Question

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

• A. $1$
• B. $-1$
• C. $4$
• D. no real value of $\alpha$

Answer 1

The correct option is D no real value of $\alpha$

$A^2$ = B$\Rightarrow \left[\begin{array}{cc}\alpha & 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}\alpha & 0\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 5& 1\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}{\alpha }^2& 0\\ \alpha +1& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 5& 1\end{array}\right]$
$\Rightarrow {\alpha }^2$ = 1, $\alpha$ + 1 = 5
$\therefore \alpha =\pm 1,\alpha =4$
There is no real value of $\alpha$ which satisfies both.