 12 + 3 \cdot (-42) = 90 – 12 – 126 = -48. \] Matice pro \(y\): \[ A_y = \begin{vmatrix} 2 & 15 & 3 \\ -1 & 1 & -1 \\ 5 & 10 & 2 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 2 \begin{vmatrix}1 & -1 \\ 10 & 2 \end{vmatrix} – 15 \begin{vmatrix} -1 & -1 \\ 5 & 2 \end{vmatrix} + 3 \begin{vmatrix} -1 & 1 \\ 5 & 10 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix}1 & -1 \\ 10 & 2 \end{vmatrix} = 1 \cdot 2 – (-1) \cdot 10 = 2 + 10 = 12, \] \[ \begin{vmatrix} -1 & -1 \\ 5 & 2 \end{vmatrix} = (-1) \cdot 2 – (-1) \cdot 5 = -2 + 5 = 3, \] \[ \begin{vmatrix} -1 & 1 \\ 5 & 10 \end{vmatrix} = (-1) \cdot 10 – 1 \cdot 5 = -10 – 5 = -15. \] Dosadíme: \[ \det(A_y) = 2 \cdot 12 – 15 \cdot 3 + 3 \cdot (-15) = 24 – 45 – 45 = -66. \] Matice pro \(z\): \[ A_z = \begin{vmatrix} 2 & 1 & 15 \\ -1 & 4 & 1 \\ 5 & -2 & 10 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 2 \begin{vmatrix}4 & 1 \\ -2 & 10 \end{vmatrix} – 1 \begin{vmatrix} -1 & 1 \\ 5 & 10 \end{vmatrix} + 15 \begin{vmatrix} -1 & 4 \\ 5 & -2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix}4 & 1 \\ -2 & 10 \end{vmatrix} = 4 \cdot 10 – 1 \cdot (-2) = 40 + 2 = 42, \] \[ \begin{vmatrix} -1 & 1 \\ 5 & 10 \end{vmatrix} = (-1) \cdot 10 – 1 \cdot 5 = -10 – 5 = -15, \] \[ \begin{vmatrix} -1 & 4 \\ 5 & -2 \end{vmatrix} = (-1) \cdot (-2) – 4 \cdot 5 = 2 – 20 = -18. \] Dosadíme: \[ \det(A_z) = 2 \cdot 42 – 1 \cdot (-15) + 15 \cdot (-18) = 84 + 15 – 270 = -171. \] Výsledky: \[ x = \frac{-48}{-45} = \frac{16}{15}, \quad y = \frac{-66}{-45} = \frac{22}{15}, \quad z = \frac{-171}{-45} = \frac{19}{5}. \]

Řešení příkladu 26:

Koeficientová matice: \[ A = \begin{vmatrix} 3 & -1 & 2 \\ 2 & 4 & -1 \\ -1 & 3 & 5 \end{vmatrix}. \] Determinant matice \(A\): \[ \det(A) = 3 \begin{vmatrix}4 & -1 \\ 3 & 5 \end{vmatrix} – (-1) \begin{vmatrix} 2 & -1 \\ -1 & 5 \end{vmatrix} + 2 \begin{vmatrix} 2 & 4 \\ -1 & 3 \end{vmatrix}. \] Vypočítáme jednotlivé determinanty 2×2: \[ \begin{vmatrix}4 & -1 \\ 3 & 5 \end{vmatrix} = 4 \cdot 5 – (-1) \cdot 3 = 20 + 3 = 23, \] \[ \begin{vmatrix} 2 & -1 \\ -1 & 5 \end{vmatrix} = 2 \cdot 5 – (-1) \cdot (-1) = 10 – 1 = 9, \] \[ \begin{vmatrix} 2 & 4 \\ -1 & 3 \end{vmatrix} = 2 \cdot 3 – 4 \cdot (-1) = 6 + 4 = 10. \] Dosadíme do vzorce pro determinant: \[ \det(A) = 3 \cdot 23 + 1 \cdot 9 + 2 \cdot 10 = 69 + 9 + 20 = 98. \] Matice \(A_x\) vznikne nahrazením první sloupce v \(A\) pravými stranami rovnic: \[ A_x = \begin{vmatrix} 7 & -1 & 2 \\ 3 & 4 & -1 \\ 10 & 3 & 5 \end{vmatrix}, \] determinant: \[ \det(A_x) = 7 \begin{vmatrix}4 & -1 \\ 3 & 5 \end{vmatrix} – (-1) \begin{vmatrix} 3 & -1 \\ 10 & 5 \end{vmatrix} + 2 \begin{vmatrix} 3 & 4 \\ 10 & 3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix}4 & -1 \\ 3 & 5 \end{vmatrix} = 23, \] \[ \begin{vmatrix} 3 & -1 \\ 10 & 5 \end{vmatrix} = 3 \cdot 5 – (-1) \cdot 10 = 15 + 10 = 25, \] \[ \begin{vmatrix} 3 & 4 \\ 10 & 3 \end{vmatrix} = 3 \cdot 3 – 4 \cdot 10 = 9 – 40 = -31. \] Dosadíme: \[ \det(A_x) = 7 \cdot 23 + 1 \cdot 25 + 2 \cdot (-31) = 161 + 25 – 62 = 124. \] Matice \(A_y\) vznikne nahrazením druhého sloupce: \[ A_y = \begin{vmatrix} 3 & 7 & 2 \\ 2 & 3 & -1 \\ -1 & 10 & 5 \end{vmatrix}, \] determinant: \[ \det(A_y) = 3 \begin{vmatrix}3 & -1 \\ 10 & 5 \end{vmatrix} – 7 \begin{vmatrix} 2 & -1 \\ -1 & 5 \end{vmatrix} + 2 \begin{vmatrix} 2 & 3 \\ -1 & 10 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix}3 & -1 \\ 10 & 5 \end{vmatrix} = 3 \cdot 5 – (-1) \cdot 10 = 15 + 10 = 25, \] \[ \begin{vmatrix} 2 & -1 \\ -1 & 5 \end{vmatrix} = 2 \cdot 5 – (-1) \cdot (-1) = 10 – 1 = 9, \] \[ \begin{vmatrix} 2 & 3 \\ -1 & 10 \end{vmatrix} = 2 \cdot 10 – 3 \cdot (-1) = 20 + 3 = 23. \] Dosadíme: \[ \det(A_y) = 3 \cdot 25 – 7 \cdot 9 + 2 \cdot 23 = 75 – 63 + 46 = 58. \] Matice \(A_z\) vznikne nahrazením třetího sloupce: \[ A_z = \begin{vmatrix} 3 & -1 & 7 \\ 2 & 4 & 3 \\ -1 & 3 & 10 \end{vmatrix}, \] determinant: \[ \det(A_z) = 3 \begin{vmatrix}4 & 3 \\ 3 & 10 \end{vmatrix} – (-1) \begin{vmatrix} 2 & 3 \\ -1 & 10 \end{vmatrix} + 7 \begin{vmatrix} 2 & 4 \\ -1 & 3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix}4 & 3 \\ 3 & 10 \end{vmatrix} = 4 \cdot 10 – 3 \cdot 3 = 40 – 9 = 31, \] \[ \begin{vmatrix} 2 & 3 \\ -1 & 10 \end{vmatrix} = 2 \cdot 10 – 3 \cdot (-1) = 20 + 3 = 23, \] \[ \begin{vmatrix} 2 & 4 \\ -1 & 3 \end{vmatrix} = 2 \cdot 3 – 4 \cdot (-1) = 6 + 4 = 10. \] Dosadíme: \[ \det(A_z) = 3 \cdot 31 + 1 \cdot 23 + 7 \cdot 10 = 93 + 23 + 70 = 186. \] Nakonec dosadíme do vzorců: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{124}{98} = \frac{62}{49}, \] \[ y = \frac{\det(A_y)}{\det(A)} = \frac{58}{98} = \frac{29}{49}, \] \[ z = \frac{\det(A_z)}{\det(A)} = \frac{186}{98} = \frac{93}{49}. \]

Řešení příkladu 27:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & 2 & 3 \\ 4 & -1 & 1 \\ 2 & 5 & -2 \end{vmatrix}. \] Determinant: \[ \det(A) = 1 \begin{vmatrix} -1 & 1 \\ 5 & -2 \end{vmatrix} – 2 \begin{vmatrix} 4 & 1 \\ 2 & -2 \end{vmatrix} + 3 \begin{vmatrix} 4 & -1 \\ 2 & 5 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 1 \\ 5 & -2 \end{vmatrix} = (-1)(-2) – 1 \cdot 5 = 2 – 5 = -3, \] \[ \begin{vmatrix} 4 & 1 \\ 2 & -2 \end{vmatrix} = 4 \cdot (-2) – 1 \cdot 2 = -8 – 2 = -10, \] \[ \begin{vmatrix} 4 & -1 \\ 2 & 5 \end{vmatrix} = 4 \cdot 5 – (-1) \cdot 2 = 20 + 2 = 22. \] Dosadíme: \[ \det(A) = 1 \cdot (-3) – 2 \cdot (-10) + 3 \cdot 22 = -3 + 20 + 66 = 83. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 14 & 2 & 3 \\ 3 & -1 & 1 \\ 1 & 5 & -2 \end{vmatrix}, \] determinant: \[ \det(A_x) = 14 \begin{vmatrix} -1 & 1 \\ 5 & -2 \end{vmatrix} – 2 \begin{vmatrix} 3 & 1 \\ 1 & -2 \end{vmatrix} + 3 \begin{vmatrix} 3 & -1 \\ 1 & 5 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 1 \\ 5 & -2 \end{vmatrix} = -3, \] \[ \begin{vmatrix} 3 & 1 \\ 1 & -2 \end{vmatrix} = 3 \cdot (-2) – 1 \cdot 1 = -6 -1 = -7, \] \[ \begin{vmatrix} 3 & -1 \\ 1 & 5 \end{vmatrix} = 3 \cdot 5 – (-1) \cdot 1 = 15 + 1 = 16. \] Dosadíme: \[ \det(A_x) = 14 \cdot (-3) – 2 \cdot (-7) + 3 \cdot 16 = -42 + 14 + 48 = 20. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 1 & 14 & 3 \\ 4 & 3 & 1 \\ 2 & 1 & -2 \end{vmatrix}, \] determinant: \[ \det(A_y) = 1 \begin{vmatrix} 3 & 1 \\ 1 & -2 \end{vmatrix} – 14 \begin{vmatrix} 4 & 1 \\ 2 & -2 \end{vmatrix} + 3 \begin{vmatrix} 4 & 3 \\ 2 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 3 & 1 \\ 1 & -2 \end{vmatrix} = -7, \] \[ \begin{vmatrix} 4 & 1 \\ 2 & -2 \end{vmatrix} = -10, \] \[ \begin{vmatrix} 4 & 3 \\ 2 & 1 \end{vmatrix} = 4 \cdot 1 – 3 \cdot 2 = 4 – 6 = -2. \] Dosadíme: \[ \det(A_y) = 1 \cdot (-7) – 14 \cdot (-10) + 3 \cdot (-2) = -7 + 140 – 6 = 127. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 1 & 2 & 14 \\ 4 & -1 & 3 \\ 2 & 5 & 1 \end{vmatrix}, \] determinant: \[ \det(A_z) = 1 \begin{vmatrix} -1 & 3 \\ 5 & 1 \end{vmatrix} – 2 \begin{vmatrix} 4 & 3 \\ 2 & 1 \end{vmatrix} + 14 \begin{vmatrix} 4 & -1 \\ 2 & 5 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 3 \\ 5 & 1 \end{vmatrix} = (-1) \cdot 1 – 3 \cdot 5 = -1 – 15 = -16, \] \[ \begin{vmatrix} 4 & 3 \\ 2 & 1 \end{vmatrix} = 4 – 6 = -2, \] \[ \begin{vmatrix} 4 & -1 \\ 2 & 5 \end{vmatrix} = 4 \cdot 5 – (-1) \cdot 2 = 20 + 2 = 22. \] Dosadíme: \[ \det(A_z) = 1 \cdot (-16) – 2 \cdot (-2) + 14 \cdot 22 = -16 + 4 + 308 = 296. \] Výsledky: \[ x = \frac{20}{83}, \quad y = \frac{127}{83}, \quad z = \frac{296}{83}. \]

Řešení příkladu 28:

