Q. 59 - 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.

$\left\{\begin{array}{l}x+2y-z=-3\\ 2x-4y+z=-7\\ -2x+2y-3z=4\end{array}\right.$

Step-by-Step Solution

Answer

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

Step 1. Given information.

The given system of equation are 

$\left\{\begin{array}{l}x+2y-z=-3\\ 2x-4y+z=-7\\ -2x+2y-3z=4\end{array}\right.$

Step 2. Calculation.

The augmented matrix of the system is: 

$\left[\begin{array}{cc}\begin{array}{ccc}1& 2y& -z\\ 2& -4& 1\\ -2& 2& -3\end{array}& \begin{array}{c}-3\\ -7\\ 4\end{array}\end{array}\right]$

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

$\left[\begin{array}{cc}\begin{array}{ccc}1& 2y& -z\\ 2& -4& 1\\ 0& -2& -2\end{array}& \begin{array}{c}-3\\ -7\\ -3\end{array}\end{array}\right]$

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

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

Perform the row operations $R_3=4r_3-r_2$:

$\left[\begin{array}{cc}\begin{array}{ccc}1& 2y& -z\\ 0& -8& 3\\ 0& 0& -11\end{array}& \begin{array}{c}-3\\ -1\\ -11\end{array}\end{array}\right]$

Perform the row operations $R_3=-\frac{1}{11}{r}_3$:

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

Use the obtained matrix to write the system of equations.

$$\begin{gathered} x+2y-z=-3 \\ -8y+3z=-1 \\ z=1 \end{gathered}$$

Step 3. Solve the equation.

Solve the equations to find the solution set.

Substitute $z=1$ into the second equation.

$$\begin{gathered} -8y+3z=-1 \\ -8y+3\left(1\right)=-1 \\ -8y=-1-3 \\ -y=-\frac{4}{8} \\ y=\frac{1}{2} \end{gathered}$$

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

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

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

Step 4. Conclusion.

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