51. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 3x + y – 2z = 7 \\ -x + 4y + z = 1 \\ 2x – 3y + 5z = -3 \end{cases} \)
Řešení příkladu 51:
Koeficientová matice: \[ A = \begin{vmatrix} 3 & 1 & -2 \\ -1 & 4 & 1 \\ 2 & -3 & 5 \end{vmatrix}. \] Determinant: \[ \det(A) = 3 \cdot \begin{vmatrix} 4 & 1 \\ -3 & 5 \end{vmatrix} – 1 \cdot \begin{vmatrix} -1 & 1 \\ 2 & 5 \end{vmatrix} + (-2) \cdot \begin{vmatrix} -1 & 4 \\ 2 & -3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & 1 \\ -3 & 5 \end{vmatrix} = 4 \cdot 5 – 1 \cdot (-3) = 20 + 3 = 23, \] \[ \begin{vmatrix} -1 & 1 \\ 2 & 5 \end{vmatrix} = (-1) \cdot 5 – 1 \cdot 2 = -5 – 2 = -7, \] \[ \begin{vmatrix} -1 & 4 \\ 2 & -3 \end{vmatrix} = (-1) \cdot (-3) – 4 \cdot 2 = 3 – 8 = -5. \] Dosadíme: \[ \det(A) = 3 \cdot 23 – 1 \cdot (-7) + (-2) \cdot (-5) = 69 + 7 + 10 = 86. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 7 & 1 & -2 \\ 1 & 4 & 1 \\ -3 & -3 & 5 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 7 \cdot \begin{vmatrix} 4 & 1 \\ -3 & 5 \end{vmatrix} – 1 \cdot \begin{vmatrix} 1 & 1 \\ -3 & 5 \end{vmatrix} + (-2) \cdot \begin{vmatrix} 1 & 4 \\ -3 & -3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & 1 \\ -3 & 5 \end{vmatrix} = 23, \] \[ \begin{vmatrix} 1 & 1 \\ -3 & 5 \end{vmatrix} = 1 \cdot 5 – 1 \cdot (-3) = 5 + 3 = 8, \] \[ \begin{vmatrix} 1 & 4 \\ -3 & -3 \end{vmatrix} = 1 \cdot (-3) – 4 \cdot (-3) = -3 + 12 = 9. \] Dosadíme: \[ \det(A_x) = 7 \cdot 23 – 1 \cdot 8 + (-2) \cdot 9 = 161 – 8 – 18 = 135. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 3 & 7 & -2 \\ -1 & 1 & 1 \\ 2 & -3 & 5 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 3 \cdot \begin{vmatrix} 1 & 1 \\ -3 & 5 \end{vmatrix} – 7 \cdot \begin{vmatrix} -1 & 1 \\ 2 & 5 \end{vmatrix} + (-2) \cdot \begin{vmatrix} -1 & 1 \\ 2 & -3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 1 & 1 \\ -3 & 5 \end{vmatrix} = 1 \cdot 5 – 1 \cdot (-3) = 5 + 3 = 8, \] \[ \begin{vmatrix} -1 & 1 \\ 2 & 5 \end{vmatrix} = -1 \cdot 5 – 1 \cdot 2 = -5 – 2 = -7, \] \[ \begin{vmatrix} -1 & 1 \\ 2 & -3 \end{vmatrix} = -1 \cdot (-3) – 1 \cdot 2 = 3 – 2 = 1. \] Dosadíme: \[ \det(A_y) = 3 \cdot 8 – 7 \cdot (-7) + (-2) \cdot 1 = 24 + 49 – 2 = 71. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 3 & 1 & 7 \\ -1 & 4 & 1 \\ 2 & -3 & -3 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 3 \cdot \begin{vmatrix} 4 & 1 \\ -3 & -3 \end{vmatrix} – 1 \cdot \begin{vmatrix} -1 & 1 \\ 2 & -3 \end{vmatrix} + 7 \cdot \begin{vmatrix} -1 & 4 \\ 2 & -3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & 1 \\ -3 & -3 \end{vmatrix} = 4 \cdot (-3) – 1 \cdot (-3) = -12 + 3 = -9, \] \[ \begin{vmatrix} -1 & 1 \\ 2 & -3 \end{vmatrix} = -1 \cdot (-3) – 1 \cdot 2 = 3 – 2 = 1, \] \[ \begin{vmatrix} -1 & 4 \\ 2 & -3 \end{vmatrix} = -1 \cdot (-3) – 4 \cdot 2 = 3 – 8 = -5. \] Dosadíme: \[ \det(A_z) = 3 \cdot (-9) – 1 \cdot 1 + 7 \cdot (-5) = -27 – 1 – 35 = -63. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{135}{86}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{71}{86}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{-63}{86}. \]
52. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 2x – y + 3z = 7 \\ x + 4y – z = 1 \\ 3x – 2y + 5z = 10 \end{cases} \)
Řešení příkladu 52:
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: \[ \begin{vmatrix} 4 & -1 \\ -2 & 5 \end{vmatrix} = 4 \cdot 5 – (-1) \cdot (-2) = 20 – 2 = 18, \] \[ \begin{vmatrix} 1 & -1 \\ 3 & 5 \end{vmatrix} = 1 \cdot 5 – (-1) \cdot 3 = 5 + 3 = 8, \] \[ \begin{vmatrix} 1 & 4 \\ 3 & -2 \end{vmatrix} = 1 \cdot (-2) – 4 \cdot 3 = -2 – 12 = -14. \] Dosadíme: \[ \det(A) = 2 \cdot 18 + 1 \cdot 8 + 3 \cdot (-14) = 36 + 8 – 42 = 2. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 7 & -1 & 3 \\ 1 & 4 & -1 \\ 10 & -2 & 5 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 7 \cdot \begin{vmatrix} 4 & -1 \\ -2 & 5 \end{vmatrix} – (-1) \cdot \begin{vmatrix} 1 & -1 \\ 10 & 5 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & 4 \\ 10 & -2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & -1 \\ -2 & 5 \end{vmatrix} = 18, \] \[ \begin{vmatrix} 1 & -1 \\ 10 & 5 \end{vmatrix} = 1 \cdot 5 – (-1) \cdot 10 = 5 + 10 = 15, \] \[ \begin{vmatrix} 1 & 4 \\ 10 & -2 \end{vmatrix} = 1 \cdot (-2) – 4 \cdot 10 = -2 – 40 = -42. \] Dosadíme: \[ \det(A_x) = 7 \cdot 18 + 1 \cdot 15 + 3 \cdot (-42) = 126 + 15 – 126 = 15. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 2 & 7 & 3 \\ 1 & 1 & -1 \\ 3 & 10 & 5 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 2 \cdot \begin{vmatrix} 1 & -1 \\ 10 & 5 \end{vmatrix} – 7 \cdot \begin{vmatrix} 1 & -1 \\ 3 & 5 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & 1 \\ 3 & 10 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 1 & -1 \\ 10 & 5 \end{vmatrix} = 1 \cdot 5 – (-1) \cdot 10 = 5 + 10 = 15, \] \[ \begin{vmatrix} 1 & -1 \\ 3 & 5 \end{vmatrix} = 1 \cdot 5 – (-1) \cdot 3 = 5 + 3 = 8, \] \[ \begin{vmatrix} 1 & 1 \\ 3 & 10 \end{vmatrix} = 1 \cdot 10 – 1 \cdot 3 = 10 – 3 = 7. \] Dosadíme: \[ \det(A_y) = 2 \cdot 15 – 7 \cdot 8 + 3 \cdot 7 = 30 – 56 + 21 = -5. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 2 & -1 & 7 \\ 1 & 4 & 1 \\ 3 & -2 & 10 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 2 \cdot \begin{vmatrix} 4 & 1 \\ -2 & 10 \end{vmatrix} – (-1) \cdot \begin{vmatrix} 1 & 1 \\ 3 & 10 \end{vmatrix} + 7 \cdot \begin{vmatrix} 1 & 4 \\ 3 & -2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & 1 \\ -2 & 10 \end{vmatrix} = 4 \cdot 10 – 1 \cdot (-2) = 40 + 2 = 42, \] \[ \begin{vmatrix} 1 & 1 \\ 3 & 10 \end{vmatrix} = 1 \cdot 10 – 1 \cdot 3 = 10 – 3 = 7, \] \[ \begin{vmatrix} 1 & 4 \\ 3 & -2 \end{vmatrix} = 1 \cdot (-2) – 4 \cdot 3 = -2 – 12 = -14. \] Dosadíme: \[ \det(A_z) = 2 \cdot 42 + 1 \cdot 7 + 7 \cdot (-14) = 84 + 7 – 98 = -7. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{15}{2}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-5}{2}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{-7}{2}. \]
53. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} x + 2y – z = 4 \\ 3x – y + 2z = 7 \\ 2x + y + 3z = 10 \end{cases} \)
Řešení příkladu 53:
Koeficientová matice: \[ A = \begin{vmatrix} 1 & 2 & -1 \\ 3 & -1 & 2 \\ 2 & 1 & 3 \end{vmatrix}. \] Determinant: \[ \det(A) = 1 \cdot \begin{vmatrix} -1 & 2 \\ 1 & 3 \end{vmatrix} – 2 \cdot \begin{vmatrix} 3 & 2 \\ 2 & 3 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 2 \\ 1 & 3 \end{vmatrix} = (-1) \cdot 3 – 2 \cdot 1 = -3 – 2 = -5, \] \[ \begin{vmatrix} 3 & 2 \\ 2 & 3 \end{vmatrix} = 3 \cdot 3 – 2 \cdot 2 = 9 – 4 = 5, \] \[ \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} = 3 \cdot 1 – (-1) \cdot 2 = 3 + 2 = 5. \] Dosadíme: \[ \det(A) = 1 \cdot (-5) – 2 \cdot 5 – 1 \cdot 5 = -5 – 10 – 5 = -20. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 4 & 2 & -1 \\ 7 & -1 & 2 \\ 10 & 1 & 3 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 4 \cdot \begin{vmatrix} -1 & 2 \\ 1 & 3 \end{vmatrix} – 2 \cdot \begin{vmatrix} 7 & 2 \\ 10 & 3 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 7 & -1 \\ 10 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 2 \\ 1 & 3 \end{vmatrix} = -5, \] \[ \begin{vmatrix} 7 & 2 \\ 10 & 3 \end{vmatrix} = 7 \cdot 3 – 2 \cdot 10 = 21 – 20 = 1, \] \[ \begin{vmatrix} 7 & -1 \\ 10 & 1 \end{vmatrix} = 7 \cdot 1 – (-1) \cdot 10 = 7 + 10 = 17. \] Dosadíme: \[ \det(A_x) = 4 \cdot (-5) – 2 \cdot 1 – 1 \cdot 17 = -20 – 2 – 17 = -39. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 1 & 4 & -1 \\ 3 & 7 & 2 \\ 2 & 10 & 3 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 1 \cdot \begin{vmatrix} 7 & 2 \\ 10 & 3 \end{vmatrix} – 4 \cdot \begin{vmatrix} 3 & 2 \\ 2 & 3 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 3 & 7 \\ 2 & 10 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 7 & 2 \\ 10 & 3 \end{vmatrix} = 1, \] \[ \begin{vmatrix} 3 & 2 \\ 2 & 3 \end{vmatrix} = 5, \] \[ \begin{vmatrix} 3 & 7 \\ 2 & 10 \end{vmatrix} = 3 \cdot 10 – 7 \cdot 2 = 30 – 14 = 16. \] Dosadíme: \[ \det(A_y) = 1 \cdot 1 – 4 \cdot 5 – 1 \cdot 16 = 1 – 20 – 16 = -35. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 1 & 2 & 4 \\ 3 & -1 & 7 \\ 2 & 1 & 10 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 1 \cdot \begin{vmatrix} -1 & 7 \\ 1 & 10 \end{vmatrix} – 2 \cdot \begin{vmatrix} 3 & 7 \\ 2 & 10 \end{vmatrix} + 4 \cdot \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} -1 & 7 \\ 1 & 10 \end{vmatrix} = (-1) \cdot 10 – 7 \cdot 1 = -10 – 7 = -17, \] \[ \begin{vmatrix} 3 & 7 \\ 2 & 10 \end{vmatrix} = 3 \cdot 10 – 7 \cdot 2 = 30 – 14 = 16, \] \[ \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} = 3 \cdot 1 – (-1) \cdot 2 = 3 + 2 = 5. \] Dosadíme: \[ \det(A_z) = 1 \cdot (-17) – 2 \cdot 16 + 4 \cdot 5 = -17 – 32 + 20 = -29. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-39}{-20} = \frac{39}{20}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-35}{-20} = \frac{35}{20} = \frac{7}{4}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{-29}{-20} = \frac{29}{20}. \]
54. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 2x + y – z = 5 \\ -x + 3y + 2z = 12 \\ 4x – y + z = 3 \end{cases} \)
Řešení příkladu 54:
Koeficientová matice: \[ A = \begin{vmatrix} 2 & 1 & -1 \\ -1 & 3 & 2 \\ 4 & -1 & 1 \end{vmatrix}. \] Determinant: \[ \det(A) = 2 \cdot \begin{vmatrix} 3 & 2 \\ -1 & 1 \end{vmatrix} – 1 \cdot \begin{vmatrix} -1 & 2 \\ 4 & 1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} -1 & 3 \\ 4 & -1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 3 & 2 \\ -1 & 1 \end{vmatrix} = 3 \cdot 1 – 2 \cdot (-1) = 3 + 2 = 5, \] \[ \begin{vmatrix} -1 & 2 \\ 4 & 1 \end{vmatrix} = (-1) \cdot 1 – 2 \cdot 4 = -1 – 8 = -9, \] \[ \begin{vmatrix} -1 & 3 \\ 4 & -1 \end{vmatrix} = (-1) \cdot (-1) – 3 \cdot 4 = 1 – 12 = -11. \] Dosadíme: \[ \det(A) = 2 \cdot 5 – 1 \cdot (-9) – 1 \cdot (-11) = 10 + 9 + 11 = 30. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 5 & 1 & -1 \\ 12 & 3 & 2 \\ 3 & -1 & 1 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 5 \cdot \begin{vmatrix} 3 & 2 \\ -1 & 1 \end{vmatrix} – 1 \cdot \begin{vmatrix} 12 & 2 \\ 3 & 1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 12 & 3 \\ 3 & -1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 3 & 2 \\ -1 & 1 \end{vmatrix} = 5, \] \[ \begin{vmatrix} 12 & 2 \\ 3 & 1 \end{vmatrix} = 12 \cdot 1 – 2 \cdot 3 = 12 – 6 = 6, \] \[ \begin{vmatrix} 12 & 3 \\ 3 & -1 \end{vmatrix} = 12 \cdot (-1) – 3 \cdot 3 = -12 – 9 = -21. \] Dosadíme: \[ \det(A_x) = 5 \cdot 5 – 1 \cdot 6 – 1 \cdot (-21) = 25 – 6 + 21 = 40. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 2 & 5 & -1 \\ -1 & 12 & 2 \\ 4 & 3 & 1 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 2 \cdot \begin{vmatrix} 12 & 2 \\ 3 & 1 \end{vmatrix} – 5 \cdot \begin{vmatrix} -1 & 2 \\ 4 & 1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} -1 & 12 \\ 4 & 3 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 12 & 2 \\ 3 & 1 \end{vmatrix} = 6, \] \[ \begin{vmatrix} -1 & 2 \\ 4 & 1 \end{vmatrix} = (-1) \cdot 1 – 2 \cdot 4 = -1 – 8 = -9, \] \[ \begin{vmatrix} -1 & 12 \\ 4 & 3 \end{vmatrix} = (-1) \cdot 3 – 12 \cdot 4 = -3 – 48 = -51. \] Dosadíme: \[ \det(A_y) = 2 \cdot 6 – 5 \cdot (-9) – 1 \cdot (-51) = 12 + 45 + 51 = 108. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 2 & 1 & 5 \\ -1 & 3 & 12 \\ 4 & -1 & 3 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 2 \cdot \begin{vmatrix} 3 & 12 \\ -1 & 3 \end{vmatrix} – 1 \cdot \begin{vmatrix} -1 & 12 \\ 4 & 3 \end{vmatrix} + 5 \cdot \begin{vmatrix} -1 & 3 \\ 4 & -1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 3 & 12 \\ -1 & 3 \end{vmatrix} = 3 \cdot 3 – 12 \cdot (-1) = 9 + 12 = 21, \] \[ \begin{vmatrix} -1 & 12 \\ 4 & 3 \end{vmatrix} = -51, \] \[ \begin{vmatrix} -1 & 3 \\ 4 & -1 \end{vmatrix} = -11. \] Dosadíme: \[ \det(A_z) = 2 \cdot 21 – 1 \cdot (-51) + 5 \cdot (-11) = 42 + 51 – 55 = 38. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{40}{30} = \frac{4}{3}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{108}{30} = \frac{18}{5}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{38}{30} = \frac{19}{15}. \]
55. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} x – 2y + 3z = 7 \\ 4x + y – z = 2 \\ -3x + 5y + 2z = 10 \end{cases} \)
Řešení příkladu 55:
