## Algebra and Trigonometry 10th Edition

$\begin{bmatrix} 1 & -\dfrac{1}{2} \\ -2 & \dfrac{3}{2} \end{bmatrix}$
$A=\begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix}$ 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: $\begin{bmatrix} 1 & -\dfrac{1}{2} \\ -2 & \dfrac{3}{2} \end{bmatrix}$