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

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(D) Given $A = \begin{bmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{bmatrix}$.First, we check if $A$ is symmetric or skew-symmetric. The transp

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$A^2 = I$

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(D) Given $A = \begin{bmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{bmatrix}$.First, we check if $A$ is symmetric or skew-symmetric. The transpose $A^T = \begin{bmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{bmatrix} = A$. Since $A^T = A$, $A$ is a symmetric matrix.Next, we calculate $A^2 = A 	imes A = \begin{bmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I$.Since $A^2 = I$, the correct option is $D$.

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

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