Q. 60 - 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.

$\left\{\begin{array}{l}x+4y-3z=-8\\ 3x-y+3z=12\\ x+y+6z=1\end{array}\right.$  

Step-by-Step Solution

Answer

The solution of the system is $x=3,y=-\frac{8}{3},z=\frac{1}{9}$ or, using ordered triplets, $\left(3,-\frac{8}{3},\frac{1}{9}\right)$.

Step 1. Given information.

The given system of equation are 

$\left\{\begin{array}{l}x+4y-3z=-8\\ 3x-y+3z=12\\ x+y+6z=1\end{array}\right.$

Step 2. Calculation.

The augmented matrix of the system is: 

$\left[\begin{array}{cc}\begin{array}{ccc}1& 4& -3\\ 3& -1& 3\\ 1& 1& 6\end{array}& \begin{array}{c}-8\\ 12\\ 1\end{array}\end{array}\right]$

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

$\left[\begin{array}{cc}\begin{array}{ccc}1& 4& -3\\ 3& -1& 3\\ 0& -3& 9\end{array}& \begin{array}{c}-8\\ 12\\ 9\end{array}\end{array}\right]$

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

$\left[\begin{array}{cc}\begin{array}{ccc}1& 4& -3\\ 0& -13& 12\\ 0& -3& 9\end{array}& \begin{array}{c}-8\\ 36\\ 9\end{array}\end{array}\right]$

Perform the row operations $R_3=13r_3-3r_2$:

$\left[\begin{array}{cc}\begin{array}{ccc}1& 4& -3\\ 0& -13& 12\\ 0& 0& 81\end{array}& \begin{array}{c}-8\\ 36\\ 9\end{array}\end{array}\right]$

Perform the row operations $R_3=\frac{1}{81}{r}_3$:

$\left[\begin{array}{cc}\begin{array}{ccc}1& 4& -3\\ 0& -13& 12\\ 0& 0& 1\end{array}& \begin{array}{c}-8\\ 36\\ \frac{1}{9}\end{array}\end{array}\right]$

Use the obtained matrix to write the system of equations.

$$\begin{gathered} x+4y-3z=-8 \\ -13y+12z=36 \\ z=\frac{1}{9} \end{gathered}$$

Step 3. Solve the equation.

Solve the equations to find the solution set.

Substitute $z=\frac{1}{9}$into the second equation.

$$\begin{gathered} -13y+12z=36 \\ -13y+12\left(\frac{1}{9}\right)=36 \\ -13y+\frac{4}{3}=36 \\ -13y=36-\frac{4}{3} \\ -13y=\frac{104}{3} \\ y=-\frac{104}{39} \\ y=-\frac{8}{3} \end{gathered}$$

Substitute  $y=-\frac{8}{3} \mathrm{and} \mathrm{z}=\frac{1}{9}$into the first equation.

$$\begin{gathered} x+4y-3z=-8 \\ x+4\left(-\frac{8}{3}\right)-3\left(\frac{1}{9}\right)=-8 \\ x-\frac{32}{3}-\frac{1}{3}=-8 \\ x-\frac{33}{3}=-8 \\ x-11=-8 \\ x=-8+11 \\ x=3 \end{gathered}$$

Therefore, the solution of the system is $x=3,y=-\frac{8}{3},z=\frac{1}{9}$or, using ordered triplets, $\left(3,-\frac{8}{3},\frac{1}{9}\right)$.

Step 4. Conclusion.

The solution of the system is $x=3,y=-\frac{8}{3},z=\frac{1}{9}$or, using ordered triplets,$\left(3,-\frac{8}{3},\frac{1}{9}\right)$.