Q. 48 - 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. 

Step-by-Step Solution

Answer

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

Step 1. Given information.

The given system of equation are  

$\left\{\begin{array}{l}2x+y=-4\\ -2y+4z=0\\ 3x-2z=-11\end{array}\right.$

Step 2. Calculation.

The augmented matrix of the system is: 

$\left[\begin{array}{cc}\begin{array}{ccc}2& 1& 0\\ 0& -2& 4\\ 3& 0& -2\end{array}& \begin{array}{c}-4\\ 0\\ -11\end{array}\end{array}\right]$

Perform the row operations $R_3=2r_3-3r_2$:

$\left[\begin{array}{cc}\begin{array}{ccc}2& 1& 0\\ 0& -2& 4\\ 0& -3& -4\end{array}& \begin{array}{c}-4\\ 0\\ -10\end{array}\end{array}\right]$

Perform the row operations $R_3=2r_3-3r_2$:

$\left[\begin{array}{cc}\begin{array}{ccc}2& 1& 0\\ 0& -2& 4\\ 0& 0& -20\end{array}& \begin{array}{c}-4\\ 0\\ -20\end{array}\end{array}\right]$

Perform the row operations $R_2=-\frac{1}{2}{r}_2; R_3=-\frac{1}{20}{r}_3$:

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

Use the obtained matrix to write the system of equations.

$$\begin{gathered} 2x+y=-4 \\ y-2z=0 \\ z=1 \end{gathered}$$

Step 3. Solve the equation.

Solve the equations to find the solution set.

Substitute $z=1$ into the second equation.

$$\begin{gathered} y-2z=0 \\ y-2\left(1\right)=0 \\ y-2=0 \\ y=2 \end{gathered}$$

Substitute $y=2$into the first equation.

$$\begin{gathered} 2x+y=-4 \\ 2x+2=-4 \\ 2x=-6 \\ x=-\frac{6}{2}=-3 \end{gathered}$$

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

Step 4. Conclusion.

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