Q. 49 - 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. 

$\left\{\begin{array}{l}x-2y+3z=7\\ 2x+y+z=4\\ -3x+2y-2z=-10\end{array}\right.$

Step-by-Step Solution

Answer

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

Step 1. Given information.

The given system of equation are  

$\left\{\begin{array}{l}x-2y+3z=7\\ 2x+y+z=4\\ -3x+2y-2z=-10\end{array}\right.$

Step 2. Calculation.

The augmented matrix of the system is:  

$\left[\begin{array}{cc}\begin{array}{ccc}1& -2& 3\\ 2& 1& 1\\ -3& 2& -2\end{array}& \begin{array}{c}7\\ 4\\ -10\end{array}\end{array}\right]$

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

$\left[\begin{array}{cc}\begin{array}{ccc}1& -2& 3\\ 0& 5& -5\\ -3& 2& -2\end{array}& \begin{array}{c}7\\ -10\\ -10\end{array}\end{array}\right]$

Perform the row operations $R_2=\frac{1}{2}{r}_2; R_3=r_3+3r_1$:

$\left[\begin{array}{cc}\begin{array}{ccc}1& -2& 3\\ 0& 1& -1\\ 0& -4& 7\end{array}& \begin{array}{c}7\\ -2\\ 11\end{array}\end{array}\right]$

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

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

Perform the row operations $R_3=\frac{1}{3}{r}_3$:

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

Use the obtained matrix to write the system of equations.

$$\begin{gathered} x-2y+3z=7 \\ y-z=-2 \\ z=1 \end{gathered}$$

Step 3. Solve the equation.

Solve the equations to find the solution set.

Substitute $z=1$into the second equation.

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

Substitute $y=-1 \mathrm{and} z=1$into first equation.

$$\begin{gathered} x-2y+3z=7 \\ x-2\left(-1\right)+3\left(1\right)=7 \\ x+2+3=7 \\ x=2 \end{gathered}$$

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

Step 4. Conclusion.

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