# How do you solve the system 9x+3y=1, 5x+y=1 using matrix equation?

Dec 8, 2016

The answer is $x = \frac{1}{3}$ and $y = - \frac{2}{3}$

#### Explanation:

Let' write the matrix corresponding to the 2 equations

$\left(\begin{matrix}9 & 3 \\ 5 & 1\end{matrix}\right) \cdot \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}1 \\ 1\end{matrix}\right)$

Let $A = \left(\begin{matrix}9 & 3 \\ 5 & 1\end{matrix}\right)$

Then

$\left(\begin{matrix}x \\ y\end{matrix}\right) = {A}^{- 1} \cdot \left(\begin{matrix}1 \\ 1\end{matrix}\right)$

We must calculate the inverse of matrix $A$

The inverse of matrix $\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$ is

$\frac{1}{a d - b c} \cdot \left(\begin{matrix}d & - b \\ - c & a\end{matrix}\right)$

So,

${A}^{- 1} = \frac{1}{- 6} \left(\begin{matrix}1 & - 3 \\ - 5 & 9\end{matrix}\right)$

$= \left(\begin{matrix}- \frac{1}{6} & \frac{1}{2} \\ \frac{5}{6} & - \frac{3}{2}\end{matrix}\right)$

Verification

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

$\left(\begin{matrix}- \frac{1}{6} & \frac{1}{2} \\ \frac{5}{6} & - \frac{3}{2}\end{matrix}\right) \cdot \left(\begin{matrix}9 & 3 \\ 5 & 1\end{matrix}\right) = \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) = I$

Now, we can solve our equation

$\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}- \frac{1}{6} & \frac{1}{2} \\ \frac{5}{6} & - \frac{3}{2}\end{matrix}\right) \cdot \left(\begin{matrix}1 \\ 1\end{matrix}\right)$

$x = - \frac{1}{6} + \frac{1}{2} = \frac{1}{3}$

$y = \frac{5}{6} - \frac{3}{2} = - \frac{2}{3}$