Question

Find $P^{-1}$, if it exists, given $P=\left[\begin{array}{cc}10& -2\\ -5& 1\end{array}\right]$

Answer 1

Given: $P=\left[\begin{array}{cc}10& -2\\ -5& 1\end{array}\right]$
We know that $P=IP$
$P=\left[\begin{array}{cc}10& -2\\ -5& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]P$
Applying $R_1\to \frac{1}{10}{R}_1$
$\Rightarrow \left[\begin{array}{cc}\frac{10}{10}& \frac{-2}{10}\\ -5& 1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{10}& \frac{0}{10}\\ 0& 1\end{array}\right]P$
$\Rightarrow \left[\begin{array}{cc}1& \frac{-1}{5}\\ -5& 1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{10}& 0\\ 0& 1\end{array}\right]P$
Applying $R_2\to R_2+5R_1$
$\Rightarrow \left[\begin{array}{cc}1& \frac{-1}{5}\\ -5+5\left(1\right)& 1+5\left(\frac{-1}{5}\right)\end{array}\right]=\left[\begin{array}{cc}\frac{1}{10}& 0\\ 0+5\left(\frac{1}{10}\right)& 1+5\left(0\right)\end{array}\right]P$
$\Rightarrow \left[\begin{array}{cc}1& \frac{-1}{5}\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}\frac{1}{10}& 0\\ \frac{1}{2}& 1\end{array}\right]P$
$\because$ Left hand side matrix involves all zeroes in the second row.
$\therefore P^{-1}$ does not exist.