# How do you solve the system 9x-6y=12, 4x+6y=-12 using matrix equation?

Feb 15, 2017

The answer is $\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}0 \\ - 2\end{matrix}\right)$

#### Explanation:

Let's rewrite the equation in matrix form

$\left(\begin{matrix}9 & - 6 \\ 4 & 6\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}12 \\ - 12\end{matrix}\right)$

Let matrix $A = \left(\begin{matrix}9 & - 6 \\ 4 & 6\end{matrix}\right)$

We need to calculate ${A}^{-} 1$, the inverse of matrix $A$

For a matrix to be invertible,

$\det A \ne 0$

$\det A = | \left(9 , - 6\right) , \left(4 , 6\right) | = 9 \cdot 6 - \left(- 6 \cdot 4\right)$

$= 54 + 24 = 78$

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

${A}^{-} 1 = \frac{1}{78} \left(\begin{matrix}6 & 6 \\ - 4 & 9\end{matrix}\right)$

$= \left(\begin{matrix}\frac{6}{78} & \frac{6}{78} \\ - \frac{4}{78} & \frac{9}{78}\end{matrix}\right) = \left(\begin{matrix}\frac{1}{13} & \frac{1}{13} \\ - \frac{2}{39} & \frac{3}{26}\end{matrix}\right)$

Therefore,

$\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}\frac{1}{13} & \frac{1}{13} \\ - \frac{2}{39} & \frac{3}{26}\end{matrix}\right) \cdot \left(\begin{matrix}12 \\ - 12\end{matrix}\right)$

$= \left(\begin{matrix}0 \\ - 2\end{matrix}\right)$