First, we write the augmented matrix.

[latex]\left[\begin{array}{ccc|c}\hfill 1& \hfill -1& \hfill 1& \hfill 8\\ \hfill 2& \hfill 3& \hfill -1& \hfill -2\\ \hfill 3& \hfill -2& \hfill -9& \hfill 9\end{array}\right][/latex]

Next, we perform row operations to obtain row-echelon form.

[latex]\begin{array}{rrrrr}\hfill -2{R}_{1}+{R}_{2}={R}_{2}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill -1& \hfill 1& \hfill 8\\ \hfill 0& \hfill 5& \hfill -3& \hfill -18\\ \hfill 3& \hfill -2& \hfill -9& \hfill 9\end{array}\right]& \hfill & \hfill & \hfill & \hfill -3{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill -1& \hfill 1& \hfill 8\\ \hfill 0& \hfill 5& \hfill -3& \hfill -18\\ \hfill 0& \hfill 1& \hfill -12& \hfill -15\end{array}\right]\end{array}[/latex]

The easiest way to obtain a 1 in row 2 of column 1 is to interchange [latex]{R}_{2}[/latex] and [latex]{R}_{3}[/latex].

[latex]\text{Interchange }{R}_{2}\text{ and }{R}_{3}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill -1& \hfill 1& \hfill 8\\ \hfill 0& \hfill 1& \hfill -12& \hfill -15\\ \hfill 0& \hfill 5& \hfill -3& \hfill -18\end{array}\right][/latex]

Then

[latex]\begin{array}{l}\\ \begin{array}{rrrrr}\hfill -5{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill -1& \hfill 1& \hfill 8\\ \hfill 0& \hfill 1& \hfill -12& \hfill -15\\ \hfill 0& \hfill 0& \hfill 57& \hfill 57\end{array}\right]& \hfill & \hfill & \hfill & \hfill -\frac{1}{57}{R}_{3}={R}_{3}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill -1& \hfill 1& \hfill 8\\ \hfill 0& \hfill 1& \hfill -12& \hfill -15\\ \hfill 0& \hfill 0& \hfill 1& \hfill 1\end{array}\right]\end{array}\end{array}[/latex]

The last matrix represents the equivalent system.

[latex]\begin{array}{l}\text{ }x-y+z=8\hfill \\ \text{ }y - 12z=-15\hfill \\ \text{ }z=1\hfill \end{array}[/latex]

Using back-substitution, we obtain the solution as [latex]\left(4,-3,1\right)[/latex].

Write the augmented matrix.

[latex]\left[\begin{array}{ccc|c}\hfill -1& \hfill -2& \hfill 1& \hfill -1\\ \hfill 2& \hfill 3& \hfill 0& \hfill 2\\ \hfill 0& \hfill 1& \hfill -2& \hfill 0\end{array}\right][/latex]

First, multiply row 1 by [latex]-1[/latex] to get a 1 in row 1, column 1. Then, perform row operations to obtain row-echelon form.

[latex]-{R}_{1}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill 2& \hfill -1& \hfill 1\\ \hfill 2& \hfill 3& \hfill 0& \hfill 2\\ \hfill 0& \hfill 1& \hfill -2& \hfill 0\end{array}\right][/latex]

[latex]{R}_{2}\leftrightarrow {R}_{3}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill 2& \hfill -1& \hfill 1\\ \hfill 0& \hfill 1& \hfill -2& \hfill 0\\ \hfill 2& \hfill 3& \hfill 0& \hfill 2\end{array}\right][/latex]

[latex]-2{R}_{1}+{R}_{3}={R}_{3}\to\left[\begin{array}{ccc|c}\hfill 1& \hfill 2& \hfill -1& \hfill 1\\ \hfill 0& \hfill 1& \hfill -2& \hfill 0\\ \hfill 0& \hfill -1& \hfill 2& \hfill 0\end{array}\right][/latex]

[latex]{R}_{2}+{R}_{3}={R}_{3}\to\left[\begin{array}{ccc|c}\hfill 1& \hfill 2& \hfill -1& \hfill 2\\ \hfill 0& \hfill 1& \hfill -2& \hfill 1\\ \hfill 0& \hfill 0& \hfill 0& \hfill 0\end{array}\right][/latex]

The last matrix represents the following system.

[latex]\begin{array}{l}\text{ }x+2y-z=1\hfill \\ \text{ }y - 2z=0\hfill \\ \text{ }0=0\hfill \end{array}[/latex]

We see by the identity [latex]0=0[/latex] that this is a dependent system with an infinite number of solutions. We then find the generic solution.

First, solve the second equation for [latex]y[/latex]:

[latex]\begin{align} y - 2z &= 0 \\ y &= 2z \end{align}[/latex]

Next, substitute [latex]y = 2z[/latex] into the first equation:

[latex]\begin{align} x + 2(2z) - z &= 1 \\ x + 4z - z &= 1 \\ x + 3z &= 1 \\ x &= 1 - 3z \end{align}[/latex]

The generic solution is [latex](1-3z,2z,z)[/latex].

Write the augmented matrix for the system of equations.

[latex]\left[\begin{array}{ccc|c}\hfill 5& \hfill 3& \hfill 9& \hfill -1\\ \hfill -2& \hfill 3& \hfill -1& \hfill -2\\ \hfill -1& \hfill -4& \hfill 5& \hfill 1\end{array}\right][/latex]

On the matrix page of the calculator, enter the augmented matrix above as the matrix variable [latex]\left[A\right][/latex].

[latex]\left[A\right]=\left[\begin{array}{rrrrrrr}\hfill 5& \hfill & \hfill 3& \hfill & \hfill 9& \hfill & \hfill -1\\ \hfill -2& \hfill & \hfill 3& \hfill & \hfill -1& \hfill & \hfill -2\\ \hfill -1& \hfill & \hfill -4& \hfill & \hfill 5& \hfill & \hfill 1\end{array}\right][/latex]

Use the ref( function in the calculator, calling up the matrix variable [latex]\left[A\right][/latex].

[latex]\text{ref}\left(\left[A\right]\right)[/latex]

Evaluate.

[latex]\begin{array}{l}\hfill \\ \left[\begin{array}{rrrr}\hfill 1& \hfill \frac{3}{5}& \hfill \frac{9}{5}& \hfill \frac{1}{5}\\ \hfill 0& \hfill 1& \hfill \frac{13}{21}& \hfill -\frac{4}{7}\\ \hfill 0& \hfill 0& \hfill 1& \hfill -\frac{24}{187}\end{array}\right]\to \begin{array}{l}x+\frac{3}{5}y+\frac{9}{5}z=-\frac{1}{5}\hfill \\ \text{ }y+\frac{13}{21}z=-\frac{4}{7}\hfill \\ \text{ }z=-\frac{24}{187}\hfill \end{array}\hfill \end{array}

[/latex]

Using back-substitution, the solution is [latex]\left(\frac{61}{187},-\frac{92}{187},-\frac{24}{187}\right)[/latex].