Chapter 4.1, Problem 38HP

To determine

To find: The solution of the given matrix.

Answer to Problem 38HP

The solution is $\left(x,y,z\right)=\left(-8,-12,-5\right)$ .

Explanation of Solution

Given:

The given matrix is given as:

  $\left[\begin{array}{cc}2x-y& 3x+4z\\ 7x-8z& 5y+12z\end{array}\right]=\left[\begin{array}{cc}9z-5x+1& 5y-2x\\ 3y-4z& 12x+2y\end{array}\right]$

The solution is given as:

  $\begin{array}{c}2x-y=9z-5x+1\\ 3x+4z=5y-2x\\ 7x-8z=3y-4z\\ 5y+12z=12x+2y\end{array}$

Rewrite the equation as:

  $\begin{array}{c}7x-y-9z=1\\ 7x-3y-4z=0\\ 5x-5y+4z=0\\ -12x+3y+12z=0\end{array}$

The solution for $x,y,z$ calculation is given as:

  $\begin{array}{c}7x-7x-y+3y-9z+4z=1\left[\text{subtract equation }\left(2\right)\text{ from equation }\left(1\right)\right]\\ 2y-5z=1\left(5\right)\\ 60y-60y-60y+15y+48z+60z=0\left[\begin{array}{l}\text{multiply equation }\left(3\right)\text{ by 12 and}\\ \text{ multiply equation }\left(4\right)\text{ by 5 and add them}\end{array}\right]\\ -45y+108z=0\left(6\right)\end{array}$

Further simplified as:

  $\begin{array}{c}90y-90y-225z+216z=45\left[\begin{array}{l}\text{multiply equation }\left(5\right)\text{ by 45 and}\\ \text{ multiply equation }\left(6\right)\text{ by 2 and add them}\end{array}\right]\\ -9z=45\\ z=-5\\ 2y-5\left(-5\right)=1\left[\text{subtitute }z=-5\text{ in equation }\left(5\right)\right]\end{array}$

Further simplified as:

  $\begin{array}{c}2y-5\left(-5\right)=1\left[\text{subtitute }z=-5\text{ in equation }\left(5\right)\right]\\ y=-12\\ 7x-\left(-12\right)-9\left(-5\right)=1\left[\text{subtitute }z=-5\text{ and }y=-12\text{ in equation }\left(1\right)\right]\\ x=-8\end{array}$

Therefore the solution is $\left(x,y,z\right)=\left(-8,-12,-5\right)$ .