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

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(A) Given $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ a & b & -1 \end{bmatrix}$.We need to find ${A^2} = A 	imes A$.${A^2} = \begin{bmatrix} 1 & 0

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Unit matrix

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(A) Given $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ a & b & -1 \end{bmatrix}$.We need to find ${A^2} = A 	imes A$.${A^2} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ a & b & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ a & b & -1 \end{bmatrix}$.Calculating the product:Row $1$: $(1 	imes 1 + 0 	imes 0 + 0 	imes a, 1 	imes 0 + 0 	imes 1 + 0 	imes b, 1 	imes 0 + 0 	imes 0 + 0 	imes -1) = (1, 0, 0)$.Row $2$: $(0 	imes 1 + 1 	imes 0 + 0 	imes a, 0 	imes 0 + 1 	imes 1 + 0 	imes b, 0 	imes 0 + 1 	imes 0 + 0 	imes -1) = (0, 1, 0)$.Row $3$: $(a 	imes 1 + b 	imes 0 + -1 	imes a, a 	imes 0 + b 	imes 1 + -1 	imes b, a 	imes 0 + b 	imes 0 + -1 	imes -1) = (a - a, b - b, 1) = (0, 0, 1)$.Thus,${A^2} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I$,which is the Unit matrix.

If $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix}$,then ${A^2} = $

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