In Problems 20–25, solve each system of equations using matrices. If the system has no solution, say that it is inconsistent.

$\left\{\begin{array}{rl}x-2z& =1\\ 2x+3y& =-3\\ 4x-3y-4z& =3\end{array}\right.$

The given system of equations is$\left\{\begin{array}{rl}x-2z& =1\\ 2x+3y& =-3\\ 4x-3y-4z& =3\end{array}\right.$

Write the augmented matrix that represents the system of equations

$\left[\begin{array}{rrrr}1& 0& -2& 1\\ 2& 3& 0& -3\\ 4& -3& -4& 3\end{array}\right]$

Perform the operations $R_2=R_2-2R_1$,

$\left[\begin{array}{cccc}1& 0& -2& 1\\ 2-2\left(1\right)& 3-2\left(0\right)& 0-2(-2)& -3-2\left(1\right)\\ 4& -3& -4& 3\end{array}\right]\to \left[\begin{array}{rrrr}1& 0& -2& 1\\ 0& 3& 4& -5\\ 4& -3& -4& 3\end{array}\right]$

Now, perform the operations $R_3=R_3-4R_1$,

$\left[\begin{array}{cccc}1& 0& -2& 1\\ 0& 3& 4& -5\\ 4-4\left(1\right)& -3-4\left(0\right)& -4-4(-2)& 3-4\left(1\right)\end{array}\right]\to \left[\begin{array}{rrrr}1& 0& -2& 1\\ 0& 3& 4& -5\\ 0& -3& 4& -1\end{array}\right]$

Now, perform the operations $R_2=\frac{R_2}{3}$,

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

Now, perform the operations $R_3=R_3+3R_2$,

$\left[\begin{array}{cccc}1& 0& -2& 1\\ \frac{{\displaystyle 0}}{{\displaystyle 3}}& \frac{{\displaystyle 3}}{{\displaystyle 3}}& \frac{{\displaystyle 4}}{{\displaystyle 3}}& -\frac{{\displaystyle 5}}{{\displaystyle 3}}\\ 0+3\left(\frac{{\displaystyle 0}}{{\displaystyle 3}}\right)& -3+3\left(\frac{{\displaystyle 3}}{{\displaystyle 3}}\right)& 4+3\left(\frac{{\displaystyle 4}}{{\displaystyle 3}}\right)& -1+3(-\frac{{\displaystyle 5}}{{\displaystyle 3}})\end{array}\right]\to \left[\begin{array}{rrrr}1& 0& -2& 1\\ 0& 1& \frac{{\displaystyle 4}}{{\displaystyle 3}}& -\frac{{\displaystyle 5}}{{\displaystyle 3}}\\ 0& 0& 8& -6\end{array}\right]$

Now, perform the operations $R_3=\frac{R_3}{8}$,

$\left[\begin{array}{rrrr}1& 0& -2& 1\\ 0& 1& \frac{{\displaystyle 4}}{{\displaystyle 3}}& -\frac{{\displaystyle 5}}{{\displaystyle 3}}\\ \frac{{\displaystyle 0}}{{\displaystyle 8}}& \frac{{\displaystyle 0}}{{\displaystyle 8}}& \frac{{\displaystyle 8}}{{\displaystyle 8}}& -\frac{{\displaystyle 6}}{{\displaystyle 8}}\end{array}\right]\to \left[\begin{array}{rrrr}1& 0& -2& 1\\ 0& 1& \frac{{\displaystyle 4}}{{\displaystyle 3}}& -\frac{{\displaystyle 5}}{{\displaystyle 3}}\\ 0& 0& 1& -\frac{{\displaystyle 3}}{{\displaystyle 4}}\end{array}\right]$

The system of equations corresponding to this matrix is

$$\begin{gathered} \left\{x-2z=1 \\ y-\frac{4}{3}z=-\frac{5}{3} \\ z=-\frac{3}{4}\right. \end{gathered}$$

Now,

role="math" localid="1647006836563" $$\begin{gathered} x-2\left(-\frac{3}{4}\right)=1 \\ x=-\frac{1}{2} \\ y-\frac{4}{3}\left(-\frac{3}{4}\right)=-\frac{5}{3} \\ y=-\frac{5}{3}-1 \\ y=-\frac{8}{3} \end{gathered}$$

Therefore, the solutions to the given equations are$x=-\frac{1}{2},y=-\frac{8}{3},z=-\frac{3}{4}$