Q. 52 - Precalculus Enhanced with Graphing Utilities

Question

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

$\left\{\begin{array}{l}bx+3y=14\\ bx+2y=10\end{array} b\ne 0\right.$

Step-by-Step Solution

Answer

The solution of the matrix is $x=\frac{2}{b}\text{ and }y=4$.

Step 1. Given Information

A matrix is given as, 

$\left\{\begin{array}{l}bx+3y=14\\ bx+2y=10\end{array} b\ne 0\right.$

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 known as:

$A^{-1}=\left[\begin{array}{cc}-\frac{2}{b}& \frac{3}{b}\\ 1& -1\end{array}\right]$

Step 3. Solution of the matrix

We know,

$X=A^{-1}B$

Therefore,

$$\begin{gathered} X=\left[\begin{array}{cc}-\frac{2}{b}& \frac{3}{b}\\ 1& -1\end{array}\right]\left[\begin{array}{l}14\\ 10\end{array}\right] \\ =\left[\begin{array}{c}-\frac{2}{b}\left(14\right)+\frac{3}{b}\left(10\right)\\ 1\left(14\right)-1\left(10\right)\end{array}\right] \\ \left[\begin{array}{l}x\\ y\end{array}\right]=\left[\begin{array}{l}\frac{2}{b}\\ 4\end{array}\right] \end{gathered}$$