# How do you solve x + 9y = 20 and 2x - 4y = 15 using matrices?

Feb 24, 2016

$x = \frac{215}{22}$

$y = \frac{25}{22}$

#### Explanation:

Writing the system in matrix form looks like this

$\left[\begin{matrix}1 & 9 \\ 2 & - 4\end{matrix}\right] \left[\begin{matrix}x \\ y\end{matrix}\right] = \left[\begin{matrix}20 \\ 15\end{matrix}\right]$

Now, the inverse of $\left[\begin{matrix}1 & 9 \\ 2 & - 4\end{matrix}\right]$ happens to be $\left[\begin{matrix}\frac{2}{11} & \frac{9}{22} \\ \frac{1}{11} & - \frac{1}{22}\end{matrix}\right]$.

${\left[\begin{matrix}1 & 9 \\ 2 & - 4\end{matrix}\right]}^{- 1} = \frac{1}{1 \times \left(- 4\right) - 9 \times 2} \cdot {\left[\begin{matrix}- 4 & - 2 \\ - 9 & 1\end{matrix}\right]}^{T}$

$= \left[\begin{matrix}\frac{2}{11} & \frac{9}{22} \\ \frac{1}{11} & - \frac{1}{22}\end{matrix}\right]$

Multiply that to the left of both sides of the first equation, you will get identity matrix on the left side, and the answer on the right.

$\left[\begin{matrix}\frac{2}{11} & \frac{9}{22} \\ \frac{1}{11} & - \frac{1}{22}\end{matrix}\right] \left[\begin{matrix}1 & 9 \\ 2 & - 4\end{matrix}\right] \left[\begin{matrix}x \\ y\end{matrix}\right] = \left[\begin{matrix}\frac{2}{11} & \frac{9}{22} \\ \frac{1}{11} & - \frac{1}{22}\end{matrix}\right] \left[\begin{matrix}20 \\ 15\end{matrix}\right]$

$\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right] \left[\begin{matrix}x \\ y\end{matrix}\right] = \left[\begin{matrix}\frac{215}{22} \\ \frac{25}{22}\end{matrix}\right]$

$\left[\begin{matrix}x \\ y\end{matrix}\right] = \left[\begin{matrix}\frac{215}{22} \\ \frac{25}{22}\end{matrix}\right]$

You can check that

$\left(\frac{215}{22}\right) + 9 \left(\frac{25}{22}\right) = 20$

$2 \left(\frac{215}{22}\right) - 4 \left(\frac{25}{22}\right) = 15$