Koeficientová matice: \[ A = \begin{vmatrix} 5 & 1 & -2 \\ 3 & -2 & 4 \\ -1 & 3 & 1 \end{vmatrix}. \] Determinant: \[ \det(A) = 5 \begin{vmatrix} -2 & 4 \\ 3 & 1 \end{vmatrix} – 1 \begin{vmatrix} 3 & 4 \\ -1 & 1 \end{vmatrix} + (-2) \begin{vmatrix} 3 & -2 \\ -1 & 3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -2 & 4 \\ 3 & 1 \end{vmatrix} = (-2)(1) – 4 \cdot 3 = -2 – 12 = -14, \] \[ \begin{vmatrix} 3 & 4 \\ -1 & 1 \end{vmatrix} = 3 \cdot 1 – 4 \cdot (-1) = 3 + 4 = 7, \] \[ \begin{vmatrix} 3 & -2 \\ -1 & 3 \end{vmatrix} = 3 \cdot 3 – (-2)(-1) = 9 – 2 = 7. \] Dosadíme: \[ \det(A) = 5 \cdot (-14) – 1 \cdot 7 – 2 \cdot 7 = -70 – 7 – 14 = -91. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 4 & 1 & -2 \\ -6 & -2 & 4 \\ 7 & 3 & 1 \end{vmatrix}, \] determinant: \[ \det(A_x) = 4 \begin{vmatrix} -2 & 4 \\ 3 & 1 \end{vmatrix} – 1 \begin{vmatrix} -6 & 4 \\ 7 & 1 \end{vmatrix} + (-2) \begin{vmatrix} -6 & -2 \\ 7 & 3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -2 & 4 \\ 3 & 1 \end{vmatrix} = -14, \] \[ \begin{vmatrix} -6 & 4 \\ 7 & 1 \end{vmatrix} = -6 \cdot 1 – 4 \cdot 7 = -6 – 28 = -34, \] \[ \begin{vmatrix} -6 & -2 \\ 7 & 3 \end{vmatrix} = -6 \cdot 3 – (-2) \cdot 7 = -18 + 14 = -4. \] Dosadíme: \[ \det(A_x) = 4 \cdot (-14) – 1 \cdot (-34) – 2 \cdot (-4) = -56 + 34 + 8 = -14. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 5 & 4 & -2 \\ 3 & -6 & 4 \\ -1 & 7 & 1 \end{vmatrix}, \] determinant: \[ \det(A_y) = 5 \begin{vmatrix} -6 & 4 \\ 7 & 1 \end{vmatrix} – 4 \begin{vmatrix} 3 & 4 \\ -1 & 1 \end{vmatrix} + (-2) \begin{vmatrix} 3 & -6 \\ -1 & 7 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -6 & 4 \\ 7 & 1 \end{vmatrix} = -6 – 28 = -34, \] \[ \begin{vmatrix} 3 & 4 \\ -1 & 1 \end{vmatrix} = 3 \cdot 1 – 4 \cdot (-1) = 3 + 4 = 7, \] \[ \begin{vmatrix} 3 & -6 \\ -1 & 7 \end{vmatrix} = 3 \cdot 7 – (-6)(-1) = 21 – 6 = 15. \] Dosadíme: \[ \det(A_y) = 5 \cdot (-34) – 4 \cdot 7 – 2 \cdot 15 = -170 – 28 – 30 = -228. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 5 & 1 & 4 \\ 3 & -2 & -6 \\ -1 & 3 & 7 \end{vmatrix}, \] determinant: \[ \det(A_z) = 5 \begin{vmatrix} -2 & -6 \\ 3 & 7 \end{vmatrix} – 1 \begin{vmatrix} 3 & -6 \\ -1 & 7 \end{vmatrix} + 4 \begin{vmatrix} 3 & -2 \\ -1 & 3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -2 & -6 \\ 3 & 7 \end{vmatrix} = (-2)(7) – (-6)(3) = -14 + 18 = 4, \] \[ \begin{vmatrix} 3 & -6 \\ -1 & 7 \end{vmatrix} = 3 \cdot 7 – (-6)(-1) = 21 – 6 = 15, \] \[ \begin{vmatrix} 3 & -2 \\ -1 & 3 \end{vmatrix} = 3 \cdot 3 – (-2)(-1) = 9 – 2 = 7. \] Dosadíme: \[ \det(A_z) = 5 \cdot 4 – 1 \cdot 15 + 4 \cdot 7 = 20 – 15 + 28 = 33. \] Výsledky: \[ x = \frac{-14}{-91} = \frac{14}{91} = \frac{2}{13}, \quad y = \frac{-228}{-91} = \frac{228}{91}, \quad z = \frac{33}{-91} = -\frac{33}{91}. \]

Řešení příkladu 29:

Koeficientová matice: \[ A = \begin{vmatrix} 2 & 3 & 1 \\ -4 & 1 & -2 \\ 1 & -2 & 3 \end{vmatrix}. \] Determinant: \[ \det(A) = 2 \begin{vmatrix} 1 & -2 \\ -2 & 3 \end{vmatrix} – 3 \begin{vmatrix} -4 & -2 \\ 1 & 3 \end{vmatrix} + 1 \begin{vmatrix} -4 & 1 \\ 1 & -2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 1 & -2 \\ -2 & 3 \end{vmatrix} = 1 \cdot 3 – (-2)(-2) = 3 – 4 = -1, \] \[ \begin{vmatrix} -4 & -2 \\ 1 & 3 \end{vmatrix} = (-4)(3) – (-2)(1) = -12 + 2 = -10, \] \[ \begin{vmatrix} -4 & 1 \\ 1 & -2 \end{vmatrix} = (-4)(-2) – 1 \cdot 1 = 8 – 1 = 7. \] Dosadíme: \[ \det(A) = 2 \cdot (-1) – 3 \cdot (-10) + 1 \cdot 7 = -2 + 30 + 7 = 35. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 10 & 3 & 1 \\ -5 & 1 & -2 \\ 7 & -2 & 3 \end{vmatrix}, \] determinant: \[ \det(A_x) = 10 \begin{vmatrix} 1 & -2 \\ -2 & 3 \end{vmatrix} – 3 \begin{vmatrix} -5 & -2 \\ 7 & 3 \end{vmatrix} + 1 \begin{vmatrix} -5 & 1 \\ 7 & -2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 1 & -2 \\ -2 & 3 \end{vmatrix} = -1, \] \[ \begin{vmatrix} -5 & -2 \\ 7 & 3 \end{vmatrix} = (-5)(3) – (-2)(7) = -15 + 14 = -1, \] \[ \begin{vmatrix} -5 & 1 \\ 7 & -2 \end{vmatrix} = (-5)(-2) – 1 \cdot 7 = 10 – 7 = 3. \] Dosadíme: \[ \det(A_x) = 10 \cdot (-1) – 3 \cdot (-1) + 1 \cdot 3 = -10 + 3 + 3 = -4. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 2 & 10 & 1 \\ -4 & -5 & -2 \\ 1 & 7 & 3 \end{vmatrix}, \] determinant: \[ \det(A_y) = 2 \begin{vmatrix} -5 & -2 \\ 7 & 3 \end{vmatrix} – 10 \begin{vmatrix} -4 & -2 \\ 1 & 3 \end{vmatrix} + 1 \begin{vmatrix} -4 & -5 \\ 1 & 7 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -5 & -2 \\ 7 & 3 \end{vmatrix} = -15 + 14 = -1, \] \[ \begin{vmatrix} -4 & -2 \\ 1 & 3 \end{vmatrix} = -12 + 2 = -10, \] \[ \begin{vmatrix} -4 & -5 \\ 1 & 7 \end{vmatrix} = (-4)(7) – (-5)(1) = -28 + 5 = -23. \] Dosadíme: \[ \det(A_y) = 2 \cdot (-1) – 10 \cdot (-10) + 1 \cdot (-23) = -2 + 100 – 23 = 75. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 2 & 3 & 10 \\ -4 & 1 & -5 \\ 1 & -2 & 7 \end{vmatrix}, \] determinant: \[ \det(A_z) = 2 \begin{vmatrix} 1 & -5 \\ -2 & 7 \end{vmatrix} – 3 \begin{vmatrix} -4 & -5 \\ 1 & 7 \end{vmatrix} + 10 \begin{vmatrix} -4 & 1 \\ 1 & -2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 1 & -5 \\ -2 & 7 \end{vmatrix} = 1 \cdot 7 – (-5)(-2) = 7 – 10 = -3, \] \[ \begin{vmatrix} -4 & -5 \\ 1 & 7 \end{vmatrix} = -28 + 5 = -23, \] \[ \begin{vmatrix} -4 & 1 \\ 1 & -2 \end{vmatrix} = (-4)(-2) – 1 \cdot 1 = 8 – 1 = 7. \] Dosadíme: \[ \det(A_z) = 2 \cdot (-3) – 3 \cdot (-23) + 10 \cdot 7 = -6 + 69 + 70 = 133. \] Výsledky: \[ x = \frac{-4}{35} = -\frac{4}{35}, \quad y = \frac{75}{35} = \frac{15}{7}, \quad z = \frac{133}{35}. \]

Řešení příkladu 30:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \\ 4 & 1 & 1 \end{vmatrix}. \] Determinant: \[ \det(A) = 1 \begin{vmatrix} -1 & 3 \\ 1 & 1 \end{vmatrix} – 2 \begin{vmatrix} 2 & 3 \\ 4 & 1 \end{vmatrix} + (-1) \begin{vmatrix} 2 & -1 \\ 4 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 3 \\ 1 & 1 \end{vmatrix} = (-1)(1) – 3 \cdot 1 = -1 – 3 = -4, \] \[ \begin{vmatrix} 2 & 3 \\ 4 & 1 \end{vmatrix} = 2 \cdot 1 – 3 \cdot 4 = 2 – 12 = -10, \] \[ \begin{vmatrix} 2 & -1 \\ 4 & 1 \end{vmatrix} = 2 \cdot 1 – (-1) \cdot 4 = 2 + 4 = 6. \] Dosadíme: \[ \det(A) = 1 \cdot (-4) – 2 \cdot (-10) – 1 \cdot 6 = -4 + 20 – 6 = 10. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 3 & 2 & -1 \\ 7 & -1 & 3 \\ 10 & 1 & 1 \end{vmatrix}, \] determinant: \[ \det(A_x) = 3 \begin{vmatrix} -1 & 3 \\ 1 & 1 \end{vmatrix} – 2 \begin{vmatrix} 7 & 3 \\ 10 & 1 \end{vmatrix} + (-1) \begin{vmatrix} 7 & -1 \\ 10 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 3 \\ 1 & 1 \end{vmatrix} = -4, \] \[ \begin{vmatrix} 7 & 3 \\ 10 & 1 \end{vmatrix} = 7 \cdot 1 – 3 \cdot 10 = 7 – 30 = -23, \] \[ \begin{vmatrix} 7 & -1 \\ 10 & 1 \end{vmatrix} = 7 \cdot 1 – (-1) \cdot 10 = 7 + 10 = 17. \] Dosadíme: \[ \det(A_x) = 3 \cdot (-4) – 2 \cdot (-23) – 1 \cdot 17 = -12 + 46 – 17 = 17. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 1 & 3 & -1 \\ 2 & 7 & 3 \\ 4 & 10 & 1 \end{vmatrix}, \] determinant: \[ \det(A_y) = 1 \begin{vmatrix} 7 & 3 \\ 10 & 1 \end{vmatrix} – 3 \begin{vmatrix} 2 & 3 \\ 4 & 1 \end{vmatrix} + (-1) \begin{vmatrix} 2 & 7 \\ 4 & 10 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 7 & 3 \\ 10 & 1 \end{vmatrix} = 7 \cdot 1 – 3 \cdot 10 = 7 – 30 = -23, \] \[ \begin{vmatrix} 2 & 3 \\ 4 & 1 \end{vmatrix} = 2 \cdot 1 – 3 \cdot 4 = 2 – 12 = -10, \] \[ \begin{vmatrix} 2 & 7 \\ 4 & 10 \end{vmatrix} = 2 \cdot 10 – 7 \cdot 4 = 20 – 28 = -8. \] Dosadíme: \[ \det(A_y) = 1 \cdot (-23) – 3 \cdot (-10) – 1 \cdot (-8) = -23 + 30 + 8 = 15. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 1 & 2 & 3 \\ 2 & -1 & 7 \\ 4 & 1 & 10 \end{vmatrix}, \] determinant: \[ \det(A_z) = 1 \begin{vmatrix} -1 & 7 \\ 1 & 10 \end{vmatrix} – 2 \begin{vmatrix} 2 & 7 \\ 4 & 10 \end{vmatrix} + 3 \begin{vmatrix} 2 & -1 \\ 4 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 7 \\ 1 & 10 \end{vmatrix} = -1 \cdot 10 – 7 \cdot 1 = -10 – 7 = -17, \] \[ \begin{vmatrix} 2 & 7 \\ 4 & 10 \end{vmatrix} = 2 \cdot 10 – 7 \cdot 4 = 20 – 28 = -8, \] \[ \begin{vmatrix} 2 & -1 \\ 4 & 1 \end{vmatrix} = 2 \cdot 1 – (-1) \cdot 4 = 2 + 4 = 6. \] Dosadíme: \[ \det(A_z) = 1 \cdot (-17) – 2 \cdot (-8) + 3 \cdot 6 = -17 + 16 + 18 = 17. \] Výsledky: \[ x = \frac{17}{10}, \quad y = \frac{15}{10} = \frac{3}{2}, \quad z = \frac{17}{10}. \]

Řešení příkladu 31:

