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

Compute $A^2$ directly. $$A^2=\begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} \alpha^2 & 0 \\ \alpha+1 & 1 \end{bmatrix}$$ Given $A^2=B$, compare corresponding entries with $\begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$. From the $(1,1)$ entry, $\alpha^2=1$, so $\alpha=\pm 1$. From the $(2,1)$ entry, $\alpha+1=5$, so $\alpha=4$. These conditions cannot hold together, so there is no real value of $\alpha$ satisfying $A^2=B$. Checking the options, this matches option $\text{D}$.