# How do you solve x + 4y = 8 and 2x + y = 9 using matrices?

Mar 4, 2017

$x = 4 , y = 1$

#### Explanation:

In preamble, we can write these equations in column vector form, ie:

$x \left(\begin{matrix}1 \\ 2\end{matrix}\right) + y \left(\begin{matrix}4 \\ 1\end{matrix}\right) = \left(\begin{matrix}8 \\ 9\end{matrix}\right)$

And so they can also be written in matrix form, as we then have a set of rules in matrix algebra which means they are the same thing:

$\left(\begin{matrix}1 & 4 \\ 2 & 1\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}8 \\ 9\end{matrix}\right)$

That is: $M m a t h b f x = m a t h b f b$

From here, we have choices. We can row reduce, but as we are looking at the simplest matrix algebra (I think), we should try to use the inverse matrix ${M}^{- 1}$, the idea being:

${M}^{- 1} M m a t h b f x = {M}^{- 1} m a t h b f b$

$\implies I m a t h b f x = {M}^{- 1} m a t h b f b$

Generally speaking, for a $2 \times 2$ matrix, $M = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$, the inverse ${M}^{- 1}$ is:

${M}^{- 1} = \frac{1}{a d - c b} \left(\begin{matrix}d & - b \\ - c & a\end{matrix}\right)$.

So we have:

$\left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = - \frac{1}{7} \left(\begin{matrix}1 & - 4 \\ - 2 & 1\end{matrix}\right) \left(\begin{matrix}8 \\ 9\end{matrix}\right)$

$\implies \left(\begin{matrix}x \\ y\end{matrix}\right) = - \frac{1}{7} \left(\begin{matrix}- 28 \\ - 7\end{matrix}\right)$

$\implies \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}4 \\ 1\end{matrix}\right)$