Chapter 11, Problem 40RE

To determine

To solve: the given system of equations using matrices.

Answer to Problem 40RE

Solution: $(\frac{1}{3},\frac{4}{5},-\frac{1}{15})$

Explanation of Solution

  $\{\begin{array}{l}x+2y-z=2\\ 2x-2y+z=-1\\ 6x+4y+3x=5\end{array}$

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

  $X=A^{-1}B$

  $A^{-1}=\frac{1}{30}\left[\begin{array}{ccc}\begin{array}{l}10\\ 0\\ -20\end{array}& \begin{array}{l}10\\ -9\\ -8\end{array}& \begin{array}{l}0\\ 3\\ 6\end{array}\end{array}\right]$

  $\begin{array}{l}\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\frac{1}{30}\left[\begin{array}{ccc}\begin{array}{l}10\\ 0\\ -20\end{array}& \begin{array}{l}10\\ -9\\ -8\end{array}& \begin{array}{l}0\\ 3\\ 6\end{array}\end{array}\right]\left[\begin{array}{l}2\\ -1\\ 5\end{array}\right]\\ \left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{l}1/3\\ 4/5\\ -1/15\end{array}\right]\end{array}$