Chapter 8, Problem 32RE

To determine

Find the solution of the system of linear equation using Gauss-Jordan elimination method.

Answer to Problem 32RE

The solutions of the system of linear equations are $x=\frac{31}{42},y=\frac{5}{14},\text{and}z=\frac{13}{84}$ .

Explanation of Solution

Given information:

The givensystem of linear equationsare

  $\begin{array}{l}4x+4y+4z=5\\ 4x-2y-8z=1\\ 5x+3y+8z=6\end{array}$

Calculation:

A matrix derived from a system of linear equations is the augmented matrix of the system. Moreover, the matrix derived from the coefficients of the system is the coefficient matrix of the system.

Let the system of linear equations are

  $\begin{array}{l}ax+by+cz=p\\ dx+ey+fz=q\end{array}$

The augmented matrix is

  $\left[\begin{array}{ccccc}a& b& c& :& p\\ d& e& f& :& q\end{array}\right]$

Write an augmented matrix for the given system of linear equations.

  $\left[\begin{array}{ccccc}4& 4& 4& :& 5\\ 4& -2& -8& :& 1\\ 5& 3& 8& :& 6\end{array}\right]$

Use Gauss-Jordan elimination method.

  $\begin{array}{l}{R}_1\leftrightarrow \frac{1}{4}{R}_1\\ \left[\begin{array}{ccccc}1& 1& 1& :& \frac{5}{4}\\ 4& -2& -8& :& 1\\ 5& 3& 8& :& 6\end{array}\right]\\ R_2\leftrightarrow R_2-4R_1\\ \left[\begin{array}{ccccc}1& 1& 1& :& \frac{5}{4}\\ 0& -6& -12& :& -4\\ 5& 3& 8& :& 6\end{array}\right]\\ R_3\leftrightarrow R_3-5R_1\\ \left[\begin{array}{ccccc}1& 1& 1& :& \frac{5}{4}\\ 0& -6& -12& :& -4\\ 0& -2& 3& :& -\frac{1}{4}\end{array}\right]\end{array}$

  $\begin{array}{l}{R}_2\leftrightarrow -\frac{1}{6}{R}_2\\ \left[\begin{array}{ccccc}1& 1& 1& :& \frac{5}{4}\\ 0& 1& 2& :& \frac{2}{3}\\ 0& -2& 3& :& -\frac{1}{4}\end{array}\right]\\ R_3\to R_3+2R_2\\ \left[\begin{array}{ccccc}1& 1& 1& :& \frac{5}{4}\\ 0& 1& 2& :& \frac{2}{3}\\ 0& 0& 7& :& \frac{13}{12}\end{array}\right]\\ \end{array}$

  $\begin{array}{l}{R}_3\leftrightarrow \frac{1}{7}{R}_3\\ \left[\begin{array}{ccccc}1& 1& 1& :& \frac{5}{4}\\ 0& 1& 2& :& \frac{2}{3}\\ 0& 0& 1& :& \frac{13}{84}\end{array}\right]\\ R_2\to R_2-2R_3\\ \left[\begin{array}{ccccc}1& 1& 1& :& \frac{5}{4}\\ 0& 1& 0& :& \frac{5}{14}\\ 0& 0& 1& :& \frac{13}{84}\end{array}\right]\\ R_1\to R_1-R_3\\ \left[\begin{array}{ccccc}1& 1& 0& :& \frac{23}{21}\\ 0& 1& 0& :& \frac{5}{14}\\ 0& 0& 1& :& \frac{13}{84}\end{array}\right]\\ R_1\to R_1-R_2\\ \left[\begin{array}{ccccc}1& 0& 0& :& \frac{31}{42}\\ 0& 1& 0& :& \frac{5}{14}\\ 0& 0& 1& :& \frac{13}{84}\end{array}\right]\end{array}$

Hence thesolutions of the system of linear equationsare $x=\frac{31}{42},y=\frac{5}{14},\text{and}z=\frac{13}{84}$ .