If A = begin{bmatrix} a & b c & -a end{bmatr | vedclass

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(C) Given $A = \begin{bmatrix} a & b \ c & -a \end{bmatrix}$.We calculate $A^2 = A 	imes A = \begin{bmatrix} a & b \ c & -a \end{bmatrix} \begin{bm

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$1 - a^2 - bc = 0$

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(C) Given $A = \begin{bmatrix} a & b \ c & -a \end{bmatrix}$.We calculate $A^2 = A 	imes A = \begin{bmatrix} a & b \ c & -a \end{bmatrix} \begin{bmatrix} a & b \ c & -a \end{bmatrix}$.Performing matrix multiplication: $A^2 = \begin{bmatrix} a^2 + bc & ab - ab \ ac - ac & bc + a^2 \end{bmatrix} = \begin{bmatrix} a^2 + bc & 0 \ 0 & a^2 + bc \end{bmatrix}$.Given that $A^2 = I$, where $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$, we equate the matrices:$\begin{bmatrix} a^2 + bc & 0 \ 0 & a^2 + bc \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.Comparing the elements, we get $a^2 + bc = 1$.Rearranging the equation, we get $1 - a^2 - bc = 0$.

If $A = \begin{bmatrix} a & b \\ c & -a \end{bmatrix}$ is such that $A^2 = I$, then . . . . . . .

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