Question

Use an augmented matrix to solve each system of equations.

$\begin{array}{l}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{8}\\ \mathbf{x}\mathbf{-}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{22}\end{array}$

Short answer

The solution of the system of equations is$\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{4}\mathbf{)}$.

Step-by-step solution

Step 1. Write the augmented matrix.

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:

$\left[\begin{array}{ccc}2& 3& 8\\ 1& -5& -22\end{array}\right]$

Step 2. Use row operations to solve the system.

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

$\left[\begin{array}{ccc}2& 3& 8\\ 1& -5& -22\end{array}\right]\stackrel{R_1\leftrightarrow R_2}{\to }\left[\begin{array}{ccc}1& -5& -22\\ 2& 3& 8\end{array}\right]$

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

$\left[\begin{array}{ccc}1& -5& -22\\ 2& 3& 8\end{array}\right]\stackrel{R_2-2R_1}{\to }\left[\begin{array}{ccc}1& -5& -22\\ 0& 13& 52\end{array}\right]$

To make the second element in the second row a 1, divide the second row by 13.

$\left[\begin{array}{ccc}1& -5& -22\\ 0& 13& 52\end{array}\right]\stackrel{\frac{1}{13}{R}_2}{\to }\left[\begin{array}{ccc}1& -5& -22\\ 0& 1& 4\end{array}\right]$

Step 3. Row reduce the matrix.

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

$\left[\begin{array}{ccc}1& -5& -22\\ 0& 1& 4\end{array}\right]\stackrel{R_1\to R_1+5R_2}{\to }\left[\begin{array}{ccc}1& 0& -2\\ 0& 1& 4\end{array}\right]$

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$(-2,4)$.