Chapter 7.4, Problem 66E

To determine

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

Answer to Problem 66E

The solution of the system of linear equation is $\underset{\_}{x=-2}$ and $\underset{\_}{y=-1}$ .

Explanation of Solution

Given information:

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

  $\begin{array}{l}5x-5y=-5\\ -2x-3y=7\end{array}$ ..

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

  $\begin{array}{l}5x-5y=-5\\ -2x-3y=7\end{array}$ ..

Represent the system in augmented matrix

  $\left[\begin{array}{cccc}5& -5& :& -5\\ -2& -3& :& 7\end{array}\right]$

Apply $R_1\to \frac{R_1}{5}$

  $\left[\begin{array}{cccc}1& -1& :& -1\\ -2& -3& :& 7\end{array}\right]$

Apply $R_2\to R_2+2R_1$

  $\left[\begin{array}{cccc}1& -1& :& -1\\ 0& -5& :& 5\end{array}\right]$

Apply $R_2\to \frac{R_2}{-5}$

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

Apply $R_1\to R_1+R_2$

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

Now represents the matrix in system of linear equation

  $\begin{array}{l}1\times x+0\times y=-2\\ 0\times x+1\times y=-1\end{array}$

  $x=-2$

  $y=-1$

Hence the solution of the system of linear equation is $\underset{\_}{x=-2}$ and $\underset{\_}{y=-1}$ .