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

Answer 1

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

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

$=\frac{1}{(25{)}^2}(\left[\begin{array}{cc}5& 0\\ a& 5\end{array}\right]\times \left[\begin{array}{cc}5& 0\\ a& 5\end{array}\right])=\frac{1}{625}\left[\begin{array}{cc}25+0& 0+0\\ 5a+5a& 0+25\end{array}\right]$

$=\frac{1}{625}\left[\begin{array}{cc}25& 0\\ 10a& 25\end{array}\right]$

$LHS=RHS$

$\left[\begin{array}{cc}1/25& 0\\ x& 1/25\end{array}\right]=\frac{1}{625}\left[\begin{array}{cc}25& 0\\ 10a& 25\end{array}\right]$

$\Rightarrow 25\times \left[\begin{array}{cc}1/25& 0\\ x& 1/25\end{array}\right]=25\times \frac{1}{25}\times \frac{1}{25}\left[\begin{array}{cc}25& 0\\ 10a& 25\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}1& 0\\ 25x& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 2/5a& 1\end{array}\right]$........ $\left(i\right)$

From the $e{q}^n\left(i\right)$

$25x=\frac{2a}{5}$

$\Rightarrow x=\frac{2a}{125}$