If A = begin{bmatrix} 0 & 0 & -5 0 & -5 & 0 | vedclass

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(A) Given $A = \begin{bmatrix} 0 & 0 & -5 \ 0 & -5 & 0 \ -5 & 0 & 0 \end{bmatrix}$.We need to calculate $A^2 = A 	imes A$.$A^2 = \begin{bmatrix} 0

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$25 I$

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(A) Given $A = \begin{bmatrix} 0 & 0 & -5 \ 0 & -5 & 0 \ -5 & 0 & 0 \end{bmatrix}$.We need to calculate $A^2 = A 	imes A$.$A^2 = \begin{bmatrix} 0 & 0 & -5 \ 0 & -5 & 0 \ -5 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -5 \ 0 & -5 & 0 \ -5 & 0 & 0 \end{bmatrix}$.Performing matrix multiplication:Row $1$: $(0)(0) + (0)(0) + (-5)(-5) = 25$,$(0)(0) + (0)(-5) + (-5)(0) = 0$,$(0)(-5) + (0)(0) + (-5)(0) = 0$.Row $2$: $(0)(0) + (-5)(0) + (0)(-5) = 0$,$(0)(0) + (-5)(-5) + (0)(0) = 25$,$(0)(-5) + (-5)(0) + (0)(0) = 0$.Row $3$: $(-5)(0) + (0)(0) + (0)(-5) = 0$,$(-5)(0) + (0)(-5) + (0)(0) = 0$,$(-5)(-5) + (0)(0) + (0)(0) = 25$.Thus,$A^2 = \begin{bmatrix} 25 & 0 & 0 \ 0 & 25 & 0 \ 0 & 0 & 25 \end{bmatrix} = 25 \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = 25 I$.Therefore,the correct option is $A$.

If $A = \begin{bmatrix} 0 & 0 & -5 \\ 0 & -5 & 0 \\ -5 & 0 & 0 \end{bmatrix}$,then $A^2 =$ . . . . . . .

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