Q. 61 - 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.

$\left\{\begin{array}{l}3x+y-z=\frac{2}{3}\\ 2x-y+z=1\\ 4x+2y=\frac{8}{3}\end{array}\right.$

Step-by-Step Solution

Answer

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

Step 1. Given information.

The given system of equation are 

$\left\{\begin{array}{l}3x+y-z=\frac{2}{3}\\ 2x-y+z=1\\ 4x+2y=\frac{8}{3}\end{array}\right.$

Step 2. Calculation.

The augmented matrix of the system is: 

$\left[\begin{array}{cc}\begin{array}{ccc}3& 1& -1\\ 2& -1& 1\\ 4& 2& 0\end{array}& \begin{array}{c}\frac{2}{3}\\ 16\\ \frac{8}{3}\end{array}\end{array}\right]$

Perform the row operations $R_2=r_2-\frac{2}{3}{r}_1, R_3=r_3-2r_2$:

$\left[\begin{array}{cc}\begin{array}{ccc}3& 1& -1\\ 0& -\frac{5}{3}& \frac{5}{3}\\ 0& 4& -2\end{array}& \begin{array}{c}\frac{2}{3}\\ \frac{5}{9}\\ \frac{6}{3}\end{array}\end{array}\right]$

Perform the row operations $R_2=\frac{3}{5}{r}_2, R_3=\frac{1}{2}{r}_3$:

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

Perform the row operations $R_3=r_3+2r_2$:

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

Use the obtained matrix to write the system of equations.

$$\begin{gathered} 3x+y-z=\frac{2}{3} \\ -y+z=\frac{1}{3} \\ z=1 \end{gathered}$$

Step 3. Solve the equation.

Solve the equations to find the solution set.

Substitute $z=1$into the second equation.

$$\begin{gathered} -y+z=\frac{1}{3} \\ -y+1=\frac{1}{3} \\ -y=\frac{1}{3}-1 \\ -y=-\frac{2}{3} \\ y=\frac{2}{3} \end{gathered}$$

Substitute  $y=\frac{2}{3} \mathrm{and} z=1$into the first equation.

$$\begin{gathered} 3x+y-z=\frac{2}{3} \\ 3x+\frac{2}{3}-1=\frac{2}{3} \\ 3x-\frac{1}{3}=\frac{2}{3} \\ 3x=\frac{2}{3}+\frac{1}{3} \\ 3x=1 \\ x=\frac{1}{3} \end{gathered}$$

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

Step 4. Conclusion.

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