Koeficientová matice: \[ A = \begin{vmatrix} 2 & -1 & 3 \\ -1 & 4 & -1 \\ 3 & 1 & 2 \end{vmatrix}. \] Determinant: \[ \det(A) = 2 \begin{vmatrix} 4 & -1 \\ 1 & 2 \end{vmatrix} – (-1) \begin{vmatrix} -1 & -1 \\ 3 & 2 \end{vmatrix} + 3 \begin{vmatrix} -1 & 4 \\ 3 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & -1 \\ 1 & 2 \end{vmatrix} = 4 \cdot 2 – (-1) \cdot 1 = 8 + 1 = 9, \] \[ \begin{vmatrix} -1 & -1 \\ 3 & 2 \end{vmatrix} = (-1) \cdot 2 – (-1) \cdot 3 = -2 + 3 = 1, \] \[ \begin{vmatrix} -1 & 4 \\ 3 & 1 \end{vmatrix} = (-1) \cdot 1 – 4 \cdot 3 = -1 – 12 = -13. \] Dosadíme: \[ \det(A) = 2 \cdot 9 + 1 \cdot 1 + 3 \cdot (-13) = 18 + 1 – 39 = -20. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 9 & -1 & 3 \\ -2 & 4 & -1 \\ 7 & 1 & 2 \end{vmatrix}, \] determinant: \[ \det(A_x) = 9 \begin{vmatrix} 4 & -1 \\ 1 & 2 \end{vmatrix} – (-1) \begin{vmatrix} -2 & -1 \\ 7 & 2 \end{vmatrix} + 3 \begin{vmatrix} -2 & 4 \\ 7 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & -1 \\ 1 & 2 \end{vmatrix} = 9, \] \[ \begin{vmatrix} -2 & -1 \\ 7 & 2 \end{vmatrix} = (-2) \cdot 2 – (-1) \cdot 7 = -4 + 7 = 3, \] \[ \begin{vmatrix} -2 & 4 \\ 7 & 1 \end{vmatrix} = (-2) \cdot 1 – 4 \cdot 7 = -2 – 28 = -30. \] Dosadíme: \[ \det(A_x) = 9 \cdot 9 + 1 \cdot 3 + 3 \cdot (-30) = 81 + 3 – 90 = -6. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 2 & 9 & 3 \\ -1 & -2 & -1 \\ 3 & 7 & 2 \end{vmatrix}, \] determinant: \[ \det(A_y) = 2 \begin{vmatrix} -2 & -1 \\ 7 & 2 \end{vmatrix} – 9 \begin{vmatrix} -1 & -1 \\ 3 & 2 \end{vmatrix} + 3 \begin{vmatrix} -1 & -2 \\ 3 & 7 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -2 & -1 \\ 7 & 2 \end{vmatrix} = 3, \] \[ \begin{vmatrix} -1 & -1 \\ 3 & 2 \end{vmatrix} = (-1) \cdot 2 – (-1) \cdot 3 = -2 + 3 = 1, \] \[ \begin{vmatrix} -1 & -2 \\ 3 & 7 \end{vmatrix} = (-1) \cdot 7 – (-2) \cdot 3 = -7 + 6 = -1. \] Dosadíme: \[ \det(A_y) = 2 \cdot 3 – 9 \cdot 1 + 3 \cdot (-1) = 6 – 9 – 3 = -6. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 2 & -1 & 9 \\ -1 & 4 & -2 \\ 3 & 1 & 7 \end{vmatrix}, \] determinant: \[ \det(A_z) = 2 \begin{vmatrix} 4 & -2 \\ 1 & 7 \end{vmatrix} – (-1) \begin{vmatrix} -1 & -2 \\ 3 & 7 \end{vmatrix} + 9 \begin{vmatrix} -1 & 4 \\ 3 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & -2 \\ 1 & 7 \end{vmatrix} = 4 \cdot 7 – (-2) \cdot 1 = 28 + 2 = 30, \] \[ \begin{vmatrix} -1 & -2 \\ 3 & 7 \end{vmatrix} = -7 + 6 = -1, \] \[ \begin{vmatrix} -1 & 4 \\ 3 & 1 \end{vmatrix} = (-1) \cdot 1 – 4 \cdot 3 = -1 – 12 = -13. \] Dosadíme: \[ \det(A_z) = 2 \cdot 30 + 1 \cdot (-1) + 9 \cdot (-13) = 60 – 1 – 117 = -58. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-6}{-20} = \frac{3}{10}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-6}{-20} = \frac{3}{10}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{-58}{-20} = \frac{29}{10}. \]

Řešení příkladu 32:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 1 & -2 & 4 \end{vmatrix}. \] Determinant: \[ \det(A) = 1 \begin{vmatrix} -1 & 3 \\ -2 & 4 \end{vmatrix} – 1 \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} + 1 \begin{vmatrix} 2 & -1 \\ 1 & -2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 3 \\ -2 & 4 \end{vmatrix} = (-1) \cdot 4 – 3 \cdot (-2) = -4 + 6 = 2, \] \[ \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} = 2 \cdot 4 – 3 \cdot 1 = 8 – 3 = 5, \] \[ \begin{vmatrix} 2 & -1 \\ 1 & -2 \end{vmatrix} = 2 \cdot (-2) – (-1) \cdot 1 = -4 + 1 = -3. \] Dosadíme: \[ \det(A) = 1 \cdot 2 – 1 \cdot 5 + 1 \cdot (-3) = 2 – 5 – 3 = -6. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 6 & 1 & 1 \\ 14 & -1 & 3 \\ 10 & -2 & 4 \end{vmatrix}, \] determinant: \[ \det(A_x) = 6 \begin{vmatrix} -1 & 3 \\ -2 & 4 \end{vmatrix} – 1 \begin{vmatrix} 14 & 3 \\ 10 & 4 \end{vmatrix} + 1 \begin{vmatrix} 14 & -1 \\ 10 & -2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 3 \\ -2 & 4 \end{vmatrix} = 2, \] \[ \begin{vmatrix} 14 & 3 \\ 10 & 4 \end{vmatrix} = 14 \cdot 4 – 3 \cdot 10 = 56 – 30 = 26, \] \[ \begin{vmatrix} 14 & -1 \\ 10 & -2 \end{vmatrix} = 14 \cdot (-2) – (-1) \cdot 10 = -28 + 10 = -18. \] Dosadíme: \[ \det(A_x) = 6 \cdot 2 – 1 \cdot 26 + 1 \cdot (-18) = 12 – 26 – 18 = -32. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 1 & 6 & 1 \\ 2 & 14 & 3 \\ 1 & 10 & 4 \end{vmatrix}, \] determinant: \[ \det(A_y) = 1 \begin{vmatrix} 14 & 3 \\ 10 & 4 \end{vmatrix} – 6 \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} + 1 \begin{vmatrix} 2 & 14 \\ 1 & 10 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 14 & 3 \\ 10 & 4 \end{vmatrix} = 26, \] \[ \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} = 2 \cdot 4 – 3 \cdot 1 = 8 – 3 = 5, \] \[ \begin{vmatrix} 2 & 14 \\ 1 & 10 \end{vmatrix} = 2 \cdot 10 – 14 \cdot 1 = 20 – 14 = 6. \] Dosadíme: \[ \det(A_y) = 1 \cdot 26 – 6 \cdot 5 + 1 \cdot 6 = 26 – 30 + 6 = 2. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 1 & 1 & 6 \\ 2 & -1 & 14 \\ 1 & -2 & 10 \end{vmatrix}, \] determinant: \[ \det(A_z) = 1 \begin{vmatrix} -1 & 14 \\ -2 & 10 \end{vmatrix} – 1 \begin{vmatrix} 2 & 14 \\ 1 & 10 \end{vmatrix} + 6 \begin{vmatrix} 2 & -1 \\ 1 & -2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 14 \\ -2 & 10 \end{vmatrix} = (-1) \cdot 10 – 14 \cdot (-2) = -10 + 28 = 18, \] \[ \begin{vmatrix} 2 & 14 \\ 1 & 10 \end{vmatrix} = 2 \cdot 10 – 14 \cdot 1 = 20 – 14 = 6, \] \[ \begin{vmatrix} 2 & -1 \\ 1 & -2 \end{vmatrix} = 2 \cdot (-2) – (-1) \cdot 1 = -4 + 1 = -3. \] Dosadíme: \[ \det(A_z) = 1 \cdot 18 – 1 \cdot 6 + 6 \cdot (-3) = 18 – 6 – 18 = -6. \] Výsledky: \[ x = \frac{-32}{-6} = \frac{16}{3}, \quad y = \frac{2}{-6} = -\frac{1}{3}, \quad z = \frac{-6}{-6} = 1. \]

Řešení příkladu 33:

Koeficientová matice: \[ A = \begin{vmatrix} 3 & -1 & 2 \\ 1 & 4 & -1 \\ 2 & -3 & 3 \end{vmatrix}. \] Determinant: \[ \det(A) = 3 \begin{vmatrix} 4 & -1 \\ -3 & 3 \end{vmatrix} – (-1) \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} + 2 \begin{vmatrix} 1 & 4 \\ 2 & -3 \end{vmatrix}. \] Výpočty determinantů podmatic: \[ \begin{vmatrix} 4 & -1 \\ -3 & 3 \end{vmatrix} = 4 \cdot 3 – (-1) \cdot (-3) = 12 – 3 = 9, \] \[ \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} = 1 \cdot 3 – (-1) \cdot 2 = 3 + 2 = 5, \] \[ \begin{vmatrix} 1 & 4 \\ 2 & -3 \end{vmatrix} = 1 \cdot (-3) – 4 \cdot 2 = -3 – 8 = -11. \] Dosadíme: \[ \det(A) = 3 \cdot 9 + 1 \cdot 5 + 2 \cdot (-11) = 27 + 5 – 22 = 10. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 5 & -1 & 2 \\ 6 & 4 & -1 \\ 4 & -3 & 3 \end{vmatrix}, \] determinant: \[ \det(A_x) = 5 \begin{vmatrix} 4 & -1 \\ -3 & 3 \end{vmatrix} – (-1) \begin{vmatrix} 6 & -1 \\ 4 & 3 \end{vmatrix} + 2 \begin{vmatrix} 6 & 4 \\ 4 & -3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & -1 \\ -3 & 3 \end{vmatrix} = 9, \] \[ \begin{vmatrix} 6 & -1 \\ 4 & 3 \end{vmatrix} = 6 \cdot 3 – (-1) \cdot 4 = 18 + 4 = 22, \] \[ \begin{vmatrix} 6 & 4 \\ 4 & -3 \end{vmatrix} = 6 \cdot (-3) – 4 \cdot 4 = -18 – 16 = -34. \] Dosadíme: \[ \det(A_x) = 5 \cdot 9 + 1 \cdot 22 + 2 \cdot (-34) = 45 + 22 – 68 = -1. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 3 & 5 & 2 \\ 1 & 6 & -1 \\ 2 & 4 & 3 \end{vmatrix}, \] determinant: \[ \det(A_y) = 3 \begin{vmatrix} 6 & -1 \\ 4 & 3 \end{vmatrix} – 5 \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} + 2 \begin{vmatrix} 1 & 6 \\ 2 & 4 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 6 & -1 \\ 4 & 3 \end{vmatrix} = 6 \cdot 3 – (-1) \cdot 4 = 18 + 4 = 22, \] \[ \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} = 1 \cdot 3 – (-1) \cdot 2 = 3 + 2 = 5, \] \[ \begin{vmatrix} 1 & 6 \\ 2 & 4 \end{vmatrix} = 1 \cdot 4 – 6 \cdot 2 = 4 – 12 = -8. \] Dosadíme: \[ \det(A_y) = 3 \cdot 22 – 5 \cdot 5 + 2 \cdot (-8) = 66 – 25 – 16 = 25. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 3 & -1 & 5 \\ 1 & 4 & 6 \\ 2 & -3 & 4 \end{vmatrix}, \] determinant: \[ \det(A_z) = 3 \begin{vmatrix} 4 & 6 \\ -3 & 4 \end{vmatrix} – (-1) \begin{vmatrix} 1 & 6 \\ 2 & 4 \end{vmatrix} + 5 \begin{vmatrix} 1 & 4 \\ 2 & -3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & 6 \\ -3 & 4 \end{vmatrix} = 4 \cdot 4 – 6 \cdot (-3) = 16 + 18 = 34, \] \[ \begin{vmatrix} 1 & 6 \\ 2 & 4 \end{vmatrix} = 1 \cdot 4 – 6 \cdot 2 = 4 – 12 = -8, \] \[ \begin{vmatrix} 1 & 4 \\ 2 & -3 \end{vmatrix} = 1 \cdot (-3) – 4 \cdot 2 = -3 – 8 = -11. \] Dosadíme: \[ \det(A_z) = 3 \cdot 34 + 1 \cdot (-8) + 5 \cdot (-11) = 102 – 8 – 55 = 39. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-1}{10}, \quad y = \frac{25}{10} = \frac{5}{2}, \quad z = \frac{39}{10}. \]

