Q56. - Algebra 1

Question

Use an augmented matrix to solve each system of equations.

$\begin{array}{l}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{5}\\ \mathbf{2}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{4}\end{array}$

Step-by-Step Solution

Answer

The solution of the system of equations is$\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{)}$.

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:

$\left[\begin{array}{ccc}1& -2& -5\\ 2& 3& 4\end{array}\right]$

Step 2. Use row operations to solve the system.

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

$\left[\begin{array}{cc}1& -2\text{ |}\\ 2& 3\text{ |}\end{array}\begin{array}{c}-5\\ 4\end{array}\right]\stackrel{R_2-2R_1}{\to }\left[\begin{array}{cc}1& -2\text{ |}\\ 0& 7\text{ |}\end{array}\begin{array}{c}-5\\ 14\end{array}\right]$

To make the second element in the second row a 1, multiply the second row by$1/7$.

$\left[\begin{array}{cc}1& -2\text{ |}\\ 0& 7\text{ |}\end{array}\begin{array}{c}-5\\ 14\end{array}\right]\stackrel{\frac{1}{7}{R}_2}{\to }\left[\begin{array}{cc}1& -2\text{ |}\\ 0& \text{ 1 |}\end{array}\begin{array}{c}-5\\ 2\end{array}\right]$

Step 3. Row reduce the matrix.

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

 $\left[\begin{array}{cc}1& -2\text{ |}\\ 0& \text{ 1 |}\end{array}\begin{array}{c}-5\\ 2\end{array}\right]\stackrel{R_1-2R_2}{\to }\left[\begin{array}{cc}1& 0\text{ |}\\ 0& \text{ 1 |}\end{array}\begin{array}{c}-1\\ 2\end{array}\right]$

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 style="max-width: none; vertical-align: -4px;" $x$is 0 and the coefficient of$y$is 1. Therefore, the solution is$(-1,2)$.