Chapter 8, Problem 88RE

To determine

The system of linear equations using the inverse matrix.

Answer to Problem 88RE

The solution is $x=-2,y=4,z=3$

Explanation of Solution

Given information:

The given linear equation as shown below,

  $\{\begin{array}{c}4x+5y-6z=-6\\ 3x+2y+2z=8\\ 2x+y+z=3\end{array}$

Formula used:

The formula used is $X=A^{-1}B$

Calculation:

Given system of equations can be written in matrix form

  $\begin{array}{l}AX=B\\ \left[\begin{array}{ccc}4& 5& -6\\ 3& 2& 2\\ 2& 1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-6\\ 8\\ 3\end{array}\right]\end{array}$

Then the solution of the system is $X=A^{-1}B$

First find the inverse of the coefficient matrix

  $A=\left[\begin{array}{ccc}4& 5& -6\\ 3& 2& 2\\ 2& 1& 1\end{array}\right]$

Consider,

  $\left[\begin{array}{ccc}A& ⋮& I\end{array}\right]=\left[\begin{array}{ccccccc}4& 5& -6& :& 1& 0& 0\\ 3& 2& 2& :& 0& 1& 0\\ 2& 1& 1& :& 0& 0& 1\end{array}\right]$

Apply the row operations to change A in to I

  $\left[\begin{array}{ccccccc}4& 5& -6& :& 1& 0& 0\\ 3& 2& 2& :& 0& 1& 0\\ 2& 1& 1& :& 0& 0& 1\end{array}\right]\begin{array}{c}{R}_2\to R_2-\frac{3}{4}{R}_1\\ R_3\to R_3-\frac{1}{2}{R}_1\end{array}\left[\begin{array}{ccccccc}4& 5& -6& :& 1& 0& 0\\ 0& -1.75& 6.5& :& -0.75& 1& 0\\ 0& -1.5& 4& :& -0.5& 0& 1\end{array}\right]$

  $\left[\begin{array}{ccccccc}4& 5& -6& :& 1& 0& 0\\ 0& -1.75& 6.5& :& -0.75& 1& 0\\ 0& -1.5& 4& :& -0.5& 0& 1\end{array}\right]R_3\to R_3-\frac{6}{7}{R}_1\left[\begin{array}{ccccccc}4& 5& -6& :& 1& 0& 0\\ 0& -1.75& 6.5& :& -0.75& 1& 0\\ 0& 0& -\frac{11}{7}& :& \frac{1}{7}& -\frac{6}{7}& 1\end{array}\right]$

  $\left[\begin{array}{ccccccc}4& 5& -6& :& 1& 0& 0\\ 0& -1.75& 6.5& :& -0.75& 1& 0\\ 0& 0& -\frac{11}{7}& :& \frac{1}{7}& -\frac{6}{7}& 1\end{array}\right]R_3\to -\frac{7}{11}{R}_3\left[\begin{array}{ccccccc}4& 5& -6& :& 1& 0& 0\\ 0& -1.75& 6.5& :& -0.75& 1& 0\\ 0& 0& 1& :& -\frac{1}{11}& \frac{6}{11}& -\frac{7}{11}\end{array}\right]$

  $\left[\begin{array}{ccccccc}4& 5& -6& :& 1& 0& 0\\ 0& -1.75& 6.5& :& -0.75& 1& 0\\ 0& 0& 1& :& -\frac{1}{11}& \frac{6}{11}& -\frac{7}{11}\end{array}\right]R_2\to R_2-6.5R_3\left[\begin{array}{ccccccc}4& 5& -6& :& 1& 0& 0\\ 0& -1.75& 0& :& -\frac{7}{44}& -\frac{25}{11}& \frac{91}{22}\\ 0& 0& 1& :& -\frac{1}{11}& \frac{6}{11}& -\frac{7}{11}\end{array}\right]$

