Answer on Question #58765 – Math – Linear Algebra
Question
Solve the set of linear equations by the matrix method:
{ a + 3 b + 2 c = 3 2 a − b − 3 c = − 8 5 a + 2 b + c = 9 \left\{ \begin{array}{l} a + 3b + 2c = 3 \\ 2a - b - 3c = -8 \\ 5a + 2b + c = 9 \end{array} \right. ⎩ ⎨ ⎧ a + 3 b + 2 c = 3 2 a − b − 3 c = − 8 5 a + 2 b + c = 9
Solution
A x = y , Ax = y, A x = y , A − 1 A x = A − 1 y , A^{-1}Ax = A^{-1}y, A − 1 A x = A − 1 y , x = A − 1 y , x = A^{-1}y, x = A − 1 y , A = ( 1 3 2 2 − 1 − 3 5 2 1 ) , x = ( a b c ) , y = ( 3 − 8 9 ) , A = \begin{pmatrix} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{pmatrix}, \quad x = \begin{pmatrix} a \\ b \\ c \end{pmatrix}, \quad y = \begin{pmatrix} 3 \\ -8 \\ 9 \end{pmatrix}, A = ⎝ ⎛ 1 2 5 3 − 1 2 2 − 3 1 ⎠ ⎞ , x = ⎝ ⎛ a b c ⎠ ⎞ , y = ⎝ ⎛ 3 − 8 9 ⎠ ⎞ , ( 1 3 2 2 − 1 − 3 5 2 1 ) ( a b c ) = ( 3 − 8 9 ) . \begin{pmatrix} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 3 \\ -8 \\ 9 \end{pmatrix}. ⎝ ⎛ 1 2 5 3 − 1 2 2 − 3 1 ⎠ ⎞ ⎝ ⎛ a b c ⎠ ⎞ = ⎝ ⎛ 3 − 8 9 ⎠ ⎞ .
The inverse of a matrix A A A
A − 1 = 1 ∣ A ∣ A + T . A^{-1} = \frac{1}{|A|} A_{+}^{T}. A − 1 = ∣ A ∣ 1 A + T .
The determinant of a matrix A A A
∣ A ∣ = ∣ 1 3 2 2 − 1 − 3 5 2 1 ∣ = 1 ⋅ ∣ − 1 − 3 2 1 ∣ − 3 ∣ 2 − 3 5 1 ∣ + 2 ∣ 2 − 1 5 2 ∣ = = 1 ⋅ ( − 1 ⋅ 1 − 2 ⋅ ( − 3 ) ) − 3 ⋅ ( 2 ⋅ 1 − 5 ⋅ ( − 3 ) ) + 2 ⋅ ( 2 ⋅ 2 − 5 ⋅ ( − 1 ) ) = = − 1 + 6 − 3 ( 2 + 15 ) + 2 ( 4 + 5 ) = 5 − 51 + 18 = − 28. \begin{aligned}
|A| &= \begin{vmatrix} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{vmatrix} = 1 \cdot \begin{vmatrix} -1 & -3 \\ 2 & 1 \end{vmatrix} - 3 \begin{vmatrix} 2 & -3 \\ 5 & 1 \end{vmatrix} + 2 \begin{vmatrix} 2 & -1 \\ 5 & 2 \end{vmatrix} = \\
&= 1 \cdot \left(-1 \cdot 1 - 2 \cdot (-3)\right) - 3 \cdot \left(2 \cdot 1 - 5 \cdot (-3)\right) + 2 \cdot \left(2 \cdot 2 - 5 \cdot (-1)\right) = \\
&= -1 + 6 - 3(2 + 15) + 2(4 + 5) = 5 - 51 + 18 = -28.
\end{aligned} ∣ A ∣ = ∣ ∣ 1 2 5 3 − 1 2 2 − 3 1 ∣ ∣ = 1 ⋅ ∣ ∣ − 1 2 − 3 1 ∣ ∣ − 3 ∣ ∣ 2 5 − 3 1 ∣ ∣ + 2 ∣ ∣ 2 5 − 1 2 ∣ ∣ = = 1 ⋅ ( − 1 ⋅ 1 − 2 ⋅ ( − 3 ) ) − 3 ⋅ ( 2 ⋅ 1 − 5 ⋅ ( − 3 ) ) + 2 ⋅ ( 2 ⋅ 2 − 5 ⋅ ( − 1 ) ) = = − 1 + 6 − 3 ( 2 + 15 ) + 2 ( 4 + 5 ) = 5 − 51 + 18 = − 28.
