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

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

The given system of equations is$$\begin{gathered} \left\{x-y-z-t=1 \\ 2x+y-z+2t=3 \\ x-2y-2z-3t=0 \\ 3x-4y+z+5t=-3\right. \end{gathered}$$

Write the augmented matrix that represents the system of equations

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

Perform the operations $R_2=R_2-2R_3$,

$\left[\begin{array}{ccccc}1& -1& -1& -1& 1\\ 2-2\left(1\right)& 1-2(-2)& -1-2(-2)& 2-2(-3)& 3-2\left(0\right)\\ 1& -2& -2& -3& 0\\ 3& -4& 1& 5& -3\end{array}\right]\to \left[\begin{array}{rrrrr}1& -1& -1& -1& 1\\ 0& 5& 3& 8& 3\\ 1& -2& -2& -3& 0\\ 3& -4& 1& 5& -3\end{array}\right]$

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

$\left[\begin{array}{ccccc}1& -1& -1& -1& 1\\ 0& 5& 3& 8& 3\\ 1-1& -2-(-1)& -2-(-1)& -3-(-1)& 0-1\\ 3& -4& 1& 5& -3\end{array}\right]\to \left[\begin{array}{rrrrr}1& -1& -1& -1& 1\\ 0& 5& 3& 8& 3\\ 0& -1& -1& -2& -1\\ 3& -4& 1& 5& -3\end{array}\right]$

Now, perform the operations $R_4=R_4-3R_1$,

$\left[\begin{array}{crrrr}1& -1& -1& -1& 1\\ 0& 5& 3& 8& 3\\ 0& -1& -1& -2& -1\\ 3-3\left(1\right)& -4-3(-1)& 1-3(-1)& 5-3(-1)& -3-3\left(1\right)\end{array}\right]\to \left[\begin{array}{rrrrr}1& -1& -1& -1& 1\\ 0& 5& 3& 8& 3\\ 0& -1& -1& -2& -1\\ 0& -1& 4& 8& -6\end{array}\right]$

Now, perform the operations $R_2=\frac{R_2}{5}$,

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

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

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

Now, perform the operations $R_4=R_4+R_2$,

$\left[\begin{array}{ccccc}1& -1& -1& -1& 1\\ 0& 1& \frac{{\displaystyle 3}}{{\displaystyle 5}}& \frac{{\displaystyle 8}}{{\displaystyle 5}}& \frac{{\displaystyle 3}}{{\displaystyle 5}}\\ 0& 0& -\frac{{\displaystyle 2}}{{\displaystyle 5}}& -\frac{{\displaystyle 2}}{{\displaystyle 5}}& -\frac{{\displaystyle 2}}{{\displaystyle 5}}\\ 0+0& 1+(-1)& \frac{{\displaystyle 3}}{{\displaystyle 5}}+4& \frac{{\displaystyle 8}}{{\displaystyle 5}}+8& \frac{{\displaystyle 3}}{{\displaystyle 5}}+(-6)\end{array}\right]\to \left[\begin{array}{rrrrr}1& -1& -1& -1& 1\\ 0& 1& \frac{{\displaystyle 3}}{{\displaystyle 5}}& \frac{{\displaystyle 8}}{{\displaystyle 5}}& \frac{{\displaystyle 3}}{{\displaystyle 5}}\\ 0& 0& -\frac{{\displaystyle 2}}{{\displaystyle 5}}& -\frac{{\displaystyle 2}}{{\displaystyle 5}}& -\frac{{\displaystyle 2}}{{\displaystyle 5}}\\ 0& 0& \frac{{\displaystyle 23}}{{\displaystyle 5}}& \frac{{\displaystyle 48}}{{\displaystyle 5}}& -\frac{{\displaystyle 27}}{{\displaystyle 5}}\end{array}\right]$

