# How do you solve 6x + 2y = -16 and 5x + 6y = 4 using matrices?

Mar 10, 2016

Use the matrix equation and the inverse matrix to find $\left[\begin{matrix}x \\ y\end{matrix}\right] = \left[\begin{matrix}- 4 \\ 4\end{matrix}\right]$

#### Explanation:

We begin by writing our system of equations in matrix form

$\left[\begin{matrix}6 & 2 \\ 5 & 6\end{matrix}\right] \left[\begin{matrix}x \\ y\end{matrix}\right] = \left[\begin{matrix}- 16 \\ 4\end{matrix}\right]$

Let $A = \left[\begin{matrix}6 & 2 \\ 5 & 6\end{matrix}\right]$

We then use the identity that a matrix, $A$ multiplied by its inverse, ${A}^{- 1}$ is the identity matrix, $I$, i.e.

${A}^{- 1} A = A {A}^{- 1} = I$

Multiplying both sides of our original matrix equation by the inverse matrix we get:

${A}^{- 1} A \left[\begin{matrix}x \\ y\end{matrix}\right] = {A}^{- 1} \left[\begin{matrix}- 16 \\ 4\end{matrix}\right]$

which simplifies to

$\left[\begin{matrix}x \\ y\end{matrix}\right] = {A}^{- 1} \left[\begin{matrix}- 16 \\ 4\end{matrix}\right]$

since any matrix or vector multiplied by the identity matrix is itself. We now need to find ${A}^{- 1}$. Use the method described here How do I find the inverse of a $2 \times 2$ matrix? to find the inverse to be:

${A}^{- 1} = \left[\begin{matrix}\frac{3}{13} & - \frac{1}{13} \\ - \frac{5}{26} & \frac{3}{13}\end{matrix}\right]$

Substitute this into the equation to find $x$ and $y$

$\left[\begin{matrix}x \\ y\end{matrix}\right] = \left[\begin{matrix}\frac{3}{13} & - \frac{1}{13} \\ - \frac{5}{26} & \frac{3}{13}\end{matrix}\right] \left[\begin{matrix}- 16 \\ 4\end{matrix}\right] = \left[\begin{matrix}- 4 \\ 4\end{matrix}\right]$

Therefore $x = - 4$ and $y = 4$