Use an augmented matrix to solve each system of equations.

$\begin{array}{l}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{11}\\ \mathbf{3}\mathbf{x}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{1}\end{array}$

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:

$\left[\begin{array}{ccc}1& -4& 11\\ 3& 5& -1\end{array}\right]$

To make the first element in the second row a 0, multiply the first row by 3 and then subtract resultant row 1 from the row 2.

$\left[\begin{array}{ccc}1& -4& 11\\ 3& 5& -1\end{array}\right]\stackrel{R_2\to R_2-3R_1}{\to }\left[\begin{array}{ccc}1& -4& 11\\ 0& 17& -34\end{array}\right]$

To make the second element in the second row a 1, divide the second row by 17.

$\left[\begin{array}{ccc}1& -4& 11\\ 0& 17& -34\end{array}\right]\stackrel{\frac{1}{17}{R}_2}{\to }\left[\begin{array}{ccc}1& -4& 11\\ 0& 1& -2\end{array}\right]$

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

$\left[\begin{array}{ccc}1& -4& 11\\ 0& 1& -2\end{array}\right]\stackrel{R_1\to R_1+4R_2}{\to }\left[\begin{array}{ccc}1& 0& 3\\ 0& 1& -2\end{array}\right]$

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 $(3,-2)$.