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(D) Given $A = \begin{bmatrix} x & 0 \ 1 & 1 \end{bmatrix}$.First,calculate $A^2 = A \cdot A$:$A^2 = \begin{bmatrix} x & 0 \ 1 & 1 \end{bmatrix} \be

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$2$

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(D) Given $A = \begin{bmatrix} x & 0 \ 1 & 1 \end{bmatrix}$.First,calculate $A^2 = A \cdot A$:$A^2 = \begin{bmatrix} x & 0 \ 1 & 1 \end{bmatrix} \begin{bmatrix} x & 0 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} x^2 + 0 & 0 + 0 \ x + 1 & 0 + 1 \end{bmatrix} = \begin{bmatrix} x^2 & 0 \ x + 1 & 1 \end{bmatrix}$.Next,calculate $A^3 = A^2 \cdot A$:$A^3 = \begin{bmatrix} x^2 & 0 \ x + 1 & 1 \end{bmatrix} \begin{bmatrix} x & 0 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} x^3 + 0 & 0 + 0 \ x(x + 1) + 1 & 0 + 1 \end{bmatrix} = \begin{bmatrix} x^3 & 0 \ x^2 + x + 1 & 1 \end{bmatrix}$.Given $A^3 = B$,we have:$\begin{bmatrix} x^3 & 0 \ x^2 + x + 1 & 1 \end{bmatrix} = \begin{bmatrix} 8 & 0 \ 7 & 1 \end{bmatrix}$.Comparing the corresponding elements:$x^3 = 8 \implies x = 2$.Also,$x^2 + x + 1 = 7 \implies x^2 + x - 6 = 0$.Factoring the quadratic equation: $(x + 3)(x - 2) = 0$,which gives $x = 2$ or $x = -3$.Since $x$ must satisfy both conditions,we take the common value,$x = 2$.

If $A = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 8 & 0 \\ 7 & 1 \end{bmatrix}$ and $A^3 = B$,then $x =$

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