Koeficientová matice: \[ A = \begin{vmatrix} 1 & -2 & 3 \\ 4 & 1 & -1 \\ -3 & 5 & 2 \end{vmatrix}. \] Determinant: \[ \det(A) = 1 \cdot \begin{vmatrix} 1 & -1 \\ 5 & 2 \end{vmatrix} – (-2) \cdot \begin{vmatrix} 4 & -1 \\ -3 & 2 \end{vmatrix} + 3 \cdot \begin{vmatrix} 4 & 1 \\ -3 & 5 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 1 & -1 \\ 5 & 2 \end{vmatrix} = 1 \cdot 2 – (-1) \cdot 5 = 2 + 5 = 7, \] \[ \begin{vmatrix} 4 & -1 \\ -3 & 2 \end{vmatrix} = 4 \cdot 2 – (-1) \cdot (-3) = 8 – 3 = 5, \] \[ \begin{vmatrix} 4 & 1 \\ -3 & 5 \end{vmatrix} = 4 \cdot 5 – 1 \cdot (-3) = 20 + 3 = 23. \] Dosadíme: \[ \det(A) = 1 \cdot 7 + 2 \cdot 5 + 3 \cdot 23 = 7 + 10 + 69 = 86. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 7 & -2 & 3 \\ 2 & 1 & -1 \\ 10 & 5 & 2 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 7 \cdot \begin{vmatrix} 1 & -1 \\ 5 & 2 \end{vmatrix} – (-2) \cdot \begin{vmatrix} 2 & -1 \\ 10 & 2 \end{vmatrix} + 3 \cdot \begin{vmatrix} 2 & 1 \\ 10 & 5 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 1 & -1 \\ 5 & 2 \end{vmatrix} = 7, \] \[ \begin{vmatrix} 2 & -1 \\ 10 & 2 \end{vmatrix} = 2 \cdot 2 – (-1) \cdot 10 = 4 + 10 = 14, \] \[ \begin{vmatrix} 2 & 1 \\ 10 & 5 \end{vmatrix} = 2 \cdot 5 – 1 \cdot 10 = 10 – 10 = 0. \] Dosadíme: \[ \det(A_x) = 7 \cdot 7 + 2 \cdot 14 + 3 \cdot 0 = 49 + 28 + 0 = 77. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 1 & 7 & 3 \\ 4 & 2 & -1 \\ -3 & 10 & 2 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 1 \cdot \begin{vmatrix} 2 & -1 \\ 10 & 2 \end{vmatrix} – 7 \cdot \begin{vmatrix} 4 & -1 \\ -3 & 2 \end{vmatrix} + 3 \cdot \begin{vmatrix} 4 & 2 \\ -3 & 10 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 2 & -1 \\ 10 & 2 \end{vmatrix} = 14, \] \[ \begin{vmatrix} 4 & -1 \\ -3 & 2 \end{vmatrix} = 4 \cdot 2 – (-1) \cdot (-3) = 8 – 3 = 5, \] \[ \begin{vmatrix} 4 & 2 \\ -3 & 10 \end{vmatrix} = 4 \cdot 10 – 2 \cdot (-3) = 40 + 6 = 46. \] Dosadíme: \[ \det(A_y) = 1 \cdot 14 – 7 \cdot 5 + 3 \cdot 46 = 14 – 35 + 138 = 117. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 1 & -2 & 7 \\ 4 & 1 & 2 \\ -3 & 5 & 10 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 1 \cdot \begin{vmatrix} 1 & 2 \\ 5 & 10 \end{vmatrix} – (-2) \cdot \begin{vmatrix} 4 & 2 \\ -3 & 10 \end{vmatrix} + 7 \cdot \begin{vmatrix} 4 & 1 \\ -3 & 5 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 1 & 2 \\ 5 & 10 \end{vmatrix} = 1 \cdot 10 – 2 \cdot 5 = 10 – 10 = 0, \] \[ \begin{vmatrix} 4 & 2 \\ -3 & 10 \end{vmatrix} = 4 \cdot 10 – 2 \cdot (-3) = 40 + 6 = 46, \] \[ \begin{vmatrix} 4 & 1 \\ -3 & 5 \end{vmatrix} = 4 \cdot 5 – 1 \cdot (-3) = 20 + 3 = 23. \] Dosadíme: \[ \det(A_z) = 1 \cdot 0 + 2 \cdot 46 + 7 \cdot 23 = 0 + 92 + 161 = 253. \] Výsledky: \[ x = \frac{77}{86}, \quad y = \frac{117}{86}, \quad z = \frac{253}{86}. \]
56. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 3x – y + 2z = 5 \\ -x + 4y – z = 1 \\ 2x + y + 3z = 10 \end{cases} \)
Řešení příkladu 56:
Koeficientová matice: \[ A = \begin{vmatrix} 3 & -1 & 2 \\ -1 & 4 & -1 \\ 2 & 1 & 3 \end{vmatrix}. \] Determinant: \[ \det(A) = 3 \cdot \begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix} – (-1) \cdot \begin{vmatrix} -1 & -1 \\ 2 & 3 \end{vmatrix} + 2 \cdot \begin{vmatrix} -1 & 4 \\ 2 & 1 \end{vmatrix}. \] Výpočty determinantů 2×2: \[ \begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix} = 4 \cdot 3 – (-1) \cdot 1 = 12 + 1 = 13, \] \[ \begin{vmatrix} -1 & -1 \\ 2 & 3 \end{vmatrix} = (-1) \cdot 3 – (-1) \cdot 2 = -3 + 2 = -1, \] \[ \begin{vmatrix} -1 & 4 \\ 2 & 1 \end{vmatrix} = (-1) \cdot 1 – 4 \cdot 2 = -1 – 8 = -9. \] Dosadíme: \[ \det(A) = 3 \cdot 13 + 1 \cdot (-1) + 2 \cdot (-9) = 39 – 1 – 18 = 20. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 5 & -1 & 2 \\ 1 & 4 & -1 \\ 10 & 1 & 3 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 5 \cdot \begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix} – (-1) \cdot \begin{vmatrix} 1 & -1 \\ 10 & 3 \end{vmatrix} + 2 \cdot \begin{vmatrix} 1 & 4 \\ 10 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix} = 13, \] \[ \begin{vmatrix} 1 & -1 \\ 10 & 3 \end{vmatrix} = 1 \cdot 3 – (-1) \cdot 10 = 3 + 10 = 13, \] \[ \begin{vmatrix} 1 & 4 \\ 10 & 1 \end{vmatrix} = 1 \cdot 1 – 4 \cdot 10 = 1 – 40 = -39. \] Dosadíme: \[ \det(A_x) = 5 \cdot 13 + 1 \cdot 13 + 2 \cdot (-39) = 65 + 13 – 78 = 0. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 3 & 5 & 2 \\ -1 & 1 & -1 \\ 2 & 10 & 3 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 3 \cdot \begin{vmatrix} 1 & -1 \\ 10 & 3 \end{vmatrix} – 5 \cdot \begin{vmatrix} -1 & -1 \\ 2 & 3 \end{vmatrix} + 2 \cdot \begin{vmatrix} -1 & 1 \\ 2 & 10 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 1 & -1 \\ 10 & 3 \end{vmatrix} = 1 \cdot 3 – (-1) \cdot 10 = 3 + 10 = 13, \] \[ \begin{vmatrix} -1 & -1 \\ 2 & 3 \end{vmatrix} = (-1) \cdot 3 – (-1) \cdot 2 = -3 + 2 = -1, \] \[ \begin{vmatrix} -1 & 1 \\ 2 & 10 \end{vmatrix} = (-1) \cdot 10 – 1 \cdot 2 = -10 – 2 = -12. \] Dosadíme: \[ \det(A_y) = 3 \cdot 13 – 5 \cdot (-1) + 2 \cdot (-12) = 39 + 5 – 24 = 20. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 3 & -1 & 5 \\ -1 & 4 & 1 \\ 2 & 1 & 10 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 3 \cdot \begin{vmatrix} 4 & 1 \\ 1 & 10 \end{vmatrix} – (-1) \cdot \begin{vmatrix} -1 & 1 \\ 2 & 10 \end{vmatrix} + 5 \cdot \begin{vmatrix} -1 & 4 \\ 2 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 4 & 1 \\ 1 & 10 \end{vmatrix} = 4 \cdot 10 – 1 \cdot 1 = 40 – 1 = 39, \] \[ \begin{vmatrix} -1 & 1 \\ 2 & 10 \end{vmatrix} = (-1) \cdot 10 – 1 \cdot 2 = -10 – 2 = -12, \] \[ \begin{vmatrix} -1 & 4 \\ 2 & 1 \end{vmatrix} = (-1) \cdot 1 – 4 \cdot 2 = -1 – 8 = -9. \] Dosadíme: \[ \det(A_z) = 3 \cdot 39 + 1 \cdot (-12) + 5 \cdot (-9) = 117 – 12 – 45 = 60. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{0}{20} = 0, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{20}{20} = 1, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{60}{20} = 3. \]
57. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 2x + 3y – z = 4 \\ -x + 5y + 2z = 7 \\ 3x – 2y + 4z = 1 \end{cases} \)
Řešení příkladu 57:
Koeficientová matice: \[ A = \begin{vmatrix} 2 & 3 & -1 \\ -1 & 5 & 2 \\ 3 & -2 & 4 \end{vmatrix}. \] Determinant: \[ \det(A) = 2 \cdot \begin{vmatrix} 5 & 2 \\ -2 & 4 \end{vmatrix} – 3 \cdot \begin{vmatrix} -1 & 2 \\ 3 & 4 \end{vmatrix} -1 \cdot \begin{vmatrix} -1 & 5 \\ 3 & -2 \end{vmatrix}. \] Výpočty determinantů 2×2: \[ \begin{vmatrix} 5 & 2 \\ -2 & 4 \end{vmatrix} = 5 \cdot 4 – 2 \cdot (-2) = 20 + 4 = 24, \] \[ \begin{vmatrix} -1 & 2 \\ 3 & 4 \end{vmatrix} = (-1) \cdot 4 – 2 \cdot 3 = -4 – 6 = -10, \] \[ \begin{vmatrix} -1 & 5 \\ 3 & -2 \end{vmatrix} = (-1) \cdot (-2) – 5 \cdot 3 = 2 – 15 = -13. \] Dosadíme: \[ \det(A) = 2 \cdot 24 – 3 \cdot (-10) – 1 \cdot (-13) = 48 + 30 + 13 = 91. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 4 & 3 & -1 \\ 7 & 5 & 2 \\ 1 & -2 & 4 \end{vmatrix}. \] Determinant: \[ \det(A_x) = 4 \cdot \begin{vmatrix} 5 & 2 \\ -2 & 4 \end{vmatrix} – 3 \cdot \begin{vmatrix} 7 & 2 \\ 1 & 4 \end{vmatrix} – 1 \cdot \begin{vmatrix} 7 & 5 \\ 1 & -2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 5 & 2 \\ -2 & 4 \end{vmatrix} = 24, \] \[ \begin{vmatrix} 7 & 2 \\ 1 & 4 \end{vmatrix} = 7 \cdot 4 – 2 \cdot 1 = 28 – 2 = 26, \] \[ \begin{vmatrix} 7 & 5 \\ 1 & -2 \end{vmatrix} = 7 \cdot (-2) – 5 \cdot 1 = -14 – 5 = -19. \] Dosadíme: \[ \det(A_x) = 4 \cdot 24 – 3 \cdot 26 – 1 \cdot (-19) = 96 – 78 + 19 = 37. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 2 & 4 & -1 \\ -1 & 7 & 2 \\ 3 & 1 & 4 \end{vmatrix}. \] Determinant: \[ \det(A_y) = 2 \cdot \begin{vmatrix} 7 & 2 \\ 1 & 4 \end{vmatrix} – 4 \cdot \begin{vmatrix} -1 & 2 \\ 3 & 4 \end{vmatrix} – 1 \cdot \begin{vmatrix} -1 & 7 \\ 3 & 1 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 7 & 2 \\ 1 & 4 \end{vmatrix} = 26, \] \[ \begin{vmatrix} -1 & 2 \\ 3 & 4 \end{vmatrix} = -10, \] \[ \begin{vmatrix} -1 & 7 \\ 3 & 1 \end{vmatrix} = (-1) \cdot 1 – 7 \cdot 3 = -1 – 21 = -22. \] Dosadíme: \[ \det(A_y) = 2 \cdot 26 – 4 \cdot (-10) – 1 \cdot (-22) = 52 + 40 + 22 = 114. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 2 & 3 & 4 \\ -1 & 5 & 7 \\ 3 & -2 & 1 \end{vmatrix}. \] Determinant: \[ \det(A_z) = 2 \cdot \begin{vmatrix} 5 & 7 \\ -2 & 1 \end{vmatrix} – 3 \cdot \begin{vmatrix} -1 & 7 \\ 3 & 1 \end{vmatrix} + 4 \cdot \begin{vmatrix} -1 & 5 \\ 3 & -2 \end{vmatrix}. \] Výpočty: \[ \begin{vmatrix} 5 & 7 \\ -2 & 1 \end{vmatrix} = 5 \cdot 1 – 7 \cdot (-2) = 5 + 14 = 19, \] \[ \begin{vmatrix} -1 & 7 \\ 3 & 1 \end{vmatrix} = -22, \] \[ \begin{vmatrix} -1 & 5 \\ 3 & -2 \end{vmatrix} = (-1)(-2) – 5 \cdot 3 = 2 – 15 = -13. \] Dosadíme: \[ \det(A_z) = 2 \cdot 19 – 3 \cdot (-22) + 4 \cdot (-13) = 38 + 66 – 52 = 52. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{37}{91}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{114}{91}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{52}{91}. \]
58. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 3x – y + 2z = 7 \\ x + 4y – z = 2 \\ 2x – 3y + 5z = 10 \end{cases} \)
Řešení příkladu 58:
Koeficientová matice: \[ A = \begin{vmatrix} 3 & -1 & 2 \\ 1 & 4 & -1 \\ 2 & -3 & 5 \end{vmatrix}. \] Determinant matice \(A\): \[ \det(A) = 3 \cdot \begin{vmatrix} 4 & -1 \\ -3 & 5 \end{vmatrix} – (-1) \cdot \begin{vmatrix} 1 & -1 \\ 2 & 5 \end{vmatrix} + 2 \cdot \begin{vmatrix} 1 & 4 \\ 2 & -3 \end{vmatrix}. \] Výpočty determinantů: \[ \begin{vmatrix} 4 & -1 \\ -3 & 5 \end{vmatrix} = 4 \cdot 5 – (-1) \cdot (-3) = 20 – 3 = 17, \] \[ \begin{vmatrix} 1 & -1 \\ 2 & 5 \end{vmatrix} = 1 \cdot 5 – (-1) \cdot 2 = 5 + 2 = 7, \] \[ \begin{vmatrix} 1 & 4 \\ 2 & -3 \end{vmatrix} = 1 \cdot (-3) – 4 \cdot 2 = -3 – 8 = -11. \] Dosadíme: \[ \det(A) = 3 \cdot 17 – (-1) \cdot 7 + 2 \cdot (-11) = 51 + 7 – 22 = 36. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 7 & -1 & 2 \\ 2 & 4 & -1 \\ 10 & -3 & 5 \end{vmatrix}. \] Determinant matice \(A_x\): \[ \det(A_x) = 7 \cdot \begin{vmatrix} 4 & -1 \\ -3 & 5 \end{vmatrix} – (-1) \cdot \begin{vmatrix} 2 & -1 \\ 10 & 5 \end{vmatrix} + 2 \cdot \begin{vmatrix} 2 & 4 \\ 10 & -3 \end{vmatrix}. \] Výpočty determinantů: \[ \begin{vmatrix} 4 & -1 \\ -3 & 5 \end{vmatrix} = 17, \] \[ \begin{vmatrix} 2 & -1 \\ 10 & 5 \end{vmatrix} = 2 \cdot 5 – (-1) \cdot 10 = 10 + 10 = 20, \] \[ \begin{vmatrix} 2 & 4 \\ 10 & -3 \end{vmatrix} = 2 \cdot (-3) – 4 \cdot 10 = -6 – 40 = -46. \] Dosadíme: \[ \det(A_x) = 7 \cdot 17 – (-1) \cdot 20 + 2 \cdot (-46) = 119 + 20 – 92 = 47. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 3 & 7 & 2 \\ 1 & 2 & -1 \\ 2 & 10 & 5 \end{vmatrix}. \] Determinant matice \(A_y\): \[ \det(A_y) = 3 \cdot \begin{vmatrix} 2 & -1 \\ 10 & 5 \end{vmatrix} – 7 \cdot \begin{vmatrix} 1 & -1 \\ 2 & 5 \end{vmatrix} + 2 \cdot \begin{vmatrix} 1 & 2 \\ 2 & 10 \end{vmatrix}. \] Výpočty determinantů: \[ \begin{vmatrix} 2 & -1 \\ 10 & 5 \end{vmatrix} = 20, \] \[ \begin{vmatrix} 1 & -1 \\ 2 & 5 \end{vmatrix} = 1 \cdot 5 – (-1) \cdot 2 = 5 + 2 = 7, \] \[ \begin{vmatrix} 1 & 2 \\ 2 & 10 \end{vmatrix} = 1 \cdot 10 – 2 \cdot 2 = 10 – 4 = 6. \] Dosadíme: \[ \det(A_y) = 3 \cdot 20 – 7 \cdot 7 + 2 \cdot 6 = 60 – 49 + 12 = 23. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 3 & -1 & 7 \\ 1 & 4 & 2 \\ 2 & -3 & 10 \end{vmatrix}. \] Determinant matice \(A_z\): \[ \det(A_z) = 3 \cdot \begin{vmatrix} 4 & 2 \\ -3 & 10 \end{vmatrix} – (-1) \cdot \begin{vmatrix} 1 & 2 \\ 2 & 10 \end{vmatrix} + 7 \cdot \begin{vmatrix} 1 & 4 \\ 2 & -3 \end{vmatrix}. \] Výpočty determinantů: \[ \begin{vmatrix} 4 & 2 \\ -3 & 10 \end{vmatrix} = 4 \cdot 10 – 2 \cdot (-3) = 40 + 6 = 46, \] \[ \begin{vmatrix} 1 & 2 \\ 2 & 10 \end{vmatrix} = 6, \] \[ \begin{vmatrix} 1 & 4 \\ 2 & -3 \end{vmatrix} = 1 \cdot (-3) – 4 \cdot 2 = -3 – 8 = -11. \] Dosadíme: \[ \det(A_z) = 3 \cdot 46 – (-1) \cdot 6 + 7 \cdot (-11) = 138 + 6 – 77 = 67. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{47}{36}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{23}{36}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{67}{36}. \]
59. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 2x + 3y – z = 5 \\ – x + 4y + 2z = 6 \\ 3x – y + 5z = 15 \end{cases} \)
Řešení příkladu 59:
