Chapter 11, Problem 44RE

To determine

Solve each system of equations using matrices.

Answer to Problem 44RE

  $\boxed{x=-\frac{17}{15},y=-\frac{1}{5},z=\frac{22}{15}and,t=-\frac{2}{15}}$

Explanation of Solution

Given information:

Solve each system of equations using matrices. If the system has no solution, say that it is inconsistent.

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

Calculation:

The given system of equations are given by the expressions given below in $x,y,z$ and $t$ .

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

If we let $A=\left[\begin{array}{cccc}1& -3& 3& -1\\ 1& 2& -1& 0\\ 1& 0& 3& 2\\ 1& 1& 5& 0\end{array}\right]$ , $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ and $B=\left[\begin{array}{c}4\\ -3\\ 3\\ 6\end{array}\right]$ , then the original equations can be written in a compact form as follows:

  $AX=B$

In order to determine the unknown variables, we need to determine the matrix $X$ . In order to obtain that, we multiply the above equation by $A^{-1}$ and solve the equation as follows:

  $\begin{array}{l}{A}^{-1}\left(AX\right)=A^{-1}B\\ \left(A^{-1}A\right)X=A^{-1}B\\ I_{4}X=A^{-1}B\\ X=A^{-1}B\end{array}$

The inverse of the matrix $A$ can now be determined using the graphing utility. The inverse of the

Matrix is $A^{-1}=\left[\begin{array}{cccc}\frac{11}{30}& \frac{13}{20}& \frac{11}{60}& -\frac{1}{5}\\ -\frac{1}{5}& \frac{1}{10}& -\frac{1}{10}& \frac{1}{5}\\ -\frac{1}{30}& -\frac{3}{20}& -\frac{1}{60}& \frac{1}{5}\\ -\frac{2}{15}& -\frac{1}{10}& \frac{13}{30}& -\frac{1}{5}\end{array}\right]$ . The solutions to the equations can now be given by

The product of the matrix $A^{-1}$ and $B$ .

  $X=\left[\begin{array}{c}4\\ -3\\ 3\\ 6\end{array}\right]\left[\begin{array}{cccc}\frac{11}{30}& \frac{13}{20}& \frac{11}{60}& -\frac{1}{5}\\ -\frac{1}{5}& \frac{1}{10}& -\frac{1}{10}& \frac{1}{5}\\ -\frac{1}{30}& -\frac{3}{20}& -\frac{1}{60}& \frac{1}{5}\\ -\frac{2}{15}& -\frac{1}{10}& \frac{13}{30}& -\frac{1}{5}\end{array}\right]$

  $=\left[\begin{array}{l}4\times \frac{11}{30}+\left(-3\right)\times \frac{13}{20}+3\times \frac{11}{60}+6\times \left(-\frac{1}{5}\right)\\ 4\times \left(-\frac{1}{5}\right)+\left(-3\right)\times \frac{1}{10}+3\times \left(-\frac{1}{10}\right)+6\times \frac{1}{5}\\ 4\times \left(-\frac{1}{30}\right)+\left(-3\right)\times \left(-\frac{3}{20}\right)+3\times \left(-\frac{1}{60}\right)+6\times \frac{1}{5}\\ 4\times \left(-\frac{2}{15}\right)+\left(-3\right)\times \left(-\frac{1}{10}\right)+3\times \frac{13}{30}+6\times \left(-\frac{1}{5}\right)\end{array}\right]$

  $=\left[\begin{array}{l}\frac{44}{30}-\frac{39}{20}+\frac{33}{60}-\frac{6}{5}\\ -\frac{4}{5}-\frac{3}{10}-\frac{3}{10}+\frac{6}{5}\\ -\frac{4}{30}+\frac{9}{20}-\frac{3}{60}+\frac{6}{5}\\ -\frac{8}{15}+\frac{3}{10}+\frac{39}{30}-\frac{6}{5}\end{array}\right]$

  $=\left[\begin{array}{c}-\frac{17}{15}\\ -\frac{1}{5}\\ \frac{22}{15}\\ -\frac{2}{15}\end{array}\right]$

Hence, the solutions to the given equations are $x=-\frac{17}{15},y=-\frac{1}{5},z=\frac{22}{15}and,t=-\frac{2}{15}$ .