If $$A = \begin{pmatrix} \alpha & 0 \\1 & 1\end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 \\5 & 1\end{pmatrix}$$and $$A^2=B$$, then value of $$\alpha$$ is​

Correct option is D

Solution:

​$$\text{Compute } A^2: \\A^2 = A \cdot A =\begin{pmatrix} \alpha & 0 \\ 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} \alpha & 0 \\ 1 & 1 \end{pmatrix} =\begin{pmatrix}\alpha^2 & 0 \\\alpha + 1 & 1\end{pmatrix} \\\text{Now equate } A^2 = B: \\\begin{pmatrix}\alpha^2 & 0 \\\alpha + 1 & 1\end{pmatrix} =\begin{pmatrix}1 & 0 \\5 & 1\end{pmatrix} \\\text{Comparing elements:} \\\alpha^2 = 1 \Rightarrow \alpha = \pm 1 \\\alpha + 1 = 5 \Rightarrow \alpha = 4 \\\text{Contradiction: } \alpha \text{ cannot satisfy both equations simultaneously.} \\\text{Therefore, no such value of } \alpha \text{ exists.} \\$$​​
Now clearly, α cannot be both ±1 and 4 at the same time, so there is no α that satisfies both equations.Now clearly, α cannot be both ±1 and 4 at the same time, so there is no α that satisfies both equations.

​$$\boxed{\text{Answer: (D) does not exist}}$$​​

​​​