College Algebra, Chapter 7, 7.1, Section 7.1, Problem 38

State whether the system of linear equations $\left\{ \begin{array}{ccccc}
& y & - 5z & = & 7 \\
3x & + 2y & & = & 12 \\
3x & & + 10z & = & 80
\end{array} \right.$ is inconsistent or dependent. If it is dependent, find the complete solution.

We transform the system into row-echelon form.

$\left[ \begin{array}{cccc}
0 & 1 & -5 & 7 \\
3 & 2 & 0 & 12 \\
3 & 0 & 10 & 80
\end{array} \right]$

$\displaystyle R_2 \leftrightarrow R_1$

$\left[ \begin{array}{cccc}
3 & 2 & 0 & 12 \\
0 & 1 & -5 & 7 \\
3 & 0 & 10 & 80
\end{array} \right]$

$\displaystyle \frac{1}{3} R_1$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{2}{3} & 0 & 4 \\
0 & 1 & -5 & 7 \\
3 & 0 & 10 & 80
\end{array} \right]$

$\displaystyle R_3 - 3R_1 \to R_3$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{2}{3} & 0 & 4 \\
0 & 1 & -5 & 7 \\
0 & -2 & 10 & 68
\end{array} \right]$

$\displaystyle \frac{-1}{2} R_3$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{2}{3} & 0 & 4 \\
0 & 1 & -5 & 7 \\
0 & 1 & -5 & -34
\end{array} \right]$

$\displaystyle R_3 - R_2 \to R_3$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{2}{3} & 0 & 4 \\
0 & 1 & -5 & 7 \\
0 & 0 & 0 & -41
\end{array} \right]$

$\displaystyle \frac{-1}{41} R_3$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{2}{3} & 0 & 4 \\
0 & 1 & -5 & 7 \\
0 & 0 & 0 & 1
\end{array} \right]$

This last matrix is in row-echelon form, so we can stop the Gaussian Elimination process. Now if we translate the last row back into equation form, we get $0x + 0y + 0z = 1$, or $0 = 1$, which is false. This means that the system is inconsistent.