If A = begin{bmatrix} 0 & 0 & -1 0 & -1 & 0 | vedclass

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

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$A = A^{-1}$

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(B) Given $A = \begin{bmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{bmatrix}$.First,we find the determinant $|A| = 0(0) - 0(0) - 1(0 - 1) = -1(-1) = 1$.Since $|A| 
eq 0$,the matrix $A$ is invertible.Now,we calculate $A^2 = \begin{bmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I$.Since $A^2 = I$,multiplying both sides by $A^{-1}$ gives $A = A^{-1}$.

If $A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$,then

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