Řešení příkladu 34:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ -3 & 4 & -1 \end{vmatrix}. \] Determinant: \[ \det(A) = 1 \begin{vmatrix} -1 & 3 \\ 4 & -1 \end{vmatrix} – 1 \begin{vmatrix} 2 & 3 \\ -3 & -1 \end{vmatrix} + 1 \begin{vmatrix} 2 & -1 \\ -3 & 4 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 3 \\ 4 & -1 \end{vmatrix} = (-1)(-1) – 3 \cdot 4 = 1 – 12 = -11, \] \[ \begin{vmatrix} 2 & 3 \\ -3 & -1 \end{vmatrix} = 2 \cdot (-1) – 3 \cdot (-3) = -2 + 9 = 7, \] \[ \begin{vmatrix} 2 & -1 \\ -3 & 4 \end{vmatrix} = 2 \cdot 4 – (-1) \cdot (-3) = 8 – 3 = 5. \] Dosadíme: \[ \det(A) = 1 \cdot (-11) – 1 \cdot 7 + 1 \cdot 5 = -11 – 7 + 5 = -13. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 6 & 1 & 1 \\ 14 & -1 & 3 \\ -2 & 4 & -1 \end{vmatrix}, \] determinant: \[ \det(A_x) = 6 \begin{vmatrix} -1 & 3 \\ 4 & -1 \end{vmatrix} – 1 \begin{vmatrix} 14 & 3 \\ -2 & -1 \end{vmatrix} + 1 \begin{vmatrix} 14 & -1 \\ -2 & 4 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 3 \\ 4 & -1 \end{vmatrix} = -11, \] \[ \begin{vmatrix} 14 & 3 \\ -2 & -1 \end{vmatrix} = 14 \cdot (-1) – 3 \cdot (-2) = -14 + 6 = -8, \] \[ \begin{vmatrix} 14 & -1 \\ -2 & 4 \end{vmatrix} = 14 \cdot 4 – (-1) \cdot (-2) = 56 – 2 = 54. \] Dosadíme: \[ \det(A_x) = 6 \cdot (-11) – 1 \cdot (-8) + 1 \cdot 54 = -66 + 8 + 54 = -4. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 1 & 6 & 1 \\ 2 & 14 & 3 \\ -3 & -2 & -1 \end{vmatrix}, \] determinant: \[ \det(A_y) = 1 \begin{vmatrix} 14 & 3 \\ -2 & -1 \end{vmatrix} – 6 \begin{vmatrix} 2 & 3 \\ -3 & -1 \end{vmatrix} + 1 \begin{vmatrix} 2 & 14 \\ -3 & -2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 14 & 3 \\ -2 & -1 \end{vmatrix} = -8, \] \[ \begin{vmatrix} 2 & 3 \\ -3 & -1 \end{vmatrix} = 2 \cdot (-1) – 3 \cdot (-3) = -2 + 9 = 7, \] \[ \begin{vmatrix} 2 & 14 \\ -3 & -2 \end{vmatrix} = 2 \cdot (-2) – 14 \cdot (-3) = -4 + 42 = 38. \] Dosadíme: \[ \det(A_y) = 1 \cdot (-8) – 6 \cdot 7 + 1 \cdot 38 = -8 – 42 + 38 = -12. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 1 & 1 & 6 \\ 2 & -1 & 14 \\ -3 & 4 & -2 \end{vmatrix}, \] determinant: \[ \det(A_z) = 1 \begin{vmatrix} -1 & 14 \\ 4 & -2 \end{vmatrix} – 1 \begin{vmatrix} 2 & 14 \\ -3 & -2 \end{vmatrix} + 6 \begin{vmatrix} 2 & -1 \\ -3 & 4 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 14 \\ 4 & -2 \end{vmatrix} = (-1)(-2) – 14 \cdot 4 = 2 – 56 = -54, \] \[ \begin{vmatrix} 2 & 14 \\ -3 & -2 \end{vmatrix} = -4 + 42 = 38, \] \[ \begin{vmatrix} 2 & -1 \\ -3 & 4 \end{vmatrix} = 2 \cdot 4 – (-1) \cdot (-3) = 8 – 3 = 5. \] Dosadíme: \[ \det(A_z) = 1 \cdot (-54) – 1 \cdot 38 + 6 \cdot 5 = -54 – 38 + 30 = -62. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-4}{-13} = \frac{4}{13}, \quad y = \frac{-12}{-13} = \frac{12}{13}, \quad z = \frac{-62}{-13} = \frac{62}{13}. \]

Řešení příkladu 35:

Koeficientová matice: \[ A = \begin{vmatrix} 2 & 3 \\ 4 & -1 \end{vmatrix} \] Determinant: \[ \det(A) = 2 \cdot (-1) – 3 \cdot 4 = -2 – 12 = -14 \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 5 & 3 \\ 1 & -1 \end{vmatrix} \] Determinant: \[ \det(A_x) = 5 \cdot (-1) – 3 \cdot 1 = -5 – 3 = -8 \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 2 & 5 \\ 4 & 1 \end{vmatrix} \] Determinant: \[ \det(A_y) = 2 \cdot 1 – 5 \cdot 4 = 2 – 20 = -18 \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-8}{-14} = \frac{4}{7}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-18}{-14} = \frac{9}{7}. \]

Řešení příkladu 36:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & -1 & 2 \\ 3 & 2 & -1 \\ 2 & -1 & 1 \end{vmatrix} \] Determinant: \[ \det(A) = 1 \begin{vmatrix} 2 & -1 \\ -1 & 1 \end{vmatrix} – (-1) \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} + 2 \begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix} \] Výpočty: \[ \begin{vmatrix} 2 & -1 \\ -1 & 1 \end{vmatrix} = 2 \cdot 1 – (-1)(-1) = 2 – 1 = 1, \] \[ \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} = 3 \cdot 1 – (-1) \cdot 2 = 3 + 2 = 5, \] \[ \begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix} = 3 \cdot (-1) – 2 \cdot 2 = -3 – 4 = -7. \] Dosadíme: \[ \det(A) = 1 \cdot 1 – (-1) \cdot 5 + 2 \cdot (-7) = 1 + 5 – 14 = -8. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 4 & -1 & 2 \\ 1 & 2 & -1 \\ 3 & -1 & 1 \end{vmatrix} \] Determinant: \[ \det(A_x) = 4 \begin{vmatrix} 2 & -1 \\ -1 & 1 \end{vmatrix} – (-1) \begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} + 2 \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} \] Výpočty: \[ \begin{vmatrix} 2 & -1 \\ -1 & 1 \end{vmatrix} = 1, \] \[ \begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} = 1 \cdot 1 – (-1) \cdot 3 = 1 + 3 = 4, \] \[ \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} = 1 \cdot (-1) – 2 \cdot 3 = -1 – 6 = -7. \] Dosadíme: \[ \det(A_x) = 4 \cdot 1 – (-1) \cdot 4 + 2 \cdot (-7) = 4 + 4 – 14 = -6. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 1 & 4 & 2 \\ 3 & 1 & -1 \\ 2 & 3 & 1 \end{vmatrix} \] Determinant: \[ \det(A_y) = 1 \begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} – 4 \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} + 2 \begin{vmatrix} 3 & 1 \\ 2 & 3 \end{vmatrix} \] Výpočty: \[ \begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} = 1 \cdot 1 – (-1) \cdot 3 = 1 + 3 = 4, \] \[ \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} = 3 \cdot 1 – (-1) \cdot 2 = 3 + 2 = 5, \] \[ \begin{vmatrix} 3 & 1 \\ 2 & 3 \end{vmatrix} = 3 \cdot 3 – 1 \cdot 2 = 9 – 2 = 7. \] Dosadíme: \[ \det(A_y) = 1 \cdot 4 – 4 \cdot 5 + 2 \cdot 7 = 4 – 20 + 14 = -2. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 1 & -1 & 4 \\ 3 & 2 & 1 \\ 2 & -1 & 3 \end{vmatrix} \] Determinant: \[ \det(A_z) = 1 \begin{vmatrix} 2 & 1 \\ -1 & 3 \end{vmatrix} – (-1) \begin{vmatrix} 3 & 1 \\ 2 & 3 \end{vmatrix} + 4 \begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix} \] Výpočty: \[ \begin{vmatrix} 2 & 1 \\ -1 & 3 \end{vmatrix} = 2 \cdot 3 – 1 \cdot (-1) = 6 + 1 = 7, \] \[ \begin{vmatrix} 3 & 1 \\ 2 & 3 \end{vmatrix} = 3 \cdot 3 – 1 \cdot 2 = 9 – 2 = 7, \] \[ \begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix} = 3 \cdot (-1) – 2 \cdot 2 = -3 – 4 = -7. \] Dosadíme: \[ \det(A_z) = 1 \cdot 7 – (-1) \cdot 7 + 4 \cdot (-7) = 7 + 7 – 28 = -14. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-6}{-8} = \frac{3}{4}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-2}{-8} = \frac{1}{4}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{-14}{-8} = \frac{7}{4}. \]

Řešení příkladu 37:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ -1 & 4 & -1 \end{vmatrix} \] Determinant: \[ \det(A) = 1 \begin{vmatrix} -1 & 3 \\ 4 & -1 \end{vmatrix} – 1 \begin{vmatrix} 2 & 3 \\ -1 & -1 \end{vmatrix} + 1 \begin{vmatrix} 2 & -1 \\ -1 & 4 \end{vmatrix} \] Výpočty: \[ \begin{vmatrix} -1 & 3 \\ 4 & -1 \end{vmatrix} = (-1)(-1) – 3 \cdot 4 = 1 – 12 = -11, \] \[ \begin{vmatrix} 2 & 3 \\ -1 & -1 \end{vmatrix} = 2 \cdot (-1) – 3 \cdot (-1) = -2 + 3 = 1, \] \[ \begin{vmatrix} 2 & -1 \\ -1 & 4 \end{vmatrix} = 2 \cdot 4 – (-1)(-1) = 8 – 1 = 7. \] Dosadíme: \[ \det(A) = 1 \cdot (-11) – 1 \cdot 1 + 1 \cdot 7 = -11 – 1 + 7 = -5. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 6 & 1 & 1 \\ 14 & -1 & 3 \\ -2 & 4 & -1 \end{vmatrix} \] Determinant: \[ \det(A_x) = 6 \begin{vmatrix} -1 & 3 \\ 4 & -1 \end{vmatrix} – 1 \begin{vmatrix} 14 & 3 \\ -2 & -1 \end{vmatrix} + 1 \begin{vmatrix} 14 & -1 \\ -2 & 4 \end{vmatrix} \] Výpočty: \[ \begin{vmatrix} -1 & 3 \\ 4 & -1 \end{vmatrix} = -11, \] \[ \begin{vmatrix} 14 & 3 \\ -2 & -1 \end{vmatrix} = 14 \cdot (-1) – 3 \cdot (-2) = -14 + 6 = -8, \] \[ \begin{vmatrix} 14 & -1 \\ -2 & 4 \end{vmatrix} = 14 \cdot 4 – (-1)(-2) = 56 – 2 = 54. \] Dosadíme: \[ \det(A_x) = 6 \cdot (-11) – 1 \cdot (-8) + 1 \cdot 54 = -66 + 8 + 54 = -4. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 1 & 6 & 1 \\ 2 & 14 & 3 \\ -1 & -2 & -1 \end{vmatrix} \] Determinant: \[ \det(A_y) = 1 \begin{vmatrix} 14 & 3 \\ -2 & -1 \end{vmatrix} – 6 \begin{vmatrix} 2 & 3 \\ -1 & -1 \end{vmatrix} + 1 \begin{vmatrix} 2 & 14 \\ -1 & -2 \end{vmatrix} \] Výpočty: \[ \begin{vmatrix} 14 & 3 \\ -2 & -1 \end{vmatrix} = -8, \] \[ \begin{vmatrix} 2 & 3 \\ -1 & -1 \end{vmatrix} = 2 \cdot (-1) – 3 \cdot (-1) = -2 + 3 = 1, \] \[ \begin{vmatrix} 2 & 14 \\ -1 & -2 \end{vmatrix} = 2 \cdot (-2) – 14 \cdot (-1) = -4 + 14 = 10. \] Dosadíme: \[ \det(A_y) = 1 \cdot (-8) – 6 \cdot 1 + 1 \cdot 10 = -8 – 6 + 10 = -4. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 1 & 1 & 6 \\ 2 & -1 & 14 \\ -1 & 4 & -2 \end{vmatrix} \] Determinant: \[ \det(A_z) = 1 \begin{vmatrix} -1 & 14 \\ 4 & -2 \end{vmatrix} – 1 \begin{vmatrix} 2 & 14 \\ -1 & -2 \end{vmatrix} + 6 \begin{vmatrix} 2 & -1 \\ -1 & 4 \end{vmatrix} \] Výpočty: \[ \begin{vmatrix} -1 & 14 \\ 4 & -2 \end{vmatrix} = (-1)(-2) – 14 \cdot 4 = 2 – 56 = -54, \] \[ \begin{vmatrix} 2 & 14 \\ -1 & -2 \end{vmatrix} = 2 \cdot (-2) – 14 \cdot (-1) = -4 + 14 = 10, \] \[ \begin{vmatrix} 2 & -1 \\ -1 & 4 \end{vmatrix} = 2 \cdot 4 – (-1)(-1) = 8 – 1 = 7. \] Dosadíme: \[ \det(A_z) = 1 \cdot (-54) – 1 \cdot 10 + 6 \cdot 7 = -54 – 10 + 42 = -22. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-4}{-5} = \frac{4}{5}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-4}{-5} = \frac{4}{5}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{-22}{-5} = \frac{22}{5}. \]

