Find the complete solution of the system
$
\left\{
\begin{array}{cccccc}
x & & +3z & & = & -1 \\
& y & & -4w & = & 5 \\
& 2y & +z & +w & = & 0 \\
2x & +y & +5z & -4w & = & 4
\end{array}
\right.
$
using Gaussian Elimination.
For this system we have
$\displaystyle \left[
\begin{array}{ccccc}
1 & 0 & 3 & 0 & -1 \\
0 & 1 & 0 & -4 & 5 \\
0 & 2 & 1 & 1 & 0 \\
2 & 1 & 5 & -4 & 4
\end{array}
\right]$
$R_4 - 2R_1 \to R_4$
$\displaystyle \left[
\begin{array}{ccccc}
1 & 0 & 3 & 0 & -1 \\
0 & 1 & 0 & -4 & 5 \\
0 & 2 & 1 & 1 & 0 \\
0 & 1 & -1 & -4 & 6
\end{array}
\right]$
$R_3 - 2 R_2 \to R_3$
$\displaystyle \left[
\begin{array}{ccccc}
1 & 0 & 3 & 0 & -1 \\
0 & 1 & 0 & -4 & 5 \\
0 & 0 & 1 & 9 & -10 \\
0 & 1 & -1 & -4 & 6
\end{array}
\right]$
$R_4 - R_2 \to R_4$
$\displaystyle \left[
\begin{array}{ccccc}
1 & 0 & 3 & 0 & -1 \\
0 & 1 & 0 & -4 & 5 \\
0 & 0 & 1 & 9 & -10 \\
0 & 0 & -1 & 0 & 1
\end{array}
\right]$
$R_4 + R_3 \to R_4$
$\displaystyle \left[
\begin{array}{ccccc}
1 & 0 & 3 & 0 & -1 \\
0 & 1 & 0 & -4 & 5 \\
0 & 0 & 1 & 9 & -10 \\
0 & 0 & 0 & 9 & -9
\end{array}
\right]$
$\displaystyle \frac{1}{9} R_4$
$\displaystyle \left[
\begin{array}{ccccc}
1 & 0 & 3 & 0 & -1 \\
0 & 1 & 0 & -4 & 5 \\
0 & 0 & 1 & 9 & -10 \\
0 & 0 & 0 & 1 & -1
\end{array}
\right]
$
The corresponding system is
$
\left\{
\begin{array}{cccccc}
x & & +3z & & = & -1 \\
& y & & -4w & = & 5 \\
& & z & +9w & = & -10 \\
& & & w & = & -1
\end{array}
\right.
$
We now use back-substitution
$
\begin{equation}
\begin{aligned}
z + 9(-1) =& -10
&& \text{Back-substitute } w = -1
\\
z =& -10 + 9
&& \text{Subtract } 9(-1) = -9
\\
z =& -1
&&
\\
\\
y - 4 (-1) =& 5
&& \text{Back-substitute } w = -1
\\
y =& 5-4
&& \text{Subtract } -4(-1)=4
\\
y =& 1
&&
\\
\\
x + 3(-1) =& -1
&& \text{Back-substitute } z = -1
\\
x =& -1+3
&& \text{Subtract } 3(-1) = -3
\\
x =& 2
&&
\end{aligned}
\end{equation}
$
The solution is
$
\begin{equation}
\begin{aligned}
x =& 2
\\
y =& 1
\\
z =& -1
\\
w =& -1
\end{aligned}
\end{equation}
$
Monday, January 5, 2015
College Algebra, Chapter 7, Review Exercises, Section Review Exercises, Problem 12
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