Use an augmented matrix to solve each system of equations.

$\begin{array}{l}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{16}\\ \mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}\end{array}$

To write the equations in augmented matrix place the coefficients of the equations and the constant terms into a matrix separated by a dashed line.

Here, the augmented matrices are:

$\left[\begin{array}{ccc}2& 4& -16\\ 1& -2& 0\end{array}\right]$

First exchange row 1 with row 2 so as to make the first element of row 1 a 1.

$\left[\begin{array}{ccc}2& 4& -16\\ 1& -2& 0\end{array}\right]\stackrel{R_1\leftrightarrow R_2}{\to }\left[\begin{array}{ccc}1& -2& 0\\ 2& 4& -16\end{array}\right]$

To make first element of row to a 0, multiply row 1 by 2 and subtract the resultant from row 1.

$\left[\begin{array}{ccc}1& -2& 0\\ 2& 4& -16\end{array}\right]\stackrel{R_2-2R_1}{\to }\left[\begin{array}{ccc}1& -2& 0\\ 0& 8& -16\end{array}\right]$

To make the second element in the second row a 1, divide the second row by 8.

$\left[\begin{array}{ccc}1& -2& 0\\ 0& 8& -16\end{array}\right]\stackrel{\frac{1}{8}{R}_2}{\to }\left[\begin{array}{ccc}1& -2& 0\\ 0& 1& -2\end{array}\right]$

Further row-reduce the augmented matrix, by making the second element in the first row a zero.

$\left[\begin{array}{ccc}1& -2& 0\\ 0& 1& -2\end{array}\right]\stackrel{R_1\to R_1+2R_2}{\to }\left[\begin{array}{ccc}1& 0& -4\\ 0& 1& -2\end{array}\right]$

Here, the first row will give the solution of $x$, because the coefficient of $y$ is 0 and the coefficient of $x$is 1. Similarly, the second row will give the solution of $y$, because the coefficient of $x$is 0 and the coefficient of $y$is 1. Therefore, the solution is$(-4,-2)$.