Q. 57 - Precalculus Enhanced with Graphing Utilities

Question

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

$\left\{\begin{array}{l}x+y-z=6\\ 3x-2y+z=-5\\ x+3y-2z=14\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the system is $x=1,y=3,z=-2$or, using ordered pair, $\left(1,3,-2\right)$.

Step 1. Given information.

The given system of equation are 
$\left\{\begin{array}{l}x+y-z=6\\ 3x-2y+z=-5\\ x+3y-2z=14\end{array}\right.$

Step 2. Calculation.

The augmented matrix of the system is: 

$\left[\begin{array}{cc}\begin{array}{ccc}1& 1& -1\\ 3& -2& 1\\ 1& 3& -2\end{array}& \begin{array}{c}6\\ -5\\ 14\end{array}\end{array}\right]$

Perform the row operations  $R_2=r_2-3r_1; R_3=r_3-r_1$:

$\left[\begin{array}{cc}\begin{array}{ccc}1& 1& -1\\ 0& -5& 4\\ 0& 2& -1\end{array}& \begin{array}{c}6\\ -23\\ 8\end{array}\end{array}\right]$

Perform the row operations $R_3=r_3+\frac{2}{5}{r}_2$:

$\left[\begin{array}{cc}\begin{array}{ccc}1& 1& -1\\ 0& -5& 4\\ 0& 0& \frac{3}{5}\end{array}& \begin{array}{c}6\\ -23\\ -\frac{6}{5}\end{array}\end{array}\right]$

Perform the row operations $R_3=\frac{5}{3}{r}_3$:

$\left[\begin{array}{cc}\begin{array}{ccc}1& 1& -1\\ 0& -5& 4\\ 0& 0& 1\end{array}& \begin{array}{c}6\\ -23\\ -2\end{array}\end{array}\right]$

Use the obtained matrix to write the system of equations. 

$$\begin{gathered} x+y-z=6 \\ -5y+4z=-23 \\ z=-2 \end{gathered}$$

Step 3. Solve the equation.

Solve the equations to find the solution set.

Substitute $z=-2$ into the second equation.

$$\begin{gathered} -5y+4z=-23 \\ -5y+4\left(-2\right)=-23 \\ -5y-8=-23 \\ 5y=-8+23 \\ 5y=15 \\ y=3 \end{gathered}$$

Substitute $y=3 \mathrm{and} z=-2$into the first equation.

$$\begin{gathered} x+y-z=6 \\ x+3-(-2)=6 \\ x+3+2=6 \\ x=6-5 \\ x=1 \end{gathered}$$

Therefore, the solution of the system is $x=1,y=3,z=-2$or, using ordered triplets, $\left(1,3,-2\right)$.

Step 4. Conclusion.

The solution of the system is $x=1,y=3,z=-2$or, using ordered triplets, $\left(1,3,-2\right)$.