Q. 54 - Precalculus Enhanced with Graphing Utilities

Question

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

$\left\{\begin{array}{rr}x+2z=& 6\\ -x+2y+3z=& -5\\ x-y=& 6\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the matrix is $x=4,y=-2\text{ and }z=1.$

Step 1. Given information

A matrix is given as, 

$\left\{\begin{array}{rr}x+2z=& 6\\ -x+2y+3z=& -5\\ x-y=& 6\end{array}\right.$

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

Step 2. Inverse of the matrix.

The inverse of the matrix is

$A^{-1}=\left[\begin{array}{ccc}3& -2& -4\\ 3& -2& -5\\ -1& 1& 2\end{array}\right]$

Step 3. Solution of the matrix

We know,

$X=A^{-1}B$

Therefore,

$$\begin{gathered} X=\left[\begin{array}{ccc}3& -2& -4\\ 3& -2& -5\\ -1& 1& 2\end{array}\right]\left[\begin{array}{c}6\\ -5\\ 6\end{array}\right] \\ =\left[\begin{array}{c}3\left(6\right)-2(-5)-4\left(6\right)\\ 3\left(6\right)-2(-5)-5\left(6\right)\\ -1\left(6\right)+1(-5)+2\left(6\right)\end{array}\right] \\ \left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}4\\ -2\\ 1\end{array}\right] \end{gathered}$$