Q. 51 - Precalculus Enhanced with Graphing Utilities

Question

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

$\left\{\begin{array}{r}2x+y=\frac{7}{a}\\ ax+ay=5\end{array} a\ne 0\right.$

Step-by-Step Solution

Answer

The solution of the matrix is $x=\frac{2}{a}\text{ and }y=\frac{3}{a}$.

Step 1. Given information

A matrix is given as,

$\left\{\begin{array}{l}2x+y=\frac{7}{a}\\ ax+ay=5\end{array} a\ne 0\right.$

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 can be found by :

$$\begin{gathered} \left[A\mid I_2\right]=\left[\begin{array}{llll}2& 1& 1& 0\\ a& a& 0& 1\end{array}\right] \\ {R}_1=\frac{1}{2}{r}_1\text{ and then }{R}_2=r_2-a{r}_1 \\ \left[\begin{array}{llll}2& 1& 1& 0\\ a& a& 0& 1\end{array}\right]\to \left[\begin{array}{llll}1& \frac{1}{2}& \frac{1}{2}& 0\\ a& a& 0& 1\end{array}\right]\to \left[\begin{array}{cccc}1& \frac{1}{2}& \frac{1}{2}& 0\\ 0& \frac{a}{2}& -\frac{a}{2}& 1\end{array}\right] \\ {R}_2=\frac{2}{a}{r}_2\text{ and }{R}_1=r_1-\frac{1}{2}{r}_2\text{. } \\ \left[\begin{array}{cccc}1& \frac{1}{2}& \frac{1}{2}& 0\\ 0& \frac{a}{2}& -\frac{a}{2}& 1\end{array}\right]\to \left[\begin{array}{cccc}1& \frac{1}{2}& \frac{1}{2}& 0\\ 0& 1& -1& \frac{2}{a}\end{array}\right]\to \left[\begin{array}{cccc}1& 0& 1& -\frac{1}{a}\\ 0& 1& -1& \frac{2}{a}\end{array}\right] \\ Therefore, \\ {A}^{-1}=\left[\begin{array}{cc}1& -\frac{1}{a}\\ -1& \frac{2}{a}\end{array}\right] \end{gathered}$$

Step 3. Solution of the matrix

We know,

$X=A^{-1}B$

Therefore,

$$\begin{gathered} \left[\begin{array}{l}x\\ y\end{array}\right]=\left[\begin{array}{cc}1& -\frac{1}{a}\\ -1& \frac{2}{a}\end{array}\right]·\left[\begin{array}{l}\frac{7}{a}\\ 5\end{array}\right] \\ =\left[\begin{array}{c}1·\frac{7}{a}+-\frac{1}{a}·5\\ -1·\frac{7}{a}+\frac{2}{a}·5\end{array}\right] \\ =\left[\begin{array}{c}\frac{7}{a}-\frac{5}{a}\\ -\frac{7}{a}+\frac{10}{a}\end{array}\right] \\ =\left[\begin{array}{l}\frac{2}{a}\\ \frac{3}{a}\end{array}\right] \\ Therefore, \\ x=\frac{2}{a}\text{ and }y=\frac{3}{a} \end{gathered}$$