Question

If A = $\left[\begin{array}{cc}\alpha & 0\\ 1& 1\end{array}\right]$ and B = $\left[\begin{array}{cc}1& 0\\ 5& 1\end{array}\right]$ and $A^2=B,$ then the value of $\alpha$ is

• A. $\alpha =\pm 1$
• B. $\alpha =4$
• C. not possible
• D. Both (a) and (b)

Answer 1

Answer: (C)

We have,
$A=\left[\begin{array}{cc}\alpha & 0\\ 1& 1\end{array}\right];B=\left[\begin{array}{cc}1& 0\\ 5& 1\end{array}\right]$
Now given $A^2=B$
$\left[\begin{array}{cc}\alpha & 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}\alpha & 0\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 5& 1\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}{\alpha }^2& 0\\ \alpha +1& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 5& 1\end{array}\right]$
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 $A^2=B$.