Thank you for reporting, we will resolve it shortly
Q.
Find $P^{-1}$,if exists and given $ p= \begin{bmatrix}10&-2\\ -5&1\end{bmatrix}$
Matrices
Solution:
We have, $P-IP$
$\therefore \begin{bmatrix}10&-2\\ -5&1\end{bmatrix}=\begin{bmatrix}1&0\\ 0&1\end{bmatrix}P$
Applying $R_{1}\rightarrow \frac{1}{10}R_{^{ }1}$,we get
$\begin{bmatrix}1&-\frac{1}{5}\\ -5&1\end{bmatrix}=\begin{bmatrix}\frac{1}{10}&0\\ 0&1\end{bmatrix}P$
Applying$ R_{2} \rightarrow R_{2} + 5R_{1}$, we get
$\begin{bmatrix}1&-\frac{1}{5}\\ 0&0\end{bmatrix}=\begin{bmatrix}\frac{1}{10}&0\\ \frac{1}{2}&1\end{bmatrix}P$
Since, all the elements in the second row of the matrix on $LHS$ are zero, therefore the given matrix is non- invertible. Hence, $P^{-1}$ does not exist.