College Algebra, Chapter 7, Review Exercises, Section Review Exercises, Problem 18

Find the complete solution of the system
$
\left\{
\begin{equation}
\begin{aligned}

x +2y +3z =& 2
\\
2x - y - 5z =& 1
\\
4x + 3y + z =& 6

\end{aligned}
\end{equation}
\right.
$
using Gauss-Jordan Elimination.

We transform the system into reduced row-echelon form

$\displaystyle \left[
\begin{array}{cccc}
1 & 2 & 3 & 2 \\
2 & -1 & -5 & 1 \\
4 & 3 & 1 & 6
\end{array}
\right]$

$R_2 - 2 R_1 \to R_2$

$\displaystyle \left[
\begin{array}{cccc}
1 & 2 & 3 & 2 \\
0 & -5 & -11 & -3 \\
4 & 3 & 1 & 6
\end{array}
\right]$

$R_3 - 4 R_1 \to R_3$

$\displaystyle \left[
\begin{array}{cccc}
1 & 2 & 3 & 2 \\
0 & -5 & -11 & -3 \\
0 & -5 & -11 & -2
\end{array}
\right]$

$\displaystyle \frac{-1}{5} R_2$

$\displaystyle \left[
\begin{array}{cccc}
1 & 2 & 3 & 2 \\
0 & 1 & \displaystyle \frac{11}{5} & \displaystyle \frac{3}{5} \\
0 & -5 & -11 & -2
\end{array}
\right]$

$R_3 + 5 R_2 \to R_3$

$\displaystyle \left[
\begin{array}{cccc}
1 & 2 & 3 & 2 \\
0 & 1 & \displaystyle \frac{11}{5} & \displaystyle \frac{3}{5} \\
0 & 0 & 0 & 1
\end{array}
\right]$

$\displaystyle R_2 - \frac{3}{5} R_3 \to R_2$

$\displaystyle \left[
\begin{array}{cccc}
1 & 2 & 3 & 2 \\
0 & 1 & \displaystyle \frac{11}{5} & 0 \\
0 & 0 & 0 & 1
\end{array}
\right]
$

$R_1 - 2 R_3 \to R_1$

$\displaystyle \left[
\begin{array}{cccc}
1 & 2 & 3 & 0 \\
0 & 1 & \displaystyle \frac{11}{5} & 0 \\
0 & 0 & 0 & 1
\end{array}
\right]$

$R_1 - 2 R_2 \to R_1$

$\displaystyle \left[
\begin{array}{cccc}
1 & 0 & \displaystyle \frac{-7}{5} & 0 \\
0 & 1 & \displaystyle \frac{11}{5} & 0 \\
0 & 0 & 0 & 1
\end{array}
\right]$

This is in reduced row echelon form. If we translate the last row back into equation, we get $0x + 0y + 0z = 1$, or $0 = 1$, which is false. This that the system has no solution or it is inconsistent.

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