Question

If $\left[\begin{array}{cc}1/25& 0\\ x& 1/25\end{array}\right]={\left[\begin{array}{cc}5& 0\\ -a& 5\end{array}\right]}^{-2}$ then the value of $x$ is

• A. $a/125$
• B. $2a/125$
• C. $2a/25$
• D. None of these

Answer 1

Answer: (B)

Let $A=\left[\begin{array}{cc}5& 0\\ -a& 5\end{array}\right]\Rightarrow adj\left(A\right)=\left[\begin{array}{cc}5& 0\\ a& 5\end{array}\right]\Rightarrow A^{-1}$
$=\frac{1}{|A|}\left[\begin{array}{cc}5& 0\\ a& 5\end{array}\right]=\frac{1}{25}\left[\begin{array}{cc}5& 0\\ a& 5\end{array}\right]$
$\Rightarrow A^{-2}={\left(A^{-1}\right)}^2=\frac{1}{25}\left[\begin{array}{cc}5& 0\\ a& 5\end{array}\right]\frac{1}{25}\left[\begin{array}{cc}5& 0\\ a& 5\end{array}\right]$
$=\frac{1}{625}\left[\begin{array}{cc}25& 0\\ 10a& 25\end{array}\right]\left[\begin{array}{cc}\frac{1}{25}& 0\\ \frac{2a}{125}& \frac{1}{25}\end{array}\right]$
Now $\left[\begin{array}{cc}1/25& 0\\ x& 125\end{array}\right]=\left[\begin{array}{cc}\frac{1}{25}& 0\\ \frac{2a}{125}& \frac{1}{25}\end{array}\right]\Rightarrow x=2a/125$