Find P^-1, if it exists, given P = […] .

The correct answer is Option (3) → The inverse does not exist ##

We have $P = IP$, i.e.,

$\begin{bmatrix} 10 & -2 \\ -5 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} P.$

or $\begin{bmatrix} 1 & \frac{-1}{5} \\ -5 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{10} & 0 \\ 0 & 1 \end{bmatrix} P \text{ (applying } R_1 \to \frac{1}{10} R_1)$

or $\begin{bmatrix} 1 & -\frac{1}{5} \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} \frac{1}{10} & 0 \\ \frac{1}{2} & 1 \end{bmatrix} P \text{ (applying } R_2 \to R_2 + 5R_1 \text{)}$

We have all zeros in the second row of the left hand side matrix of the above equation. Therefore, $P^{-1}$ does not exist.