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(D) Given $A = \begin{bmatrix} x & 1 \ 1 & 0 \end{bmatrix}$.Since $A^2 = I$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bma

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(D) Given $A = \begin{bmatrix} x & 1 \ 1 & 0 \end{bmatrix}$.Since $A^2 = I$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$,we have:$A^2 = \begin{bmatrix} x & 1 \ 1 & 0 \end{bmatrix} \begin{bmatrix} x & 1 \ 1 & 0 \end{bmatrix} = \begin{bmatrix} x^2 + 1 & x \ x & 1 \end{bmatrix}$.Equating this to the identity matrix:$\begin{bmatrix} x^2 + 1 & x \ x & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.Comparing the corresponding elements,we get $x = 0$ and $x^2 + 1 = 1$.From $x^2 + 1 = 1$,we get $x^2 = 0$,which implies $x = 0$.Thus,the value of $x$ is $0$.

If $A = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}$ and $A^2$ is the identity matrix,then $x =$

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