Chapter 8, Problem 27RE

To determine

Find the solution of the system of linear equation using Gaussian elimination with back-substitution method.

Answer to Problem 27RE

The solutions of the system are $x=1,y=0,z=4\text{and}w=3$ .

Explanation of Solution

Given information:

The givensystem of linear equationsare

  $\begin{array}{l}2x+y+z=6\\ -2y+3z-w=9\\ 3x+3y-2z-2w=-11\\ x+z+3w=14\end{array}$

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.

  $\left[\begin{array}{cccccc}2& 1& 1& 0& :& 6\\ 0& -2& 3& -1& :& 9\\ 3& 3& -2& -2& :& -11\\ 1& 0& 1& 3& :& 14\end{array}\right]$

Use Gaussian elimination method.

  $\begin{array}{l}{R}_1\leftrightarrow R_4\\ \left[\begin{array}{cccccc}1& 0& 1& 3& :& 14\\ 0& -2& 3& -1& :& 9\\ 3& 3& -2& -2& :& -11\\ 2& 1& 1& 0& :& 6\end{array}\right]\\ R_3\leftrightarrow R_3-3R_1\\ \left[\begin{array}{cccccc}1& 0& 1& 3& :& 14\\ 0& -2& 3& -1& :& 9\\ 0& 3& -5& -11& :& -53\\ 2& 1& 1& 0& :& 6\end{array}\right]\\ R_4\leftrightarrow R_4-2R_1\\ \left[\begin{array}{cccccc}1& 0& 1& 3& :& 14\\ 0& -2& 3& -1& :& 9\\ 0& 3& -5& -11& :& -53\\ 0& 1& -1& -6& :& -22\end{array}\right]\\ R_2\leftrightarrow R_4\\ \left[\begin{array}{cccccc}1& 0& 1& 3& :& 14\\ 0& 1& -1& -6& :& -22\\ 0& 3& -5& -11& :& -53\\ 0& -2& 3& -1& :& 9\end{array}\right]\end{array}$

  $\begin{array}{l}{R}_3\leftrightarrow R_3-3R_2\\ \left[\begin{array}{cccccc}1& 0& 1& 3& :& 14\\ 0& 1& -1& -6& :& -22\\ 0& 0& -2& 7& :& 13\\ 0& -2& 3& -1& :& 9\end{array}\right]\\ R_4\leftrightarrow R_4+2R_2\\ \left[\begin{array}{cccccc}1& 0& 1& 3& :& 14\\ 0& 1& -1& -6& :& -22\\ 0& 0& -2& 7& :& 13\\ 0& 0& 1& -13& :& -35\end{array}\right]\\ R_3\leftrightarrow R_4\\ \left[\begin{array}{cccccc}1& 0& 1& 3& :& 14\\ 0& 1& -1& -6& :& -22\\ 0& 0& 1& -13& :& -35\\ 0& 0& -2& 7& :& 13\end{array}\right]\end{array}$

  $\begin{array}{l}{R}_4\leftrightarrow R_4+2R_3\\ \left[\begin{array}{cccccc}1& 0& 1& 3& :& 14\\ 0& 1& -1& -6& :& -22\\ 0& 0& 1& -13& :& -35\\ 0& 0& 0& -19& :& -57\end{array}\right]\\ -\frac{1}{19}{R}_4\leftrightarrow R_4\\ \left[\begin{array}{cccccc}1& 0& 1& 3& :& 14\\ 0& 1& -1& -6& :& -22\\ 0& 0& 1& -13& :& -35\\ 0& 0& 0& 1& :& 3\end{array}\right]\end{array}$

Now the system of linear equations are

  $\begin{array}{l}x+z+3w=14---\left(i\right)\\ y-z-6w=-22---\left(ii\right)\\ z-13w=-35---\left(iii\right)\\ w=3---\left(iv\right)\end{array}$

Substitute the value of $w=3$ in equation $\left(iii\right)$ .

  $\begin{array}{l}z-13w=-35\\ \Rightarrow z-13\cdot 3=-35\\ \Rightarrow z-39=-35\\ \Rightarrow z-39+39=-35+39\\ \Rightarrow z=4\end{array}$

Substitute the value of $z=4$ and $w=3$ in equation $\left(i\right)$ and $\left(ii\right)$ .

  $\begin{array}{l}x+z+3w=14\\ \Rightarrow x+4+3\cdot 3=14\\ \Rightarrow x+13=14\\ \Rightarrow x+13-13=14-13\\ \Rightarrow x=1\end{array}$

  $\begin{array}{l}y-z-6w=-22\\ \Rightarrow y-4-6\cdot 3=-22\\ \Rightarrow y-22=-22\\ \Rightarrow y-22+22=-22+22\\ \Rightarrow y=0\end{array}$

Hence the solutions of the system are $x=1,y=0,z=4\text{and}w=3$ .