Answer
$\begin{bmatrix} 0 & \dfrac{1}{2} \\ -\dfrac{1}{2} & 1\end{bmatrix} $
Work Step by Step
A matrix $A=\begin{bmatrix} a_{11} & a_{12} \\a_{21} &a_{22} \end{bmatrix} $ is said to be invertible when the determinant of a matrix $det=a_{11}a_{22}-a_{12}a_{21} \ne 0$ . Then, the inverse of a matrix $A$ can be computed as: $A^{-1}=\dfrac{1}{a_{11}a_{22}-a_{12}a_{21} }\begin{bmatrix} a_{22} & -a_{12} \\ -a_{21} &a_{11} \end{bmatrix} $ and when $a_{11}a_{22}-a_{12}a_{21} =0$; then a matrix $A$ is not invertible.
Now, we have: $det=(4)(0)-(-2)(2)=4$
This shows that matrix $A$ is invertible and its inverse can be calculated as:
$A^{-1}=\dfrac{1}{4} \begin{bmatrix} 0 & 2 \\ -2 & 4\end{bmatrix} =\begin{bmatrix} 0 & \dfrac{1}{2} \\ -\dfrac{1}{2} & 1\end{bmatrix} $