Koeficientová matice: \[ A = \begin{vmatrix} 2 & 3 & -1 \\ -1 & 4 & 2 \\ 3 & -1 & 5 \end{vmatrix}. \] Determinant matice \(A\): \[ \det(A) = 2 \cdot \begin{vmatrix} 4 & 2 \\ -1 & 5 \end{vmatrix} – 3 \cdot \begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix} + (-1) \cdot \begin{vmatrix} -1 & 4 \\ 3 & -1 \end{vmatrix}. \] Výpočty determinantů: \[ \begin{vmatrix} 4 & 2 \\ -1 & 5 \end{vmatrix} = 4 \cdot 5 – 2 \cdot (-1) = 20 + 2 = 22, \] \[ \begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix} = (-1) \cdot 5 – 2 \cdot 3 = -5 – 6 = -11, \] \[ \begin{vmatrix} -1 & 4 \\ 3 & -1 \end{vmatrix} = (-1) \cdot (-1) – 4 \cdot 3 = 1 – 12 = -11. \] Dosadíme: \[ \det(A) = 2 \cdot 22 – 3 \cdot (-11) + (-1) \cdot (-11) = 44 + 33 + 11 = 88. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 5 & 3 & -1 \\ 6 & 4 & 2 \\ 15 & -1 & 5 \end{vmatrix}. \] Determinant matice \(A_x\): \[ \det(A_x) = 5 \cdot \begin{vmatrix} 4 & 2 \\ -1 & 5 \end{vmatrix} – 3 \cdot \begin{vmatrix} 6 & 2 \\ 15 & 5 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 6 & 4 \\ 15 & -1 \end{vmatrix}. \] Výpočty determinantů: \[ \begin{vmatrix} 4 & 2 \\ -1 & 5 \end{vmatrix} = 22, \] \[ \begin{vmatrix} 6 & 2 \\ 15 & 5 \end{vmatrix} = 6 \cdot 5 – 2 \cdot 15 = 30 – 30 = 0, \] \[ \begin{vmatrix} 6 & 4 \\ 15 & -1 \end{vmatrix} = 6 \cdot (-1) – 4 \cdot 15 = -6 – 60 = -66. \] Dosadíme: \[ \det(A_x) = 5 \cdot 22 – 3 \cdot 0 + (-1) \cdot (-66) = 110 + 0 + 66 = 176. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 2 & 5 & -1 \\ -1 & 6 & 2 \\ 3 & 15 & 5 \end{vmatrix}. \] Determinant matice \(A_y\): \[ \det(A_y) = 2 \cdot \begin{vmatrix} 6 & 2 \\ 15 & 5 \end{vmatrix} – 5 \cdot \begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix} + (-1) \cdot \begin{vmatrix} -1 & 6 \\ 3 & 15 \end{vmatrix}. \] Výpočty determinantů: \[ \begin{vmatrix} 6 & 2 \\ 15 & 5 \end{vmatrix} = 30 – 30 = 0, \] \[ \begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix} = -11, \] \[ \begin{vmatrix} -1 & 6 \\ 3 & 15 \end{vmatrix} = (-1) \cdot 15 – 6 \cdot 3 = -15 – 18 = -33. \] Dosadíme: \[ \det(A_y) = 2 \cdot 0 – 5 \cdot (-11) + (-1) \cdot (-33) = 0 + 55 + 33 = 88. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 2 & 3 & 5 \\ -1 & 4 & 6 \\ 3 & -1 & 15 \end{vmatrix}. \] Determinant matice \(A_z\): \[ \det(A_z) = 2 \cdot \begin{vmatrix} 4 & 6 \\ -1 & 15 \end{vmatrix} – 3 \cdot \begin{vmatrix} -1 & 6 \\ 3 & 15 \end{vmatrix} + 5 \cdot \begin{vmatrix} -1 & 4 \\ 3 & -1 \end{vmatrix}. \] Výpočty determinantů: \[ \begin{vmatrix} 4 & 6 \\ -1 & 15 \end{vmatrix} = 4 \cdot 15 – 6 \cdot (-1) = 60 + 6 = 66, \] \[ \begin{vmatrix} -1 & 6 \\ 3 & 15 \end{vmatrix} = -33, \] \[ \begin{vmatrix} -1 & 4 \\ 3 & -1 \end{vmatrix} = (-1) \cdot (-1) – 4 \cdot 3 = 1 – 12 = -11. \] Dosadíme: \[ \det(A_z) = 2 \cdot 66 – 3 \cdot (-33) + 5 \cdot (-11) = 132 + 99 – 55 = 176. \] Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{176}{88} = 2, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{88}{88} = 1, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{176}{88} = 2. \]
60. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} x + 2y + 3z = 14 \\ 2x + y – z = 1 \\ x – y + z = 4 \end{cases} \)
Řešení příkladu 60:
Koeficientová matice: \[ A = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 1 & -1 \\ 1 & -1 & 1 \end{vmatrix}. \] Výpočet determinantu: \[ \det(A) = 1 \cdot \begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix} – 2 \cdot \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix}. \] \[ = 1 \cdot (1 – 1) – 2 \cdot (2 \cdot 1 – (-1) \cdot 1) + 3 \cdot (2 \cdot (-1) – 1 \cdot 1) \] \[ = 0 – 2 \cdot (2 + 1) + 3 \cdot (-2 – 1) = 0 – 6 – 9 = -15. \] Matice \(A_x\): \[ A_x = \begin{vmatrix} 14 & 2 & 3 \\ 1 & 1 & -1 \\ 4 & -1 & 1 \end{vmatrix}, \quad \det(A_x) = 14 \cdot \begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix} – 2 \cdot \begin{vmatrix} 1 & -1 \\ 4 & 1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & 1 \\ 4 & -1 \end{vmatrix}. \] \[ = 14(1 – 1) – 2(1 + 4) + 3(-1 – 4) = 0 – 10 – 15 = -25. \] Matice \(A_y\): \[ A_y = \begin{vmatrix} 1 & 14 & 3 \\ 2 & 1 & -1 \\ 1 & 4 & 1 \end{vmatrix}, \quad \det(A_y) = 1 \cdot \begin{vmatrix} 1 & -1 \\ 4 & 1 \end{vmatrix} – 14 \cdot \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 2 & 1 \\ 1 & 4 \end{vmatrix}. \] \[ = 1(1 + 4) – 14(2 \cdot 1 + 1 \cdot 1) + 3(8 – 1) = 5 – 14 \cdot 3 + 3 \cdot 7 = 5 – 42 + 21 = -16. \] Matice \(A_z\): \[ A_z = \begin{vmatrix} 1 & 2 & 14 \\ 2 & 1 & 1 \\ 1 & -1 & 4 \end{vmatrix}, \quad \det(A_z) = 1 \cdot \begin{vmatrix} 1 & 1 \\ -1 & 4 \end{vmatrix} – 2 \cdot \begin{vmatrix} 2 & 1 \\ 1 & 4 \end{vmatrix} + 14 \cdot \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix}. \] \[ = 1(4 + 1) – 2(8 – 1) + 14(-2 – 1) = 5 – 14 – 42 = -51. \] Výsledky: \[ x = \frac{-25}{-15} = \frac{5}{3}, \quad y = \frac{-16}{-15} = \frac{16}{15}, \quad z = \frac{-51}{-15} = \frac{17}{5}. \]
61. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 2x + y – z = 5 \\ x – 2y + 3z = -4 \\ 3x + y + 2z = 10 \end{cases} \)
Řešení příkladu 61:
Koeficientová matice: \[ A = \begin{vmatrix} 2 & 1 & -1 \\ 1 & -2 & 3 \\ 3 & 1 & 2 \end{vmatrix}. \]
Determinant: \[ \det(A) = 2 \cdot \begin{vmatrix} -2 & 3 \\ 1 & 2 \end{vmatrix} – 1 \cdot \begin{vmatrix} 1 & 3 \\ 3 & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix}. \]
Výpočty: \[ \begin{vmatrix} -2 & 3 \\ 1 & 2 \end{vmatrix} = (-2)(2) – (3)(1) = -4 – 3 = -7, \] \[ \begin{vmatrix} 1 & 3 \\ 3 & 2 \end{vmatrix} = 1 \cdot 2 – 3 \cdot 3 = 2 – 9 = -7, \] \[ \begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} = 1 \cdot 1 – (-2) \cdot 3 = 1 + 6 = 7. \]
Dosadíme: \[ \det(A) = 2 \cdot (-7) – 1 \cdot (-7) – 1 \cdot 7 = -14 + 7 – 7 = -14. \]
Matice \(A_x\): \[ A_x = \begin{vmatrix} 5 & 1 & -1 \\ -4 & -2 & 3 \\ 10 & 1 & 2 \end{vmatrix}. \]
\[ \det(A_x) = 5 \cdot \begin{vmatrix} -2 & 3 \\ 1 & 2 \end{vmatrix} – 1 \cdot \begin{vmatrix} -4 & 3 \\ 10 & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} -4 & -2 \\ 10 & 1 \end{vmatrix}. \]
Výpočty: \[ \begin{vmatrix} -2 & 3 \\ 1 & 2 \end{vmatrix} = -4 – 3 = -7, \] \[ \begin{vmatrix} -4 & 3 \\ 10 & 2 \end{vmatrix} = -4 \cdot 2 – 3 \cdot 10 = -8 – 30 = -38, \] \[ \begin{vmatrix} -4 & -2 \\ 10 & 1 \end{vmatrix} = -4 \cdot 1 – (-2) \cdot 10 = -4 + 20 = 16. \]
Dosadíme: \[ \det(A_x) = 5 \cdot (-7) – (-38) – 16 = -35 + 38 – 16 = -13. \]
Matice \(A_y\): \[ A_y = \begin{vmatrix} 2 & 5 & -1 \\ 1 & -4 & 3 \\ 3 & 10 & 2 \end{vmatrix}. \]
\[ \det(A_y) = 2 \cdot \begin{vmatrix} -4 & 3 \\ 10 & 2 \end{vmatrix} – 5 \cdot \begin{vmatrix} 1 & 3 \\ 3 & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 1 & -4 \\ 3 & 10 \end{vmatrix}. \]
Výpočty: \[ \begin{vmatrix} -4 & 3 \\ 10 & 2 \end{vmatrix} = -8 – 30 = -38, \] \[ \begin{vmatrix} 1 & 3 \\ 3 & 2 \end{vmatrix} = 2 – 9 = -7, \] \[ \begin{vmatrix} 1 & -4 \\ 3 & 10 \end{vmatrix} = 10 + 12 = 22. \]
Dosadíme: \[ \det(A_y) = 2 \cdot (-38) – 5 \cdot (-7) – 22 = -76 + 35 – 22 = -63. \]
Matice \(A_z\): \[ A_z = \begin{vmatrix} 2 & 1 & 5 \\ 1 & -2 & -4 \\ 3 & 1 & 10 \end{vmatrix}. \]
\[ \det(A_z) = 2 \cdot \begin{vmatrix} -2 & -4 \\ 1 & 10 \end{vmatrix} – 1 \cdot \begin{vmatrix} 1 & -4 \\ 3 & 10 \end{vmatrix} + 5 \cdot \begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix}. \]
Výpočty: \[ \begin{vmatrix} -2 & -4 \\ 1 & 10 \end{vmatrix} = -20 + 4 = -16, \] \[ \begin{vmatrix} 1 & -4 \\ 3 & 10 \end{vmatrix} = 10 + 12 = 22, \] \[ \begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} = 1 + 6 = 7. \]
Dosadíme: \[ \det(A_z) = 2 \cdot (-16) – 22 + 5 \cdot 7 = -32 – 22 + 35 = -19. \]
Výsledky: \[ x = \frac{-13}{-14} = \frac{13}{14}, \quad y = \frac{-63}{-14} = \frac{63}{14} = \frac{9}{2}, \quad z = \frac{-19}{-14} = \frac{19}{14}. \]
62. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 2x – y + 3z = 9 \\ x + 4y – z = 2 \\ 3x + 2y + z = 10 \end{cases} \)
Řešení příkladu 62:
Koeficientová matice: \[ A = \begin{vmatrix} 2 & -1 & 3 \\ 1 & 4 & -1 \\ 3 & 2 & 1 \end{vmatrix}. \]
Výpočet determinantu: \[ \det(A) = 2 \cdot \begin{vmatrix} 4 & -1 \\ 2 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix}. \]
\[ \begin{vmatrix} 4 & -1 \\ 2 & 1 \end{vmatrix} = 4 \cdot 1 – (-1) \cdot 2 = 4 + 2 = 6, \] \[ \begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} = 1 \cdot 1 – (-1) \cdot 3 = 1 + 3 = 4, \] \[ \begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix} = 1 \cdot 2 – 4 \cdot 3 = 2 – 12 = -10. \]
\[ \det(A) = 2 \cdot 6 – 1 \cdot 4 + 3 \cdot (-10) = 12 – 4 – 30 = -22. \]
Matice \( A_x \): \[ A_x = \begin{vmatrix} 9 & -1 & 3 \\ 2 & 4 & -1 \\ 10 & 2 & 1 \end{vmatrix}. \]
\[ \det(A_x) = 9 \cdot \begin{vmatrix} 4 & -1 \\ 2 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & -1 \\ 10 & 1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 2 & 4 \\ 10 & 2 \end{vmatrix}. \]
\[ \begin{vmatrix} 4 & -1 \\ 2 & 1 \end{vmatrix} = 4 \cdot 1 – (-1) \cdot 2 = 4 + 2 = 6, \] \[ \begin{vmatrix} 2 & -1 \\ 10 & 1 \end{vmatrix} = 2 \cdot 1 – (-1) \cdot 10 = 2 + 10 = 12, \] \[ \begin{vmatrix} 2 & 4 \\ 10 & 2 \end{vmatrix} = 2 \cdot 2 – 4 \cdot 10 = 4 – 40 = -36. \]
\[ \det(A_x) = 9 \cdot 6 – 1 \cdot 12 + 3 \cdot (-36) = 54 – 12 – 108 = -66. \]
Matice \( A_y \): \[ A_y = \begin{vmatrix} 2 & 9 & 3 \\ 1 & 2 & -1 \\ 3 & 10 & 1 \end{vmatrix}. \]
\[ \det(A_y) = 2 \cdot \begin{vmatrix} 2 & -1 \\ 10 & 1 \end{vmatrix} – 9 \cdot \begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & 2 \\ 3 & 10 \end{vmatrix}. \]
\[ \begin{vmatrix} 2 & -1 \\ 10 & 1 \end{vmatrix} = 2 \cdot 1 – (-1) \cdot 10 = 2 + 10 = 12, \] \[ \begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} = 1 \cdot 1 – (-1) \cdot 3 = 1 + 3 = 4, \] \[ \begin{vmatrix} 1 & 2 \\ 3 & 10 \end{vmatrix} = 1 \cdot 10 – 2 \cdot 3 = 10 – 6 = 4. \]
\[ \det(A_y) = 2 \cdot 12 – 9 \cdot 4 + 3 \cdot 4 = 24 – 36 + 12 = 0. \]
Matice \( A_z \): \[ A_z = \begin{vmatrix} 2 & -1 & 9 \\ 1 & 4 & 2 \\ 3 & 2 & 10 \end{vmatrix}. \]
\[ \det(A_z) = 2 \cdot \begin{vmatrix} 4 & 2 \\ 2 & 10 \end{vmatrix} – (-1) \cdot \begin{vmatrix} 1 & 2 \\ 3 & 10 \end{vmatrix} + 9 \cdot \begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix}. \]
\[ \begin{vmatrix} 4 & 2 \\ 2 & 10 \end{vmatrix} = 4 \cdot 10 – 2 \cdot 2 = 40 – 4 = 36, \] \[ \begin{vmatrix} 1 & 2 \\ 3 & 10 \end{vmatrix} = 1 \cdot 10 – 2 \cdot 3 = 10 – 6 = 4, \] \[ \begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix} = 1 \cdot 2 – 4 \cdot 3 = 2 – 12 = -10. \]
\[ \det(A_z) = 2 \cdot 36 + 1 \cdot 4 + 9 \cdot (-10) = 72 + 4 – 90 = -14. \]
Výsledky: \[ x = \frac{-66}{-22} = 3, \quad y = \frac{0}{-22} = 0, \quad z = \frac{-14}{-22} = \frac{7}{11}. \]
63. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 2x + 3y – z = 7 \\ 4x – y + 2z = 1 \\ -3x + 2y + z = 4 \end{cases} \)
Řešení příkladu 63:
Koeficientová matice: \[ A = \begin{vmatrix} 2 & 3 & -1 \\ 4 & -1 & 2 \\ -3 & 2 & 1 \end{vmatrix}. \]
Determinant: \[ \det(A) = 2 \cdot \begin{vmatrix} -1 & 2 \\ 2 & 1 \end{vmatrix} – 3 \cdot \begin{vmatrix} 4 & 2 \\ -3 & 1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 4 & -1 \\ -3 & 2 \end{vmatrix}. \]
Výpočty: \[ \begin{vmatrix} -1 & 2 \\ 2 & 1 \end{vmatrix} = (-1)(1) – (2)(2) = -1 – 4 = -5, \] \[ \begin{vmatrix} 4 & 2 \\ -3 & 1 \end{vmatrix} = 4 \cdot 1 – 2 \cdot (-3) = 4 + 6 = 10, \] \[ \begin{vmatrix} 4 & -1 \\ -3 & 2 \end{vmatrix} = 4 \cdot 2 – (-1)(-3) = 8 – 3 = 5. \]
Dosadíme: \[ \det(A) = 2 \cdot (-5) – 3 \cdot 10 – 1 \cdot 5 = -10 – 30 – 5 = -45. \]
Matice \(A_x\): \[ A_x = \begin{vmatrix} 7 & 3 & -1 \\ 1 & -1 & 2 \\ 4 & 2 & 1 \end{vmatrix}. \]
Výpočty: \[ \det(A_x) = 7 \cdot \begin{vmatrix} -1 & 2 \\ 2 & 1 \end{vmatrix} – 3 \cdot \begin{vmatrix} 1 & 2 \\ 4 & 1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 1 & -1 \\ 4 & 2 \end{vmatrix}, \] \[ \begin{vmatrix} -1 & 2 \\ 2 & 1 \end{vmatrix} = -1 \cdot 1 – 2 \cdot 2 = -1 – 4 = -5, \] \[ \begin{vmatrix} 1 & 2 \\ 4 & 1 \end{vmatrix} = 1 \cdot 1 – 2 \cdot 4 = 1 – 8 = -7, \] \[ \begin{vmatrix} 1 & -1 \\ 4 & 2 \end{vmatrix} = 1 \cdot 2 – (-1) \cdot 4 = 2 + 4 = 6. \]
Dosadíme: \[ \det(A_x) = 7 \cdot (-5) – 3 \cdot (-7) – 6 = -35 + 21 – 6 = -20. \]
Matice \(A_y\): \[ A_y = \begin{vmatrix} 2 & 7 & -1 \\ 4 & 1 & 2 \\ -3 & 4 & 1 \end{vmatrix}. \]
Výpočty: \[ \det(A_y) = 2 \cdot \begin{vmatrix} 1 & 2 \\ 4 & 1 \end{vmatrix} – 7 \cdot \begin{vmatrix} 4 & 2 \\ -3 & 1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 4 & 1 \\ -3 & 4 \end{vmatrix}, \] \[ \begin{vmatrix} 1 & 2 \\ 4 & 1 \end{vmatrix} = 1 – 8 = -7, \] \[ \begin{vmatrix} 4 & 2 \\ -3 & 1 \end{vmatrix} = 4 \cdot 1 – 2 \cdot (-3) = 4 + 6 = 10, \] \[ \begin{vmatrix} 4 & 1 \\ -3 & 4 \end{vmatrix} = 4 \cdot 4 – 1 \cdot (-3) = 16 + 3 = 19. \]
Dosadíme: \[ \det(A_y) = 2 \cdot (-7) – 7 \cdot 10 – 19 = -14 – 70 – 19 = -103. \]
Matice \(A_z\): \[ A_z = \begin{vmatrix} 2 & 3 & 7 \\ 4 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix}. \]
