Chapter 2.5, Problem 41E

To determine

To find: solution of the given matrix equation.

Answer to Problem 41E

  $x=\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$9$}\right.\text{ , }y=\raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\text{ and }z=\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

Explanation of Solution

Given information: The given matrix equations is shown below.

  $\left[\begin{array}{ccc}\begin{array}{l}-6\\ 9\\ 3\end{array}& \begin{array}{l}5\\ -2\\ 1\end{array}& \begin{array}{l}3\\ -1\\ 1\end{array}\end{array}\right]\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{l}-9\\ 5\\ -1\end{array}\right]$

The inverse coefficient matrix is as follows:

  $A^{-1}=-\frac{1}{9}\left[\begin{array}{ccc}\begin{array}{l}-1\\ -12\\ 15\end{array}& \begin{array}{l}-2\\ -15\\ 21\end{array}& \begin{array}{l}1\\ 21\\ -33\end{array}\end{array}\right]$

Calculation:

Multiply equation from left with $A^{-1}$ . Since $A\cdot A^{-1}=I$ , so only the variable matrix will remain on the left side.

  $\begin{array}{c}\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=-\frac{1}{9}\left[\begin{array}{ccc}\begin{array}{l}-1\\ -12\\ 15\end{array}& \begin{array}{l}-2\\ -15\\ 21\end{array}& \begin{array}{l}1\\ 21\\ -33\end{array}\end{array}\right]\cdot \left[\begin{array}{l}-9\\ 5\\ -1\end{array}\right]\\ =-\frac{1}{9}\left[\begin{array}{l}9-10-1\\ 108-75-21\\ -135+105+33\end{array}\right]\\ =-\frac{1}{9}\left[\begin{array}{l}-2\\ 12\\ 3\end{array}\right]\\ =\left[\begin{array}{l}2/9\\ -4/3\\ -1/3\end{array}\right]\end{array}$

Thus, the value of x, y and z is $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$9$}\right.$ , $\raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ and $\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ respectively.