Řešení příkladu 38:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & 2 & -1 \\ 3 & -1 & 4 \\ 2 & 1 & 3 \end{vmatrix} \] Determinant: \[ \det(A) = 1 \begin{vmatrix} -1 & 4 \\ 1 & 3 \end{vmatrix} – 2 \begin{vmatrix} 3 & 4 \\ 2 & 3 \end{vmatrix} + (-1) \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} \] Výpočty: \[ \begin{vmatrix} -1 & 4 \\ 1 & 3 \end{vmatrix} = (-1)(3) – 4 \cdot 1 = -3 – 4 = -7, \] \[ \begin{vmatrix} 3 & 4 \\ 2 & 3 \end{vmatrix} = 3 \cdot 3 – 4 \cdot 2 = 9 – 8 = 1, \] \[ \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} = 3 \cdot 1 – (-1)(2) = 3 + 2 = 5. \] Dosadíme: \[ \det(A) = 1 \cdot (-7) – 2 \cdot 1 + (-1) \cdot 5 = -7 – 2 – 5 = -14. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 0 & 2 & -1 \\ 10 & -1 & 4 \\ 13 & 1 & 3 \end{vmatrix} \] Determinant: \[ \det(A_x) = 0 \begin{vmatrix} -1 & 4 \\ 1 & 3 \end{vmatrix} – 2 \begin{vmatrix} 10 & 4 \\ 13 & 3 \end{vmatrix} + (-1) \begin{vmatrix} 10 & -1 \\ 13 & 1 \end{vmatrix} \] Výpočty: \[ \begin{vmatrix} -1 & 4 \\ 1 & 3 \end{vmatrix} = -7, \] \[ \begin{vmatrix} 10 & 4 \\ 13 & 3 \end{vmatrix} = 10 \cdot 3 – 4 \cdot 13 = 30 – 52 = -22, \] \[ \begin{vmatrix} 10 & -1 \\ 13 & 1 \end{vmatrix} = 10 \cdot 1 – (-1) \cdot 13 = 10 + 13 = 23. \] Dosadíme: \[ \det(A_x) = 0 \cdot (-7) – 2 \cdot (-22) + (-1) \cdot 23 = 0 + 44 – 23 = 21. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 1 & 0 & -1 \\ 3 & 10 & 4 \\ 2 & 13 & 3 \end{vmatrix} \] Determinant: \[ \det(A_y) = 1 \begin{vmatrix} 10 & 4 \\ 13 & 3 \end{vmatrix} – 0 \begin{vmatrix} 3 & 4 \\ 2 & 3 \end{vmatrix} + (-1) \begin{vmatrix} 3 & 10 \\ 2 & 13 \end{vmatrix} \] Výpočty: \[ \begin{vmatrix} 10 & 4 \\ 13 & 3 \end{vmatrix} = -22, \] \[ \begin{vmatrix} 3 & 10 \\ 2 & 13 \end{vmatrix} = 3 \cdot 13 – 10 \cdot 2 = 39 – 20 = 19. \] Dosadíme: \[ \det(A_y) = 1 \cdot (-22) – 0 + (-1) \cdot 19 = -22 – 19 = -41. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 1 & 2 & 0 \\ 3 & -1 & 10 \\ 2 & 1 & 13 \end{vmatrix} \] Determinant: \[ \det(A_z) = 1 \begin{vmatrix} -1 & 10 \\ 1 & 13 \end{vmatrix} – 2 \begin{vmatrix} 3 & 10 \\ 2 & 13 \end{vmatrix} + 0 \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} \] Výpočty: \[ \begin{vmatrix} -1 & 10 \\ 1 & 13 \end{vmatrix} = (-1)(13) – 10 \cdot 1 = -13 – 10 = -23, \] \[ \begin{vmatrix} 3 & 10 \\ 2 & 13 \end{vmatrix} = 3 \cdot 13 – 10 \cdot 2 = 39 – 20 = 19. \] Dosadíme: \[ \det(A_z) = 1 \cdot (-23) – 2 \cdot 19 + 0 = -23 – 38 = -61. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{21}{-14} = -\frac{3}{2}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-41}{-14} = \frac{41}{14}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{-61}{-14} = \frac{61}{14}. \]

Řešení příkladu 39:

Nejprve sestavíme koeficientovou matici soustavy: \[ A = \begin{vmatrix} 2 & -1 & 3 \\ -1 & 4 & -2 \\ 3 & 2 & 1 \end{vmatrix}. \] Vypočítáme determinant matice \(A\): \[ \det(A) = 2 \begin{vmatrix} 4 & -2 \\ 2 & 1 \end{vmatrix} – (-1) \begin{vmatrix} -1 & -2 \\ 3 & 1 \end{vmatrix} + 3 \begin{vmatrix} -1 & 4 \\ 3 & 2 \end{vmatrix}. \] Nyní jednotlivé determinanty 2×2: \[ \begin{vmatrix} 4 & -2 \\ 2 & 1 \end{vmatrix} = 4 \cdot 1 – (-2) \cdot 2 = 4 + 4 = 8, \] \[ \begin{vmatrix} -1 & -2 \\ 3 & 1 \end{vmatrix} = (-1) \cdot 1 – (-2) \cdot 3 = -1 + 6 = 5, \] \[ \begin{vmatrix} -1 & 4 \\ 3 & 2 \end{vmatrix} = (-1) \cdot 2 – 4 \cdot 3 = -2 – 12 = -14. \] Dosadíme zpět: \[ \det(A) = 2 \cdot 8 + 1 \cdot 5 + 3 \cdot (-14) = 16 + 5 – 42 = -21. \] Protože \(\det(A) \neq 0\), soustava má jedinečné řešení. — Vypočítáme determinant matice \(A_x\), kde první sloupec nahradíme pravými stranami rovnic: \[ A_x = \begin{vmatrix} 7 & -1 & 3 \\ -1 & 4 & -2 \\ 10 & 2 & 1 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 7 \begin{vmatrix} 4 & -2 \\ 2 & 1 \end{vmatrix} – (-1) \begin{vmatrix} -1 & -2 \\ 10 & 1 \end{vmatrix} + 3 \begin{vmatrix} -1 & 4 \\ 10 & 2 \end{vmatrix}. \] Vypočítáme determinanty 2×2: \[ \begin{vmatrix} 4 & -2 \\ 2 & 1 \end{vmatrix} = 8 \quad (\text{z předchozího výpočtu}), \] \[ \begin{vmatrix} -1 & -2 \\ 10 & 1 \end{vmatrix} = (-1) \cdot 1 – (-2) \cdot 10 = -1 + 20 = 19, \] \[ \begin{vmatrix} -1 & 4 \\ 10 & 2 \end{vmatrix} = (-1) \cdot 2 – 4 \cdot 10 = -2 – 40 = -42. \] Dosadíme: \[ \det(A_x) = 7 \cdot 8 + 1 \cdot 19 + 3 \cdot (-42) = 56 + 19 – 126 = -51. \] — Matice \(A_y\), druhý sloupec nahradíme pravými stranami: \[ A_y = \begin{vmatrix} 2 & 7 & 3 \\ -1 & -1 & -2 \\ 3 & 10 & 1 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 2 \begin{vmatrix} -1 & -2 \\ 10 & 1 \end{vmatrix} – 7 \begin{vmatrix} -1 & -2 \\ 3 & 1 \end{vmatrix} + 3 \begin{vmatrix} -1 & -1 \\ 3 & 10 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & -2 \\ 10 & 1 \end{vmatrix} = 19 \quad (\text{předchozí výpočet}), \] \[ \begin{vmatrix} -1 & -2 \\ 3 & 1 \end{vmatrix} = (-1) \cdot 1 – (-2) \cdot 3 = -1 + 6 = 5, \] \[ \begin{vmatrix} -1 & -1 \\ 3 & 10 \end{vmatrix} = (-1) \cdot 10 – (-1) \cdot 3 = -10 + 3 = -7. \] Dosadíme: \[ \det(A_y) = 2 \cdot 19 – 7 \cdot 5 + 3 \cdot (-7) = 38 – 35 – 21 = -18. \] — Matice \(A_z\), třetí sloupec nahradíme pravými stranami: \[ A_z = \begin{vmatrix} 2 & -1 & 7 \\ -1 & 4 & -1 \\ 3 & 2 & 10 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 2 \begin{vmatrix} 4 & -1 \\ 2 & 10 \end{vmatrix} – (-1) \begin{vmatrix} -1 & -1 \\ 3 & 10 \end{vmatrix} + 7 \begin{vmatrix} -1 & 4 \\ 3 & 2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & -1 \\ 2 & 10 \end{vmatrix} = 4 \cdot 10 – (-1) \cdot 2 = 40 + 2 = 42, \] \[ \begin{vmatrix} -1 & -1 \\ 3 & 10 \end{vmatrix} = -7 \quad (\text{předchozí výpočet}), \] \[ \begin{vmatrix} -1 & 4 \\ 3 & 2 \end{vmatrix} = -14 \quad (\text{předchozí výpočet}). \] Dosadíme: \[ \det(A_z) = 2 \cdot 42 + 1 \cdot (-7) + 7 \cdot (-14) = 84 – 7 – 98 = -21. \] — Nakonec spočítáme řešení: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-51}{-21} = \frac{51}{21} = \frac{17}{7}, \] \[ y = \frac{\det(A_y)}{\det(A)} = \frac{-18}{-21} = \frac{18}{21} = \frac{6}{7}, \] \[ z = \frac{\det(A_z)}{\det(A)} = \frac{-21}{-21} = 1. \] Tedy \[ x = \frac{17}{7}, \quad y = \frac{6}{7}, \quad z = 1. \]

Řešení příkladu 40:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \\ 3 & 1 & 2 \end{vmatrix}. \] Determinant spočítáme podle pravidla pro 3×3 matice: \[ \det(A) = 1 \cdot \begin{vmatrix} -1 & 3 \\ 1 & 2 \end{vmatrix} – 2 \cdot \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix}. \] Vypočítáme jednotlivé determinanty 2×2: \[ \begin{vmatrix} -1 & 3 \\ 1 & 2 \end{vmatrix} = (-1) \cdot 2 – 3 \cdot 1 = -2 – 3 = -5, \] \[ \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} = 2 \cdot 2 – 3 \cdot 3 = 4 – 9 = -5, \] \[ \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} = 2 \cdot 1 – (-1) \cdot 3 = 2 + 3 = 5. \] Dosadíme: \[ \det(A) = 1 \cdot (-5) – 2 \cdot (-5) + (-1) \cdot 5 = -5 + 10 – 5 = 0. \] Jelikož \(\det(A) = 0\), soustava může být buď neřešitelná, nebo mít nekonečně mnoho řešení. Dále analyzujeme soustavu.

Řešení příkladu 41:

Koeficientová matice: \[ A = \begin{vmatrix} 2 & -1 & 1 \\ 1 & 3 & -2 \\ 4 & -1 & 5 \end{vmatrix}. \] Determinant: \[ \det(A) = 2 \cdot \begin{vmatrix} 3 & -2 \\ -1 & 5 \end{vmatrix} – (-1) \cdot \begin{vmatrix} 1 & -2 \\ 4 & 5 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 3 \\ 4 & -1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 3 & -2 \\ -1 & 5 \end{vmatrix} = 3 \cdot 5 – (-2) \cdot (-1) = 15 – 2 = 13, \] \[ \begin{vmatrix} 1 & -2 \\ 4 & 5 \end{vmatrix} = 1 \cdot 5 – (-2) \cdot 4 = 5 + 8 = 13, \] \[ \begin{vmatrix} 1 & 3 \\ 4 & -1 \end{vmatrix} = 1 \cdot (-1) – 3 \cdot 4 = -1 – 12 = -13. \] Dosadíme: \[ \det(A) = 2 \cdot 13 + 1 \cdot 13 + 1 \cdot (-13) = 26 + 13 – 13 = 26. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 5 & -1 & 1 \\ -4 & 3 & -2 \\ 12 & -1 & 5 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 5 \cdot \begin{vmatrix} 3 & -2 \\ -1 & 5 \end{vmatrix} – (-1) \cdot \begin{vmatrix} -4 & -2 \\ 12 & 5 \end{vmatrix} + 1 \cdot \begin{vmatrix} -4 & 3 \\ 12 & -1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 3 & -2 \\ -1 & 5 \end{vmatrix} = 13, \] \[ \begin{vmatrix} -4 & -2 \\ 12 & 5 \end{vmatrix} = (-4) \cdot 5 – (-2) \cdot 12 = -20 + 24 = 4, \] \[ \begin{vmatrix} -4 & 3 \\ 12 & -1 \end{vmatrix} = (-4) \cdot (-1) – 3 \cdot 12 = 4 – 36 = -32. \] Dosadíme: \[ \det(A_x) = 5 \cdot 13 + 1 \cdot 4 + 1 \cdot (-32) = 65 + 4 – 32 = 37. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 2 & 5 & 1 \\ 1 & -4 & -2 \\ 4 & 12 & 5 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 2 \cdot \begin{vmatrix} -4 & -2 \\ 12 & 5 \end{vmatrix} – 5 \cdot \begin{vmatrix} 1 & -2 \\ 4 & 5 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -4 \\ 4 & 12 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -4 & -2 \\ 12 & 5 \end{vmatrix} = 4, \] \[ \begin{vmatrix} 1 & -2 \\ 4 & 5 \end{vmatrix} = 1 \cdot 5 – (-2) \cdot 4 = 5 + 8 = 13, \] \[ \begin{vmatrix} 1 & -4 \\ 4 & 12 \end{vmatrix} = 1 \cdot 12 – (-4) \cdot 4 = 12 + 16 = 28. \] Dosadíme: \[ \det(A_y) = 2 \cdot 4 – 5 \cdot 13 + 1 \cdot 28 = 8 – 65 + 28 = -29. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 2 & -1 & 5 \\ 1 & 3 & -4 \\ 4 & -1 & 12 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 2 \cdot \begin{vmatrix} 3 & -4 \\ -1 & 12 \end{vmatrix} – (-1) \cdot \begin{vmatrix} 1 & -4 \\ 4 & 12 \end{vmatrix} + 5 \cdot \begin{vmatrix} 1 & 3 \\ 4 & -1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 3 & -4 \\ -1 & 12 \end{vmatrix} = 3 \cdot 12 – (-4) \cdot (-1) = 36 – 4 = 32, \] \[ \begin{vmatrix} 1 & -4 \\ 4 & 12 \end{vmatrix} = 1 \cdot 12 – (-4) \cdot 4 = 12 + 16 = 28, \] \[ \begin{vmatrix} 1 & 3 \\ 4 & -1 \end{vmatrix} = 1 \cdot (-1) – 3 \cdot 4 = -1 – 12 = -13. \] Dosadíme: \[ \det(A_z) = 2 \cdot 32 + 1 \cdot 28 + 5 \cdot (-13) = 64 + 28 – 65 = 27. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{37}{26}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-29}{26}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{27}{26}. \]

Řešení příkladu 42:

