Q. 60 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 41– 60, use the inverses found in Problems 31–40 to solve each system of equation 

$\left\{\begin{array}{r}3x+3y+z=1\\ x+2y+z=0\\ 2x-y+z=4\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the matrix is $x=1, y=-1\text{ and }z=1.$

Step 1. Given information

A matrix is given as, 

$\left\{\begin{array}{r}3x+3y+z=1\\ x+2y+z=0\\ 2x-y+z=4\end{array}\right.$

We need to find the solution of the matrix using inverse of the matrix. 

Step 2. Solution of the matrix

The inverse of the matrix is,

$$\begin{gathered} A^{-1}=\frac{1}{7}\left[\begin{array}{ccc}3& -4& 1\\ 1& 1& -2\\ -5& 9& 3\end{array}\right] \\ X=\frac{1}{7}\left[\begin{array}{ccc}3& -4& 1\\ 1& 1& -2\\ -5& 9& 3\end{array}\right]\left[\begin{array}{l}1\\ 0\\ 4\end{array}\right] \\ =\frac{1}{7}\left[\begin{array}{c}3\left(1\right)-4\left(0\right)+1\left(4\right)\\ 1\left(1\right)+1\left(0\right)-2\left(4\right)\\ -5\left(1\right)+9\left(0\right)+3\left(4\right)\end{array}\right] \\ \left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right] \end{gathered}$$