Chapter 2.5, Problem 39E

To determine

To find: solution of the given system of equations using matrix equation.

Answer to Problem 39E

  $\begin{array}{l}x=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\\ y=\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\end{array}$

Explanation of Solution

Given information:

The given equations are as follows:

  $\begin{array}{l}9x+3y=1\\ 5x+y=1\end{array}$

Calculation:

Set up the matrix for the given equations as follows:

  $\begin{array}{l}\begin{array}{ccc}A& B& \text{ C}\end{array}\\ \left[\begin{array}{cc}\begin{array}{l}9\\ 5\end{array}& \begin{array}{l}3\\ \text{ 1}\end{array}\end{array}\right]\left[\begin{array}{l}x\\ y\end{array}\right]=\left[\begin{array}{l}1\\ 1\end{array}\right]\end{array}$

For $A=\left[\begin{array}{cc}\begin{array}{l}a\\ c\end{array}& \begin{array}{l}b\\ d\end{array}\end{array}\right]$ , the inverse formula is $A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}\begin{array}{l}d\\ -c\end{array}& \begin{array}{l}-b\\ a\end{array}\end{array}\right]$ .

Put the values in the above formula and evaluate the inverse for the given matrix as follows:

  $\begin{array}{c}{\left[\begin{array}{cc}\begin{array}{l}9\\ 5\end{array}& \begin{array}{l}3\\ \text{ 1}\end{array}\end{array}\right]}^{-1}=\frac{1}{9-15}\left[\begin{array}{cc}\begin{array}{l}1\\ -5\end{array}& \begin{array}{l}-3\\ 9\end{array}\end{array}\right]\\ =\frac{1}{-6}\left[\begin{array}{cc}\begin{array}{l}1\\ -5\end{array}& \begin{array}{l}-3\\ 9\end{array}\end{array}\right]\end{array}$

Multiply both the sides of the matrix by the inverse as follows:

  $\frac{1}{-6}\left[\begin{array}{cc}\begin{array}{l}1\\ -5\end{array}& \begin{array}{l}-3\\ 9\end{array}\end{array}\right]\left[\begin{array}{cc}\begin{array}{l}9\\ 5\end{array}& \begin{array}{l}3\\ \text{ 1}\end{array}\end{array}\right]\left[\begin{array}{l}x\\ y\end{array}\right]=\frac{1}{-6}\left[\begin{array}{cc}\begin{array}{l}1\\ -5\end{array}& \begin{array}{l}-3\\ 9\end{array}\end{array}\right]\left[\begin{array}{l}1\\ 1\end{array}\right]$

Now solve each side to get,

  $\begin{array}{l}\left[\begin{array}{l}x\\ y\end{array}\right]=\frac{1}{-6}\left[\begin{array}{l}1-3\\ -5+9\end{array}\right]\\ =\frac{1}{-6}\left[\begin{array}{c}-2\\ 4\end{array}\right]\\ =\left[\begin{array}{l}1/3\\ -2/3\end{array}\right]\end{array}$

Thus, the value of x and y is $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ and $\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ respectively.