Koeficientová matice: \[ A = \begin{vmatrix} 3 & 1 & -2 \\ 1 & -4 & 1 \\ 2 & 3 & -1 \end{vmatrix}. \] Determinant: \[ \det(A) = 3 \cdot \begin{vmatrix} -4 & 1 \\ 3 & -1 \end{vmatrix} – 1 \cdot \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} + (-2) \cdot \begin{vmatrix} 1 & -4 \\ 2 & 3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -4 & 1 \\ 3 & -1 \end{vmatrix} = (-4) \cdot (-1) – 1 \cdot 3 = 4 – 3 = 1, \] \[ \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = 1 \cdot (-1) – 1 \cdot 2 = -1 – 2 = -3, \] \[ \begin{vmatrix} 1 & -4 \\ 2 & 3 \end{vmatrix} = 1 \cdot 3 – (-4) \cdot 2 = 3 + 8 = 11. \] Dosadíme: \[ \det(A) = 3 \cdot 1 – 1 \cdot (-3) – 2 \cdot 11 = 3 + 3 – 22 = -16. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 5 & 1 & -2 \\ -2 & -4 & 1 \\ 4 & 3 & -1 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 5 \cdot \begin{vmatrix} -4 & 1 \\ 3 & -1 \end{vmatrix} – 1 \cdot \begin{vmatrix} -2 & 1 \\ 4 & -1 \end{vmatrix} + (-2) \cdot \begin{vmatrix} -2 & -4 \\ 4 & 3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -4 & 1 \\ 3 & -1 \end{vmatrix} = 1, \] \[ \begin{vmatrix} -2 & 1 \\ 4 & -1 \end{vmatrix} = (-2) \cdot (-1) – 1 \cdot 4 = 2 – 4 = -2, \] \[ \begin{vmatrix} -2 & -4 \\ 4 & 3 \end{vmatrix} = (-2) \cdot 3 – (-4) \cdot 4 = -6 + 16 = 10. \] Dosadíme: \[ \det(A_x) = 5 \cdot 1 – 1 \cdot (-2) – 2 \cdot 10 = 5 + 2 – 20 = -13. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 3 & 5 & -2 \\ 1 & -2 & 1 \\ 2 & 4 & -1 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 3 \cdot \begin{vmatrix} -2 & 1 \\ 4 & -1 \end{vmatrix} – 5 \cdot \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} + (-2) \cdot \begin{vmatrix} 1 & -2 \\ 2 & 4 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -2 & 1 \\ 4 & -1 \end{vmatrix} = (-2) \cdot (-1) – 1 \cdot 4 = 2 – 4 = -2, \] \[ \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = 1 \cdot (-1) – 1 \cdot 2 = -1 – 2 = -3, \] \[ \begin{vmatrix} 1 & -2 \\ 2 & 4 \end{vmatrix} = 1 \cdot 4 – (-2) \cdot 2 = 4 + 4 = 8. \] Dosadíme: \[ \det(A_y) = 3 \cdot (-2) – 5 \cdot (-3) – 2 \cdot 8 = -6 + 15 – 16 = -7. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 3 & 1 & 5 \\ 1 & -4 & -2 \\ 2 & 3 & 4 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 3 \cdot \begin{vmatrix} -4 & -2 \\ 3 & 4 \end{vmatrix} – 1 \cdot \begin{vmatrix} 1 & -2 \\ 2 & 4 \end{vmatrix} + 5 \cdot \begin{vmatrix} 1 & -4 \\ 2 & 3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -4 & -2 \\ 3 & 4 \end{vmatrix} = (-4) \cdot 4 – (-2) \cdot 3 = -16 + 6 = -10, \] \[ \begin{vmatrix} 1 & -2 \\ 2 & 4 \end{vmatrix} = 1 \cdot 4 – (-2) \cdot 2 = 4 + 4 = 8, \] \[ \begin{vmatrix} 1 & -4 \\ 2 & 3 \end{vmatrix} = 1 \cdot 3 – (-4) \cdot 2 = 3 + 8 = 11. \] Dosadíme: \[ \det(A_z) = 3 \cdot (-10) – 1 \cdot 8 + 5 \cdot 11 = -30 – 8 + 55 = 17. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-13}{-16} = \frac{13}{16}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-7}{-16} = \frac{7}{16}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{17}{-16} = -\frac{17}{16}. \]

Řešení příkladu 43:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ -1 & 4 & -1 \end{vmatrix}. \] Determinant: \[ \det(A) = 1 \cdot \begin{vmatrix} -1 & 3 \\ 4 & -1 \end{vmatrix} – 1 \cdot \begin{vmatrix} 2 & 3 \\ -1 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & -1 \\ -1 & 4 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 3 \\ 4 & -1 \end{vmatrix} = (-1) \cdot (-1) – 3 \cdot 4 = 1 – 12 = -11, \] \[ \begin{vmatrix} 2 & 3 \\ -1 & -1 \end{vmatrix} = 2 \cdot (-1) – 3 \cdot (-1) = -2 + 3 = 1, \] \[ \begin{vmatrix} 2 & -1 \\ -1 & 4 \end{vmatrix} = 2 \cdot 4 – (-1) \cdot (-1) = 8 – 1 = 7. \] Dosadíme: \[ \det(A) = 1 \cdot (-11) – 1 \cdot 1 + 1 \cdot 7 = -11 – 1 + 7 = -5. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 6 & 1 & 1 \\ 14 & -1 & 3 \\ -2 & 4 & -1 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 6 \cdot \begin{vmatrix} -1 & 3 \\ 4 & -1 \end{vmatrix} – 1 \cdot \begin{vmatrix} 14 & 3 \\ -2 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 14 & -1 \\ -2 & 4 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 3 \\ 4 & -1 \end{vmatrix} = -11, \] \[ \begin{vmatrix} 14 & 3 \\ -2 & -1 \end{vmatrix} = 14 \cdot (-1) – 3 \cdot (-2) = -14 + 6 = -8, \] \[ \begin{vmatrix} 14 & -1 \\ -2 & 4 \end{vmatrix} = 14 \cdot 4 – (-1) \cdot (-2) = 56 – 2 = 54. \] Dosadíme: \[ \det(A_x) = 6 \cdot (-11) – 1 \cdot (-8) + 1 \cdot 54 = -66 + 8 + 54 = -4. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 1 & 6 & 1 \\ 2 & 14 & 3 \\ -1 & -2 & -1 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 1 \cdot \begin{vmatrix} 14 & 3 \\ -2 & -1 \end{vmatrix} – 6 \cdot \begin{vmatrix} 2 & 3 \\ -1 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 14 \\ -1 & -2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 14 & 3 \\ -2 & -1 \end{vmatrix} = -8, \] \[ \begin{vmatrix} 2 & 3 \\ -1 & -1 \end{vmatrix} = -1, \] \[ \begin{vmatrix} 2 & 14 \\ -1 & -2 \end{vmatrix} = 2 \cdot (-2) – 14 \cdot (-1) = -4 + 14 = 10. \] Dosadíme: \[ \det(A_y) = 1 \cdot (-8) – 6 \cdot (-1) + 1 \cdot 10 = -8 + 6 + 10 = 8. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 1 & 1 & 6 \\ 2 & -1 & 14 \\ -1 & 4 & -2 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 1 \cdot \begin{vmatrix} -1 & 14 \\ 4 & -2 \end{vmatrix} – 1 \cdot \begin{vmatrix} 2 & 14 \\ -1 & -2 \end{vmatrix} + 6 \cdot \begin{vmatrix} 2 & -1 \\ -1 & 4 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 14 \\ 4 & -2 \end{vmatrix} = (-1) \cdot (-2) – 14 \cdot 4 = 2 – 56 = -54, \] \[ \begin{vmatrix} 2 & 14 \\ -1 & -2 \end{vmatrix} = 2 \cdot (-2) – 14 \cdot (-1) = -4 + 14 = 10, \] \[ \begin{vmatrix} 2 & -1 \\ -1 & 4 \end{vmatrix} = 2 \cdot 4 – (-1) \cdot (-1) = 8 – 1 = 7. \] Dosadíme: \[ \det(A_z) = 1 \cdot (-54) – 1 \cdot 10 + 6 \cdot 7 = -54 – 10 + 42 = -22. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-4}{-5} = \frac{4}{5}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{8}{-5} = -\frac{8}{5}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{-22}{-5} = \frac{22}{5}. \]

Řešení příkladu 44:

Koeficientová matice: \[ A = \begin{vmatrix} 2 & -1 & 1 \\ -1 & 3 & 2 \\ 3 & 1 & -4 \end{vmatrix}. \] Determinant: \[ \det(A) = 2 \cdot \begin{vmatrix} 3 & 2 \\ 1 & -4 \end{vmatrix} – (-1) \cdot \begin{vmatrix} -1 & 2 \\ 3 & -4 \end{vmatrix} + 1 \cdot \begin{vmatrix} -1 & 3 \\ 3 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 3 & 2 \\ 1 & -4 \end{vmatrix} = 3 \cdot (-4) – 2 \cdot 1 = -12 – 2 = -14, \] \[ \begin{vmatrix} -1 & 2 \\ 3 & -4 \end{vmatrix} = (-1) \cdot (-4) – 2 \cdot 3 = 4 – 6 = -2, \] \[ \begin{vmatrix} -1 & 3 \\ 3 & 1 \end{vmatrix} = (-1) \cdot 1 – 3 \cdot 3 = -1 – 9 = -10. \] Dosadíme: \[ \det(A) = 2 \cdot (-14) – (-1) \cdot (-2) + 1 \cdot (-10) = -28 – 2 – 10 = -40. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 3 & -1 & 1 \\ 7 & 3 & 2 \\ -2 & 1 & -4 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 3 \cdot \begin{vmatrix} 3 & 2 \\ 1 & -4 \end{vmatrix} – (-1) \cdot \begin{vmatrix} 7 & 2 \\ -2 & -4 \end{vmatrix} + 1 \cdot \begin{vmatrix} 7 & 3 \\ -2 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 3 & 2 \\ 1 & -4 \end{vmatrix} = -14, \] \[ \begin{vmatrix} 7 & 2 \\ -2 & -4 \end{vmatrix} = 7 \cdot (-4) – 2 \cdot (-2) = -28 + 4 = -24, \] \[ \begin{vmatrix} 7 & 3 \\ -2 & 1 \end{vmatrix} = 7 \cdot 1 – 3 \cdot (-2) = 7 + 6 = 13. \] Dosadíme: \[ \det(A_x) = 3 \cdot (-14) – (-1) \cdot (-24) + 1 \cdot 13 = -42 – 24 + 13 = -53. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 2 & 3 & 1 \\ -1 & 7 & 2 \\ 3 & -2 & -4 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 2 \cdot \begin{vmatrix} 7 & 2 \\ -2 & -4 \end{vmatrix} – 3 \cdot \begin{vmatrix} -1 & 2 \\ 3 & -4 \end{vmatrix} + 1 \cdot \begin{vmatrix} -1 & 7 \\ 3 & -2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 7 & 2 \\ -2 & -4 \end{vmatrix} = -24, \] \[ \begin{vmatrix} -1 & 2 \\ 3 & -4 \end{vmatrix} = (-1) \cdot (-4) – 2 \cdot 3 = 4 – 6 = -2, \] \[ \begin{vmatrix} -1 & 7 \\ 3 & -2 \end{vmatrix} = (-1) \cdot (-2) – 7 \cdot 3 = 2 – 21 = -19. \] Dosadíme: \[ \det(A_y) = 2 \cdot (-24) – 3 \cdot (-2) + 1 \cdot (-19) = -48 + 6 – 19 = -61. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 2 & -1 & 3 \\ -1 & 3 & 7 \\ 3 & 1 & -2 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 2 \cdot \begin{vmatrix} 3 & 7 \\ 1 & -2 \end{vmatrix} – (-1) \cdot \begin{vmatrix} -1 & 7 \\ 3 & -2 \end{vmatrix} + 3 \cdot \begin{vmatrix} -1 & 3 \\ 3 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 3 & 7 \\ 1 & -2 \end{vmatrix} = 3 \cdot (-2) – 7 \cdot 1 = -6 – 7 = -13, \] \[ \begin{vmatrix} -1 & 7 \\ 3 & -2 \end{vmatrix} = (-1) \cdot (-2) – 7 \cdot 3 = 2 – 21 = -19, \] \[ \begin{vmatrix} -1 & 3 \\ 3 & 1 \end{vmatrix} = (-1) \cdot 1 – 3 \cdot 3 = -1 – 9 = -10. \] Dosadíme: \[ \det(A_z) = 2 \cdot (-13) – (-1) \cdot (-19) + 3 \cdot (-10) = -26 – 19 – 30 = -75. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-53}{-40} = \frac{53}{40}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-61}{-40} = \frac{61}{40}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{-75}{-40} = \frac{75}{40} = \frac{15}{8}. \]

Řešení příkladu 45:

