Q. 62 - 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.

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

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

$\left\{\begin{array}{l}x+y=6\\ 2x-y+z=1\\ x+2y+z=\frac{8}{3}\end{array}\right.$

Step 2. Calculation.

The augmented matrix of the system is:  

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

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

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

Perform the row operations $R_2=r_2-2r_1$:

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

Perform the row operations $R_3=3r_3+r_2$:

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

Perform the row operations $R_3=\frac{1}{4}{r}_3$:

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

Use the obtained matrix to write the system of equations.

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

Step 3. Solve the equation.

Solve the equations to find the solution set.

Substitute $z=1$into the second equation.

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

Substitute  $y=\frac{2}{3}$into the first equation.

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

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,$\left(\frac{1}{3},\frac{2}{3},1\right)$.