Question

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

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

Short answer

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

Step-by-step solution

Step 1. Given Information

A matrix is given as

$\left\{\begin{array}{l}x+2z=2\\ -x+2y+3z=-\frac{3}{2}\\ x-y=2\end{array}\right.$

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

Step 2. Inverse of a 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,

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