## Algebra and Trigonometry 10th Edition

$\begin{bmatrix} 3/2 & -2 \\ -7/2 & 5 \end{bmatrix}$
The general form of a matrix of order $2 \times 2$ is: $det \ A=\begin{bmatrix} p & q \\ r & s\end{bmatrix}=ps-qr$ Now, $det \ A =\begin{bmatrix} 10 & 4 \\ 7 & 3 \end{bmatrix}=30-28=2 \ne 0$ We can determine the inverse of the matrix $A$ as follows: $A^{-1}=\dfrac{1}{det \ A} \begin{bmatrix} s & -q \\ -r & p\end{bmatrix}$ $A^{-1}=\dfrac{1}{2} \begin{bmatrix} 3 & -4 \\ -7 & 10 \end{bmatrix}$ Our answer is: $\begin{bmatrix} 3/2 & -2 \\ -7/2 & 5 \end{bmatrix}$