## Precalculus (10th Edition)

$\begin{bmatrix} \dfrac{1}{2}&-1\\-\dfrac{1}{6}&\dfrac{2}{3}\end{bmatrix}$
We are given the matrix: $A=\begin{bmatrix}4&6\\1&3\end{bmatrix}$ In order to compute $A^{-1}$ use the formula: $A^{-1}=\dfrac{1}{det A}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$ where $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ and $det A=ad-bc$. First compute $detA$: $detA=4(3)-1(6)=6\not=0$ As $detA\not=0$, the inverse of $A$ exists. Determine $A^{-1}$: $A^{-1}=\dfrac{1}{6}\begin{bmatrix} 3&-6\\-1&4\end{bmatrix}=\begin{bmatrix} \dfrac{1}{2}&-1\\-\dfrac{1}{6}&\dfrac{2}{3}\end{bmatrix}$