Q. 59 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 41– 60, use the inverses found in Problems 31–40 to solve each system of equations

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

Step-by-Step Solution

Answer

The solution of the matrix is $x=\frac{1}{3},y=1,\text{ and }z=\frac{2}{3}.$

Step 1. Given information

A matrix is given as, 

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

We need to find the solution of the matrix using inverse of the matrix. 

Step 2. Calculating Inverse of the matrix.

The inverse of the matrix can be found by : 

$$\begin{gathered} \left[A\mid I_3\right]=\left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 3& 2& -1& 0& 1& 0\\ 3& 1& 2& 0& 0& 1\end{array}\right] \\ {R}_2=\frac{r_2}{3}\text{ and }{R}_3=\frac{r_3}{3} \\ \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 3& 2& -1& 0& 1& 0\\ 3& 1& 2& 0& 0& 1\end{array}\right]\to \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ \frac{3}{3}& \frac{2}{3}& \frac{-1}{3}& \frac{0}{3}& \frac{1}{3}& \frac{0}{3}\\ \frac{3}{3}& \frac{1}{3}& \frac{2}{3}& \frac{0}{3}& \frac{0}{3}& \frac{1}{3}\end{array}\right]\to \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 1& \frac{2}{3}& -\frac{1}{3}& 0& \frac{1}{3}& 0\\ 1& \frac{1}{3}& \frac{2}{3}& 0& 0& \frac{1}{3}\end{array}\right] \\ {R}_2=r_2-r_1 \\ \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 1& \frac{2}{3}& -\frac{1}{3}& 0& \frac{1}{3}& 0\\ 1& \frac{1}{3}& \frac{2}{3}& 0& 0& \frac{1}{3}\end{array}\right]\to \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 1-1& \frac{2}{3}-1& -\frac{1}{3}-1& 0-1& \frac{1}{3}-0& 0-0\\ 1& \frac{1}{3}& \frac{2}{3}& 0& 0& \frac{1}{3}\end{array}\right]\to \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 0& -\frac{1}{3}& -\frac{4}{3}& -1& \frac{1}{3}& 0\\ 1& \frac{1}{3}& \frac{2}{3}& 0& 0& \frac{1}{3}\end{array}\right] \\ {R}_3=r_3-r_1 \\ \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 0& -\frac{1}{3}& -\frac{4}{3}& -1& \frac{1}{3}& 0\\ 1& \frac{1}{3}& \frac{2}{3}& 0& 0& \frac{1}{3}\end{array}\right]\to \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 0& -\frac{1}{3}& -\frac{4}{3}& -1& \frac{1}{3}& 0\\ 1-1& \frac{1}{3}-1& \frac{2}{3}-1& 0-1& 0-0& \frac{1}{3}-0\end{array}\right]\to \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 0& -\frac{1}{3}& -\frac{4}{3}& -1& \frac{1}{3}& 0\\ 0& -\frac{2}{3}& -\frac{1}{3}& -1& 0& \frac{1}{3}\end{array}\right] \end{gathered}$$

