In Problems 37–72, solve each system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent.

$\left\{\begin{array}{l}\begin{array}{c}x-3y+z=1\\ 2x-y-4z=0\end{array}\\ \begin{array}{c}x-3y+2z=1\\ x-2y=5\end{array}\end{array}\right.$

The given system of equations is

$\left\{\begin{array}{l}\begin{array}{c}x-3y+z=1\\ 2x-y-4z=0\end{array}\\ \begin{array}{c}x-3y+2z=1\\ x-2y=5\end{array}\end{array}\right.$

Row operation :

1. Interchange any two rows.
2. Replace a row by a nonzero multiple of that row.
3. Replace a row by the sum of that row and a constant nonzero multiple of some other row.

The augmented matrix of the system is:

$\left[\begin{array}{cc}\begin{array}{ccc}1& -3& 1\\ 2& -1& -4\\ 1& -3& 2\\ 1& -2& 0\end{array}& \begin{array}{c}\begin{array}{c}\begin{array}{c}1\\ 0\end{array}\\ \begin{array}{c}1\\ 5\end{array}\end{array}\end{array}\end{array}\right]$

Perform the row operations$R_2=r_2-2r_1$:

role="math" localid="1648197781539" $\left[\begin{array}{cc}\begin{array}{ccc}1& -3& 1\\ 0& 5& -6\\ 1& -3& 2\\ 1& -2& 0\end{array}& \begin{array}{c}\begin{array}{c}\begin{array}{c}1\\ -2\end{array}\\ \begin{array}{c}1\\ 5\end{array}\end{array}\end{array}\end{array}\right]$

Perform the row operations$R_3=r_3-r_1$:

role="math" localid="1648197791537" $\left[\begin{array}{cc}\begin{array}{ccc}1& -3& 1\\ 0& 5& -6\\ 0& 0& 1\\ 1& -2& 0\end{array}& \begin{array}{c}\begin{array}{c}\begin{array}{c}1\\ -2\end{array}\\ \begin{array}{c}0\\ 5\end{array}\end{array}\end{array}\end{array}\right]$

Perform the row operations$R_4=r_4-r_1$:

role="math" localid="1648197813528" $\left[\begin{array}{cc}\begin{array}{ccc}1& -3& 1\\ 0& 5& -6\\ 0& 0& 1\\ 0& 1& -1\end{array}& \begin{array}{c}\begin{array}{c}\begin{array}{c}1\\ -2\end{array}\\ \begin{array}{c}0\\ 4\end{array}\end{array}\end{array}\end{array}\right]$

Perform the row operations$R_2=\frac{r_2}{5}$:

$\left[\begin{array}{cc}\begin{array}{ccc}1& -3& 1\\ 0& 1& -\frac{6}{5}\\ 0& 0& 1\\ 0& 1& -1\end{array}& \begin{array}{c}\begin{array}{c}\begin{array}{c}1\\ -\frac{2}{5}\end{array}\\ \begin{array}{c}0\\ 4\end{array}\end{array}\end{array}\end{array}\right]$

Perform the row operations$R_1=r_1+3r_2$:

$\left[\begin{array}{cc}\begin{array}{ccc}1& 0& -\frac{13}{5}\\ 0& 1& -\frac{6}{5}\\ 0& 0& 1\\ 0& 1& -1\end{array}& \begin{array}{c}\begin{array}{c}\begin{array}{c}-\frac{1}{5}\\ -\frac{2}{5}\end{array}\\ \begin{array}{c}0\\ 4\end{array}\end{array}\end{array}\end{array}\right]$

Perform the row operations$R_4=r_4-r_2$:

$\left[\begin{array}{cc}\begin{array}{ccc}1& 0& -\frac{13}{5}\\ 0& 1& -\frac{6}{5}\\ 0& 0& 1\\ 0& 1& \frac{1}{5}\end{array}& \begin{array}{c}\begin{array}{c}\begin{array}{c}-\frac{1}{5}\\ -\frac{2}{5}\end{array}\\ \begin{array}{c}0\\ \frac{22}{5}\end{array}\end{array}\end{array}\end{array}\right]$

Perform the row operations$R_1=r_1+\frac{13}{5}{r}_3$:

$\left[\begin{array}{cc}\begin{array}{ccc}1& 0& 0\\ 0& 1& -\frac{6}{5}\\ 0& 0& 1\\ 0& 1& \frac{1}{5}\end{array}& \begin{array}{c}\begin{array}{c}\begin{array}{c}-\frac{1}{5}\\ -\frac{2}{5}\end{array}\\ \begin{array}{c}0\\ \frac{22}{5}\end{array}\end{array}\end{array}\end{array}\right]$

Perform the row operations$R_2=r_2+\frac{6}{5}{r}_3$:

$\left[\begin{array}{cc}\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\\ 0& 1& \frac{1}{5}\end{array}& \begin{array}{c}\begin{array}{c}\begin{array}{c}-\frac{1}{5}\\ -\frac{2}{5}\end{array}\\ \begin{array}{c}0\\ \frac{22}{5}\end{array}\end{array}\end{array}\end{array}\right]$

Perform the row operations$R_4=r_4-\frac{1}{5}{r}_3$:

$\left[\begin{array}{cc}\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}& \begin{array}{c}\begin{array}{c}\begin{array}{c}-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\\ -\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\end{array}\\ \begin{array}{c}0\\ \raisebox{1ex}{$22$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\end{array}\end{array}\end{array}\end{array}\right]$

This matrix is in row echelon form.

Use the obtained matrix to write the system of equations.

$$\begin{gathered} x=-\frac{1}{5}...\left(1\right) \\ y=-\frac{2}{5}...\left(21\right) \\ z=0...\left(3\right) \end{gathered}$$

The bottom row of the obtained matrix is equivalent to the equation

$0·x+0·y+0·z=\frac{22}{5}$

which has no solution.

Thus, the original system is inconsistent.

The given system of equations is inconsistent.