Koeficientová matice: \[ A = \begin{vmatrix} 2 & -1 & 3 \\ -1 & 4 & -2 \\ 3 & 1 & 1 \end{vmatrix}. \] Determinant: \[ \det(A) = 2 \cdot \begin{vmatrix} 4 & -2 \\ 1 & 1 \end{vmatrix} – (-1) \cdot \begin{vmatrix} -1 & -2 \\ 3 & 1 \end{vmatrix} + 3 \cdot \begin{vmatrix} -1 & 4 \\ 3 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & -2 \\ 1 & 1 \end{vmatrix} = 4 \cdot 1 – (-2) \cdot 1 = 4 + 2 = 6, \] \[ \begin{vmatrix} -1 & -2 \\ 3 & 1 \end{vmatrix} = (-1) \cdot 1 – (-2) \cdot 3 = -1 + 6 = 5, \] \[ \begin{vmatrix} -1 & 4 \\ 3 & 1 \end{vmatrix} = (-1) \cdot 1 – 4 \cdot 3 = -1 – 12 = -13. \] Dosadíme: \[ \det(A) = 2 \cdot 6 – (-1) \cdot 5 + 3 \cdot (-13) = 12 + 5 – 39 = -22. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 9 & -1 & 3 \\ -1 & 4 & -2 \\ 10 & 1 & 1 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 9 \cdot \begin{vmatrix} 4 & -2 \\ 1 & 1 \end{vmatrix} – (-1) \cdot \begin{vmatrix} -1 & -2 \\ 10 & 1 \end{vmatrix} + 3 \cdot \begin{vmatrix} -1 & 4 \\ 10 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & -2 \\ 1 & 1 \end{vmatrix} = 6, \] \[ \begin{vmatrix} -1 & -2 \\ 10 & 1 \end{vmatrix} = (-1) \cdot 1 – (-2) \cdot 10 = -1 + 20 = 19, \] \[ \begin{vmatrix} -1 & 4 \\ 10 & 1 \end{vmatrix} = (-1) \cdot 1 – 4 \cdot 10 = -1 – 40 = -41. \] Dosadíme: \[ \det(A_x) = 9 \cdot 6 – (-1) \cdot 19 + 3 \cdot (-41) = 54 + 19 – 123 = -50. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 2 & 9 & 3 \\ -1 & -1 & -2 \\ 3 & 10 & 1 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 2 \cdot \begin{vmatrix} -1 & -2 \\ 10 & 1 \end{vmatrix} – 9 \cdot \begin{vmatrix} -1 & -2 \\ 3 & 1 \end{vmatrix} + 3 \cdot \begin{vmatrix} -1 & -1 \\ 3 & 10 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & -2 \\ 10 & 1 \end{vmatrix} = 19, \] \[ \begin{vmatrix} -1 & -2 \\ 3 & 1 \end{vmatrix} = (-1) \cdot 1 – (-2) \cdot 3 = -1 + 6 = 5, \] \[ \begin{vmatrix} -1 & -1 \\ 3 & 10 \end{vmatrix} = (-1) \cdot 10 – (-1) \cdot 3 = -10 + 3 = -7. \] Dosadíme: \[ \det(A_y) = 2 \cdot 19 – 9 \cdot 5 + 3 \cdot (-7) = 38 – 45 – 21 = -28. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 2 & -1 & 9 \\ -1 & 4 & -1 \\ 3 & 1 & 10 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 2 \cdot \begin{vmatrix} 4 & -1 \\ 1 & 10 \end{vmatrix} – (-1) \cdot \begin{vmatrix} -1 & -1 \\ 3 & 10 \end{vmatrix} + 9 \cdot \begin{vmatrix} -1 & 4 \\ 3 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & -1 \\ 1 & 10 \end{vmatrix} = 4 \cdot 10 – (-1) \cdot 1 = 40 + 1 = 41, \] \[ \begin{vmatrix} -1 & -1 \\ 3 & 10 \end{vmatrix} = -7, \] \[ \begin{vmatrix} -1 & 4 \\ 3 & 1 \end{vmatrix} = -13. \] Dosadíme: \[ \det(A_z) = 2 \cdot 41 – (-1) \cdot (-7) + 9 \cdot (-13) = 82 – 7 – 117 = -42. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-50}{-22} = \frac{25}{11}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-28}{-22} = \frac{14}{11}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{-42}{-22} = \frac{21}{11}. \]

Řešení příkladu 46:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & 2 & -1 \\ 4 & -1 & 5 \\ -2 & 3 & 2 \end{vmatrix}. \] Determinant: \[ \det(A) = 1 \cdot \begin{vmatrix} -1 & 5 \\ 3 & 2 \end{vmatrix} – 2 \cdot \begin{vmatrix} 4 & 5 \\ -2 & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 4 & -1 \\ -2 & 3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 5 \\ 3 & 2 \end{vmatrix} = (-1) \cdot 2 – 5 \cdot 3 = -2 – 15 = -17, \] \[ \begin{vmatrix} 4 & 5 \\ -2 & 2 \end{vmatrix} = 4 \cdot 2 – 5 \cdot (-2) = 8 + 10 = 18, \] \[ \begin{vmatrix} 4 & -1 \\ -2 & 3 \end{vmatrix} = 4 \cdot 3 – (-1) \cdot (-2) = 12 – 2 = 10. \] Dosadíme: \[ \det(A) = 1 \cdot (-17) – 2 \cdot 18 + (-1) \cdot 10 = -17 – 36 – 10 = -63. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 3 & 2 & -1 \\ 10 & -1 & 5 \\ 1 & 3 & 2 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 3 \cdot \begin{vmatrix} -1 & 5 \\ 3 & 2 \end{vmatrix} – 2 \cdot \begin{vmatrix} 10 & 5 \\ 1 & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 10 & -1 \\ 1 & 3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 5 \\ 3 & 2 \end{vmatrix} = -17, \] \[ \begin{vmatrix} 10 & 5 \\ 1 & 2 \end{vmatrix} = 10 \cdot 2 – 5 \cdot 1 = 20 – 5 = 15, \] \[ \begin{vmatrix} 10 & -1 \\ 1 & 3 \end{vmatrix} = 10 \cdot 3 – (-1) \cdot 1 = 30 + 1 = 31. \] Dosadíme: \[ \det(A_x) = 3 \cdot (-17) – 2 \cdot 15 + (-1) \cdot 31 = -51 – 30 – 31 = -112. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 1 & 3 & -1 \\ 4 & 10 & 5 \\ -2 & 1 & 2 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 1 \cdot \begin{vmatrix} 10 & 5 \\ 1 & 2 \end{vmatrix} – 3 \cdot \begin{vmatrix} 4 & 5 \\ -2 & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 4 & 10 \\ -2 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 10 & 5 \\ 1 & 2 \end{vmatrix} = 15, \] \[ \begin{vmatrix} 4 & 5 \\ -2 & 2 \end{vmatrix} = 18, \] \[ \begin{vmatrix} 4 & 10 \\ -2 & 1 \end{vmatrix} = 4 \cdot 1 – 10 \cdot (-2) = 4 + 20 = 24. \] Dosadíme: \[ \det(A_y) = 1 \cdot 15 – 3 \cdot 18 + (-1) \cdot 24 = 15 – 54 – 24 = -63. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 1 & 2 & 3 \\ 4 & -1 & 10 \\ -2 & 3 & 1 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 1 \cdot \begin{vmatrix} -1 & 10 \\ 3 & 1 \end{vmatrix} – 2 \cdot \begin{vmatrix} 4 & 10 \\ -2 & 1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 4 & -1 \\ -2 & 3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 10 \\ 3 & 1 \end{vmatrix} = (-1) \cdot 1 – 10 \cdot 3 = -1 – 30 = -31, \] \[ \begin{vmatrix} 4 & 10 \\ -2 & 1 \end{vmatrix} = 24, \] \[ \begin{vmatrix} 4 & -1 \\ -2 & 3 \end{vmatrix} = 4 \cdot 3 – (-1) \cdot (-2) = 12 – 2 = 10. \] Dosadíme: \[ \det(A_z) = 1 \cdot (-31) – 2 \cdot 24 + 3 \cdot 10 = -31 – 48 + 30 = -49. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-112}{-63} = \frac{112}{63} = \frac{16}{9}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-63}{-63} = 1, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{-49}{-63} = \frac{49}{63} = \frac{7}{9}. \]

Řešení příkladu 46:

Koeficientová matice: \[ A = \begin{vmatrix} 2 & -1 & 3 \\ 1 & 4 & -1 \\ 3 & -2 & 5 \end{vmatrix}. \] Determinant: \[ \det(A) = 2 \cdot \begin{vmatrix} 4 & -1 \\ -2 & 5 \end{vmatrix} – (-1) \cdot \begin{vmatrix} 1 & -1 \\ 3 & 5 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & 4 \\ 3 & -2 \end{vmatrix}. \] Výpočty determinantů 2×2: \[ \begin{vmatrix} 4 & -1 \\ -2 & 5 \end{vmatrix} = 4 \cdot 5 – (-1) \cdot (-2) = 20 – 2 = 18, \] \[ \begin{vmatrix} 1 & -1 \\ 3 & 5 \end{vmatrix} = 1 \cdot 5 – (-1) \cdot 3 = 5 + 3 = 8, \] \[ \begin{vmatrix} 1 & 4 \\ 3 & -2 \end{vmatrix} = 1 \cdot (-2) – 4 \cdot 3 = -2 – 12 = -14. \] Dosadíme: \[ \det(A) = 2 \cdot 18 + 1 \cdot 8 + 3 \cdot (-14) = 36 + 8 – 42 = 2. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 7 & -1 & 3 \\ 3 & 4 & -1 \\ 10 & -2 & 5 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 7 \cdot \begin{vmatrix} 4 & -1 \\ -2 & 5 \end{vmatrix} – (-1) \cdot \begin{vmatrix} 3 & -1 \\ 10 & 5 \end{vmatrix} + 3 \cdot \begin{vmatrix} 3 & 4 \\ 10 & -2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & -1 \\ -2 & 5 \end{vmatrix} = 18, \] \[ \begin{vmatrix} 3 & -1 \\ 10 & 5 \end{vmatrix} = 3 \cdot 5 – (-1) \cdot 10 = 15 + 10 = 25, \] \[ \begin{vmatrix} 3 & 4 \\ 10 & -2 \end{vmatrix} = 3 \cdot (-2) – 4 \cdot 10 = -6 – 40 = -46. \] Dosadíme: \[ \det(A_x) = 7 \cdot 18 + 1 \cdot 25 + 3 \cdot (-46) = 126 + 25 – 138 = 13. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 2 & 7 & 3 \\ 1 & 3 & -1 \\ 3 & 10 & 5 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 2 \cdot \begin{vmatrix} 3 & -1 \\ 10 & 5 \end{vmatrix} – 7 \cdot \begin{vmatrix} 1 & -1 \\ 3 & 5 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & 3 \\ 3 & 10 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 3 & -1 \\ 10 & 5 \end{vmatrix} = 3 \cdot 5 – (-1) \cdot 10 = 15 + 10 = 25, \] \[ \begin{vmatrix} 1 & -1 \\ 3 & 5 \end{vmatrix} = 1 \cdot 5 – (-1) \cdot 3 = 5 + 3 = 8, \] \[ \begin{vmatrix} 1 & 3 \\ 3 & 10 \end{vmatrix} = 1 \cdot 10 – 3 \cdot 3 = 10 – 9 = 1. \] Dosadíme: \[ \det(A_y) = 2 \cdot 25 – 7 \cdot 8 + 3 \cdot 1 = 50 – 56 + 3 = -3. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 2 & -1 & 7 \\ 1 & 4 & 3 \\ 3 & -2 & 10 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 2 \cdot \begin{vmatrix} 4 & 3 \\ -2 & 10 \end{vmatrix} – (-1) \cdot \begin{vmatrix} 1 & 3 \\ 3 & 10 \end{vmatrix} + 7 \cdot \begin{vmatrix} 1 & 4 \\ 3 & -2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & 3 \\ -2 & 10 \end{vmatrix} = 4 \cdot 10 – 3 \cdot (-2) = 40 + 6 = 46, \] \[ \begin{vmatrix} 1 & 3 \\ 3 & 10 \end{vmatrix} = 1 \cdot 10 – 3 \cdot 3 = 10 – 9 = 1, \] \[ \begin{vmatrix} 1 & 4 \\ 3 & -2 \end{vmatrix} = 1 \cdot (-2) – 4 \cdot 3 = -2 – 12 = -14. \] Dosadíme: \[ \det(A_z) = 2 \cdot 46 + 1 \cdot 1 + 7 \cdot (-14) = 92 + 1 – 98 = -5. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{13}{2}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-3}{2}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{-5}{2}. \]

Řešení příkladu 47:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & 2 & -1 \\ 3 & -1 & 4 \\ 2 & 1 & 3 \end{vmatrix}. \] Determinant: \[ \det(A) = 1 \cdot \begin{vmatrix} -1 & 4 \\ 1 & 3 \end{vmatrix} – 2 \cdot \begin{vmatrix} 3 & 4 \\ 2 & 3 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 4 \\ 1 & 3 \end{vmatrix} = (-1) \cdot 3 – 4 \cdot 1 = -3 – 4 = -7, \] \[ \begin{vmatrix} 3 & 4 \\ 2 & 3 \end{vmatrix} = 3 \cdot 3 – 4 \cdot 2 = 9 – 8 = 1, \] \[ \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} = 3 \cdot 1 – (-1) \cdot 2 = 3 + 2 = 5. \] Dosadíme: \[ \det(A) = 1 \cdot (-7) – 2 \cdot 1 + (-1) \cdot 5 = -7 – 2 – 5 = -14. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 4 & 2 & -1 \\ 10 & -1 & 4 \\ 13 & 1 & 3 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 4 \cdot \begin{vmatrix} -1 & 4 \\ 1 & 3 \end{vmatrix} – 2 \cdot \begin{vmatrix} 10 & 4 \\ 13 & 3 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 10 & -1 \\ 13 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 4 \\ 1 & 3 \end{vmatrix} = -7, \] \[ \begin{vmatrix} 10 & 4 \\ 13 & 3 \end{vmatrix} = 10 \cdot 3 – 4 \cdot 13 = 30 – 52 = -22, \] \[ \begin{vmatrix} 10 & -1 \\ 13 & 1 \end{vmatrix} = 10 \cdot 1 – (-1) \cdot 13 = 10 + 13 = 23. \] Dosadíme: \[ \det(A_x) = 4 \cdot (-7) – 2 \cdot (-22) + (-1) \cdot 23 = -28 + 44 – 23 = -7. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 1 & 4 & -1 \\ 3 & 10 & 4 \\ 2 & 13 & 3 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 1 \cdot \begin{vmatrix} 10 & 4 \\ 13 & 3 \end{vmatrix} – 4 \cdot \begin{vmatrix} 3 & 4 \\ 2 & 3 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 3 & 10 \\ 2 & 13 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 10 & 4 \\ 13 & 3 \end{vmatrix} = -22, \] \[ \begin{vmatrix} 3 & 4 \\ 2 & 3 \end{vmatrix} = 1, \] \[ \begin{vmatrix} 3 & 10 \\ 2 & 13 \end{vmatrix} = 3 \cdot 13 – 10 \cdot 2 = 39 – 20 = 19. \] Dosadíme: \[ \det(A_y) = 1 \cdot (-22) – 4 \cdot 1 + (-1) \cdot 19 = -22 – 4 – 19 = -45. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 1 & 2 & 4 \\ 3 & -1 & 10 \\ 2 & 1 & 13 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 1 \cdot \begin{vmatrix} -1 & 10 \\ 1 & 13 \end{vmatrix} – 2 \cdot \begin{vmatrix} 3 & 10 \\ 2 & 13 \end{vmatrix} + 4 \cdot \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 10 \\ 1 & 13 \end{vmatrix} = (-1) \cdot 13 – 10 \cdot 1 = -13 – 10 = -23, \] \[ \begin{vmatrix} 3 & 10 \\ 2 & 13 \end{vmatrix} = 3 \cdot 13 – 10 \cdot 2 = 39 – 20 = 19, \] \[ \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} = 3 \cdot 1 – (-1) \cdot 2 = 3 + 2 = 5. \] Dosadíme: \[ \det(A_z) = 1 \cdot (-23) – 2 \cdot 19 + 4 \cdot 5 = -23 – 38 + 20 = -41. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-7}{-14} = \frac{1}{2}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-45}{-14} = \frac{45}{14}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{-41}{-14} = \frac{41}{14}. \]

