Answer on Question #79911 – Math – Linear Algebra
Question
Find inverse of the matrix B B B
B = [ − 1 − 3 0 2 4 0 − 1 − 1 2 ] B = \begin{bmatrix} -1 & -3 & 0 \\ 2 & 4 & 0 \\ -1 & -1 & 2 \end{bmatrix} B = ⎣ ⎡ − 1 2 − 1 − 3 4 − 1 0 0 2 ⎦ ⎤ Solution
0. det B = ∣ − 1 − 3 0 2 4 0 − 1 − 1 2 ∣ = 2 ∣ − 1 − 3 2 4 ∣ = 2 ⋅ ( − 4 + 6 ) = 4 ≠ 0 \det B = \begin{vmatrix} -1 & -3 & 0 \\ 2 & 4 & 0 \\ -1 & -1 & 2 \end{vmatrix} = 2 \begin{vmatrix} -1 & -3 \\ 2 & 4 \end{vmatrix} = 2 \cdot (-4 + 6) = 4 \neq 0 det B = ∣ ∣ − 1 2 − 1 − 3 4 − 1 0 0 2 ∣ ∣ = 2 ∣ ∣ − 1 2 − 3 4 ∣ ∣ = 2 ⋅ ( − 4 + 6 ) = 4 = 0
1. B − 1 = 1 det B adj B = 1 2 ⋅ ( − 4 + 6 ) [ 8 ∗ ∗ ∗ − 2 ∗ ∗ ∗ 2 ] = 1 4 [ 8 6 0 − 4 − 2 0 2 2 2 ] B^{-1} = \frac{1}{\det B} \operatorname{adj} B = \frac{1}{2 \cdot (-4 + 6)} \begin{bmatrix} 8 & * & * \\ * & -2 & * \\ * & * & 2 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 8 & 6 & 0 \\ -4 & -2 & 0 \\ 2 & 2 & 2 \end{bmatrix} B − 1 = d e t B 1 adj B = 2 ⋅ ( − 4 + 6 ) 1 ⎣ ⎡ 8 ∗ ∗ ∗ − 2 ∗ ∗ ∗ 2 ⎦ ⎤ = 4 1 ⎣ ⎡ 8 − 4 2 6 − 2 2 0 0 2 ⎦ ⎤
Recall that the adjoint of a matrix is the transpose of its cofactor matrix:
adj B = [ Cofactor of b i j ] T = [ ( − 1 ) i + j ( Minor of b i j ) ] T = [ ( − 1 ) i + j ( Minor of b j i ) ] = [ b 22 b 33 − b 23 b 32 − ( b 12 b 33 − b 13 b 32 ) − ( b 12 b 23 − b 13 b 22 ) − ( b 21 b 33 − b 23 b 31 ) b 11 b 33 − b 13 b 31 − ( b 11 b 23 − b 13 b 21 ) − ( b 21 b 32 − b 22 b 31 ) − ( b 11 b 32 − b 12 b 31 ) b 11 b 22 − b 12 b 21 ] \begin{aligned}
\operatorname{adj} B &= \left[ \text{Cofactor of } b_{ij} \right]^T = \left[ (-1)^{i+j} (\text{Minor of } b_{ij}) \right]^T = \left[ (-1)^{i+j} (\text{Minor of } b_{ji}) \right] \\
&= \begin{bmatrix} b_{22}b_{33} - b_{23}b_{32} & -(b_{12}b_{33} - b_{13}b_{32}) & -(b_{12}b_{23} - b_{13}b_{22}) \\ -(b_{21}b_{33} - b_{23}b_{31}) & b_{11}b_{33} - b_{13}b_{31} & -(b_{11}b_{23} - b_{13}b_{21}) \\ -(b_{21}b_{32} - b_{22}b_{31}) & -(b_{11}b_{32} - b_{12}b_{31}) & b_{11}b_{22} - b_{12}b_{21} \end{bmatrix}
\end{aligned} adj B = [ Cofactor of b ij ] T = [ ( − 1 ) i + j ( Minor of b ij ) ] T = [ ( − 1 ) i + j ( Minor of b ji ) ] = ⎣ ⎡ b 22 b 33 − b 23 b 32 − ( b 21 b 33 − b 23 b 31 ) − ( b 21 b 32 − b 22 b 31 ) − ( b 12 b 33 − b 13 b 32 ) b 11 b 33 − b 13 b 31 − ( b 11 b 32 − b 12 b 31 ) − ( b 12 b 23 − b 13 b 22 ) − ( b 11 b 23 − b 13 b 21 ) b 11 b 22 − b 12 b 21 ⎦ ⎤
