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

AMUAMU 2017

Solution:

We have,
$A=\begin{bmatrix}\alpha & 0 \\1 & 1\end{bmatrix} ; B=\begin{bmatrix}1 & 0 \\5 & 1\end{bmatrix}$
Now given $ A^{2}=B$
$\begin{bmatrix}\alpha & 0 \\ 1 & 1\end{bmatrix}\begin{bmatrix}\alpha & 0 \\ 1 & 1\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 5 & 1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}\alpha^{2} & 0 \\ \alpha+1 & 1\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 5 & 1\end{bmatrix}$
On comparing both sides, we get
$\Rightarrow \alpha^{2}=1 \text { and } \alpha+1=5 $
$\Rightarrow \alpha=\pm 1 \text { and } \alpha=4 .$
$\because$ Both the values cannot occur simultaneously.
$\therefore$ No value of $\alpha$ is possible for which $A^{2}=B$.