# How do you solve 9x-5y=-44 and 4x-3y=-18 using matrices?

Jul 10, 2016

The answer (in matrix form) is: $\left(\begin{matrix}1 & 0 & - 6 \\ 0 & 1 & 2\end{matrix}\right)$.

#### Explanation:

We can translate the given equations into matrix notation by transcribing the coefficients to elements of a 2x3 matrix:

$\left(\begin{matrix}9 & - 5 & - 44 \\ 4 & - 3 & - 18\end{matrix}\right)$

Divide the second row by 4 to get a one in the "x column."

$\left(\begin{matrix}9 & - 5 & - 44 \\ 1 & - \frac{3}{4} & - \frac{9}{2}\end{matrix}\right)$

Add -9 times the second row to the top row to get a zero in the "x column." We'll also revert the second row back to its previous form by multiplying by 4 again.

$\left(\begin{matrix}0 & \frac{7}{4} & - \frac{7}{2} \\ 4 & - 3 & - 18\end{matrix}\right)$

Multiply the top row by $\frac{4}{7}$ to get a 1 in the "y column."

$\left(\begin{matrix}0 & 1 & - 2 \\ 4 & - 3 & - 18\end{matrix}\right)$

We now have an answer for y. In order to solve for x, we add 3 times the first row to the second row.

$\left(\begin{matrix}0 & 1 & - 2 \\ 4 & 0 & - 24\end{matrix}\right)$

Then divide the second row by 4.

$\left(\begin{matrix}0 & 1 & - 2 \\ 1 & 0 & - 6\end{matrix}\right)$

And we finish by reversing the rows since it's traditional to show your final solution in the form of an identity matrix and an auxiliary column.

$\left(\begin{matrix}1 & 0 & - 6 \\ 0 & 1 & - 2\end{matrix}\right)$

This is equivalent to the set of equations:
$x = - 6$
$y = - 2$