Řešení příkladu 48:

Koeficientová matice: \[ A = \begin{vmatrix} 2 & -1 & 3 \\ -1 & 4 & -1 \\ 3 & 2 & 1 \end{vmatrix}. \] Determinant: \[ \det(A) = 2 \cdot \begin{vmatrix} 4 & -1 \\ 2 & 1 \end{vmatrix} – (-1) \cdot \begin{vmatrix} -1 & -1 \\ 3 & 1 \end{vmatrix} + 3 \cdot \begin{vmatrix} -1 & 4 \\ 3 & 2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & -1 \\ 2 & 1 \end{vmatrix} = 4 \cdot 1 – (-1) \cdot 2 = 4 + 2 = 6, \] \[ \begin{vmatrix} -1 & -1 \\ 3 & 1 \end{vmatrix} = (-1) \cdot 1 – (-1) \cdot 3 = -1 + 3 = 2, \] \[ \begin{vmatrix} -1 & 4 \\ 3 & 2 \end{vmatrix} = (-1) \cdot 2 – 4 \cdot 3 = -2 – 12 = -14. \] Dosadíme: \[ \det(A) = 2 \cdot 6 – (-1) \cdot 2 + 3 \cdot (-14) = 12 + 2 – 42 = -28. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 9 & -1 & 3 \\ 2 & 4 & -1 \\ 7 & 2 & 1 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 9 \cdot \begin{vmatrix} 4 & -1 \\ 2 & 1 \end{vmatrix} – (-1) \cdot \begin{vmatrix} 2 & -1 \\ 7 & 1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 2 & 4 \\ 7 & 2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & -1 \\ 2 & 1 \end{vmatrix} = 6, \] \[ \begin{vmatrix} 2 & -1 \\ 7 & 1 \end{vmatrix} = 2 \cdot 1 – (-1) \cdot 7 = 2 + 7 = 9, \] \[ \begin{vmatrix} 2 & 4 \\ 7 & 2 \end{vmatrix} = 2 \cdot 2 – 4 \cdot 7 = 4 – 28 = -24. \] Dosadíme: \[ \det(A_x) = 9 \cdot 6 – (-1) \cdot 9 + 3 \cdot (-24) = 54 + 9 – 72 = -9. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 2 & 9 & 3 \\ -1 & 2 & -1 \\ 3 & 7 & 1 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 2 \cdot \begin{vmatrix} 2 & -1 \\ 7 & 1 \end{vmatrix} – 9 \cdot \begin{vmatrix} -1 & -1 \\ 3 & 1 \end{vmatrix} + 3 \cdot \begin{vmatrix} -1 & 2 \\ 3 & 7 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 2 & -1 \\ 7 & 1 \end{vmatrix} = 2 \cdot 1 – (-1) \cdot 7 = 2 + 7 = 9, \] \[ \begin{vmatrix} -1 & -1 \\ 3 & 1 \end{vmatrix} = 2, \] \[ \begin{vmatrix} -1 & 2 \\ 3 & 7 \end{vmatrix} = (-1) \cdot 7 – 2 \cdot 3 = -7 – 6 = -13. \] Dosadíme: \[ \det(A_y) = 2 \cdot 9 – 9 \cdot 2 + 3 \cdot (-13) = 18 – 18 – 39 = -39. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 2 & -1 & 9 \\ -1 & 4 & 2 \\ 3 & 2 & 7 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 2 \cdot \begin{vmatrix} 4 & 2 \\ 2 & 7 \end{vmatrix} – (-1) \cdot \begin{vmatrix} -1 & 2 \\ 3 & 7 \end{vmatrix} + 9 \cdot \begin{vmatrix} -1 & 4 \\ 3 & 2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & 2 \\ 2 & 7 \end{vmatrix} = 4 \cdot 7 – 2 \cdot 2 = 28 – 4 = 24, \] \[ \begin{vmatrix} -1 & 2 \\ 3 & 7 \end{vmatrix} = (-1) \cdot 7 – 2 \cdot 3 = -7 – 6 = -13, \] \[ \begin{vmatrix} -1 & 4 \\ 3 & 2 \end{vmatrix} = (-1) \cdot 2 – 4 \cdot 3 = -2 – 12 = -14. \] Dosadíme: \[ \det(A_z) = 2 \cdot 24 – (-1) \cdot (-13) + 9 \cdot (-14) = 48 – 13 – 126 = -91. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-9}{-28} = \frac{9}{28}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-39}{-28} = \frac{39}{28}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{-91}{-28} = \frac{91}{28}. \]

Řešení příkladu 49:

Koeficientová matice: \[ A = \begin{vmatrix} 1 & 2 & -1 \\ 3 & -1 & 2 \\ 2 & 1 & 1 \end{vmatrix}. \] Determinant: \[ \det(A) = 1 \cdot \begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix} – 2 \cdot \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix} = (-1) \cdot 1 – 2 \cdot 1 = -1 – 2 = -3, \] \[ \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} = 3 \cdot 1 – 2 \cdot 2 = 3 – 4 = -1, \] \[ \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} = 3 \cdot 1 – (-1) \cdot 2 = 3 + 2 = 5. \] Dosadíme: \[ \det(A) = 1 \cdot (-3) – 2 \cdot (-1) + (-1) \cdot 5 = -3 + 2 – 5 = -6. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 4 & 2 & -1 \\ 5 & -1 & 2 \\ 6 & 1 & 1 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 4 \cdot \begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix} – 2 \cdot \begin{vmatrix} 5 & 2 \\ 6 & 1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 5 & -1 \\ 6 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix} = -3, \] \[ \begin{vmatrix} 5 & 2 \\ 6 & 1 \end{vmatrix} = 5 \cdot 1 – 2 \cdot 6 = 5 – 12 = -7, \] \[ \begin{vmatrix} 5 & -1 \\ 6 & 1 \end{vmatrix} = 5 \cdot 1 – (-1) \cdot 6 = 5 + 6 = 11. \] Dosadíme: \[ \det(A_x) = 4 \cdot (-3) – 2 \cdot (-7) + (-1) \cdot 11 = -12 + 14 – 11 = -9. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 1 & 4 & -1 \\ 3 & 5 & 2 \\ 2 & 6 & 1 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 1 \cdot \begin{vmatrix} 5 & 2 \\ 6 & 1 \end{vmatrix} – 4 \cdot \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 3 & 5 \\ 2 & 6 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 5 & 2 \\ 6 & 1 \end{vmatrix} = -7, \] \[ \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} = 3 \cdot 1 – 2 \cdot 2 = 3 – 4 = -1, \] \[ \begin{vmatrix} 3 & 5 \\ 2 & 6 \end{vmatrix} = 3 \cdot 6 – 5 \cdot 2 = 18 – 10 = 8. \] Dosadíme: \[ \det(A_y) = 1 \cdot (-7) – 4 \cdot (-1) + (-1) \cdot 8 = -7 + 4 – 8 = -11. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 1 & 2 & 4 \\ 3 & -1 & 5 \\ 2 & 1 & 6 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 1 \cdot \begin{vmatrix} -1 & 5 \\ 1 & 6 \end{vmatrix} – 2 \cdot \begin{vmatrix} 3 & 5 \\ 2 & 6 \end{vmatrix} + 4 \cdot \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 5 \\ 1 & 6 \end{vmatrix} = (-1) \cdot 6 – 5 \cdot 1 = -6 – 5 = -11, \] \[ \begin{vmatrix} 3 & 5 \\ 2 & 6 \end{vmatrix} = 8, \] \[ \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} = 3 \cdot 1 – (-1) \cdot 2 = 3 + 2 = 5. \] Dosadíme: \[ \det(A_z) = 1 \cdot (-11) – 2 \cdot 8 + 4 \cdot 5 = -11 – 16 + 20 = -7. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-9}{-6} = \frac{3}{2}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-11}{-6} = \frac{11}{6}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{-7}{-6} = \frac{7}{6}. \]

Řešení příkladu 50:

Koeficientová matice: \[ A = \begin{vmatrix} 2 & -1 & 1 \\ 1 & 3 & -2 \\ 4 & -1 & 5 \end{vmatrix}. \] Determinant: \[ \det(A) = 2 \cdot \begin{vmatrix} 3 & -2 \\ -1 & 5 \end{vmatrix} – (-1) \cdot \begin{vmatrix} 1 & -2 \\ 4 & 5 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 3 \\ 4 & -1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 3 & -2 \\ -1 & 5 \end{vmatrix} = 3 \cdot 5 – (-2) \cdot (-1) = 15 – 2 = 13, \] \[ \begin{vmatrix} 1 & -2 \\ 4 & 5 \end{vmatrix} = 1 \cdot 5 – (-2) \cdot 4 = 5 + 8 = 13, \] \[ \begin{vmatrix} 1 & 3 \\ 4 & -1 \end{vmatrix} = 1 \cdot (-1) – 3 \cdot 4 = -1 – 12 = -13. \] Dosadíme: \[ \det(A) = 2 \cdot 13 + 1 \cdot 13 + 1 \cdot (-13) = 26 + 13 – 13 = 26. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 5 & -1 & 1 \\ -4 & 3 & -2 \\ 10 & -1 & 5 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 5 \cdot \begin{vmatrix} 3 & -2 \\ -1 & 5 \end{vmatrix} – (-1) \cdot \begin{vmatrix} -4 & -2 \\ 10 & 5 \end{vmatrix} + 1 \cdot \begin{vmatrix} -4 & 3 \\ 10 & -1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 3 & -2 \\ -1 & 5 \end{vmatrix} = 13, \] \[ \begin{vmatrix} -4 & -2 \\ 10 & 5 \end{vmatrix} = (-4) \cdot 5 – (-2) \cdot 10 = -20 + 20 = 0, \] \[ \begin{vmatrix} -4 & 3 \\ 10 & -1 \end{vmatrix} = (-4) \cdot (-1) – 3 \cdot 10 = 4 – 30 = -26. \] Dosadíme: \[ \det(A_x) = 5 \cdot 13 + 1 \cdot 0 + 1 \cdot (-26) = 65 + 0 – 26 = 39. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 2 & 5 & 1 \\ 1 & -4 & -2 \\ 4 & 10 & 5 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 2 \cdot \begin{vmatrix} -4 & -2 \\ 10 & 5 \end{vmatrix} – 5 \cdot \begin{vmatrix} 1 & -2 \\ 4 & 5 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -4 \\ 4 & 10 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -4 & -2 \\ 10 & 5 \end{vmatrix} = 0, \] \[ \begin{vmatrix} 1 & -2 \\ 4 & 5 \end{vmatrix} = 1 \cdot 5 – (-2) \cdot 4 = 5 + 8 = 13, \] \[ \begin{vmatrix} 1 & -4 \\ 4 & 10 \end{vmatrix} = 1 \cdot 10 – (-4) \cdot 4 = 10 + 16 = 26. \] Dosadíme: \[ \det(A_y) = 2 \cdot 0 – 5 \cdot 13 + 1 \cdot 26 = 0 – 65 + 26 = -39. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 2 & -1 & 5 \\ 1 & 3 & -4 \\ 4 & -1 & 10 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 2 \cdot \begin{vmatrix} 3 & -4 \\ -1 & 10 \end{vmatrix} – (-1) \cdot \begin{vmatrix} 1 & -4 \\ 4 & 10 \end{vmatrix} + 5 \cdot \begin{vmatrix} 1 & 3 \\ 4 & -1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 3 & -4 \\ -1 & 10 \end{vmatrix} = 3 \cdot 10 – (-4) \cdot (-1) = 30 – 4 = 26, \] \[ \begin{vmatrix} 1 & -4 \\ 4 & 10 \end{vmatrix} = 26, \] \[ \begin{vmatrix} 1 & 3 \\ 4 & -1 \end{vmatrix} = 1 \cdot (-1) – 3 \cdot 4 = -1 – 12 = -13. \] Dosadíme: \[ \det(A_z) = 2 \cdot 26 + 1 \cdot 26 + 5 \cdot (-13) = 52 + 26 – 65 = 13. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{39}{26} = \frac{3}{2}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-39}{26} = -\frac{3}{2}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{13}{26} = \frac{1}{2}. \]