The minor matrix is
M = ( M 11 M 12 M 13 M 21 M 22 M 23 M 31 M 32 M 33 ) . M = \begin{pmatrix} M_{11} & M_{12} & M_{13} \\ M_{21} & M_{22} & M_{23} \\ M_{31} & M_{32} & M_{33} \end{pmatrix}. M = ⎝ ⎛ M 11 M 21 M 31 M 12 M 22 M 32 M 13 M 23 M 33 ⎠ ⎞ .
The cofactor matrix is
A ∗ = ( M 11 − M 12 M 13 − M 21 M 22 − M 23 M 31 − M 32 M 33 ) . A _ {*} = \left( \begin{array}{c c c} M _ {1 1} & - M _ {1 2} & M _ {1 3} \\ - M _ {2 1} & M _ {2 2} & - M _ {2 3} \\ M _ {3 1} & - M _ {3 2} & M _ {3 3} \end{array} \right). A ∗ = ⎝ ⎛ M 11 − M 21 M 31 − M 12 M 22 − M 32 M 13 − M 23 M 33 ⎠ ⎞ .
Minors are
M 11 = ∣ − 1 − 3 2 1 ∣ = − 1 ⋅ 1 − 2 ⋅ ( − 3 ) = − 1 + 6 = 5 , M _ {1 1} = \left| \begin{array}{c c} - 1 & - 3 \\ 2 & 1 \end{array} \right| = - 1 \cdot 1 - 2 \cdot (- 3) = - 1 + 6 = 5, M 11 = ∣ ∣ − 1 2 − 3 1 ∣ ∣ = − 1 ⋅ 1 − 2 ⋅ ( − 3 ) = − 1 + 6 = 5 , M 12 = ∣ 2 − 3 5 1 ∣ = 2 ⋅ 1 − 5 ⋅ ( − 3 ) = 2 + 15 = 17 , M _ {1 2} = \left| \begin{array}{c c} 2 & - 3 \\ 5 & 1 \end{array} \right| = 2 \cdot 1 - 5 \cdot (- 3) = 2 + 1 5 = 1 7, M 12 = ∣ ∣ 2 5 − 3 1 ∣ ∣ = 2 ⋅ 1 − 5 ⋅ ( − 3 ) = 2 + 15 = 17 , M 13 = ∣ 2 − 1 5 2 ∣ = 2 ⋅ 2 − 5 ⋅ ( − 1 ) = 9 , M _ {1 3} = \left| \begin{array}{c c} 2 & - 1 \\ 5 & 2 \end{array} \right| = 2 \cdot 2 - 5 \cdot (- 1) = 9, M 13 = ∣ ∣ 2 5 − 1 2 ∣ ∣ = 2 ⋅ 2 − 5 ⋅ ( − 1 ) = 9 , M 21 = ∣ 3 2 2 1 ∣ = 3 ⋅ 1 − 2 ⋅ 2 = − 1 , M _ {2 1} = \left| \begin{array}{c c} 3 & 2 \\ 2 & 1 \end{array} \right| = 3 \cdot 1 - 2 \cdot 2 = - 1, M 21 = ∣ ∣ 3 2 2 1 ∣ ∣ = 3 ⋅ 1 − 2 ⋅ 2 = − 1 , M 22 = ∣ 1 2 5 1 ∣ = 1 ⋅ 1 − 5 ⋅ 2 = − 9 , M _ {2 2} = \left| \begin{array}{c c} 1 & 2 \\ 5 & 1 \end{array} \right| = 1 \cdot 1 - 5 \cdot 2 = - 9, M 22 = ∣ ∣ 1 5 2 1 ∣ ∣ = 1 ⋅ 1 − 5 ⋅ 2 = − 9 , M 23 = ∣ 1 3 5 2 ∣ = 1 ⋅ 2 − 5 ⋅ 3 = − 13 , M _ {2 3} = \left| \begin{array}{c c} 1 & 3 \\ 5 & 2 \end{array} \right| = 1 \cdot 2 - 5 \cdot 3 = - 1 3, M 23 = ∣ ∣ 1 5 3 2 ∣ ∣ = 1 ⋅ 2 − 5 ⋅ 3 = − 13 , M 31 = ∣ 3 2 − 1 − 3 ∣ = 3 ⋅ ( − 3 ) − ( − 1 ) ⋅ 2 = − 7 , M _ {3 1} = \left| \begin{array}{c c} 3 & 2 \\ - 1 & - 3 \end{array} \right| = 3 \cdot (- 3) - (- 1) \cdot 2 = - 7, M 31 = ∣ ∣ 3 − 1 2 − 3 ∣ ∣ = 3 ⋅ ( − 3 ) − ( − 1 ) ⋅ 2 = − 7 , M 32 = ∣ 1 2 2 − 3 ∣ = 1 ⋅ ( − 3 ) − 2 ⋅ 2 = − 7 , M _ {3 2} = \left| \begin{array}{c c} 1 & 2 \\ 2 & - 3 \end{array} \right| = 1 \cdot (- 3) - 2 \cdot 2 = - 7, M 32 = ∣ ∣ 1 2 2 − 3 ∣ ∣ = 1 ⋅ ( − 3 ) − 2 ⋅ 2 = − 7 , M 33 = ∣ 1 3 2 − 1 ∣ = 1 ⋅ ( − 1 ) − 2 ⋅ 3 = − 7. M _ {3 3} = \left| \begin{array}{c c} 1 & 3 \\ 2 & - 1 \end{array} \right| = 1 \cdot (- 1) - 2 \cdot 3 = - 7. M 33 = ∣ ∣ 1 2 3 − 1 ∣ ∣ = 1 ⋅ ( − 1 ) − 2 ⋅ 3 = − 7.
The cofactor matrix will be
A ∗ = ( 5 − 17 9 1 − 9 13 − 7 7 − 7 ) . A _ {*} = \left( \begin{array}{c c c} 5 & - 1 7 & 9 \\ 1 & - 9 & 1 3 \\ - 7 & 7 & - 7 \end{array} \right). A ∗ = ⎝ ⎛ 5 1 − 7 − 17 − 9 7 9 13 − 7 ⎠ ⎞ .
The transpose of matrix A ∗ A_{*} A ∗
A ∗ T = ( 5 1 − 7 − 17 − 9 7 9 13 − 7 ) . A _ {*} ^ {T} = \left( \begin{array}{c c c} 5 & 1 & - 7 \\ - 1 7 & - 9 & 7 \\ 9 & 1 3 & - 7 \end{array} \right). A ∗ T = ⎝ ⎛ 5 − 17 9 1 − 9 13 − 7 7 − 7 ⎠ ⎞ .
The inverse of a matrix A A A
A − 1 = − 1 28 ⋅ ( 5 1 − 7 − 17 − 9 7 9 13 − 7 ) = ( − 5 28 − 1 28 1 4 17 28 9 28 − 1 4 − 9 28 − 13 28 1 4 ) . A ^ {- 1} = - \frac {1}{2 8} \cdot \left( \begin{array}{c c c} 5 & 1 & - 7 \\ - 1 7 & - 9 & 7 \\ 9 & 1 3 & - 7 \end{array} \right) = \left( \begin{array}{c c c} - \frac {5}{2 8} & - \frac {1}{2 8} & \frac {1}{4} \\ \frac {1 7}{2 8} & \frac {9}{2 8} & - \frac {1}{4} \\ - \frac {9}{2 8} & - \frac {1 3}{2 8} & \frac {1}{4} \end{array} \right). A − 1 = − 28 1 ⋅ ⎝ ⎛ 5 − 17 9 1 − 9 13 − 7 7 − 7 ⎠ ⎞ = ⎝ ⎛ − 28 5 28 17 − 28 9 − 28 1 28 9 − 28 13 4 1 − 4 1 4 1 ⎠ ⎞ .