Výpočty: \[ \det(A_z) = 2 \cdot \begin{vmatrix} -1 & 1 \\ 2 & 4 \end{vmatrix} – 3 \cdot \begin{vmatrix} 4 & 1 \\ -3 & 4 \end{vmatrix} + 7 \cdot \begin{vmatrix} 4 & -1 \\ -3 & 2 \end{vmatrix}, \] \[ \begin{vmatrix} -1 & 1 \\ 2 & 4 \end{vmatrix} = -1 \cdot 4 – 1 \cdot 2 = -4 – 2 = -6, \] \[ \begin{vmatrix} 4 & 1 \\ -3 & 4 \end{vmatrix} = 4 \cdot 4 – 1 \cdot (-3) = 16 + 3 = 19, \] \[ \begin{vmatrix} 4 & -1 \\ -3 & 2 \end{vmatrix} = 4 \cdot 2 – (-1) \cdot (-3) = 8 – 3 = 5. \]
Dosadíme: \[ \det(A_z) = 2 \cdot (-6) – 3 \cdot 19 + 7 \cdot 5 = -12 – 57 + 35 = -34. \]
Výsledky: \[ x = \frac{-20}{-45} = \frac{4}{9}, \quad y = \frac{-103}{-45} = \frac{103}{45}, \quad z = \frac{-34}{-45} = \frac{34}{45}. \]
64. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 2x + y – z = 3 \\ x – 2y + 4z = 1 \\ 3x + y + z = 7 \end{cases} \)
Řešení příkladu 64:
Koeficientová matice: \[ A = \begin{vmatrix} 2 & 1 & -1 \\ 1 & -2 & 4 \\ 3 & 1 & 1 \end{vmatrix}. \]
Determinant: \[ \det(A) = 2 \cdot \begin{vmatrix} -2 & 4 \\ 1 & 1 \end{vmatrix} – 1 \cdot \begin{vmatrix} 1 & 4 \\ 3 & 1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix}. \]
Výpočty: \[ \begin{vmatrix} -2 & 4 \\ 1 & 1 \end{vmatrix} = (-2)(1) – (4)(1) = -2 – 4 = -6, \] \[ \begin{vmatrix} 1 & 4 \\ 3 & 1 \end{vmatrix} = 1 \cdot 1 – 4 \cdot 3 = 1 – 12 = -11, \] \[ \begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} = 1 \cdot 1 – (-2)(3) = 1 + 6 = 7. \]
Dosazení: \[ \det(A) = 2 \cdot (-6) – 1 \cdot (-11) + (-1) \cdot 7 = -12 + 11 – 7 = -8. \]
Matice \(A_x\): \[ A_x = \begin{vmatrix} 3 & 1 & -1 \\ 1 & -2 & 4 \\ 7 & 1 & 1 \end{vmatrix}. \]
Determinant: \[ \det(A_x) = 3 \cdot \begin{vmatrix} -2 & 4 \\ 1 & 1 \end{vmatrix} – 1 \cdot \begin{vmatrix} 1 & 4 \\ 7 & 1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 1 & -2 \\ 7 & 1 \end{vmatrix}. \]
Výpočty: \[ \begin{vmatrix} -2 & 4 \\ 1 & 1 \end{vmatrix} = -6, \quad \begin{vmatrix} 1 & 4 \\ 7 & 1 \end{vmatrix} = 1 – 28 = -27, \quad \begin{vmatrix} 1 & -2 \\ 7 & 1 \end{vmatrix} = 1 + 14 = 15. \]
Dosazení: \[ \det(A_x) = 3 \cdot (-6) – 1 \cdot (-27) + (-1) \cdot 15 = -18 + 27 – 15 = -6. \]
Matice \(A_y\): \[ A_y = \begin{vmatrix} 2 & 3 & -1 \\ 1 & 1 & 4 \\ 3 & 7 & 1 \end{vmatrix}. \]
Determinant: \[ \det(A_y) = 2 \cdot \begin{vmatrix} 1 & 4 \\ 7 & 1 \end{vmatrix} – 3 \cdot \begin{vmatrix} 1 & 4 \\ 3 & 1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 1 & 1 \\ 3 & 7 \end{vmatrix}. \]
Výpočty: \[ \begin{vmatrix} 1 & 4 \\ 7 & 1 \end{vmatrix} = 1 – 28 = -27, \quad \begin{vmatrix} 1 & 4 \\ 3 & 1 \end{vmatrix} = 1 – 12 = -11, \quad \begin{vmatrix} 1 & 1 \\ 3 & 7 \end{vmatrix} = 7 – 3 = 4. \]
Dosazení: \[ \det(A_y) = 2 \cdot (-27) – 3 \cdot (-11) + (-1) \cdot 4 = -54 + 33 – 4 = -25. \]
Matice \(A_z\): \[ A_z = \begin{vmatrix} 2 & 1 & 3 \\ 1 & -2 & 1 \\ 3 & 1 & 7 \end{vmatrix}. \]
Determinant: \[ \det(A_z) = 2 \cdot \begin{vmatrix} -2 & 1 \\ 1 & 7 \end{vmatrix} – 1 \cdot \begin{vmatrix} 1 & 1 \\ 3 & 7 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix}. \]
Výpočty: \[ \begin{vmatrix} -2 & 1 \\ 1 & 7 \end{vmatrix} = -14 – 1 = -15, \quad \begin{vmatrix} 1 & 1 \\ 3 & 7 \end{vmatrix} = 7 – 3 = 4, \quad \begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} = 1 + 6 = 7. \]
Dosazení: \[ \det(A_z) = 2 \cdot (-15) – 1 \cdot 4 + 3 \cdot 7 = -30 – 4 + 21 = -13. \]
Výsledky: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{-6}{-8} = \frac{3}{4}, \quad y = \frac{\det(A_y)}{\det(A)} = \frac{-25}{-8} = \frac{25}{8}, \quad z = \frac{\det(A_z)}{\det(A)} = \frac{-13}{-8} = \frac{13}{8}. \]
65. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} x + 2y + z = 4 \\ 2x – y + 3z = 1 \\ 3x + y – z = 2 \end{cases} \)
Řešení příkladu 65:
Koeficientová matice: \[ A = \begin{vmatrix} 1 & 2 & 1 \\ 2 & -1 & 3 \\ 3 & 1 & -1 \end{vmatrix} \]
Vypočítáme determinant: \[ \det(A) = 1 \cdot \begin{vmatrix} -1 & 3 \\ 1 & -1 \end{vmatrix} – 2 \cdot \begin{vmatrix} 2 & 3 \\ 3 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} \]
\(
\begin{vmatrix} -1 & 3 \\ 1 & -1 \end{vmatrix} = (-1)(-1) – (3)(1) = 1 – 3 = -2
\)
\(
\begin{vmatrix} 2 & 3 \\ 3 & -1 \end{vmatrix} = 2 \cdot (-1) – 3 \cdot 3 = -2 – 9 = -11
\)
\(
\begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} = 2 \cdot 1 – (-1) \cdot 3 = 2 + 3 = 5
\)
\( \det(A) = 1 \cdot (-2) – 2 \cdot (-11) + 1 \cdot 5 = -2 + 22 + 5 = 25 \)
Matice \(A_x\) (sloupec x nahrazen pravou stranou): \[ A_x = \begin{vmatrix} 4 & 2 & 1 \\ 1 & -1 & 3 \\ 2 & 1 & -1 \end{vmatrix} \]
\( \det(A_x) = 4 \cdot \begin{vmatrix} -1 & 3 \\ 1 & -1 \end{vmatrix} – 2 \cdot \begin{vmatrix} 1 & 3 \\ 2 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} \)
\( \begin{vmatrix} -1 & 3 \\ 1 & -1 \end{vmatrix} = -2,\quad \begin{vmatrix} 1 & 3 \\ 2 & -1 \end{vmatrix} = 1 \cdot (-1) – 3 \cdot 2 = -1 – 6 = -7,\quad \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} = 1 \cdot 1 – (-1) \cdot 2 = 1 + 2 = 3 \)
\( \det(A_x) = 4 \cdot (-2) – 2 \cdot (-7) + 1 \cdot 3 = -8 + 14 + 3 = 9 \)
Matice \(A_y\): \[ A_y = \begin{vmatrix} 1 & 4 & 1 \\ 2 & 1 & 3 \\ 3 & 2 & -1 \end{vmatrix} \]
\( \det(A_y) = 1 \cdot \begin{vmatrix} 1 & 3 \\ 2 & -1 \end{vmatrix} – 4 \cdot \begin{vmatrix} 2 & 3 \\ 3 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} \)
\(
\begin{vmatrix} 1 & 3 \\ 2 & -1 \end{vmatrix} = 1 \cdot (-1) – 3 \cdot 2 = -1 – 6 = -7
\)
\(
\begin{vmatrix} 2 & 3 \\ 3 & -1 \end{vmatrix} = 2 \cdot (-1) – 3 \cdot 3 = -2 – 9 = -11
\)
\(
\begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} = 2 \cdot 2 – 1 \cdot 3 = 4 – 3 = 1
\)
\( \det(A_y) = 1 \cdot (-7) – 4 \cdot (-11) + 1 \cdot 1 = -7 + 44 + 1 = 38 \)
Matice \(A_z\): \[ A_z = \begin{vmatrix} 1 & 2 & 4 \\ 2 & -1 & 1 \\ 3 & 1 & 2 \end{vmatrix} \]
\( \det(A_z) = 1 \cdot \begin{vmatrix} -1 & 1 \\ 1 & 2 \end{vmatrix} – 2 \cdot \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} + 4 \cdot \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} \)
\(
\begin{vmatrix} -1 & 1 \\ 1 & 2 \end{vmatrix} = -1 \cdot 2 – 1 \cdot 1 = -2 – 1 = -3
\)
\(
\begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} = 2 \cdot 2 – 1 \cdot 3 = 4 – 3 = 1
\)
\(
\begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} = 2 \cdot 1 – (-1) \cdot 3 = 2 + 3 = 5
\)
\( \det(A_z) = 1 \cdot (-3) – 2 \cdot 1 + 4 \cdot 5 = -3 – 2 + 20 = 15 \)
Výsledky: \[ x = \frac{9}{25},\quad y = \frac{38}{25},\quad z = \frac{15}{25} = \frac{3}{5} \]
66. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 2x + 3y – z = 1 \\ x – y + 4z = 7 \\ 3x + y + 2z = 5 \end{cases} \)
Řešení příkladu 66:
Koeficientová matice: \[ A = \begin{vmatrix} 2 & 3 & -1 \\ 1 & -1 & 4 \\ 3 & 1 & 2 \end{vmatrix} \]
Vypočítáme determinant: \[ \det(A) = 2 \cdot \begin{vmatrix} -1 & 4 \\ 1 & 2 \end{vmatrix} – 3 \cdot \begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} \]
\(
\begin{vmatrix} -1 & 4 \\ 1 & 2 \end{vmatrix} = (-1)(2) – (4)(1) = -2 – 4 = -6
\)
\(
\begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix} = 1 \cdot 2 – 4 \cdot 3 = 2 – 12 = -10
\)
\(
\begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} = 1 \cdot 1 – (-1)(3) = 1 + 3 = 4
\)
\( \det(A) = 2 \cdot (-6) – 3 \cdot (-10) – 1 \cdot 4 = -12 + 30 – 4 = 14 \)
Matice \(A_x\): \[ \begin{vmatrix} 1 & 3 & -1 \\ 7 & -1 & 4 \\ 5 & 1 & 2 \end{vmatrix} \]
\( \det(A_x) = 1 \cdot \begin{vmatrix} -1 & 4 \\ 1 & 2 \end{vmatrix} – 3 \cdot \begin{vmatrix} 7 & 4 \\ 5 & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 7 & -1 \\ 5 & 1 \end{vmatrix} \)
\(
\begin{vmatrix} -1 & 4 \\ 1 & 2 \end{vmatrix} = -1 \cdot 2 – 4 \cdot 1 = -2 – 4 = -6
\)
\(
\begin{vmatrix} 7 & 4 \\ 5 & 2 \end{vmatrix} = 7 \cdot 2 – 4 \cdot 5 = 14 – 20 = -6
\)
\(
\begin{vmatrix} 7 & -1 \\ 5 & 1 \end{vmatrix} = 7 \cdot 1 – (-1) \cdot 5 = 7 + 5 = 12
\)
\( \det(A_x) = 1 \cdot (-6) – 3 \cdot (-6) – 1 \cdot 12 = -6 + 18 – 12 = 0 \)
Matice \(A_y\): \[ \begin{vmatrix} 2 & 1 & -1 \\ 1 & 7 & 4 \\ 3 & 5 & 2 \end{vmatrix} \]
\( \det(A_y) = 2 \cdot \begin{vmatrix} 7 & 4 \\ 5 & 2 \end{vmatrix} – 1 \cdot \begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 1 & 7 \\ 3 & 5 \end{vmatrix} \)
\( \begin{vmatrix} 7 & 4 \\ 5 & 2 \end{vmatrix} = 14 – 20 = -6,\quad \begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix} = 1 \cdot 2 – 4 \cdot 3 = 2 – 12 = -10,\quad \begin{vmatrix} 1 & 7 \\ 3 & 5 \end{vmatrix} = 1 \cdot 5 – 7 \cdot 3 = 5 – 21 = -16 \)
\( \det(A_y) = 2 \cdot (-6) – 1 \cdot (-10) – 1 \cdot (-16) = -12 + 10 + 16 = 14 \)
Matice \(A_z\): \[ \begin{vmatrix} 2 & 3 & 1 \\ 1 & -1 & 7 \\ 3 & 1 & 5 \end{vmatrix} \]
\( \det(A_z) = 2 \cdot \begin{vmatrix} -1 & 7 \\ 1 & 5 \end{vmatrix} – 3 \cdot \begin{vmatrix} 1 & 7 \\ 3 & 5 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} \)
\( \begin{vmatrix} -1 & 7 \\ 1 & 5 \end{vmatrix} = -1 \cdot 5 – 7 \cdot 1 = -5 – 7 = -12,\quad \begin{vmatrix} 1 & 7 \\ 3 & 5 \end{vmatrix} = 1 \cdot 5 – 7 \cdot 3 = 5 – 21 = -16,\quad \begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} = 1 \cdot 1 – (-1)(3) = 1 + 3 = 4 \)
\( \det(A_z) = 2 \cdot (-12) – 3 \cdot (-16) + 1 \cdot 4 = -24 + 48 + 4 = 28 \)
Výsledky: \[ x = \frac{0}{14} = 0,\quad y = \frac{14}{14} = 1,\quad z = \frac{28}{14} = 2 \]
67. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} x + 2y + 3z = 14 \\ 2x – y + z = 5 \\ x + y – z = 1 \end{cases} \)
Řešení příkladu 67:
Koeficientová matice: \[ A = \begin{vmatrix} 1 & 2 & 3 \\ 2 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} \]
Determinant: \[ \det(A) = 1 \cdot \begin{vmatrix} -1 & 1 \\ 1 & -1 \end{vmatrix} – 2 \cdot \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} \]
\(
\begin{vmatrix} -1 & 1 \\ 1 & -1 \end{vmatrix} = (-1)(-1) – (1)(1) = 1 – 1 = 0 \)
\(
\begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} = (2)(-1) – (1)(1) = -2 – 1 = -3 \)
\(
\begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} = (2)(1) – (-1)(1) = 2 + 1 = 3
\)
\( \det(A) = 1 \cdot 0 – 2 \cdot (-3) + 3 \cdot 3 = 0 + 6 + 9 = 15 \)
Matice \(A_x\): \[ \begin{vmatrix} 14 & 2 & 3 \\ 5 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} \]
\( \det(A_x) = 14 \cdot \begin{vmatrix} -1 & 1 \\ 1 & -1 \end{vmatrix} – 2 \cdot \begin{vmatrix} 5 & 1 \\ 1 & -1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 5 & -1 \\ 1 & 1 \end{vmatrix} \)
\( \begin{vmatrix} -1 & 1 \\ 1 & -1 \end{vmatrix} = 0,\quad \begin{vmatrix} 5 & 1 \\ 1 & -1 \end{vmatrix} = -5 – 1 = -6,\quad \begin{vmatrix} 5 & -1 \\ 1 & 1 \end{vmatrix} = 5 + 1 = 6 \)
\( \det(A_x) = 14 \cdot 0 – 2 \cdot (-6) + 3 \cdot 6 = 0 + 12 + 18 = 30 \)
Matice \(A_y\): \[ \begin{vmatrix} 1 & 14 & 3 \\ 2 & 5 & 1 \\ 1 & 1 & -1 \end{vmatrix} \]
\( \det(A_y) = 1 \cdot \begin{vmatrix} 5 & 1 \\ 1 & -1 \end{vmatrix} – 14 \cdot \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 2 & 5 \\ 1 & 1 \end{vmatrix} \)
\( \begin{vmatrix} 5 & 1 \\ 1 & -1 \end{vmatrix} = -5 – 1 = -6,\quad \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} = -2 – 1 = -3,\quad \begin{vmatrix} 2 & 5 \\ 1 & 1 \end{vmatrix} = 2 – 5 = -3 \)
\( \det(A_y) = 1 \cdot (-6) – 14 \cdot (-3) + 3 \cdot (-3) = -6 + 42 – 9 = 27 \)
Matice \(A_z\): \[ \begin{vmatrix} 1 & 2 & 14 \\ 2 & -1 & 5 \\ 1 & 1 & 1 \end{vmatrix} \]
\( \det(A_z) = 1 \cdot \begin{vmatrix} -1 & 5 \\ 1 & 1 \end{vmatrix} – 2 \cdot \begin{vmatrix} 2 & 5 \\ 1 & 1 \end{vmatrix} + 14 \cdot \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} \)
\( \begin{vmatrix} -1 & 5 \\ 1 & 1 \end{vmatrix} = -1 – 5 = -6,\quad \begin{vmatrix} 2 & 5 \\ 1 & 1 \end{vmatrix} = 2 – 5 = -3,\quad \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} = 2 + 1 = 3 \)
\( \det(A_z) = 1 \cdot (-6) – 2 \cdot (-3) + 14 \cdot 3 = -6 + 6 + 42 = 42 \)
Výsledky: \[ x = \frac{30}{15} = 2,\quad y = \frac{27}{15} = \frac{9}{5},\quad z = \frac{42}{15} = \frac{14}{5} \]
68. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} x + 2y – z = 3 \\ 2x – y + z = 0 \\ 3x + y + 2z = 7 \end{cases} \)
Řešení příkladu 68:
Koeficientová matice \( A \): \[ A = \begin{vmatrix} 1 & 2 & -1 \\ 2 & -1 & 1 \\ 3 & 1 & 2 \end{vmatrix} \]
Výpočet determinantu \( \det(A) \): \[ \det(A) = 1 \cdot \begin{vmatrix} -1 & 1 \\ 1 & 2 \end{vmatrix} – 2 \cdot \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} \]
\( \begin{vmatrix} -1 & 1 \\ 1 & 2 \end{vmatrix} = (-1)(2) – (1)(1) = -2 – 1 = -3 \\ \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} = 2(2) – 1(3) = 4 – 3 = 1 \\ \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} = 2(1) – (-1)(3) = 2 + 3 = 5 \)
\( \det(A) = 1 \cdot (-3) – 2 \cdot 1 + (-1) \cdot 5 = -3 – 2 – 5 = -10 \)
Vektor pravých stran \( b = (3, 0, 7) \).