Step 3. Inverse of the Matrix

$$\begin{gathered} R_2=-3r_2 \\ \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 0& -\frac{1}{3}& -\frac{4}{3}& -1& \frac{1}{3}& 0\\ 0& -\frac{2}{3}& -\frac{1}{3}& -1& 0& \frac{1}{3}\end{array}\right]\to \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 0(-3)& -\frac{1}{3}(-3)& -\frac{4}{3}(-3)& -1(-3)& \frac{1}{3}(-3)& 0(-3)\\ 0& -\frac{2}{3}& -\frac{1}{3}& -1& 0& \frac{1}{3}\end{array}\right]\to \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 0& 1& 4& 3& -1& 0\\ 0& -\frac{2}{3}& -\frac{1}{3}& -1& 0& \frac{1}{3}\end{array}\right] \\ {R}_1=r_1-r_2 \\ \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 0& 1& 4& 3& -1& 0\\ 0& -\frac{2}{3}& -\frac{1}{3}& -1& 0& \frac{1}{3}\end{array}\right]\to \left[\begin{array}{cccccc}1-0& 1-1& 1-4& 1-3& 0+1& 0-0\\ 0& 1& 4& 3& -1& 0\\ 0& -\frac{2}{3}& -\frac{1}{3}& -1& 0& \frac{1}{3}\end{array}\right]\to \left[\begin{array}{cccccc}1& 0& -3& -2& 1& 0\\ 0& 1& 4& 3& -1& 0\\ 0& -\frac{2}{3}& -\frac{1}{3}& -1& 0& \frac{1}{3}\end{array}\right] \\ {R}_3=r_3+\frac{2}{3}{r}_2 \\ \left[\begin{array}{cccccc}1& 0& -3& -2& 1& 0\\ 0& 1& 4& 3& -1& 0\\ 0& -\frac{2}{3}& -\frac{1}{3}& -1& 0& \frac{1}{3}\end{array}\right]\to \left[\begin{array}{cccccc}1& 0& -3& -2& 1& 0\\ 0& 1& 4& 3& -1& 0\\ 0+0& -\frac{2}{3}+\frac{2}{3}& -\frac{1}{3}+\frac{8}{3}& -1+2& 0-\frac{2}{3}& \frac{1}{3}+0\end{array}\right]\to \left[\begin{array}{cccccc}1& 0& -3& -2& 1& 0\\ 0& 1& 4& 3& -1& 0\\ 0& 0& \frac{7}{3}& 1& -\frac{2}{3}& \frac{1}{3}\end{array}\right] \\ {R}_3=r_3·\frac{3}{7} \\ \left[\begin{array}{cccccc}1& 0& -3& -2& 1& 0\\ 0& 1& 4& 3& -1& 0\\ 0& 0& \frac{7}{3}& 1& -\frac{2}{3}& \frac{1}{3}\end{array}\right]\to \left[\begin{array}{cccccc}1& 0& -3& -2& 1& 0\\ 0& 1& 4& 3& -1& 0\\ 0\left(\frac{3}{7}\right)& 0\left(\frac{3}{7}\right)& \frac{7}{3}\left(\frac{3}{7}\right)& 1\left(\frac{3}{7}\right)& -\frac{2}{3}\left(\frac{3}{7}\right)& \frac{1}{3}\left(\frac{3}{7}\right)\end{array}\right]\to \left[\begin{array}{cccccc}1& 0& -3& -2& 1& 0\\ 0& 1& 4& 3& -1& 0\\ 0& 0& 1& \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right] \\ {R}_1=r_1+3r_3 \\ \left[\begin{array}{cccccc}1& 0& -3& -2& 1& 0\\ 0& 1& 4& 3& -1& 0\\ 0& 0& 1& \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right]\to \left[\begin{array}{rrrrrr}1+0& 0+0& -3+3& -2+\frac{9}{7}& 1-\frac{6}{7}& 0+\frac{3}{7}\\ 0& 1& 4& 3& -1& 0\\ 0& 0& 1& \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right]\to \left[\begin{array}{cccccc}1& 0& 0& -\frac{5}{7}& \frac{1}{7}& \frac{3}{7}\\ 0& 1& 4& 3& -1& 0\\ 0& 0& 1& \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right] \\ {R}_1=r_2-4r_3 \\ \left[\begin{array}{cccccc}1& 0& 0& -\frac{5}{7}& \frac{1}{7}& \frac{3}{7}\\ 0& 1& 4& 3& -1& 0\\ 0& 0& 1& \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right]\to \left[\begin{array}{cccccc}1& 0& 0& -\frac{5}{7}& \frac{1}{7}& \frac{3}{7}\\ 0-0& 1-0& 4-4& 3-\frac{12}{7}& -1+\frac{8}{7}& 0-\frac{4}{7}\\ 0& 0& 1& \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right]\to \left[\begin{array}{cccccc}1& 0& 0& -\frac{5}{7}& \frac{1}{7}& \frac{3}{7}\\ 0& 1& 0& \frac{9}{7}& \frac{1}{7}& -\frac{4}{7}\\ 0& 1& \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}& \end{array}\right] \\ {A}^{-1}=\left[\begin{array}{ccc}-\frac{5}{7}& \frac{1}{7}& \frac{3}{7}\\ \frac{9}{7}& \frac{1}{7}& -\frac{4}{7}\\ \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right] \end{gathered}$$

Step 4. Solution of the matrix

We know,

$$\begin{gathered} X=A^{-1}B \\ X=\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=A^{-1}B \\ =\left[\begin{array}{ccc}-\frac{5}{7}& \frac{1}{7}& \frac{3}{7}\\ \frac{9}{7}& \frac{1}{7}& -\frac{4}{7}\\ \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right]·\left[\begin{array}{c}2\\ \frac{7}{3}\\ \frac{10}{3}\end{array}\right] \\ X=\left[\begin{array}{c}-\frac{5}{7}\left(2\right)+\frac{1}{7}\left(\frac{7}{3}\right)+\frac{3}{7}\left(\frac{10}{3}\right)\\ \frac{9}{7}\left(2\right)[phone]\frac{\mathrm{}}{\mathrm{}}\left(\frac{\mathrm{}}{\mathrm{}}\right)\frac{\mathrm{}}{\mathrm{}}\left(\frac{\mathrm{}}{\mathrm{}}\right)\\ \frac{\mathrm{}}{\mathrm{}}\left(2\right)-\frac{2}{7}\left(\frac{7}{3}\right)+\frac{1}{7}\left(\frac{10}{3}\right)\end{array}\right] \\ =\left[\begin{array}{c}-\frac{10}{7}+\frac{1}{3}+\frac{10}{7}\\ \frac{18}{7}+\frac{1}{3}-\frac{40}{21}\\ \frac{6}{7}-\frac{2}{3}+\frac{10}{21}\end{array}\right] \\ =\left[\begin{array}{l}\frac{1}{3}\\ 1\\ \frac{2}{3}\end{array}\right] \end{gathered}$$