# Using matrices when solving system of equations

Matrices could be used to solve systems of equations but first one must master to find the inverse of a matrice, C-1.

A matrix C will have an inverse C-1 if and only if the determinant of C is not equal to zero, then we call the matrix invertible.

$if\: c=\begin{bmatrix} a & b\\ c & d \end{bmatrix} \; and \; \begin{vmatrix} a & b\\ c & d \end{vmatrix}\neq 0\\ \\ \\ then\: C^{-1}=\frac{1}{ad-bc}\begin{bmatrix} d & -b\\ -c & a \end{bmatrix}$

We will now in an example show how to solve systems of equations using matrices and the inverse of matrices.

Example

Consider the following simultaneous equations (this example is also shown in our video lesson)

$\left\{\begin{matrix} 3x+y=5\\ 2x-y=0\\ \end{matrix}\right.$

Provided that we know how to multiply matrices we realize that our equations could be written as

$\begin{bmatrix} 3 & 1\\ 2 & -1 \end{bmatrix}\cdot \begin{bmatrix} x\\ y\\ \end{bmatrix}=\begin{bmatrix} 5\\ 0 \end{bmatrix}$

First we find the inverse of the coefficient matrix:

$C^{-1}=\frac{1}{3\cdot -1-1\cdot 2}\begin{bmatrix} -1 & -1\\ -2& 3 \end{bmatrix}=$

$=-\frac{1}{5}\begin{bmatrix} -1 & -1\\ -2& 3 \end{bmatrix}$

The next step is to multiply both sides of our matrix equation by the inverse matrix:

$-\frac{1}{5}\begin{bmatrix} -1 & -1\\ -2& 3 \end{bmatrix} \begin{bmatrix} 3 & 1\\ 2 & -1 \end{bmatrix}\cdot \begin{bmatrix} x\\ y\\ \end{bmatrix}= -\frac{1}{5}\begin{bmatrix} -1 & -1\\ -2& 3 \end{bmatrix} \begin{bmatrix} 5\\ 0 \end{bmatrix}$

$-\frac{1}{5}\begin{bmatrix} -5 & 0\\ 0 & -5 \end{bmatrix}\cdot \begin{bmatrix} x\\ y \end{bmatrix}=-\frac{1}{5}\begin{bmatrix} -5\\ -10 \end{bmatrix}$

$\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\cdot \begin{bmatrix} x\\ y \end{bmatrix}=\begin{bmatrix} 1\\ 2 \end{bmatrix}$

Our solution is (1,2), the easiest way to check if we are right is to plug our values into our original equations.

## Video lesson

The example above in videoformat.