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

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(A) Given $A = \begin{bmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{bmatrix}$.We evaluate the options:$(i)$ Calculate $A^2$:$A^2 = \begin{bmatr

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$A^2 = I$

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(A) Given $A = \begin{bmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{bmatrix}$.We evaluate the options:$(i)$ Calculate $A^2$:$A^2 = \begin{bmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I$.Thus,option $A$ is correct.$(ii)$ $(-1)I = \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{bmatrix} 
eq A$.$(iii)$ Calculate the determinant $|A|$:$|A| = 0(0 - 0) - 0(0 - 0) - 1(0 - 1) = 1$.Since $|A| 
eq 0$,$A^{-1}$ exists.$(iv)$ $A$ is clearly not a zero matrix as it contains non-zero elements.

Let $A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$. The only correct statement about the matrix $A$ is:

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