College Algebra, Chapter 7, 7.1, Section 7.1, Problem 42

Find the complete solution of the system

$
\left\{
\begin{equation}
\begin{aligned}

-4x - y + 36z =& 24
\\
x - 2y + 9z =& 3
\\
-2x + y + 6z =& 6

\end{aligned}
\end{equation}
\right.
$


We first write the augmented matrix of the system and using Gauss-Jordan Elimination.

$\left[ \begin{array}{cccc}
-4 & -1 & 36 & 24 \\
1 & -2 & 9 & 3 \\
-2 & 1 & 6 & 6
\end{array} \right]$

$\displaystyle \frac{-1}{4} R_1$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{1}{4} & -9 & -6 \\
1 & -2 & 9 & 3 \\
-2 & 1 & 6 & 6
\end{array} \right]$

$\displaystyle R_3 + 2 R_1 \to R_3$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{1}{4} & -9 & -6 \\
1 & -2 & 9 & 3 \\
0 & \displaystyle \frac{3}{2} & -12 & -6
\end{array} \right]$

$\displaystyle \frac{2}{3} R_3$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{1}{4} & -9 & -6 \\
1 & -2 & 9 & 3 \\
0 & 1 & -8 & -4
\end{array} \right]$

$\displaystyle R_2 - R_1 \to R_2$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{1}{4} & -9 & -6 \\
0 & \displaystyle \frac{-9}{4} & 18 & 9 \\
0 & 1 & -8 & -4
\end{array} \right]$

$\displaystyle \frac{-4}{9} R_2$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{1}{4} & -9 & -6 \\
0 & 1 & -8 & -4 \\
0 & 1 & -8 & -4
\end{array} \right]$

$\displaystyle R_3 - R_2 \to R_3$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{1}{4} & -9 & -6 \\
0 & 1 & -8 & -4 \\
0 & 0 & 0 & 0
\end{array} \right]$

$\displaystyle R_1 - \frac{1}{4} R_2 \to R_1$

$\left[ \begin{array}{cccc}
1 & 0 & -7 & -5 \\
0 & 1 & -8 & -4 \\
0 & 0 & 0 & 0
\end{array} \right]$


This is in reduced row-echelon form since the last row represents the equation $0 = 0$, we may discard it. So the last matrix corresponds to the system


$
\left\{
\begin{array}{ccccc}
x & & - 7z & = & -5 \\
& y & - 8z & = & -4
\end{array}
\right.
$


To obtain the complete solution, we solve for the leading variables $x$ and $y$ in terms of the nonleading variable $z$ and we let $z$ be any real numbers. Thus, the complete solution is


$
\begin{equation}
\begin{aligned}

x =& 7t - 5
\\
y =& 8t - 4
\\
z =& t

\end{aligned}
\end{equation}
$


where $t$ is any real number.