Now, perform the operations $R_3=-\frac{5}{2}{R}_3$,

role="math" localid="1647011325202" $\left[\begin{array}{ccccc}1& -1& -1& -1& 1\\ 0& 1& \frac{{\displaystyle 3}}{{\displaystyle 5}}& \frac{{\displaystyle 8}}{{\displaystyle 5}}& \frac{{\displaystyle 3}}{{\displaystyle 5}}\\ -\frac{{\displaystyle 5}}{{\displaystyle 2}}\cdot 0& -\frac{{\displaystyle 5}}{{\displaystyle 2}}\cdot 0& -\frac{{\displaystyle 5}}{{\displaystyle 2}}(-\frac{{\displaystyle 2}}{{\displaystyle 5}})& -\frac{{\displaystyle 5}}{{\displaystyle 2}}(-\frac{{\displaystyle 2}}{{\displaystyle 5}})& -\frac{{\displaystyle 5}}{{\displaystyle 2}}(-\frac{{\displaystyle 2}}{{\displaystyle 5}})\\ 0& 0& \frac{{\displaystyle 23}}{{\displaystyle 5}}& \frac{{\displaystyle 48}}{{\displaystyle 5}}& -\frac{{\displaystyle 27}}{{\displaystyle 5}}\end{array}\right]\to \left[\begin{array}{ccccc}1& -1& -1& -1& 1\\ 0& 1& \frac{{\displaystyle 3}}{{\displaystyle 5}}& \frac{{\displaystyle 8}}{{\displaystyle 5}}& \frac{{\displaystyle 3}}{{\displaystyle 5}}\\ 0& 0& 1& 1& 1\\ 0& 0& \frac{{\displaystyle 23}}{{\displaystyle 5}}& \frac{{\displaystyle 48}}{{\displaystyle 5}}& -\frac{{\displaystyle 27}}{{\displaystyle 5}}\end{array}\right]$

Now, perform the operations $R_4=\frac{5}{48}{R}_4$,

$\left[\begin{array}{ccccc}1& -1& -1& -1& 1\\ 0& 1& \frac{{\displaystyle 3}}{{\displaystyle 5}}& \frac{{\displaystyle 8}}{{\displaystyle 5}}& \frac{{\displaystyle 3}}{{\displaystyle 5}}\\ 0& 0& 1& 1& 1\\ \frac{5}{48}·0& \frac{5}{48}·0& \frac{5}{48}·\frac{{\displaystyle 23}}{{\displaystyle 5}}& \frac{5}{48}·\frac{{\displaystyle 48}}{{\displaystyle 5}}& \frac{5}{48}·\left(-\frac{{\displaystyle 27}}{{\displaystyle 5}}\right)\end{array}\right]\to \left[\begin{array}{ccccc}1& -1& -1& -1& 1\\ 0& 1& \frac{{\displaystyle 3}}{{\displaystyle 5}}& \frac{{\displaystyle 8}}{{\displaystyle 5}}& \frac{{\displaystyle 3}}{{\displaystyle 5}}\\ 0& 0& 1& 1& 1\\ 0& 0& \frac{{\displaystyle 23}}{{\displaystyle 48}}& 1& -\frac{{\displaystyle 27}}{{\displaystyle 48}}\end{array}\right]$

The system of equations corresponding to this matrix is

$$\begin{gathered} \left\{x-y-z-t=1 \\ y+\frac{3}{5}z+\frac{8}{5}t=\frac{3}{5} \\ z+t=1 \\ \frac{23}{48}z+t=-\frac{27}{48}\right. \end{gathered}$$

role="math" localid="1647012045858" $$\begin{gathered} z=1-t \\ \frac{23}{48}\left(1-t\right)+t=-\frac{27}{48} \\ \frac{25}{48}t=-\frac{50}{48} \\ t=-2 \\ z=1-\left(-2\right) \\ z=3 \end{gathered}$$

$$\begin{gathered} y+\frac{3}{5}\left(3\right)+\frac{8}{5}\left(-2\right)=\frac{3}{5} \\ y=2 \end{gathered}$$

$$\begin{gathered} x-2-3-\left(-2\right)=1 \\ x=4 \end{gathered}$$

Therefore, the solutions are$x=4,y=2,z=3,t=-2$