  $\left[\begin{array}{ccccccc}4& 5& -6& :& 1& 0& 0\\ 0& -1.75& 0& :& -\frac{7}{44}& -\frac{25}{11}& \frac{91}{22}\\ 0& 0& 1& :& -\frac{1}{11}& \frac{6}{11}& -\frac{7}{11}\end{array}\right]\frac{-1}{1.75}{R}_2\left[\begin{array}{ccccccc}4& 5& -6& :& 1& 0& 0\\ 0& 1& 0& :& \frac{1}{11}& -\frac{16}{11}& -\frac{26}{11}\\ 0& 0& 1& :& -\frac{1}{11}& \frac{6}{11}& -\frac{7}{11}\end{array}\right]$

  $\left[\begin{array}{ccccccc}4& 5& -6& :& 1& 0& 0\\ 0& 1& 0& :& \frac{1}{11}& -\frac{16}{11}& -\frac{26}{11}\\ 0& 0& 1& :& -\frac{1}{11}& \frac{6}{11}& -\frac{7}{11}\end{array}\right]R_1\to R_1-5R_2\left[\begin{array}{ccccccc}4& 0& -6& :& \frac{6}{11}& -\frac{80}{11}& \frac{130}{11}\\ 0& 1& 0& :& \frac{1}{11}& \frac{16}{11}& -\frac{26}{11}\\ 0& 0& 1& :& -\frac{1}{11}& \frac{6}{11}& -\frac{7}{11}\end{array}\right]$

  $\left[\begin{array}{ccccccc}4& 0& -6& :& \frac{6}{11}& -\frac{80}{11}& \frac{130}{11}\\ 0& 1& 0& :& \frac{1}{11}& \frac{16}{11}& -\frac{26}{11}\\ 0& 0& 1& :& -\frac{1}{11}& \frac{6}{11}& -\frac{7}{11}\end{array}\right]R_1\to R_1+6R_3\left[\begin{array}{ccccccc}4& 0& 0& :& 0& -\frac{44}{11}& 8\\ 0& 1& 0& :& \frac{1}{11}& \frac{16}{11}& -\frac{26}{11}\\ 0& 0& 1& :& -\frac{1}{11}& \frac{6}{11}& -\frac{7}{11}\end{array}\right]$

  $\left[\begin{array}{ccccccc}4& 0& 0& :& 0& -\frac{44}{11}& 8\\ 0& 1& 0& :& \frac{1}{11}& \frac{16}{11}& -\frac{26}{11}\\ 0& 0& 1& :& -\frac{1}{11}& \frac{6}{11}& -\frac{7}{11}\end{array}\right]\frac{1}{4}{R}_1\left[\begin{array}{ccccccc}1& 0& 0& :& 0& -1& 2\\ 0& 1& 0& :& \frac{1}{11}& \frac{16}{11}& -\frac{26}{11}\\ 0& 0& 1& :& -\frac{1}{11}& \frac{6}{11}& -\frac{7}{11}\end{array}\right]$

Now from the above operations

  $A^{-1}=\left[\begin{array}{ccc}0& -1& 2\\ \frac{1}{11}& \frac{16}{11}& -\frac{26}{11}\\ -\frac{1}{11}& \frac{6}{11}& -\frac{7}{11}\end{array}\right]$

Now solution of the system of equation is

  $\begin{array}{l}X=A^{-1}B\\ =\left[\begin{array}{ccc}0& -1& 2\\ \frac{1}{11}& \frac{16}{11}& -\frac{26}{11}\\ -\frac{1}{11}& \frac{6}{11}& -\frac{7}{11}\end{array}\right]\left[\begin{array}{c}-6\\ 8\\ 3\end{array}\right]\\ =\left[\begin{array}{c}-8+6\\ \begin{array}{l}\frac{-6}{11}+\frac{128}{11}-\frac{78}{11}\\ \frac{6}{11}+\frac{48}{11}-\frac{21}{11}\end{array}\end{array}\right]\\ X=\left[\begin{array}{c}-2\\ 4\\ 3\end{array}\right]\end{array}$

Hence the solution is $x=-2,y=4,z=3$

Conclusion:

The solution is $x=-2,y=4,z=3$