Matica \( A_x \): \[ A_x = \begin{vmatrix} 3 & 2 & -1 \\ 0 & -1 & 1 \\ 7 & 1 & 2 \end{vmatrix} \]
\( \det(A_x) = 3 \cdot \begin{vmatrix} -1 & 1 \\ 1 & 2 \end{vmatrix} – 2 \cdot \begin{vmatrix} 0 & 1 \\ 7 & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 0 & -1 \\ 7 & 1 \end{vmatrix} \)
\( \begin{vmatrix} -1 & 1 \\ 1 & 2 \end{vmatrix} = -1 \cdot 2 – 1 \cdot 1 = -2 – 1 = -3 \\ \begin{vmatrix} 0 & 1 \\ 7 & 2 \end{vmatrix} = 0 \cdot 2 – 1 \cdot 7 = -7 \\ \begin{vmatrix} 0 & -1 \\ 7 & 1 \end{vmatrix} = 0 \cdot 1 – (-1)(7) = 7 \)
\( \det(A_x) = 3(-3) – 2(-7) – 1(7) = -9 + 14 – 7 = -2 \)
Matica \( A_y \): \[ A_y = \begin{vmatrix} 1 & 3 & -1 \\ 2 & 0 & 1 \\ 3 & 7 & 2 \end{vmatrix} \]
\( \det(A_y) = 1 \cdot \begin{vmatrix} 0 & 1 \\ 7 & 2 \end{vmatrix} – 3 \cdot \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 2 & 0 \\ 3 & 7 \end{vmatrix} \)
\( \begin{vmatrix} 0 & 1 \\ 7 & 2 \end{vmatrix} = 0 \cdot 2 – 1 \cdot 7 = -7 \\ \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} = 2 \cdot 2 – 1 \cdot 3 = 4 – 3 = 1 \\ \begin{vmatrix} 2 & 0 \\ 3 & 7 \end{vmatrix} = 2 \cdot 7 – 0 \cdot 3 = 14 \)
\( \det(A_y) = 1(-7) – 3(1) – 1(14) = -7 – 3 – 14 = -24 \)
Matica \( A_z \): \[ A_z = \begin{vmatrix} 1 & 2 & 3 \\ 2 & -1 & 0 \\ 3 & 1 & 7 \end{vmatrix} \]
\( \det(A_z) = 1 \cdot \begin{vmatrix} -1 & 0 \\ 1 & 7 \end{vmatrix} – 2 \cdot \begin{vmatrix} 2 & 0 \\ 3 & 7 \end{vmatrix} + 3 \cdot \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} \)
\( \begin{vmatrix} -1 & 0 \\ 1 & 7 \end{vmatrix} = -1 \cdot 7 – 0 \cdot 1 = -7 \\ \begin{vmatrix} 2 & 0 \\ 3 & 7 \end{vmatrix} = 2 \cdot 7 – 0 \cdot 3 = 14 \\ \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} = 2 \cdot 1 – (-1)(3) = 2 + 3 = 5 \)
\( \det(A_z) = 1(-7) – 2(14) + 3(5) = -7 – 28 + 15 = -20 \)
Výsledky: \[ x = \frac{-2}{-10} = \frac{1}{5},\quad y = \frac{-24}{-10} = \frac{12}{5},\quad z = \frac{-20}{-10} = 2 \]
69. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 2x + 3y + z = 4 \\ x – y + 2z = 1 \\ 3x + 2y – z = 0 \end{cases} \)
Řešení příkladu 69:
Koeficientová matice \( A \): \[ A = \begin{vmatrix} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 3 & 2 & -1 \end{vmatrix} \]
Výpočet determinantu \( \det(A) \): \[ \det(A) = 2 \cdot \begin{vmatrix} -1 & 2 \\ 2 & -1 \end{vmatrix} – 3 \cdot \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix} \]
\( \begin{vmatrix} -1 & 2 \\ 2 & -1 \end{vmatrix} = (-1)(-1) – (2)(2) = 1 – 4 = -3 \\ \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} = 1(-1) – 2(3) = -1 – 6 = -7 \\ \begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix} = 1(2) – (-1)(3) = 2 + 3 = 5 \)
\( \det(A) = 2(-3) – 3(-7) + 1(5) = -6 + 21 + 5 = 20 \)
Vektor pravých stran \( b = (4, 1, 0) \).
Matica \( A_x \): \[ A_x = \begin{vmatrix} 4 & 3 & 1 \\ 1 & -1 & 2 \\ 0 & 2 & -1 \end{vmatrix} \]
\( \det(A_x) = 4 \cdot \begin{vmatrix} -1 & 2 \\ 2 & -1 \end{vmatrix} – 3 \cdot \begin{vmatrix} 1 & 2 \\ 0 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -1 \\ 0 & 2 \end{vmatrix} \)
\( \begin{vmatrix} -1 & 2 \\ 2 & -1 \end{vmatrix} = -3 \\ \begin{vmatrix} 1 & 2 \\ 0 & -1 \end{vmatrix} = 1(-1) – 2(0) = -1 \\ \begin{vmatrix} 1 & -1 \\ 0 & 2 \end{vmatrix} = 1(2) – (-1)(0) = 2 \)
\( \det(A_x) = 4(-3) – 3(-1) + 1(2) = -12 + 3 + 2 = -7 \)
Matica \( A_y \): \[ A_y = \begin{vmatrix} 2 & 4 & 1 \\ 1 & 1 & 2 \\ 3 & 0 & -1 \end{vmatrix} \]
\( \det(A_y) = 2 \cdot \begin{vmatrix} 1 & 2 \\ 0 & -1 \end{vmatrix} – 4 \cdot \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} \)
\( \begin{vmatrix} 1 & 2 \\ 0 & -1 \end{vmatrix} = -1 \\ \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} = 1(-1) – 2(3) = -1 – 6 = -7 \\ \begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} = 1(0) – 1(3) = -3 \)
\( \det(A_y) = 2(-1) – 4(-7) + 1(-3) = -2 + 28 – 3 = 23 \)
Matica \( A_z \): \[ A_z = \begin{vmatrix} 2 & 3 & 4 \\ 1 & -1 & 1 \\ 3 & 2 & 0 \end{vmatrix} \]
\( \det(A_z) = 2 \cdot \begin{vmatrix} -1 & 1 \\ 2 & 0 \end{vmatrix} – 3 \cdot \begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} + 4 \cdot \begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix} \)
\( \begin{vmatrix} -1 & 1 \\ 2 & 0 \end{vmatrix} = -1(0) – 1(2) = -2 \\ \begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} = 1(0) – 1(3) = -3 \\ \begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix} = 1(2) – (-1)(3) = 2 + 3 = 5 \)
\( \det(A_z) = 2(-2) – 3(-3) + 4(5) = -4 + 9 + 20 = 25 \)
Výsledky: \[ x = \frac{-7}{20},\quad y = \frac{23}{20},\quad z = \frac{25}{20} = \frac{5}{4} \]
70. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} x + 2y + 3z = 14 \\ 2x + y + z = 10 \\ x – y + 2z = 5 \end{cases} \)
Řešení příkladu 70:
Matice \( A \): \[ A = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{vmatrix} \]
\( \det(A) = 1 \cdot \begin{vmatrix} 1 & 1 \\ -1 & 2 \end{vmatrix} – 2 \cdot \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} + 3 \cdot \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} \)
\( = 1(1 \cdot 2 – (-1) \cdot 1) – 2(2 \cdot 2 – 1 \cdot 1) + 3(2 \cdot (-1) – 1 \cdot 1) = 1(2 + 1) – 2(4 – 1) + 3(-2 – 1) \)
= 3 – 2 \cdot 3 – 9 = 3 – 6 – 9 = -12
Matica \( A_x \): \[ \begin{vmatrix} 14 & 2 & 3 \\ 10 & 1 & 1 \\ 5 & -1 & 2 \end{vmatrix} \]
\( \det(A_x) = 14 \cdot \begin{vmatrix} 1 & 1 \\ -1 & 2 \end{vmatrix} – 2 \cdot \begin{vmatrix} 10 & 1 \\ 5 & 2 \end{vmatrix} + 3 \cdot \begin{vmatrix} 10 & 1 \\ 5 & -1 \end{vmatrix} \)
= 14(1 \cdot 2 – (-1) \cdot 1) – 2(10 \cdot 2 – 1 \cdot 5) + 3(10 \cdot (-1) – 1 \cdot 5) = 14(3) – 2(20 – 5) + 3(-10 – 5)
= 42 – 2 \cdot 15 – 3 \cdot 15 = 42 – 30 – 45 = -33
Matica \( A_y \): \[ \begin{vmatrix} 1 & 14 & 3 \\ 2 & 10 & 1 \\ 1 & 5 & 2 \end{vmatrix} \]
\( \det(A_y) = 1 \cdot \begin{vmatrix} 10 & 1 \\ 5 & 2 \end{vmatrix} – 14 \cdot \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} + 3 \cdot \begin{vmatrix} 2 & 10 \\ 1 & 5 \end{vmatrix} \)
= 1(20 – 5) – 14(4 – 1) + 3(10 – 10) = 15 – 14 \cdot 3 + 0 = 15 – 42 = -27
Matica \( A_z \): \[ \begin{vmatrix} 1 & 2 & 14 \\ 2 & 1 & 10 \\ 1 & -1 & 5 \end{vmatrix} \]
\( \det(A_z) = 1 \cdot \begin{vmatrix} 1 & 10 \\ -1 & 5 \end{vmatrix} – 2 \cdot \begin{vmatrix} 2 & 10 \\ 1 & 5 \end{vmatrix} + 14 \cdot \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} \)
= 1(5 + 10) – 2(10 – 10) + 14(-2 – 1) = 15 – 0 – 42 = -27
\( x = \frac{-33}{-12} = \frac{11}{4},\quad y = \frac{-27}{-12} = \frac{9}{4},\quad z = \frac{-27}{-12} = \frac{9}{4} \)
71. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} x + y + z = 6 \\ 2x – y + 3z = 14 \\ x + 4y – z = 2 \end{cases} \)
Řešení příkladu 71:
Matice \( A \): \[ \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 1 & 4 & -1 \end{vmatrix} \]
\( \det(A) = 1 \cdot (-1 \cdot -1 – 3 \cdot 4) – 1 \cdot (2 \cdot -1 – 3 \cdot 1) + 1 \cdot (2 \cdot 4 – (-1) \cdot 1) = 1(1 – 12) – 1(-2 – 3) + 1(8 + 1) = -11 + 5 + 9 = 3 \)
Matica \( A_x \): \[ \begin{vmatrix} 6 & 1 & 1 \\ 14 & -1 & 3 \\ 2 & 4 & -1 \end{vmatrix} \]
\( \det(A_x) = 6(-1 \cdot -1 – 3 \cdot 4) – 1(14 \cdot -1 – 3 \cdot 2) + 1(14 \cdot 4 – (-1) \cdot 2) = 6(1 – 12) – ( -14 – 6) + (56 + 2) = 6(-11) + 20 + 58 = -66 + 78 = 12 \)
Matica \( A_y \): \[ \begin{vmatrix} 1 & 6 & 1 \\ 2 & 14 & 3 \\ 1 & 2 & -1 \end{vmatrix} \]
\( \det(A_y) = 1(14 \cdot -1 – 3 \cdot 2) – 6(2 \cdot -1 – 3 \cdot 1) + 1(2 \cdot 2 – 14 \cdot 1) = 1(-14 – 6) – 6(-2 – 3) + (4 – 14) = -20 + 30 – 10 = 0 \)
Matica \( A_z \): \[ \begin{vmatrix} 1 & 1 & 6 \\ 2 & -1 & 14 \\ 1 & 4 & 2 \end{vmatrix} \]
\( \det(A_z) = 1(-1 \cdot 2 – 14 \cdot 4) – 1(2 \cdot 2 – 14 \cdot 1) + 6(2 \cdot 4 – (-1) \cdot 1) = 1(-2 – 56) – (4 – 14) + 6(8 + 1) = -58 + 10 + 54 = 6 \)
\( x = \frac{12}{3} = 4,\quad y = \frac{0}{3} = 0,\quad z = \frac{6}{3} = 2 \)
72. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} x + 2y – z = 3 \\ 2x – y + z = 0 \\ 3x + y + 2z = 7 \end{cases} \)
Řešení příkladu 72:
Matice \( A \): \[ \begin{vmatrix} 1 & 2 & -1 \\ 2 & -1 & 1 \\ 3 & 1 & 2 \end{vmatrix} \]
\( \det(A) = 1(-1 \cdot 2 – 1 \cdot 1) – 2(2 \cdot 2 – 1 \cdot 3) + (-1)(2 \cdot 1 – (-1) \cdot 3) = 1(-2 – 1) – 2(4 – 3) – 1(2 + 3) = -3 – 2 – 5 = -10 \)
Matica \( A_x \): \[ \begin{vmatrix} 3 & 2 & -1 \\ 0 & -1 & 1 \\ 7 & 1 & 2 \end{vmatrix} \]
\( \det(A_x) = 3(-1 \cdot 2 – 1 \cdot 1) – 2(0 \cdot 2 – 1 \cdot 7) + (-1)(0 \cdot 1 – (-1) \cdot 7) = 3(-2 – 1) – 2(-7) – (0 + 7) = -9 + 14 – 7 = -2 \)
Matica \( A_y \): \[ \begin{vmatrix} 1 & 3 & -1 \\ 2 & 0 & 1 \\ 3 & 7 & 2 \end{vmatrix} \]
\( \det(A_y) = 1(0 \cdot 2 – 1 \cdot 7) – 3(2 \cdot 2 – 1 \cdot 3) + (-1)(2 \cdot 7 – 0 \cdot 3) = 1(0 – 7) – 3(4 – 3) – 1(14) = -7 – 3 – 14 = -24 \)
Matica \( A_z \): \[ \begin{vmatrix} 1 & 2 & 3 \\ 2 & -1 & 0 \\ 3 & 1 & 7 \end{vmatrix} \]
\( \det(A_z) = 1(-1 \cdot 7 – 0 \cdot 1) – 2(2 \cdot 7 – 0 \cdot 3) + 3(2 \cdot 1 – (-1) \cdot 3) = 1(-7) – 2(14) + 3(2 + 3) = -7 – 28 + 15 = -20 \)
\( x = \frac{-2}{-10} = \frac{1}{5},\quad y = \frac{-24}{-10} = \frac{12}{5},\quad z = \frac{-20}{-10} = 2 \)
73. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 2x – y + 3z = 9 \\ x + y – z = 2 \\ 3x + 2y + z = 10 \end{cases} \)
Řešení příkladu 73:
Matice \( A \): \[ \begin{vmatrix} 2 & -1 & 3 \\ 1 & 1 & -1 \\ 3 & 2 & 1 \end{vmatrix} \]
\( \det(A) = 2(1 \cdot 1 – (-1) \cdot 2) + 1(1 \cdot 1 – (-1) \cdot 3) + 3(1 \cdot 2 – 1 \cdot 3) = 2(1 + 2) + (-1)(1 + 3) + 3(2 – 3) = 6 – 4 – 3 = -1 \)
Matica \( A_x \): \[ \begin{vmatrix} 9 & -1 & 3 \\ 2 & 1 & -1 \\ 10 & 2 & 1 \end{vmatrix} \]
\( \det(A_x) = 9(1 \cdot 1 – (-1) \cdot 2) + 1(2 \cdot 1 – (-1) \cdot 10) + 3(2 \cdot 2 – 1 \cdot 10) = 9(1 + 2) – 1(2 + 10) + 3(4 – 10) = 27 – 12 – 18 = -3 \)
Matica \( A_y \): \[ \begin{vmatrix} 2 & 9 & 3 \\ 1 & 2 & -1 \\ 3 & 10 & 1 \end{vmatrix} \]
\( \det(A_y) = 2(2 \cdot 1 – (-1) \cdot 10) – 9(1 \cdot 1 – (-1) \cdot 3) + 3(1 \cdot 10 – 2 \cdot 3) = 2(2 + 10) – 9(1 + 3) + 3(10 – 6) = 24 – 36 + 12 = 0 \)
Matica \( A_z \): \[ \begin{vmatrix} 2 & -1 & 9 \\ 1 & 1 & 2 \\ 3 & 2 & 10 \end{vmatrix} \]
\( \det(A_z) = 2(1 \cdot 10 – 2 \cdot 2) – (-1)(1 \cdot 10 – 2 \cdot 3) + 9(1 \cdot 2 – 1 \cdot 3) = 2(10 – 4) + (10 – 6) + 9(-1) = 12 + 4 – 9 = 7 \)
\( x = \frac{-3}{-1} = 3,\quad y = \frac{0}{-1} = 0,\quad z = \frac{7}{-1} = -7 \)
74. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 3x + y – 2z = 5 \\ 2x – 2y + 4z = -2 \\ – x + \frac{1}{2}y – z = 0 \end{cases} \)
Řešení příkladu 74:
Matice \( A \): \[ \begin{vmatrix} 3 & 1 & -2 \\ 2 & -2 & 4 \\ -1 & \frac{1}{2} & -1 \end{vmatrix} \]
\( \det(A) = 3 \cdot (-2 \cdot -1 – 4 \cdot \frac{1}{2}) – 1 \cdot (2 \cdot -1 – 4 \cdot -1) + (-2) \cdot (2 \cdot \frac{1}{2} – (-2) \cdot -1) \)
\( = 3(2 – 2) – 1(-2 + 4) – 2(1 – 2) = 3 \cdot 0 – 1 \cdot 2 – 2 \cdot (-1) = 0 – 2 + 2 = 0 \)
Determinant je 0, takže Cramerovo pravidlo nelze použít, soustava může být neřešitelná nebo mít nekonečně mnoho řešení.
75. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} x – y + z = 4 \\ 2x + y – 3z = 1 \\ x + 3y + 2z = 7 \end{cases} \)
Řešení příkladu 75:
Matice \( A \): \[ \begin{vmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 3 & 2 \end{vmatrix} \]
\( \det(A) = 1(1 \cdot 2 – (-3) \cdot 3) – (-1)(2 \cdot 2 – (-3) \cdot 1) + 1(2 \cdot 3 – 1 \cdot 1) = 1(2 + 9) + 1(4 + 3) + 1(6 – 1) = 11 + 7 + 5 = 23 \)
Matice \( A_x \): \[ \begin{vmatrix} 4 & -1 & 1 \\ 1 & 1 & -3 \\ 7 & 3 & 2 \end{vmatrix} \]
\( \det(A_x) = 4(1 \cdot 2 – (-3) \cdot 3) – (-1)(1 \cdot 2 – (-3) \cdot 7) + 1(1 \cdot 3 – 1 \cdot 7) = 4(2 + 9) + 1(2 + 21) + 1(3 – 7) = 44 + 23 – 4 = 63 \)
Matice \( A_y \): \[ \begin{vmatrix} 1 & 4 & 1 \\ 2 & 1 & -3 \\ 1 & 7 & 2 \end{vmatrix} \]
\( \det(A_y) = 1(1 \cdot 2 – (-3) \cdot 7) – 4(2 \cdot 2 – (-3) \cdot 1) + 1(2 \cdot 7 – 1 \cdot 1) = 1(2 + 21) – 4(4 + 3) + 1(14 – 1) = 23 – 28 + 13 = 8 \)
Matice \( A_z \): \[ \begin{vmatrix} 1 & -1 & 4 \\ 2 & 1 & 1 \\ 1 & 3 & 7 \end{vmatrix} \]
\( \det(A_z) = 1(1 \cdot 7 – 1 \cdot 3) – (-1)(2 \cdot 7 – 1 \cdot 1) + 4(2 \cdot 3 – 1 \cdot 1) = 1(7 – 3) + 1(14 – 1) + 4(6 – 1) = 4 + 13 + 20 = 37 \)
\( x = \frac{63}{23},\quad y = \frac{8}{23},\quad z = \frac{37}{23} \)
76. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 4x + 2y + z = 7 \\ -2x + y + 5z = 3 \\ x – y + 2z = 4 \end{cases} \)
Řešení příkladu 76:
Matice \( A \): \[ \begin{vmatrix} 4 & 2 & 1 \\ -2 & 1 & 5 \\ 1 & -1 & 2 \end{vmatrix} \]
\( \det(A) = 4(1 \cdot 2 – 5 \cdot -1) – 2(-2 \cdot 2 – 5 \cdot 1) + 1(-2 \cdot -1 – 1 \cdot 1) = 4(2 + 5) – 2(-4 – 5) + 1(2 – 1) = 4 \cdot 7 + 2 \cdot 9 + 1 = 28 + 18 + 1 = 47 \)
Matice \( A_x \): \[ \begin{vmatrix} 7 & 2 & 1 \\ 3 & 1 & 5 \\ 4 & -1 & 2 \end{vmatrix} \]
\( \det(A_x) = 7(1 \cdot 2 – 5 \cdot -1) – 2(3 \cdot 2 – 5 \cdot 4) + 1(3 \cdot -1 – 1 \cdot 4) = 7(2 + 5) – 2(6 – 20) + ( -3 – 4) = 49 – 2(-14) – 7 = 49 + 28 – 7 = 70 \)
Matice \( A_y \): \[ \begin{vmatrix} 4 & 7 & 1 \\ -2 & 3 & 5 \\ 1 & 4 & 2 \end{vmatrix} \]
\( \det(A_y) = 4(3 \cdot 2 – 5 \cdot 4) – 7(-2 \cdot 2 – 5 \cdot 1) + 1(-2 \cdot 4 – 3 \cdot 1) = 4(6 – 20) – 7(-4 – 5) + (-8 – 3) = 4(-14) + 7 \cdot 9 – 11 = -56 + 63 – 11 = -4 \)
Matice \( A_z \): \[ \begin{vmatrix} 4 & 2 & 7 \\ -2 & 1 & 3 \\ 1 & -1 & 4 \end{vmatrix} \]
\( \det(A_z) = 4(1 \cdot 4 – 3 \cdot -1) – 2(-2 \cdot 4 – 3 \cdot 1) + 7(-2 \cdot -1 – 1 \cdot 1) = 4(4 + 3) – 2(-8 – 3) + 7(2 – 1) = 4 \cdot 7 + 2 \cdot 11 + 7 = 28 + 22 + 7 = 57 \)
\( x = \frac{70}{47},\quad y = \frac{-4}{47},\quad z = \frac{57}{47} \)
77. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} x + 2y + 3z = 14 \\ 3x – y + z = 5 \\ 2x + y – z = 7 \end{cases} \)
Řešení příkladu 77:
Matice \( A \): \[ \begin{vmatrix} 1 & 2 & 3 \\ 3 & -1 & 1 \\ 2 & 1 & -1 \end{vmatrix} \]
\( \det(A) = 1((-1)(-1) – 1 \cdot 1) – 2(3 \cdot -1 – 1 \cdot 2) + 3(3 \cdot 1 – (-1) \cdot 2) = 1(1 – 1) – 2(-3 – 2) + 3(3 + 2) = 0 + 10 + 15 = 25 \)
Matice \( A_x \): \[ \begin{vmatrix} 14 & 2 & 3 \\ 5 & -1 & 1 \\ 7 & 1 & -1 \end{vmatrix} \]
\( \det(A_x) = 14((-1)(-1) – 1 \cdot 1) – 2(5 \cdot -1 – 1 \cdot 7) + 3(5 \cdot 1 – (-1) \cdot 7) = 14(1 – 1) – 2(-5 – 7) + 3(5 + 7) = 0 + 24 + 36 = 60 \)
Matice \( A_y \): \[ \begin{vmatrix} 1 & 14 & 3 \\ 3 & 5 & 1 \\ 2 & 7 & -1 \end{vmatrix} \]
\( \det(A_y) = 1(5 \cdot -1 – 1 \cdot 7) – 14(3 \cdot -1 – 1 \cdot 2) + 3(3 \cdot 7 – 5 \cdot 2) = 1(-5 – 7) – 14(-3 – 2) + 3(21 – 10) = -12 + 70 + 33 = 91 \)
Matice \( A_z \): \[ \begin{vmatrix} 1 & 2 & 14 \\ 3 & -1 & 5 \\ 2 & 1 & 7 \end{vmatrix} \]
\( \det(A_z) = 1((-1) \cdot 7 – 5 \cdot 1) – 2(3 \cdot 7 – 5 \cdot 2) + 14(3 \cdot 1 – (-1) \cdot 2) = 1(-7 – 5) – 2(21 – 10) + 14(3 + 2) = -12 – 22 + 70 = 36 \)
\( x = \frac{60}{25} = \frac{12}{5}, \quad y = \frac{91}{25}, \quad z = \frac{36}{25} \)
78. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 2x – y + 3z = 9 \\ x + 4y – 2z = 1 \\ 3x – 2y + z = 7 \end{cases} \)
Řešení příkladu 78:
Matice \( A \): \[ \begin{vmatrix} 2 & -1 & 3 \\ 1 & 4 & -2 \\ 3 & -2 & 1 \end{vmatrix} \]
\( \det(A) = 2(4 \cdot 1 – (-2) \cdot -2) – (-1)(1 \cdot 1 – (-2) \cdot 3) + 3(1 \cdot -2 – 4 \cdot 3) = 2(4 – 4) + 1(1 + 6) + 3(-2 – 12) = 0 + 7 – 42 = -35 \)
Matice \( A_x \): \[ \begin{vmatrix} 9 & -1 & 3 \\ 1 & 4 & -2 \\ 7 & -2 & 1 \end{vmatrix} \]
\( \det(A_x) = 9(4 \cdot 1 – (-2) \cdot -2) – (-1)(1 \cdot 1 – (-2) \cdot 7) + 3(1 \cdot -2 – 4 \cdot 7) = 9(4 – 4) + 1(1 + 14) + 3(-2 – 28) = 0 + 15 – 90 = -75 \)
Matice \( A_y \): \[ \begin{vmatrix} 2 & 9 & 3 \\ 1 & 1 & -2 \\ 3 & 7 & 1 \end{vmatrix} \]
\( \det(A_y) = 2(1 \cdot 1 – (-2) \cdot 7) – 9(1 \cdot 1 – (-2) \cdot 3) + 3(1 \cdot 7 – 1 \cdot 3) = 2(1 + 14) – 9(1 + 6) + 3(7 – 3) = 30 – 63 + 12 = -21 \)
Matice \( A_z \): \[ \begin{vmatrix} 2 & -1 & 9 \\ 1 & 4 & 1 \\ 3 & -2 & 7 \end{vmatrix} \]
\( \det(A_z) = 2(4 \cdot 7 – 1 \cdot -2) – (-1)(1 \cdot 7 – 1 \cdot 3) + 9(1 \cdot -2 – 4 \cdot 3) = 2(28 + 2) + 1(7 – 3) + 9(-2 – 12) = 2 \cdot 30 + 4 – 126 = 60 + 4 – 126 = -62 \)
\( x = \frac{-75}{-35} = \frac{15}{7}, \quad y = \frac{-21}{-35} = \frac{3}{5}, \quad z = \frac{-62}{-35} = \frac{62}{35} \)
79. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 3x + 2y – z = 5 \\ -2x + y + 4z = 3 \\ x – 3y + 2z = -4 \end{cases} \)
Řešení příkladu 79:
Matice \( A \): \[ \begin{vmatrix} 3 & 2 & -1 \\ -2 & 1 & 4 \\ 1 & -3 & 2 \end{vmatrix} \]
\( \det(A) = 3(1 \cdot 2 – 4 \cdot -3) – 2(-2 \cdot 2 – 4 \cdot 1) + (-1)(-2 \cdot -3 – 1 \cdot 1) = 3(2 + 12) – 2(-4 – 4) -1(6 – 1) = 3 \cdot 14 + 2 \cdot 8 – 5 = 42 + 16 – 5 = 53 \)
Matice \( A_x \): \[ \begin{vmatrix} 5 & 2 & -1 \\ 3 & 1 & 4 \\ -4 & -3 & 2 \end{vmatrix} \]
\( \det(A_x) = 5(1 \cdot 2 – 4 \cdot -3) – 2(3 \cdot 2 – 4 \cdot -4) + (-1)(3 \cdot -3 – 1 \cdot -4) = 5(2 + 12) – 2(6 + 16) -1(-9 + 4) = 5 \cdot 14 – 2 \cdot 22 + 5 = 70 – 44 + 5 = 31 \)
Matice \( A_y \): \[ \begin{vmatrix} 3 & 5 & -1 \\ -2 & 3 & 4 \\ 1 & -4 & 2 \end{vmatrix} \]
\( \det(A_y) = 3(3 \cdot 2 – 4 \cdot -4) – 5(-2 \cdot 2 – 4 \cdot 1) + (-1)(-2 \cdot -4 – 3 \cdot 1) = 3(6 + 16) – 5(-4 – 4) -1(8 – 3) = 3 \cdot 22 + 5 \cdot 8 – 5 = 66 + 40 – 5 = 101 \)
Matice \( A_z \): \[ \begin{vmatrix} 3 & 2 & 5 \\ -2 & 1 & 3 \\ 1 & -3 & -4 \end{vmatrix} \]
\( \det(A_z) = 3(1 \cdot -4 – 3 \cdot -3) – 2(-2 \cdot -4 – 3 \cdot 1) + 5(-2 \cdot -3 – 1 \cdot 1) = 3(-4 + 9) – 2(8 – 3) + 5(6 – 1) = 3 \cdot 5 – 2 \cdot 5 + 5 \cdot 5 = 15 – 10 + 25 = 30 \)
\( x = \frac{31}{53}, \quad y = \frac{101}{53}, \quad z = \frac{30}{53} \)
80. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} x + y + z = 6 \\ 2x – y + 3z = 14 \\ -x + 4y – z = -2 \end{cases} \)
Řešení příkladu 80:
Matice \( A \): \[ \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ -1 & 4 & -1 \end{vmatrix} \]
\( \det(A) = 1((-1)(-1) – 3 \cdot 4) – 1(2 \cdot -1 – 3 \cdot -1) + 1(2 \cdot 4 – (-1) \cdot -1) = 1(1 – 12) – 1(-2 + 3) + 1(8 – 1) = -11 – 1 + 7 = -5 \)
Matice \( A_x \): \[ \begin{vmatrix} 6 & 1 & 1 \\ 14 & -1 & 3 \\ -2 & 4 & -1 \end{vmatrix} \]
\( \det(A_x) = 6((-1)(-1) – 3 \cdot 4) – 1(14 \cdot -1 – 3 \cdot -2) + 1(14 \cdot 4 – (-1) \cdot -2) = 6(1 – 12) – 1(-14 + 6) + 1(56 – 2) = 6(-11) – (-8) + 54 = -66 + 8 + 54 = -4 \)
Matice \( A_y \): \[ \begin{vmatrix} 1 & 6 & 1 \\ 2 & 14 & 3 \\ -1 & -2 & -1 \end{vmatrix} \]
\( \det(A_y) = 1(14 \cdot -1 – 3 \cdot -2) – 6(2 \cdot -1 – 3 \cdot -1) + 1(2 \cdot -2 – 14 \cdot -1) = 1(-14 + 6) – 6(-2 + 3) + 1(-4 + 14) = -8 – 6 + 10 = -4 \)
Matice \( A_z \): \[ \begin{vmatrix} 1 & 1 & 6 \\ 2 & -1 & 14 \\ -1 & 4 & -2 \end{vmatrix} \]
\( \det(A_z) = 1((-1)(-2) – 14 \cdot 4) – 1(2 \cdot -2 – 14 \cdot -1) + 6(2 \cdot 4 – (-1) \cdot -1) = 1(2 – 56) – 1(-4 + 14) + 6(8 – 1) = -54 – 10 + 42 = -22 \)
\( x = \frac{-4}{-5} = \frac{4}{5}, \quad y = \frac{-4}{-5} = \frac{4}{5}, \quad z = \frac{-22}{-5} = \frac{22}{5} \)
81. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 4x – y + 2z = 7 \\ -3x + 5y – z = -8 \\ 2x + 3y + 4z = 10 \end{cases} \)
Řešení příkladu 81:
Matice \( A \): \[ \begin{vmatrix} 4 & -1 & 2 \\ -3 & 5 & -1 \\ 2 & 3 & 4 \end{vmatrix} \]
\( \det(A) = 4(5 \cdot 4 – (-1) \cdot 3) – (-1)(-3 \cdot 4 – (-1) \cdot 2) + 2(-3 \cdot 3 – 5 \cdot 2) = 4(20 + 3) – (-1)(-12 + 2) + 2(-9 – 10) = 4 \cdot 23 – (-1) \cdot (-10) + 2(-19) = 92 – 10 – 38 = 44 \)
Matice \( A_x \): \[ \begin{vmatrix} 7 & -1 & 2 \\ -8 & 5 & -1 \\ 10 & 3 & 4 \end{vmatrix} \]
\( \det(A_x) = 7(5 \cdot 4 – (-1) \cdot 3) – (-1)(-8 \cdot 4 – (-1) \cdot 10) + 2(-8 \cdot 3 – 5 \cdot 10) = 7(20 + 3) – (-1)(-32 + 10) + 2(-24 – 50) = 7 \cdot 23 – (-1) \cdot (-22) + 2(-74) = 161 – 22 – 148 = -9 \)
Matice \( A_y \): \[ \begin{vmatrix} 4 & 7 & 2 \\ -3 & -8 & -1 \\ 2 & 10 & 4 \end{vmatrix} \]
\( \det(A_y) = 4(-8 \cdot 4 – (-1) \cdot 10) – 7(-3 \cdot 4 – (-1) \cdot 2) + 2(-3 \cdot 10 – (-8) \cdot 2) = 4(-32 + 10) – 7(-12 + 2) + 2(-30 + 16) = 4(-22) – 7(-10) + 2(-14) = -88 + 70 – 28 = -46 \)
Matice \( A_z \): \[ \begin{vmatrix} 4 & -1 & 7 \\ -3 & 5 & -8 \\ 2 & 3 & 10 \end{vmatrix} \]
\( \det(A_z) = 4(5 \cdot 10 – (-8) \cdot 3) – (-1)(-3 \cdot 10 – (-8) \cdot 2) + 7(-3 \cdot 3 – 5 \cdot 2) = 4(50 + 24) – (-1)(-30 + 16) + 7(-9 – 10) = 4 \cdot 74 – (-1) \cdot (-14) + 7(-19) = 296 – 14 – 133 = 149 \)
\( x = \frac{-9}{44}, \quad y = \frac{-46}{44} = \frac{-23}{22}, \quad z = \frac{149}{44} \)
82. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 2x + y – 3z = 4 \\ -x + 4y + 2z = 7 \\ 3x – 2y + z = -1 \end{cases} \)
Řešení příkladu 82:
Matice \( A \): \[ \begin{vmatrix} 2 & 1 & -3 \\ -1 & 4 & 2 \\ 3 & -2 & 1 \end{vmatrix} \]
\( \det(A) = 2(4 \cdot 1 – 2 \cdot (-2)) – 1(-1 \cdot 1 – 2 \cdot 3) + (-3)(-1 \cdot (-2) – 4 \cdot 3) = 2(4 + 4) – 1(-1 – 6) -3(2 – 12) = 2 \cdot 8 – 1 \cdot (-7) – 3 \cdot (-10) = 16 + 7 + 30 = 53 \)
Matice \( A_x \): \[ \begin{vmatrix} 4 & 1 & -3 \\ 7 & 4 & 2 \\ -1 & -2 & 1 \end{vmatrix} \]
\( \det(A_x) = 4(4 \cdot 1 – 2 \cdot (-2)) – 1(7 \cdot 1 – 2 \cdot (-1)) + (-3)(7 \cdot (-2) – 4 \cdot (-1)) = 4(4 + 4) – 1(7 + 2) – 3(-14 + 4) = 4 \cdot 8 – 9 – 3 \cdot (-10) = 32 – 9 + 30 = 53 \)
Matice \( A_y \): \[ \begin{vmatrix} 2 & 4 & -3 \\ -1 & 7 & 2 \\ 3 & -1 & 1 \end{vmatrix} \]
\( \det(A_y) = 2(7 \cdot 1 – 2 \cdot (-1)) – 4(-1 \cdot 1 – 2 \cdot 3) + (-3)(-1 \cdot (-1) – 7 \cdot 3) = 2(7 + 2) – 4(-1 – 6) – 3(1 – 21) = 2 \cdot 9 + 4 \cdot 7 – 3 \cdot (-20) = 18 + 28 + 60 = 106 \)
Matice \( A_z \): \[ \begin{vmatrix} 2 & 1 & 4 \\ -1 & 4 & 7 \\ 3 & -2 & -1 \end{vmatrix} \]
\( \det(A_z) = 2(4 \cdot (-1) – 7 \cdot (-2)) – 1(-1 \cdot (-1) – 7 \cdot 3) + 4(-1 \cdot (-2) – 4 \cdot 3) = 2(-4 + 14) – 1(1 – 21) + 4(2 – 12) = 2 \cdot 10 – 1 \cdot (-20) + 4 \cdot (-10) = 20 + 20 – 40 = 0 \)
\( x = \frac{\det(A_x)}{\det(A)} = \frac{53}{53} = 1, \quad y = \frac{106}{53} = 2, \quad z = \frac{0}{53} = 0 \)
83. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} x + 2y + z = 6 \\ 2x – y + 3z = 14 \\ -x + 3y – 2z = -2 \end{cases} \)
Řešení příkladu 83:
Matice \( A \): \[ \begin{vmatrix} 1 & 2 & 1 \\ 2 & -1 & 3 \\ -1 & 3 & -2 \end{vmatrix} \]
\( \det(A) = 1((-1)(-2) – 3 \cdot 3) – 2(2 \cdot (-2) – 3 \cdot (-1)) + 1(2 \cdot 3 – (-1) \cdot (-1)) = 1(2 – 9) – 2(-4 + 3) + 1(6 – 1) = 1(-7) – 2(-1) + 5 = -7 + 2 + 5 = 0 \)
Determinant je nula, což znamená, že soustava nemá jednoznačné řešení nebo je nekonzistentní.
84. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 4x – y + z = 5 \\ -2x + 3y + 2z = 3 \\ x + 2y – 3z = -4 \end{cases} \)
Řešení příkladu 84:
Matice \( A \): \[ \begin{vmatrix} 4 & -1 & 1 \\ -2 & 3 & 2 \\ 1 & 2 & -3 \end{vmatrix} \]
\( \det(A) = 4(3 \cdot (-3) – 2 \cdot 2) – (-1)(-2 \cdot (-3) – 2 \cdot 1) + 1(-2 \cdot 2 – 3 \cdot 1) = 4(-9 – 4) – (-1)(6 – 2) + 1(-4 – 3) = 4(-13) + 4 – 7 = -52 + 4 -7 = -55 \)
Matice \( A_x \): \[ \begin{vmatrix} 5 & -1 & 1 \\ 3 & 3 & 2 \\ -4 & 2 & -3 \end{vmatrix} \]
\( \det(A_x) = 5(3 \cdot (-3) – 2 \cdot 2) – (-1)(3 \cdot (-3) – 2 \cdot (-4)) + 1(3 \cdot 2 – 3 \cdot (-4)) = 5(-9 – 4) – (-1)(-9 + 8) + 1(6 + 12) = 5(-13) – (-1)(-1) + 18 = -65 – 1 + 18 = -48 \)
Matice \( A_y \): \[ \begin{vmatrix} 4 & 5 & 1 \\ -2 & 3 & 2 \\ 1 & -4 & -3 \end{vmatrix} \]
\( \det(A_y) = 4(3 \cdot (-3) – 2 \cdot (-4)) – 5(-2 \cdot (-3) – 2 \cdot 1) + 1(-2 \cdot (-4) – 3 \cdot 1) = 4(-9 + 8) – 5(6 – 2) + 1(8 – 3) = 4(-1) – 5 \cdot 4 + 5 = -4 – 20 + 5 = -19 \)
Matice \( A_z \): \[ \begin{vmatrix} 4 & -1 & 5 \\ -2 & 3 & 3 \\ 1 & 2 & -4 \end{vmatrix} \]
\( \det(A_z) = 4(3 \cdot (-4) – 3 \cdot 2) – (-1)(-2 \cdot (-4) – 3 \cdot 1) + 5(-2 \cdot 2 – 3 \cdot 1) = 4(-12 – 6) – (-1)(8 – 3) + 5(-4 – 3) = 4(-18) + 5 – 35 = -72 + 5 – 35 = -102 \)
\( x = \frac{-48}{-55} = \frac{48}{55}, \quad y = \frac{-19}{-55} = \frac{19}{55}, \quad z = \frac{-102}{-55} = \frac{102}{55} \)
85. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} x – 2y + 3z = 9 \\ 3x + y – z = 4 \\ 2x – y + 4z = 11 \end{cases} \)
Řešení příkladu 85:
Matice \( A \): \[ \begin{vmatrix} 1 & -2 & 3 \\ 3 & 1 & -1 \\ 2 & -1 & 4 \end{vmatrix} \]
\( \det(A) = 1(1 \cdot 4 – (-1) \cdot (-1)) – (-2)(3 \cdot 4 – (-1) \cdot 2) + 3(3 \cdot (-1) – 1 \cdot 2) = 1(4 – 1) – (-2)(12 – (-2)) + 3(-3 – 2) = 3 + 2 \cdot 14 – 15 = 3 + 28 – 15 = 16 \)
Matice \( A_x \): \[ \begin{vmatrix} 9 & -2 & 3 \\ 4 & 1 & -1 \\ 11 & -1 & 4 \end{vmatrix} \]
\( \det(A_x) = 9(1 \cdot 4 – (-1) \cdot (-1)) – (-2)(4 \cdot 4 – (-1) \cdot 11) + 3(4 \cdot (-1) – 1 \cdot 11) = 9(4 – 1) – (-2)(16 – (-11)) + 3(-4 – 11) = 27 + 2 \cdot 27 – 45 = 27 + 54 – 45 = 36 \)
Matice \( A_y \): \[ \begin{vmatrix} 1 & 9 & 3 \\ 3 & 4 & -1 \\ 2 & 11 & 4 \end{vmatrix} \]
\( \det(A_y) = 1(4 \cdot 4 – (-1) \cdot 11) – 9(3 \cdot 4 – (-1) \cdot 2) + 3(3 \cdot 11 – 4 \cdot 2) = 1(16 + 11) – 9(12 + 2) + 3(33 – 8) = 27 – 126 + 75 = -24 \)
Matice \( A_z \): \[ \begin{vmatrix} 1 & -2 & 9 \\ 3 & 1 & 4 \\ 2 & -1 & 11 \end{vmatrix} \]
\( \det(A_z) = 1(1 \cdot 11 – 4 \cdot (-1)) – (-2)(3 \cdot 11 – 4 \cdot 2) + 9(3 \cdot (-1) – 1 \cdot 2) = 1(11 + 4) – (-2)(33 – 8) + 9(-3 – 2) = 15 + 2 \cdot 25 – 45 = 15 + 50 – 45 = 20 \)
\( x = \frac{36}{16} = \frac{9}{4}, \quad y = \frac{-24}{16} = -\frac{3}{2}, \quad z = \frac{20}{16} = \frac{5}{4} \)
86. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 5x + 2y – z = 3 \\ x – 3y + 4z = -2 \\ 2x + y + 3z = 7 \end{cases} \)
Řešení příkladu 86:
Matice \( A \): \[ \begin{vmatrix} 5 & 2 & -1 \\ 1 & -3 & 4 \\ 2 & 1 & 3 \end{vmatrix} \]
\( \det(A) = 5(-3 \cdot 3 – 4 \cdot 1) – 2(1 \cdot 3 – 4 \cdot 2) + (-1)(1 \cdot 1 – (-3) \cdot 2) = 5(-9 – 4) – 2(3 – 8) – 1(1 + 6) = 5(-13) – 2(-5) – 7 = -65 + 10 – 7 = -62 \)
Matice \( A_x \): \[ \begin{vmatrix} 3 & 2 & -1 \\ -2 & -3 & 4 \\ 7 & 1 & 3 \end{vmatrix} \]
\( \det(A_x) = 3(-3 \cdot 3 – 4 \cdot 1) – 2(-2 \cdot 3 – 4 \cdot 7) + (-1)(-2 \cdot 1 – (-3) \cdot 7) = 3(-9 – 4) – 2(-6 – 28) – 1(-2 + 21) = 3(-13) + 2 \cdot 34 – 19 = -39 + 68 – 19 = 10 \)
Matice \( A_y \): \[ \begin{vmatrix} 5 & 3 & -1 \\ 1 & -2 & 4 \\ 2 & 7 & 3 \end{vmatrix} \]
\( \det(A_y) = 5(-2 \cdot 3 – 4 \cdot 7) – 3(1 \cdot 3 – 4 \cdot 2) + (-1)(1 \cdot 7 – (-2) \cdot 2) = 5(-6 – 28) – 3(3 – 8) – 1(7 + 4) = 5(-34) – 3(-5) – 11 = -170 + 15 – 11 = -166 \)
Matice \( A_z \): \[ \begin{vmatrix} 5 & 2 & 3 \\ 1 & -3 & -2 \\ 2 & 1 & 7 \end{vmatrix} \]
\( \det(A_z) = 5(-3 \cdot 7 – (-2) \cdot 1) – 2(1 \cdot 7 – (-2) \cdot 2) + 3(1 \cdot 1 – (-3) \cdot 2) = 5(-21 + 2) – 2(7 + 4) + 3(1 + 6) = 5(-19) – 2 \cdot 11 + 3 \cdot 7 = -95 – 22 + 21 = -96 \)
\( x = \frac{10}{-62} = -\frac{5}{31}, \quad y = \frac{-166}{-62} = \frac{83}{31}, \quad z = \frac{-96}{-62} = \frac{48}{31} \)
87. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 2x + y – z = 4 \\ -x + 3y + 2z = 1 \\ 4x – y + 5z = 10 \end{cases} \)
Řešení příkladu 87:
Matice \( A \): \[ \begin{vmatrix} 2 & 1 & -1 \\ -1 & 3 & 2 \\ 4 & -1 & 5 \end{vmatrix} \]
\( \det(A) = 2(3 \cdot 5 – 2 \cdot (-1)) – 1(-1 \cdot 5 – 2 \cdot 4) + (-1)(-1 \cdot (-1) – 3 \cdot 4) = 2(15 + 2) – 1(-5 – 8) – 1(1 – 12) = 2 \cdot 17 + 13 – (-11) = 34 + 13 + 11 = 58 \)
Matice \( A_x \): \[ \begin{vmatrix} 4 & 1 & -1 \\ 1 & 3 & 2 \\ 10 & -1 & 5 \end{vmatrix} \]
\( \det(A_x) = 4(3 \cdot 5 – 2 \cdot (-1)) – 1(1 \cdot 5 – 2 \cdot 10) + (-1)(1 \cdot (-1) – 3 \cdot 10) = 4(15 + 2) – 1(5 – 20) – 1(-1 – 30) = 4 \cdot 17 + 15 + 31 = 68 + 15 + 31 = 114 \)
Matice \( A_y \): \[ \begin{vmatrix} 2 & 4 & -1 \\ -1 & 1 & 2 \\ 4 & 10 & 5 \end{vmatrix} \]
\( \det(A_y) = 2(1 \cdot 5 – 2 \cdot 10) – 4(-1 \cdot 5 – 2 \cdot 4) + (-1)(-1 \cdot 10 – 1 \cdot 4) = 2(5 – 20) – 4(-5 – 8) – 1(-10 – 4) = 2(-15) + 4 \cdot 13 + 14 = -30 + 52 + 14 = 36 \)
Matice \( A_z \): \[ \begin{vmatrix} 2 & 1 & 4 \\ -1 & 3 & 1 \\ 4 & -1 & 10 \end{vmatrix} \]
\( \det(A_z) = 2(3 \cdot 10 – 1 \cdot (-1)) – 1(-1 \cdot 10 – 1 \cdot 4) + 4(-1 \cdot (-1) – 3 \cdot 4) = 2(30 + 1) – 1(-10 – 4) + 4(1 – 12) = 2 \cdot 31 + 14 + 4(-11) = 62 + 14 – 44 = 32 \)
\( x = \frac{114}{58} = \frac{57}{29}, \quad y = \frac{36}{58} = \frac{18}{29}, \quad z = \frac{32}{58} = \frac{16}{29} \)
88. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} x + 2y + 3z = 9 \\ 4x – y + z = 8 \\ -2x + 5y – z = -3 \end{cases} \)
Řešení příkladu 88:
Matice \( A \): \[ \begin{vmatrix} 1 & 2 & 3 \\ 4 & -1 & 1 \\ -2 & 5 & -1 \end{vmatrix} \]
\( \det(A) = 1((-1) \cdot (-1) – 1 \cdot 5) – 2(4 \cdot (-1) – 1 \cdot (-2)) + 3(4 \cdot 5 – (-1) \cdot (-2)) = 1(1 – 5) – 2(-4 + 2) + 3(20 – 2) = 1(-4) – 2(-2) + 3 \cdot 18 = -4 + 4 + 54 = 54 \)
Matice \( A_x \): \[ \begin{vmatrix} 9 & 2 & 3 \\ 8 & -1 & 1 \\ -3 & 5 & -1 \end{vmatrix} \]
\( \det(A_x) = 9((-1) \cdot (-1) – 1 \cdot 5) – 2(8 \cdot (-1) – 1 \cdot (-3)) + 3(8 \cdot 5 – (-1) \cdot (-3)) = 9(1 – 5) – 2(-8 + 3) + 3(40 – 3) = 9(-4) – 2(-5) + 3 \cdot 37 = -36 + 10 + 111 = 85 \)
Matice \( A_y \): \[ \begin{vmatrix} 1 & 9 & 3 \\ 4 & 8 & 1 \\ -2 & -3 & -1 \end{vmatrix} \]
\( \det(A_y) = 1(8 \cdot (-1) – 1 \cdot (-3)) – 9(4 \cdot (-1) – 1 \cdot (-2)) + 3(4 \cdot (-3) – 8 \cdot (-2)) = 1(-8 + 3) – 9(-4 + 2) + 3(-12 + 16) = -5 – 9(-2) + 3 \cdot 4 = -5 + 18 + 12 = 25 \)
Matice \( A_z \): \[ \begin{vmatrix} 1 & 2 & 9 \\ 4 & -1 & 8 \\ -2 & 5 & -3 \end{vmatrix} \]
\( \det(A_z) = 1((-1) \cdot (-3) – 8 \cdot 5) – 2(4 \cdot (-3) – 8 \cdot (-2)) + 9(4 \cdot 5 – (-1) \cdot (-2)) = 1(3 – 40) – 2(-12 + 16) + 9(20 – 2) = -37 – 2 \cdot 4 + 9 \cdot 18 = -37 – 8 + 162 = 117 \)
\( x = \frac{85}{54}, \quad y = \frac{25}{54}, \quad z = \frac{117}{54} = \frac{39}{18} = \frac{13}{6} \)
89. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 3x – y + 4z = 12 \\ 2x + 5y – z = 7 \\ -x + 3y + 2z = 5 \end{cases} \)
Řešení příkladu 89:
Matice \( A \): \[ \begin{vmatrix} 3 & -1 & 4 \\ 2 & 5 & -1 \\ -1 & 3 & 2 \end{vmatrix} \]
\( \det(A) = 3(5 \cdot 2 – (-1) \cdot 3) – (-1)(2 \cdot 2 – (-1) \cdot (-1)) + 4(2 \cdot 3 – 5 \cdot (-1)) = 3(10 + 3) + 1(4 – 1) + 4(6 + 5) = 3 \cdot 13 + 3 + 4 \cdot 11 = 39 + 3 + 44 = 86 \)
Matice \( A_x \): \[ \begin{vmatrix} 12 & -1 & 4 \\ 7 & 5 & -1 \\ 5 & 3 & 2 \end{vmatrix} \]
\( \det(A_x) = 12(5 \cdot 2 – (-1) \cdot 3) – (-1)(7 \cdot 2 – (-1) \cdot 5) + 4(7 \cdot 3 – 5 \cdot 5) = 12(10 + 3) + 1(14 + 5) + 4(21 – 25) = 12 \cdot 13 + 19 + 4(-4) = 156 + 19 – 16 = 159 \)
Matice \( A_y \): \[ \begin{vmatrix} 3 & 12 & 4 \\ 2 & 7 & -1 \\ -1 & 5 & 2 \end{vmatrix} \]
\( \det(A_y) = 3(7 \cdot 2 – (-1) \cdot 5) – 12(2 \cdot 2 – (-1) \cdot (-1)) + 4(2 \cdot 5 – 7 \cdot (-1)) = 3(14 + 5) – 12(4 – 1) + 4(10 + 7) = 3 \cdot 19 – 12 \cdot 3 + 4 \cdot 17 = 57 – 36 + 68 = 89 \)
Matice \( A_z \): \[ \begin{vmatrix} 3 & -1 & 12 \\ 2 & 5 & 7 \\ -1 & 3 & 5 \end{vmatrix} \]
\( \det(A_z) = 3(5 \cdot 5 – 7 \cdot 3) – (-1)(2 \cdot 5 – 7 \cdot (-1)) + 12(2 \cdot 3 – 5 \cdot (-1)) = 3(25 – 21) + 1(10 + 7) + 12(6 + 5) = 3 \cdot 4 + 17 + 12 \cdot 11 = 12 + 17 + 132 = 161 \)
\( x = \frac{159}{86}, \quad y = \frac{89}{86}, \quad z = \frac{161}{86} \)
90. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 5x – 2y + z = 4 \\ 3x + y – 4z = -2 \\ -x + 4y + 3z = 7 \end{cases} \)
Řešení příkladu 90:
Matice \( A \): \[ \begin{vmatrix} 5 & -2 & 1 \\ 3 & 1 & -4 \\ -1 & 4 & 3 \end{vmatrix} \]
\( \det(A) = 5(1 \cdot 3 – (-4) \cdot 4) – (-2)(3 \cdot 3 – (-4) \cdot (-1)) + 1(3 \cdot 4 – 1 \cdot (-1)) = 5(3 + 16) + 2(9 – 4) + (12 + 1) = 5 \cdot 19 + 2 \cdot 5 + 13 = 95 + 10 + 13 = 118 \)
Matice \( A_x \): \[ \begin{vmatrix} 4 & -2 & 1 \\ -2 & 1 & -4 \\ 7 & 4 & 3 \end{vmatrix} \]
\( \det(A_x) = 4(1 \cdot 3 – (-4) \cdot 4) – (-2)(-2 \cdot 3 – (-4) \cdot 7) + 1(-2 \cdot 4 – 1 \cdot 7) = 4(3 + 16) – 2(-6 + 28) + ( -8 – 7) = 4 \cdot 19 – 2 \cdot 22 – 15 = 76 – 44 – 15 = 17 \)
Matice \( A_y \): \[ \begin{vmatrix} 5 & 4 & 1 \\ 3 & -2 & -4 \\ -1 & 7 & 3 \end{vmatrix} \]
\( \det(A_y) = 5((-2) \cdot 3 – (-4) \cdot 7) – 4(3 \cdot 3 – (-4) \cdot (-1)) + 1(3 \cdot 7 – (-2) \cdot (-1)) = 5(-6 + 28) – 4(9 – 4) + (21 – 2) = 5 \cdot 22 – 4 \cdot 5 + 19 = 110 – 20 + 19 = 109 \)
Matice \( A_z \): \[ \begin{vmatrix} 5 & -2 & 4 \\ 3 & 1 & -2 \\ -1 & 4 & 7 \end{vmatrix} \]
\( \det(A_z) = 5(1 \cdot 7 – (-2) \cdot 4) – (-2)(3 \cdot 7 – (-2) \cdot (-1)) + 4(3 \cdot 4 – 1 \cdot (-1)) = 5(7 + 8) + 2(21 – 2) + 4(12 + 1) = 5 \cdot 15 + 2 \cdot 19 + 4 \cdot 13 = 75 + 38 + 52 = 165 \)
\( x = \frac{17}{118}, \quad y = \frac{109}{118}, \quad z = \frac{165}{118} \)
91. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} x + y + z = 6 \\ 2x – y + 3z = 14 \\ -x + 4y – z = 2 \end{cases} \)
Řešení příkladu 91:
Matice \( A \): \[ \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ -1 & 4 & -1 \end{vmatrix} \]
\( \det(A) = 1((-1)(-1) – 3 \cdot 4) – 1(2 \cdot (-1) – 3 \cdot (-1)) + 1(2 \cdot 4 – (-1) \cdot (-1)) = 1(1 – 12) – 1(-2 + 3) + 1(8 – 1) = -11 – 1 + 7 = -5 \)
Matice \( A_x \): \[ \begin{vmatrix} 6 & 1 & 1 \\ 14 & -1 & 3 \\ 2 & 4 & -1 \end{vmatrix} \]
\( \det(A_x) = 6((-1)(-1) – 3 \cdot 4) – 1(14 \cdot (-1) – 3 \cdot 2) + 1(14 \cdot 4 – (-1) \cdot 2) = 6(1 – 12) – 1(-14 – 6) + 1(56 + 2) = 6(-11) + 20 + 58 = -66 + 78 = 12 \)
Matice \( A_y \): \[ \begin{vmatrix} 1 & 6 & 1 \\ 2 & 14 & 3 \\ -1 & 2 & -1 \end{vmatrix} \]
\( \det(A_y) = 1(14 \cdot (-1) – 3 \cdot 2) – 6(2 \cdot (-1) – 3 \cdot (-1)) + 1(2 \cdot 2 – 14 \cdot (-1)) = 1(-14 – 6) – 6(-2 + 3) + 1(4 + 14) = -20 – 6 + 18 = -8 \)
Matice \( A_z \): \[ \begin{vmatrix} 1 & 1 & 6 \\ 2 & -1 & 14 \\ -1 & 4 & 2 \end{vmatrix} \]
\( \det(A_z) = 1((-1) \cdot 2 – 14 \cdot 4) – 1(2 \cdot 2 – 14 \cdot (-1)) + 6(2 \cdot 4 – (-1) \cdot (-1)) = 1(-2 – 56) – 1(4 + 14) + 6(8 – 1) = -58 – 18 + 42 = -34 \)