The solution of the system is
x = ( a b c ) = ( − 5 28 − 1 28 1 4 17 28 9 28 − 1 4 − 9 28 − 13 28 1 4 ) ( 3 − 8 9 ) = ( − 5 ⋅ 3 28 + − 1 ⋅ ( − 8 ) 28 + 1 ⋅ 9 4 17 ⋅ 3 28 + 9 ⋅ ( − 8 ) 28 + ( − 1 ) ⋅ 9 4 − 9 ⋅ 3 28 + − 13 ⋅ ( − 8 ) 28 + 1 ⋅ 9 28 ) = ( − 15 28 + 8 28 + 9 4 51 28 − 72 28 − 9 4 − 27 28 + 104 28 + 9 4 ) = = ( − 15 + 8 + 9 ⋅ 7 28 51 − 72 − 9 ⋅ 7 28 − 27 + 104 + 7 ⋅ 9 28 ) = ( 56 28 − 84 28 140 28 ) = ( 2 − 3 5 ) , \begin{array}{l} x = \left( \begin{array}{c} a \\ b \\ c \end{array} \right) = \left( \begin{array}{c c c} - \frac {5}{2 8} & - \frac {1}{2 8} & \frac {1}{4} \\ \frac {1 7}{2 8} & \frac {9}{2 8} & - \frac {1}{4} \\ - \frac {9}{2 8} & - \frac {1 3}{2 8} & \frac {1}{4} \end{array} \right) \left( \begin{array}{c} 3 \\ - 8 \\ 9 \end{array} \right) = \left( \begin{array}{c} \frac {- 5 \cdot 3}{2 8} + \frac {- 1 \cdot (- 8)}{2 8} + \frac {1 \cdot 9}{4} \\ \frac {1 7 \cdot 3}{2 8} + \frac {9 \cdot (- 8)}{2 8} + \frac {(- 1) \cdot 9}{4} \\ - \frac {9 \cdot 3}{2 8} + \frac {- 1 3 \cdot (- 8)}{2 8} + \frac {1 \cdot 9}{2 8} \end{array} \right) \\ = \left( \begin{array}{c} - \frac {1 5}{2 8} + \frac {8}{2 8} + \frac {9}{4} \\ \frac {5 1}{2 8} - \frac {7 2}{2 8} - \frac {9}{4} \\ - \frac {2 7}{2 8} + \frac {1 0 4}{2 8} + \frac {9}{4} \end{array} \right) = \\ = \left( \begin{array}{c} \frac {- 1 5 + 8 + 9 \cdot 7}{2 8} \\ \frac {5 1 - 7 2 - 9 \cdot 7}{2 8} \\ \frac {- 2 7 + 1 0 4 + 7 \cdot 9}{2 8} \end{array} \right) = \left( \begin{array}{c} \frac {5 6}{2 8} \\ \frac {- 8 4}{2 8} \\ \frac {1 4 0}{2 8} \end{array} \right) = \left( \begin{array}{c} 2 \\ - 3 \\ 5 \end{array} \right), \\ \end{array} x = ⎝ ⎛ a b c ⎠ ⎞ = ⎝ ⎛ − 28 5 28 17 − 28 9 − 28 1 28 9 − 28 13 4 1 − 4 1 4 1 ⎠ ⎞ ⎝ ⎛ 3 − 8 9 ⎠ ⎞ = ⎝ ⎛ 28 − 5 ⋅ 3 + 28 − 1 ⋅ ( − 8 ) + 4 1 ⋅ 9 28 17 ⋅ 3 + 28 9 ⋅ ( − 8 ) + 4 ( − 1 ) ⋅ 9 − 28 9 ⋅ 3 + 28 − 13 ⋅ ( − 8 ) + 28 1 ⋅ 9 ⎠ ⎞ = ⎝ ⎛ − 28 15 + 28 8 + 4 9 28 51 − 28 72 − 4 9 − 28 27 + 28 104 + 4 9 ⎠ ⎞ = = ⎝ ⎛ 28 − 15 + 8 + 9 ⋅ 7 28 51 − 72 − 9 ⋅ 7 28 − 27 + 104 + 7 ⋅ 9 ⎠ ⎞ = ⎝ ⎛ 28 56 28 − 84 28 140 ⎠ ⎞ = ⎝ ⎛ 2 − 3 5 ⎠ ⎞ ,
that is, a = 2 , b = − 3 , c = 5 a = 2, b = -3, c = 5 a = 2 , b = − 3 , c = 5
Answer: a = 2 a = 2 a = 2 b = − 3 b = -3 b = − 3 c = 5 c = 5 c = 5
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#339914 on Dec 2023