Question

If $A=\left[\begin{array}{cc}\alpha & 0\\ 1& 1\end{array}\right]$ and $B=\left[\begin{array}{cc}1& 0\\ 3& 1\end{array}\right],$ Then the value of $\alpha$ for which $A^2=B$ is

• A. 1
• B. -1
• C. $i$
• D. no real values of $\alpha$

Answer 1

The correct option is D

no real values of $\alpha$

$A=\left[\begin{array}{cc}\alpha & 0\\ 1& 1\end{array}\right],A^2=\left[\begin{array}{cc}\alpha & 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}\alpha & 0\\ 1& 1\end{array}\right],=\left[\begin{array}{cc}{\alpha }^2& 0\\ \alpha +1& 1\end{array}\right],B=\left[\begin{array}{cc}1& 0\\ 3& 1\end{array}\right]$

$If{A}^2=Bi){\alpha }^2=1\Rightarrow \alpha =\pm 1ii)\alpha +1=3\Rightarrow \alpha =2$

There is no real value of $\alpha$ which satisfies both equations simultaneously.