Q62. - Algebra 1

Question

Use an augmented matrix to solve each system of equations.

$\begin{array}{l}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{26}\\ \mathbf{2}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{42}\end{array}$

Step-by-Step Solution

Answer

The solution of the system of equations is $\mathbf{(}\mathbf{-}\mathbf{6}\mathbf{,}\mathbf{10}\mathbf{)}$.

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:

 $\left[\begin{array}{ccc}1& -2& |-26\\ 2& -3& |-42\end{array}\right]$

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 then subtract the resultant row 1 from the row 2.

 $\left[\begin{array}{ccc}1& -2& |-26\\ 2& -3& |-42\end{array}\right]\stackrel{R_2\to R_2-2R_1}{\to }\left[\begin{array}{ccc}1& -2& |-26\\ 0& 1& | 10 \end{array}\right]$

Step 3. Row reduce the matrix.

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

$\left[\begin{array}{ccc}1& -2& |-26\\ 0& 1& | 10\end{array}\right]\stackrel{R_1\to R_1+2R_2}{\to }\left[\begin{array}{ccc}1& 0& |-6\\ 0& 1& | 10\end{array}\right]$ 

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 $(-6,10)$.