2. p ( λ ) = det ( λ I − B ) = ( − 1 ) 3 det ( B − λ I ) = − det [ − 1 − λ − 3 0 2 4 − λ 0 − 1 − 1 2 − λ ] = − ( 2 − λ ) ( 6 + λ 2 − 3 λ − 4 ) = ( λ − 2 ) ( λ 2 − 3 λ + 2 ) = λ 3 − 5 λ 2 + 8 λ − 4 p(\lambda) = \det(\lambda I - B) = (-1)^3 \det(B - \lambda I) = -\det\begin{bmatrix} -1 - \lambda & -3 & 0 \\ 2 & 4 - \lambda & 0 \\ -1 & -1 & 2 - \lambda \end{bmatrix} = -(2 - \lambda) (6 + \lambda^2 - 3\lambda - 4) = (\lambda - 2)(\lambda^2 - 3\lambda + 2) = \lambda^3 - 5\lambda^2 + 8\lambda - 4 p ( λ ) = det ( λ I − B ) = ( − 1 ) 3 det ( B − λ I ) = − det ⎣ ⎡ − 1 − λ 2 − 1 − 3 4 − λ − 1 0 0 2 − λ ⎦ ⎤ = − ( 2 − λ ) ( 6 + λ 2 − 3 λ − 4 ) = ( λ − 2 ) ( λ 2 − 3 λ + 2 ) = λ 3 − 5 λ 2 + 8 λ − 4
p ( B ) = 0 p(B) = 0 p ( B ) = 0
4 B − 1 = B 2 − 5 B + 8 I , where 8 I = [ 8 0 0 0 8 0 0 0 8 ] 4B^{-1} = B^2 - 5B + 8I, \text{ where } 8I = \begin{bmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{bmatrix} 4 B − 1 = B 2 − 5 B + 8 I , where 8 I = ⎣ ⎡ 8 0 0 0 8 0 0 0 8 ⎦ ⎤ B 2 = [ − 5 − 9 0 6 10 0 − 3 − 3 4 ] , − 5 B + 8 I = [ 13 15 0 − 10 − 12 0 5 5 − 2 ] , 4 B − 1 = [ 8 6 0 − 4 − 2 0 2 2 2 ] B^2 = \begin{bmatrix} -5 & -9 & 0 \\ 6 & 10 & 0 \\ -3 & -3 & 4 \end{bmatrix}, \quad -5B + 8I = \begin{bmatrix} 13 & 15 & 0 \\ -10 & -12 & 0 \\ 5 & 5 & -2 \end{bmatrix}, \quad 4B^{-1} = \begin{bmatrix} 8 & 6 & 0 \\ -4 & -2 & 0 \\ 2 & 2 & 2 \end{bmatrix} B 2 = ⎣ ⎡ − 5 6 − 3 − 9 10 − 3 0 0 4 ⎦ ⎤ , − 5 B + 8 I = ⎣ ⎡ 13 − 10 5 15 − 12 5 0 0 − 2 ⎦ ⎤ , 4 B − 1 = ⎣ ⎡ 8 − 4 2 6 − 2 2 0 0 2 ⎦ ⎤ B − 1 = 1 4 [ 8 6 0 − 4 − 2 0 2 2 2 ] = 1 2 [ 4 3 0 − 2 − 1 0 1 1 1 ] . B^{-1} = \frac{1}{4} \begin{bmatrix} 8 & 6 & 0 \\ -4 & -2 & 0 \\ 2 & 2 & 2 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 4 & 3 & 0 \\ -2 & -1 & 0 \\ 1 & 1 & 1 \end{bmatrix}. B − 1 = 4 1 ⎣ ⎡ 8 − 4 2 6 − 2 2 0 0 2 ⎦ ⎤ = 2 1 ⎣ ⎡ 4 − 2 1 3 − 1 1 0 0 1 ⎦ ⎤ .
Answer:
The inverse is B − 1 = 1 2 [ 4 3 0 − 2 − 1 0 1 1 1 ] B^{-1} = \frac{1}{2}\begin{bmatrix} 4 & 3 & 0 \\ -2 & -1 & 0 \\ 1 & 1 & 1 \end{bmatrix} B − 1 = 2 1 ⎣ ⎡ 4 − 2 1 3 − 1 1 0 0 1 ⎦ ⎤
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#339914 on Dec 2023