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(D) Given $A = \begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$.Calculate $A^2$:$A^2 = \begin

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No real values

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(D) Given $A = \begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$.Calculate $A^2$:$A^2 = \begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix} \begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} \alpha^2 + 0 & 0 \ \alpha + 1 & 1 \end{bmatrix} = \begin{bmatrix} \alpha^2 & 0 \ \alpha + 1 & 1 \end{bmatrix}$.Since $A^2 = B$,we equate the corresponding elements:$1$) $\alpha^2 = 1 \Rightarrow \alpha = \pm 1$$2$) $\alpha + 1 = 5 \Rightarrow \alpha = 4$There is no common value of $\alpha$ that satisfies both equations simultaneously.Therefore,there are no real values of $\alpha$ for which $A^2 = B$.

If $A = \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$,then the value of $\alpha$ for which $A^2 = B$ is

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