The inverse of matrix A = begin{bmatrix} 0 & | vedclass

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(A) Given $A = \begin{bmatrix} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{bmatrix}$.First,we find the determinant $|A| = 0(0-0) - 1(1-0) + 0(0-0) = -1$.

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$A$

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(A) Given $A = \begin{bmatrix} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{bmatrix}$.First,we find the determinant $|A| = 0(0-0) - 1(1-0) + 0(0-0) = -1$.Since $|A| 
eq 0$,the inverse $A^{-1}$ exists.We find the cofactor matrix $C$ where $C_{ij} = (-1)^{i+j} M_{ij}$.$C_{11} = +(0-0) = 0, C_{12} = -(1-0) = -1, C_{13} = +(0-0) = 0$.$C_{21} = -(1-0) = -1, C_{22} = +(0-0) = 0, C_{23} = -(0-0) = 0$.$C_{31} = +(0-0) = 0, C_{32} = -(0-0) = 0, C_{33} = +(0-1) = -1$.Thus,$adj(A) = C^T = \begin{bmatrix} 0 & -1 & 0 \ -1 & 0 & 0 \ 0 & 0 & -1 \end{bmatrix}$.Finally,$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-1} \begin{bmatrix} 0 & -1 & 0 \ -1 & 0 & 0 \ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{bmatrix} = A$.

The inverse of matrix $A = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ is

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