# How do you solve the system 4x-y=1, x+2y=7 using matrix equation?

Feb 14, 2017

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

#### Explanation:

In matrix form it's this:

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

You can the go through the formality of getting the inverse of the matrix (assuming of course it has one). If we start with the determinant:

$\det \left(\begin{matrix}4 & - 1 \\ 1 & 2\end{matrix}\right) = 4 \left(2\right) - 1 \left(- 1\right) = 9$ so the matrix is invertible!

And we know that the inverse is:

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

And we can say that:

$\textcolor{red}{\frac{1}{9} \left(\begin{matrix}2 & 1 \\ - 1 & 4\end{matrix}\right)} \textcolor{g r e e n}{\left(\begin{matrix}4 & - 1 \\ 1 & 2\end{matrix}\right)} \left(\begin{matrix}x \\ y\end{matrix}\right) = \textcolor{red}{\frac{1}{9} \left(\begin{matrix}2 & 1 \\ - 1 & 4\end{matrix}\right)} \left(\begin{matrix}1 \\ 7\end{matrix}\right)$

And because the red and the green terms combine as the identity matrix:

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

$\implies \left(\begin{matrix}x \\ y\end{matrix}\right) = \frac{1}{9} \left(\begin{matrix}9 \\ 27\end{matrix}\right) = \left(\begin{matrix}1 \\ 3\end{matrix}\right)$

Generally speaking solving systems is easier using Row Reduction but in the case of a 2x2 it is also pretty easy to work through this method.