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. $\frac{1}{125}$
• B. $\frac{2a}{125}$
• C. $\frac{2a}{25}$
• D. $\frac{2a}{625}$

Answer 1

The correct option is B $\frac{2a}{125}$

Let, $A=\left[\begin{array}{cc}5& 0\\ -a& 5\end{array}\right]$

$\Rightarrow adj\left(A\right)=\left[\begin{array}{cc}5& 0\\ a& 5\end{array}\right]$
$|A|=25-0=25$
$\Rightarrow A^{-1}=\frac{1}{|A|}\left[\begin{array}{cc}5& 0\\ a& 5\end{array}\right]=\frac{1}{25}\left[\begin{array}{cc}5& 0\\ a& 5\end{array}\right]$

$\Rightarrow A^{-2}=(A^{-1}{)}^2=\frac{1}{25}\left[\begin{array}{cc}5& 0\\ a& 5\end{array}\right]\frac{1}{25}\left[\begin{array}{cc}5& 0\\ a& 5\end{array}\right]$

$\Rightarrow A^{-2}=\frac{1}{625}\left[\begin{array}{cc}25& 0\\ 10a& 25\end{array}\right]=\left[\begin{array}{cc}\frac{1}{25}& 0\\ \frac{2a}{125}& \frac{1}{25}\end{array}\right]$

Now, according to given condition, we have:

$\left[\begin{array}{cc}1/25& 0\\ x& 1/25\end{array}\right]=\left[\begin{array}{cc}\frac{1}{25}& 0\\ \frac{2a}{125}& \frac{1}{25}\end{array}\right]$

$\Rightarrow x=2a/125$