Q57. - Algebra 1

Question

Use an augmented matrix to solve each system of equations.

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

Step-by-Step Solution

Answer

The solution of the system of equations is$\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{-}\mathbf{2}\mathbf{)}$.

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:

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

Step 2. Use row operations to solve the system.

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

$\begin{array}{r}\left[\begin{array}{ccc}1& -1& 5\\ 3& 3& 3\end{array}\right]\stackrel{\frac{1}{3}{R}_2}{\to }\left[\begin{array}{ccc}1& -1& 5\\ 1& 1& 1\end{array}\right]\\ \stackrel{R_2-R_1}{\to }\left[\begin{array}{ccc}1& -1& 5\\ 0& 2& -4\end{array}\right]\end{array}$

To make the second element in the second row a 1, divide the second row by 2.

$\left[\begin{array}{ccc}1& -1& 5\\ 0& 2& -4\end{array}\right]\stackrel{\frac{1}{2}{R}_2}{\to }\left[\begin{array}{ccc}1& -1& 5\\ 0& 1& -2\end{array}\right]$

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}{ccc}1& -1& 5\\ 0& 1& -2\end{array}\right]\stackrel{R_1+R_2}{\to }\left[\begin{array}{ccc}1& 0& 3\\ 0& 1& -2\end{array}\right]$$\left[\begin{array}{ccc}1& -1& 5\\ 0& 1& -2\end{array}\right]\stackrel{R_1+R_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 $(3,-2)$.