Question

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

• A. $1$
• B. $-1$
• C. $4$
• D. Does not exists

Answer 1

The correct option is D Does not exists

Given $A=\left[\begin{array}{cc}\alpha & 0\\ 1& 1\end{array}\right],B=\left[\begin{array}{cc}1& 0\\ 5& 1\end{array}\right]$
$A^2=\left[\begin{array}{cc}\alpha & 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}\alpha & 0\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}{\alpha }^2& 0\\ \alpha +1& 1\end{array}\right]$
Also, given $A^2=B$
$\Rightarrow \left[\begin{array}{cc}{\alpha }^2& 0\\ \alpha +1& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 5& 1\end{array}\right]$
On comparing ${\alpha }^2=1$ and $\alpha +1=5$
No value of $\alpha$ satisfies above equations