If A = begin{bmatrix} alpha & 0 1 & 1 end{bm | vedclass

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(D) Given $A = \begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$.We calculate $A^2$ as follows

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No real values

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(D) Given $A = \begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$.We calculate $A^2$ as follows:$A^2 = A 	imes A = \begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix} \begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} \alpha^2 + 0 & 0 + 0 \ \alpha + 1 & 0 + 1 \end{bmatrix} = \begin{bmatrix} \alpha^2 & 0 \ \alpha + 1 & 1 \end{bmatrix}$.We are given that $A^2 = B$,so:$\begin{bmatrix} \alpha^2 & 0 \ \alpha + 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$.Comparing the corresponding elements,we get:$1$) $\alpha^2 = 1 \implies \alpha = \pm 1$$2$) $\alpha + 1 = 5 \implies \alpha = 4$Since there is no value of $\alpha$ that satisfies both equations simultaneously,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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