Chapter 7.4, Problem 69E

To determine

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

Answer to Problem 69E

The system of linear equation have an infinite number of solution of the form

  $\underset{\_}{x=\frac{3}{2}-\frac{3}{2}a}$ , $\underset{\_}{y=\frac{1}{3}+\frac{1}{3}a}$ and $\underset{\_}{z=a}$ where $a$ is any real number.

Explanation of Solution

Given information:

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

  $\begin{array}{l}2x+3z=3\\ 4x-3y+7z=5\\ 8x-9y+15z=9\end{array}$ ..

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

  $\begin{array}{l}2x+3z=3\\ 4x-3y+7z=5\\ 8x-9y+15z=9\end{array}$ ..

Represent the system in augmented matrix

  $\left[\begin{array}{ccccc}2& 0& 3& :& 3\\ 4& -3& 7& :& 5\\ 8& -9& 15& :& 9\end{array}\right]$

Apply $R_1\to \frac{R_1}{2}$

  $\left[\begin{array}{ccccc}1& 0& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& :& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 4& -3& 7& :& 5\\ 8& -9& 15& :& 9\end{array}\right]$

Apply $R_2\to R_2-4R_1$

  $R_3\to R_3-8R_1$

  $\left[\begin{array}{ccccc}1& 0& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& :& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 0& -3& 1& :& -1\\ 0& -9& 3& :& -3\end{array}\right]$

Apply $R_3\to R_3-3R_2$

  $\left[\begin{array}{ccccc}1& 0& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& :& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 0& -3& 1& :& -1\\ 0& 0& 0& :& 0\end{array}\right]$

Now represents the matrix in system of linear equation

  $\begin{array}{l}1\times x+0\times y+\frac{3}{2}\times z=\frac{3}{2}\\ 0\times x-3\times y+1\times z=-1\\ 0\times x+0\times y+0\times z=0\end{array}$

  $\begin{array}{l}x+\frac{3}{2}\times z=\frac{3}{2}\\ -3y+z=-1\end{array}$

Solving $x$ and $y$ in term of $z$ ,

  $\begin{array}{l}x=\frac{3}{2}-\frac{3}{2}z\\ y=\frac{1}{3}+\frac{1}{3}z\end{array}$

Let $z=a$ where $a$ is any real number.

  $x=\frac{3}{2}-\frac{3}{2}a$

  $y=\frac{1}{3}+\frac{1}{3}a$

Solution set has the form $\left(\frac{3}{2}-\frac{3}{2}a,\frac{1}{3}+\frac{1}{3}a,a\right)$ which represents an infinite number of solutions.

Hence the system of linear equation have an infinite number of solution of the form $\underset{\_}{x=\frac{3}{2}-\frac{3}{2}a}$ , $\underset{\_}{y=\frac{1}{3}+\frac{1}{3}a}$ and $\underset{\_}{z=a}$ where $a$ is any real number.