Answer
$\begin{bmatrix} 1 & -\dfrac{1}{2} \\ -2 & \dfrac{3}{2} \end{bmatrix}$
Work Step by Step
$A=\begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix}$
Therefore, the general form of a matrix of order $ 2 \times 2$ is:
$\begin{bmatrix} p & q \\ r & s\end{bmatrix}=ps-qr$
Now, $ps-qr=\begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix}=6-4=2 \ne 0$
$A^{-1}=\dfrac{1}{2} \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix}$
Our answer is: $\begin{bmatrix} 1 & -\dfrac{1}{2} \\ -2 & \dfrac{3}{2} \end{bmatrix}$
