Q. 78 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 75– 82, algebraically solve each system of equations using any method you wish.

$\left\{\begin{array}{l}2x+3y-z=-2\\ 4x+3z=6\\ 6y-2z=2\end{array}\right.$

Step-by-Step Solution

Answer

The values of unknown variables are $x=-1.5,y=1.67 and z=4$

Step 1. Given information

The given equations $\left\{\begin{array}{l}2x+3y-z=-2\\ 4x+3z=6\\ 6y-2z=2\end{array}\right.$

Step 2. Finding the values of x,y

Consider the equations $2x+3y-z=-2,4x+3z=6$ and $6y-2z=2$.To solve these equations,we first convert them into matrix form and then transform it into reduced echelon form

$$\begin{gathered} \left[\begin{array}{ccc}2& 3& -1\\ 4& 0& 3\\ 0& 6& -2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-2\\ 6\\ 2\end{array}\right] \left\{R_1=\frac{R_1}{2}\right\} \\ \left[\begin{array}{ccc}1& \frac{3}{2}& -\frac{1}{2}\\ 4& 0& 3\\ 0& 6& -2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-1\\ 6\\ 2\end{array}\right] \left\{R_2=R_2-4R_1\right\} \\ \left[\begin{array}{ccc}1& \frac{3}{2}& -\frac{1}{2}\\ 0& -6& 5\\ 0& 6& -2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-1\\ 10\\ 2\end{array}\right] \left\{R_2=-\frac{R_2}{6}\right\} \\ \left[\begin{array}{ccc}1& \frac{3}{2}& -\frac{1}{2}\\ 0& 1& -\frac{5}{6}\\ 0& 6& -2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-1\\ -\frac{5}{3}\\ 2\end{array}\right] \left\{R_1=R_1-\frac{3}{2}{R}_2;R_3=R_3-6R_2\right\} \\ \left[\begin{array}{ccc}1& 0& \frac{3}{4}\\ 0& 1& -\frac{5}{6}\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}\frac{3}{2}\\ -\frac{5}{3}\\ 4\end{array}\right] \left\{R_1=R_1-\frac{3}{4}{R}_3;R_2=R_2+\frac{5}{6}{R}_3\right\} \\ \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-1.5\\ 1.67\\ 4\end{array}\right] \\ \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-1.5\\ 1.67\\ 4\end{array}\right] \end{gathered}$$

Thus,we have solved the equations and the values of unknown variables are $x=-1.5,y=1.67$ and $z=4$