\( x = \frac{12}{-5} = -\frac{12}{5}, \quad y = \frac{-8}{-5} = \frac{8}{5}, \quad z = \frac{-34}{-5} = \frac{34}{5} \)
92. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 2x + y – z = 3 \\ -x + 4y + 2z = 1 \\ 3x – 2y + 5z = 7 \end{cases} \)
Řešení příkladu 92:
Matice \( A \): \[ \begin{vmatrix} 2 & 1 & -1 \\ -1 & 4 & 2 \\ 3 & -2 & 5 \end{vmatrix} \]
\( \det(A) = 2(4 \cdot 5 – 2 \cdot (-2)) – 1(-1 \cdot 5 – 2 \cdot 3) + (-1)(-1 \cdot (-2) – 4 \cdot 3) = 2(20 + 4) – 1(-5 – 6) – 1(2 – 12) = 2 \cdot 24 + 11 – 1(-10) = 48 + 11 + 10 = 69 \)
Matice \( A_x \): \[ \begin{vmatrix} 3 & 1 & -1 \\ 1 & 4 & 2 \\ 7 & -2 & 5 \end{vmatrix} \]
\( \det(A_x) = 3(4 \cdot 5 – 2 \cdot (-2)) – 1(1 \cdot 5 – 2 \cdot 7) + (-1)(1 \cdot (-2) – 4 \cdot 7) = 3(20 + 4) – 1(5 – 14) – 1(-2 – 28) = 3 \cdot 24 – 1(-9) – 1(-30) = 72 + 9 + 30 = 111 \)
Matice \( A_y \): \[ \begin{vmatrix} 2 & 3 & -1 \\ -1 & 1 & 2 \\ 3 & 7 & 5 \end{vmatrix} \]
\( \det(A_y) = 2(1 \cdot 5 – 2 \cdot 7) – 3(-1 \cdot 5 – 2 \cdot 3) + (-1)(-1 \cdot 7 – 1 \cdot 3) = 2(5 – 14) – 3(-5 – 6) – 1(-7 – 3) = 2(-9) + 33 – 1(-10) = -18 + 33 + 10 = 25 \)
Matice \( A_z \): \[ \begin{vmatrix} 2 & 1 & 3 \\ -1 & 4 & 1 \\ 3 & -2 & 7 \end{vmatrix} \]
\( \det(A_z) = 2(4 \cdot 7 – 1 \cdot (-2)) – 1(-1 \cdot 7 – 1 \cdot 3) + 3(-1 \cdot (-2) – 4 \cdot 3) = 2(28 + 2) – 1(-7 – 3) + 3(2 – 12) = 2 \cdot 30 + 10 + 3(-10) = 60 + 10 – 30 = 40 \)
\( x = \frac{111}{69}, \quad y = \frac{25}{69}, \quad z = \frac{40}{69} \)
93. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} x + 2y + 3z = 6 \\ 2x – y + z = 3 \\ 3x + y – 2z = 4 \end{cases} \)
Řešení příkladu 93:
Matice \( A \): \[ \begin{vmatrix} 1 & 2 & 3 \\ 2 & -1 & 1 \\ 3 & 1 & -2 \end{vmatrix} \]
\( \det(A) = 1((-1)(-2) – 1 \cdot 1) – 2(2 \cdot (-2) – 1 \cdot 3) + 3(2 \cdot 1 – (-1) \cdot 3) = 1(2 – 1) – 2(-4 – 3) + 3(2 + 3) = 1(1) – 2(-7) + 3(5) = 1 + 14 + 15 = 30 \)
Matice \( A_x \): \[ \begin{vmatrix} 6 & 2 & 3 \\ 3 & -1 & 1 \\ 4 & 1 & -2 \end{vmatrix} \]
\( \det(A_x) = 6((-1)(-2) – 1 \cdot 1) – 2(3 \cdot (-2) – 1 \cdot 4) + 3(3 \cdot 1 – (-1) \cdot 4) = 6(2 – 1) – 2(-6 – 4) + 3(3 + 4) = 6(1) – 2(-10) + 3(7) = 6 + 20 + 21 = 47 \)
Matice \( A_y \): \[ \begin{vmatrix} 1 & 6 & 3 \\ 2 & 3 & 1 \\ 3 & 4 & -2 \end{vmatrix} \]
\( \det(A_y) = 1(3 \cdot (-2) – 1 \cdot 4) – 6(2 \cdot (-2) – 1 \cdot 3) + 3(2 \cdot 4 – 3 \cdot 3) = 1(-6 – 4) – 6(-4 – 3) + 3(8 – 9) = -10 + 6 \cdot 7 + 3(-1) = -10 + 42 – 3 = 29 \)
Matice \( A_z \): \[ \begin{vmatrix} 1 & 2 & 6 \\ 2 & -1 & 3 \\ 3 & 1 & 4 \end{vmatrix} \]
\( \det(A_z) = 1((-1) \cdot 4 – 3 \cdot 1) – 2(2 \cdot 4 – 3 \cdot 3) + 6(2 \cdot 1 – (-1) \cdot 3) = 1(-4 – 3) – 2(8 – 9) + 6(2 + 3) = -7 – 2(-1) + 6 \cdot 5 = -7 + 2 + 30 = 25 \)
\( x = \frac{47}{30}, \quad y = \frac{29}{30}, \quad z = \frac{25}{30} = \frac{5}{6} \)
94. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 5x – 2y + z = 1 \\ 3x + y – 4z = -2 \\ -x + 4y + 3z = 7 \end{cases} \)
Řešení příkladu 94:
Matice \( A \): \[ \begin{vmatrix} 5 & -2 & 1 \\ 3 & 1 & -4 \\ -1 & 4 & 3 \end{vmatrix} \]
\( \det(A) = 5(1 \cdot 3 – (-4) \cdot 4) – (-2)(3 \cdot 3 – (-4) \cdot (-1)) + 1(3 \cdot 4 – 1 \cdot (-1)) = 5(3 + 16) + 2(9 – 4) + 1(12 + 1) = 5 \cdot 19 + 2 \cdot 5 + 13 = 95 + 10 + 13 = 118 \)
Matice \( A_x \): \[ \begin{vmatrix} 1 & -2 & 1 \\ -2 & 1 & -4 \\ 7 & 4 & 3 \end{vmatrix} \]
\( \det(A_x) = 1(1 \cdot 3 – (-4) \cdot 4) – (-2)(-2 \cdot 3 – (-4) \cdot 7) + 1(-2 \cdot 4 – 1 \cdot 7) = 1(3 + 16) – (-2)(-6 + 28) + 1(-8 – 7) = 19 – (-2)(22) – 15 = 19 – 44 – 15 = -40 \)
Matice \( A_y \): \[ \begin{vmatrix} 5 & 1 & 1 \\ 3 & -2 & -4 \\ -1 & 7 & 3 \end{vmatrix} \]
\( \det(A_y) = 5(-2 \cdot 3 – (-4) \cdot 7) – 1(3 \cdot 3 – (-4) \cdot (-1)) + 1(3 \cdot 7 – (-2) \cdot (-1)) = 5(-6 + 28) – 1(9 – 4) + 1(21 – 2) = 5 \cdot 22 – 5 + 19 = 110 – 5 + 19 = 124 \)
Matice \( A_z \): \[ \begin{vmatrix} 5 & -2 & 1 \\ 3 & 1 & -2 \\ -1 & 4 & 7 \end{vmatrix} \]
\( \det(A_z) = 5(1 \cdot 7 – (-2) \cdot 4) – (-2)(3 \cdot 7 – (-2) \cdot (-1)) + 1(3 \cdot 4 – 1 \cdot (-1)) = 5(7 + 8) + 2(21 – 2) + 1(12 + 1) = 5 \cdot 15 + 2 \cdot 19 + 13 = 75 + 38 + 13 = 126 \)
\( x = \frac{-40}{118}, \quad y = \frac{124}{118}, \quad z = \frac{126}{118} \)
95. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} x – y + 2z = 5 \\ 2x + 3y – z = 4 \\ -3x + y + z = -1 \end{cases} \)
Řešení příkladu 95:
Matice \( A \): \[ \begin{vmatrix} 1 & -1 & 2 \\ 2 & 3 & -1 \\ -3 & 1 & 1 \end{vmatrix} \]
\( \det(A) = 1(3 \cdot 1 – (-1) \cdot 1) – (-1)(2 \cdot 1 – (-1) \cdot (-3)) + 2(2 \cdot 1 – 3 \cdot (-3)) = 1(3 + 1) + 1(2 – 3) + 2(2 + 9) = 4 + (-1) + 2 \cdot 11 = 4 – 1 + 22 = 25 \)
Matice \( A_x \): \[ \begin{vmatrix} 5 & -1 & 2 \\ 4 & 3 & -1 \\ -1 & 1 & 1 \end{vmatrix} \]
\( \det(A_x) = 5(3 \cdot 1 – (-1) \cdot 1) – (-1)(4 \cdot 1 – (-1) \cdot (-1)) + 2(4 \cdot 1 – 3 \cdot (-1)) = 5(3 + 1) + 1(4 – 1) + 2(4 + 3) = 20 + 3 + 14 = 37 \)
Matice \( A_y \): \[ \begin{vmatrix} 1 & 5 & 2 \\ 2 & 4 & -1 \\ -3 & -1 & 1 \end{vmatrix} \]
\( \det(A_y) = 1(4 \cdot 1 – (-1) \cdot (-1)) – 5(2 \cdot 1 – (-1) \cdot (-3)) + 2(2 \cdot (-1) – 4 \cdot (-3)) = 1(4 – 1) – 5(2 + 3) + 2(-2 + 12) = 3 – 25 + 20 = -2 \)
Matice \( A_z \): \[ \begin{vmatrix} 1 & -1 & 5 \\ 2 & 3 & 4 \\ -3 & 1 & -1 \end{vmatrix} \]
\( \det(A_z) = 1(3 \cdot (-1) – 4 \cdot 1) – (-1)(2 \cdot (-1) – 4 \cdot (-3)) + 5(2 \cdot 1 – 3 \cdot (-3)) = 1(-3 – 4) + 1(-2 + 12) + 5(2 + 9) = -7 + 10 + 55 = 58 \)
\( x = \frac{37}{25}, \quad y = \frac{-2}{25}, \quad z = \frac{58}{25} \)
96. Řešte soustavu rovnic pomocí Cramerova pravidla: \( \begin{cases} 2x + 3y – z = 4 \\ – x + 4y + 5z = 7 \\ 3x – y + 2z = 1 \end{cases} \)
Řešení příkladu 96:
Matice \( A \): \[ \begin{vmatrix} 2 & 3 & -1 \\ -1 & 4 & 5 \\ 3 & -1 & 2 \end{vmatrix} \]
\( \det(A) = 2(4 \cdot 2 – 5 \cdot (-1)) – 3(-1 \cdot 2 – 5 \cdot 3) + (-1)(-1 \cdot (-1) – 4 \cdot 3) = 2(8 + 5) – 3(-2 – 15) – 1(1 – 12) = 2 \cdot 13 – 3 \cdot (-17) – 1 \cdot (-11) = 26 + 51 + 11 = 88 \)
Matice \( A_x \): \[ \begin{vmatrix} 4 & 3 & -1 \\ 7 & 4 & 5 \\ 1 & -1 & 2 \end{vmatrix} \]
\( \det(A_x) = 4(4 \cdot 2 – 5 \cdot (-1)) – 3(7 \cdot 2 – 5 \cdot 1) + (-1)(7 \cdot (-1) – 4 \cdot 1) = 4(8 + 5) – 3(14 – 5) – 1(-7 – 4) = 4 \cdot 13 – 3 \cdot 9 + 11 = 52 – 27 + 11 = 36 \)
Matice \( A_y \): \[ \begin{vmatrix} 2 & 4 & -1 \\ -1 & 7 & 5 \\ 3 & 1 & 2 \end{vmatrix} \]
\( \det(A_y) = 2(7 \cdot 2 – 5 \cdot 1) – 4(-1 \cdot 2 – 5 \cdot 3) + (-1)(-1 \cdot 1 – 7 \cdot 3) = 2(14 – 5) – 4(-2 – 15) – 1(-1 – 21) = 2 \cdot 9 + 4 \cdot 17 + 22 = 18 + 68 + 22 = 108 \)
Matice \( A_z \): \[ \begin{vmatrix} 2 & 3 & 4 \\ -1 & 4 & 7 \\ 3 & -1 & 1 \end{vmatrix} \]
\( \det(A_z) = 2(4 \cdot 1 – 7 \cdot (-1)) – 3(-1 \cdot 1 – 7 \cdot 3) + 4(-1 \cdot (-1) – 4 \cdot 3) = 2(4 + 7) – 3(-1 – 21) + 4(1 – 12) = 2 \cdot 11 + 3 \cdot 22 + 4 \cdot (-11) = 22 + 66 – 44 = 44 \)
\( x = \frac{36}{88} = \frac{9}{22}, \quad y = \frac{108}{88} = \frac{27}{22}, \quad z = \frac{44}{88} = \frac{1}{2} \)
97. Najděte determinant matice a rozhodněte, zda je matice invertovatelná: \[ \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} \]
Řešení příkladu 97:
Výpočet determinantu: \[ \det = 1(5 \cdot 9 – 6 \cdot 8) – 2(4 \cdot 9 – 6 \cdot 7) + 3(4 \cdot 8 – 5 \cdot 7) = 1(45 – 48) – 2(36 – 42) + 3(32 – 35) = 1(-3) – 2(-6) + 3(-3) = -3 + 12 – 9 = 0 \]
Protože determinant je \(0\), matice není invertovatelná.
98. Vyřešte soustavu rovnic: \( \begin{cases} x + 2y + 3z = 6 \\ 2x + 3y + 4z = 9 \\ 3x + y + 2z = 7 \end{cases} \) pomocí Cramerova pravidla.
Řešení příkladu 98:
Matice \( A \): \[ \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 1 & 2 \end{vmatrix} \]
\( \det(A) = 1(3 \cdot 2 – 4 \cdot 1) – 2(2 \cdot 2 – 4 \cdot 3) + 3(2 \cdot 1 – 3 \cdot 3) = 1(6 – 4) – 2(4 – 12) + 3(2 – 9) = 2 – 2(-8) + 3(-7) = 2 + 16 – 21 = -3 \)
Matice \( A_x \): \[ \begin{vmatrix} 6 & 2 & 3 \\ 9 & 3 & 4 \\ 7 & 1 & 2 \end{vmatrix} \]
\( \det(A_x) = 6(3 \cdot 2 – 4 \cdot 1) – 2(9 \cdot 2 – 4 \cdot 7) + 3(9 \cdot 1 – 3 \cdot 7) = 6(6 – 4) – 2(18 – 28) + 3(9 – 21) = 6 \cdot 2 – 2(-10) + 3(-12) = 12 + 20 – 36 = -4 \)
Matice \( A_y \): \[ \begin{vmatrix} 1 & 6 & 3 \\ 2 & 9 & 4 \\ 3 & 7 & 2 \end{vmatrix} \]
\( \det(A_y) = 1(9 \cdot 2 – 4 \cdot 7) – 6(2 \cdot 2 – 4 \cdot 3) + 3(2 \cdot 7 – 9 \cdot 3) = 1(18 – 28) – 6(4 – 12) + 3(14 – 27) = -10 – 6(-8) + 3(-13) = -10 + 48 – 39 = -1 \)
Matice \( A_z \): \[ \begin{vmatrix} 1 & 2 & 6 \\ 2 & 3 & 9 \\ 3 & 1 & 7 \end{vmatrix} \]
\( \det(A_z) = 1(3 \cdot 7 – 9 \cdot 1) – 2(2 \cdot 7 – 9 \cdot 3) + 6(2 \cdot 1 – 3 \cdot 3) = 1(21 – 9) – 2(14 – 27) + 6(2 – 9) = 12 – 2(-13) + 6(-7) = 12 + 26 – 42 = -4 \)
\( x = \frac{-4}{-3} = \frac{4}{3}, \quad y = \frac{-1}{-3} = \frac{1}{3}, \quad z = \frac{-4}{-3} = \frac{4}{3} \)
99. Vyřešte soustavu rovnic: \( \begin{cases} 2x + y + z = 5 \\ x + 3y + 2z = 8 \\ 3x + 2y + 4z = 11 \end{cases} \) pomocí Cramerova pravidla.
Řešení příkladu 99:
Matice \( A \): \[ \begin{vmatrix} 2 & 1 & 1 \\ 1 & 3 & 2 \\ 3 & 2 & 4 \end{vmatrix} \]
\( \det(A) = 2(3 \cdot 4 – 2 \cdot 2) – 1(1 \cdot 4 – 2 \cdot 3) + 1(1 \cdot 2 – 3 \cdot 3) = 2(12 – 4) – 1(4 – 6) + 1(2 – 9) = 16 + 2 – 7 = 11 \)
Matice \( A_x \): \[ \begin{vmatrix} 5 & 1 & 1 \\ 8 & 3 & 2 \\ 11 & 2 & 4 \end{vmatrix} \]
\( \det(A_x) = 5(3 \cdot 4 – 2 \cdot 2) – 1(8 \cdot 4 – 2 \cdot 11) + 1(8 \cdot 2 – 3 \cdot 11) = 5(12 – 4) – 1(32 – 22) + 1(16 – 33) = 40 – 10 – 17 = 13 \)
Matice \( A_y \): \[ \begin{vmatrix} 2 & 5 & 1 \\ 1 & 8 & 2 \\ 3 & 11 & 4 \end{vmatrix} \]
\( \det(A_y) = 2(8 \cdot 4 – 2 \cdot 11) – 5(1 \cdot 4 – 2 \cdot 3) + 1(1 \cdot 11 – 8 \cdot 3) = 2(32 – 22) – 5(4 – 6) + 1(11 – 24) = 20 + 10 – 13 = 17 \)
Matice \( A_z \): \[ \begin{vmatrix} 2 & 1 & 5 \\ 1 & 3 & 8 \\ 3 & 2 & 11 \end{vmatrix} \]
\( \det(A_z) = 2(3 \cdot 11 – 8 \cdot 2) – 1(1 \cdot 11 – 8 \cdot 3) + 5(1 \cdot 2 – 3 \cdot 3) = 2(33 – 16) – 1(11 – 24) + 5(2 – 9) = 34 + 13 – 35 = 12 \)
\( x = \frac{13}{11}, \quad y = \frac{17}{11}, \quad z = \frac{12}{11} \)
100. Vyřešte soustavu rovnic: \( \begin{cases} x + 2y + z = 4 \\ 2x + y + 3z = 7 \\ 3x + y + 2z = 8 \end{cases} \) pomocí Cramerova pravidla.
Řešení příkladu 100:
Matice \( A \): \[ \begin{vmatrix} 1 & 2 & 1 \\ 2 & 1 & 3 \\ 3 & 1 & 2 \end{vmatrix} \]
\( \det(A) = 1(1 \cdot 2 – 3 \cdot 1) – 2(2 \cdot 2 – 3 \cdot 3) + 1(2 \cdot 1 – 1 \cdot 3) = 1(2 – 3) – 2(4 – 9) + 1(2 – 3) = -1 – 2(-5) – 1 = -1 + 10 – 1 = 8 \)
Matice \( A_x \): \[ \begin{vmatrix} 4 & 2 & 1 \\ 7 & 1 & 3 \\ 8 & 1 & 2 \end{vmatrix} \]
\( \det(A_x) = 4(1 \cdot 2 – 3 \cdot 1) – 2(7 \cdot 2 – 3 \cdot 8) + 1(7 \cdot 1 – 1 \cdot 8) = 4(2 – 3) – 2(14 – 24) + (7 – 8) = -4 – 2(-10) -1 = -4 + 20 -1 = 15 \)
Matice \( A_y \): \[ \begin{vmatrix} 1 & 4 & 1 \\ 2 & 7 & 3 \\ 3 & 8 & 2 \end{vmatrix} \]
\( \det(A_y) = 1(7 \cdot 2 – 3 \cdot 8) – 4(2 \cdot 2 – 3 \cdot 3) + 1(2 \cdot 8 – 7 \cdot 3) = 1(14 – 24) – 4(4 – 9) + 1(16 – 21) = -10 – 4(-5) -5 = -10 + 20 -5 = 5 \)
Matice \( A_z \): \[ \begin{vmatrix} 1 & 2 & 4 \\ 2 & 1 & 7 \\ 3 & 1 & 8 \end{vmatrix} \]
\( \det(A_z) = 1(1 \cdot 8 – 7 \cdot 1) – 2(2 \cdot 8 – 7 \cdot 3) + 4(2 \cdot 1 – 1 \cdot 3) = 1(8 – 7) – 2(16 – 21) + 4(2 – 3) = 1 – 2(-5) + 4(-1) = 1 + 10 -4 = 7 \)
\( x = \frac{15}{8}, \quad y = \frac{5}{8}, \quad z = \frac{7}{8} \)