College Algebra, Chapter 7, 7.1, Section 7.1, Problem 50

Find the complete solution of the system
$
\left\{
\begin{array}{cccccc}
x & - 3y & + 2z & + w & = & -2 \\
x & - 2y & & - 2w & = & -10 \\
& & z & + 5w & = & 15 \\
3x & & + 2z & + w & = & -3
\end{array}
\right.
$


We transform the system into reduced row-echelon form

$\left[ \begin{array}{ccccc}
1 & -3 & 2 & 1 & -2 \\
1 & -2 & 0 & -2 & -10 \\
0 & 0 & 1 & 5 & 15 \\
3 & 0 & 2 & 1 & -3
\end{array} \right]$

$\displaystyle R_2 - R_1 \to R_2$

$\left[ \begin{array}{ccccc}
1 & -3 & 2 & 1 & -2 \\
0 & 1 & -2 & -3 & -8 \\
0 & 0 & 1 & 5 & 15 \\
3 & 0 & 2 & 1 & -3
\end{array} \right]$

$\displaystyle R_4 - 3R_1 \to R_4$

$\left[ \begin{array}{ccccc}
1 & -3 & 2 & 1 & -2 \\
0 & 1 & -2 & -3 & -8 \\
0 & 0 & 1 & 5 & 15 \\
0 & 9 & -4 & -2 & 3
\end{array} \right]$

$\displaystyle R_4 - 9 R_2 \to R_4$

$\left[ \begin{array}{ccccc}
1 & -3 & 2 & 1 & -2 \\
0 & 1 & -2 & -3 & -8 \\
0 & 0 & 1 & 5 & 15 \\
0 & 0 & 14 & 25 & 75
\end{array} \right]$

$\displaystyle R_4 - 14 R_3 \to R_4$

$\left[ \begin{array}{ccccc}
1 & -3 & 2 & 1 & -2 \\
0 & 1 & -2 & -3 & -8 \\
0 & 0 & 1 & 5 & 15 \\
0 & 0 & 0 & -45 & -135
\end{array} \right]$

$\displaystyle \frac{-1}{45} R_4$

$\left[ \begin{array}{ccccc}
1 & -3 & 2 & 1 & -2 \\
0 & 1 & -2 & -3 & -8 \\
0 & 0 & 1 & 5 & 15 \\
0 & 0 & 0 & 1 & 3
\end{array} \right]$

$\displaystyle R_3 - 5 R_4 \to R_3$

$\left[ \begin{array}{ccccc}
1 & -3 & 2 & 1 & -2 \\
0 & 1 & -2 & -3 & -8 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 3
\end{array} \right]$

$\displaystyle R_2 + 3 R_4 \to R_2$

$\left[ \begin{array}{ccccc}
1 & -3 & 2 & 1 & -2 \\
0 & 1 & -2 & 0 & 1 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 3
\end{array} \right]$

$\displaystyle R_1 - R_4 \to R_1$

$\left[ \begin{array}{ccccc}
1 & -3 & 2 & 0 & -5 \\
0 & 1 & -2 & 0 & 1 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 3
\end{array} \right]$

$\displaystyle R_2 + 2R_3 \to R_2$

$\left[ \begin{array}{ccccc}
1 & -3 & 2 & 0 & -5 \\
0 & 1 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 3
\end{array} \right]$

$\displaystyle R_1 - 2 R_3 \to R_1$

$\left[ \begin{array}{ccccc}
1 & -3 & 0 & 0 & -5 \\
0 & 1 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 3
\end{array} \right]$

$\displaystyle R_1 + 3 R_2 \to R_1$

$\left[ \begin{array}{ccccc}
1 & 0 & 0 & 0 & -2 \\
0 & 1 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 3
\end{array} \right]$


We now have an equivalent matrix in reduced row-echelon form, and the system of equations is


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

x =& -2
\\
\\
y =& 1
\\
\\
z =& 0
\\
\\
w =& 3

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


We can write the solution as the ordered quadruple $(-2, 1, 0, 3)$.