Question

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

• A. $1$
• B. $-1$
• C. $4$
• D. No real value

Answer 1

The correct option is D No real value

Given : $A=\left[\begin{array}{cc}\alpha & 0\\ 1& 1\end{array}\right]$ and $B=\left[\begin{array}{cc}1& 0\\ 5& 1\end{array}\right]$

$A^2=\left[\begin{array}{cc}\alpha & 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}\alpha & 0\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}{\alpha }^2& 0\\ \alpha +1& 1\end{array}\right]$

Now $A^2=B$
$\left[\begin{array}{cc}{\alpha }^2& 0\\ \alpha +1& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 5& 1\end{array}\right]$

$\Rightarrow {\alpha }^2=1$ and $\alpha +1=5$

Clearly, no real value of $\alpha$ exist.