# How do you solve 4x + 5y = -7  and 3x - 6y = 24 using matrices?

##### 1 Answer
Mar 9, 2017

The solution is $\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}2 \\ - 3\end{matrix}\right)$

#### Explanation:

We rewrite the equatins in matrix form

$\left(\begin{matrix}4 & 5 \\ 3 & - 6\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}- 7 \\ 24\end{matrix}\right)$

Let $A = \left(\begin{matrix}4 & 5 \\ 3 & - 6\end{matrix}\right)$

We need the inverse of matrix $A$

First, we calculate the determinant of matrix $A$

$\det A = | \left(4 , 5\right) , \left(3 , - 6\right) | = - 24 - 15 = - 39$

As, $\det A \ne 0$, the matrix $A$ is invertible

${A}^{-} 1 = - \frac{1}{39} \left(\begin{matrix}- 6 & - 5 \\ - 3 & 4\end{matrix}\right)$

$= \left(\begin{matrix}\frac{6}{39} & \frac{5}{39} \\ \frac{3}{39} & - \frac{4}{39}\end{matrix}\right)$

Therefore,

$\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}\frac{6}{39} & \frac{5}{39} \\ \frac{3}{39} & - \frac{4}{39}\end{matrix}\right) \left(\begin{matrix}- 7 \\ 24\end{matrix}\right)$

$= \left(\begin{matrix}- \frac{42}{39} + \frac{120}{39} \\ - \frac{21}{39} - \frac{96}{39}\end{matrix}\right)$

$= \left(\begin{matrix}\frac{78}{39} \\ - \frac{117}{39}\end{matrix}\right)$

$= \left(\begin{matrix}2 \\ - 3\end{matrix}\right)$