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Question

(B) Let $A = \begin{bmatrix} 1 & 0 & 0 \ 3 & 3 & 0 \ 5 & 2 & -1 \end{bmatrix}$.First,calculate the determinant $|A| = 1(3(-1) - 0(2)) - 0 + 0 = -3$.

Vedclass · Accepted Answer

$-\frac{1}{3} \begin{bmatrix} -3 & 0 & 0 \ 3 & -1 & 0 \ -9 & -2 & 3 \end{bmatrix}$

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(B) Let $A = \begin{bmatrix} 1 & 0 & 0 \ 3 & 3 & 0 \ 5 & 2 & -1 \end{bmatrix}$.First,calculate the determinant $|A| = 1(3(-1) - 0(2)) - 0 + 0 = -3$.Next,find the matrix of cofactors $C_{ij}$:$C_{11} = +((-1)(3) - 0) = -3$,$C_{12} = -(3(-1) - 0) = 3$,$C_{13} = +(3(2) - 3(5)) = -9$.$C_{21} = -(0(-1) - 0(2)) = 0$,$C_{22} = +(1(-1) - 0(5)) = -1$,$C_{23} = -(1(2) - 0(5)) = -2$.$C_{31} = +(0(0) - 3(0)) = 0$,$C_{32} = -(1(0) - 0(3)) = 0$,$C_{33} = +(1(3) - 0(3)) = 3$.Thus,$	ext{Adj}(A) = \begin{bmatrix} -3 & 0 & 0 \ 3 & -1 & 0 \ -9 & -2 & 3 \end{bmatrix}$.Finally,$A^{-1} = \frac{1}{|A|} 	ext{Adj}(A) = \frac{1}{-3} \begin{bmatrix} -3 & 0 & 0 \ 3 & -1 & 0 \ -9 & -2 & 3 \end{bmatrix}$.

The inverse of the matrix $\begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$ is

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