Chapter 7.4, Problem 65E

To determine

To solve: the system of linear equation using Gauss Jordan elimination.

Answer to Problem 65E

The solution of the system of linear equation is $\underset{\_}{x=-1}$ and $\underset{\_}{y=-4}$ .

Explanation of Solution

Given information:

System of linear equation which need to be solved using Gauss Jordan elimination is as follows

  $\begin{array}{l}-2x+6y=-22\\ x+2y=-9\end{array}$ ..

Formula used:

According to Gauss Jordan elimination represent system of linear equation in augmented matrix and using row transformations, convert the obtained matrix into identity matrix to get the solution.

Calculation:

System of linear equation is as follows

  $\begin{array}{l}-2x+6y=-22\\ x+2y=-9\end{array}$ ..

Represent the system in augmented matrix

  $\left[\begin{array}{cccc}-2& 6& :& -22\\ 1& 2& :& -9\end{array}\right]$

Apply

  $R_1\leftrightarrow R_2$

  $\left[\begin{array}{cccc}1& 2& :& -9\\ -2& 6& :& -22\end{array}\right]$

Apply

  $R_2\to R_2+2R_1$

  $\left[\begin{array}{cccc}1& 2& :& -9\\ 0& 10& :& -40\end{array}\right]$

Apply

  $R_2\to \frac{R_2}{10}$

  $\left[\begin{array}{cccc}1& 2& :& -9\\ 0& 1& :& -4\end{array}\right]$

Apply

  $R_1\to R_1-2R_2$

  $\left[\begin{array}{cccc}1& 0& :& -1\\ 0& 1& :& -4\end{array}\right]$

Now represents the matrix in system of linear equation

  $\begin{array}{l}1\times x+0\times y=-1\\ 0\times x+1\times y=-4\end{array}$

  $x=-1$

  $y=-4$

Hence the solution of the system of linear equation is $\underset{\_}{x=-1}$ and $\underset{\_}{y=-4}$ .