Answer
$\begin{bmatrix} 0 &\frac{1}{2} \\ -1 & \frac{1}{2} \end{bmatrix}$
Work Step by Step
Let us suppose that $P$ is the given matrix.
First we check if the inverse exists:
$det P=\begin{vmatrix} 1 & -1 \\ 2 & 0 \end{vmatrix}=1(0)-2(-1)=2$
Since $det P\not=0$, the inverse exists.
We will apply the formula for the inverse of $ 2\times 2$ matrix to obtain:
$P^{-1}=\dfrac{1}{det P} \begin{bmatrix} 0 & 1 \\ -2 & 1 \end{bmatrix}$
$=\dfrac{1}{2} \begin{bmatrix} 0 & 1 \\ -2 & 1 \end{bmatrix}$
$=\begin{bmatrix} 0 &\frac{1}{2} \\ -1 & \frac{1}{2} \end{bmatrix}$
