Chapter 8, Problem 28RE

To determine

values of x, y and z by Gauss Jordan elimination method.

Answer to Problem 28RE

  $x=-2,y=-1,z=1$

Explanation of Solution

Given information: A system of equations is given as follows-

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

Calculation:

The given system of equation in matrix form is as follows-

  $\left[\begin{array}{ccc}1& -3& 1\\ 3& -1& -1\\ -1& 1& -3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2\\ -6\\ -2\end{array}\right]$

Let A= $\left[\begin{array}{ccc}1& -3& 1\\ 3& -1& -1\\ -1& 1& -3\end{array}\right]$ and B= $\left[\begin{array}{c}2\\ -6\\ -2\end{array}\right]$

Its augmented matrix form is,

C= [A: B] = $\left[\begin{array}{ccccc}1& -3& 1& :& 2\\ 3& -1& -1& :& -6\\ -1& 1& -3& :& -2\end{array}\right]$

Here our main motive is to bring part A of the augmented matrix in the form $\left[\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right]$ where a, b, c are constant numbers.

  $C\sim \left[\begin{array}{ccccc}1& -3& 1& :& 2\\ 3& -1& -1& :& -6\\ -1& 1& -3& :& -2\end{array}\right]$

By applying operations $\begin{array}{l}{R}_2\to R_2-3R_1\\ R_3\to R_3+R_1\end{array}$ we get,

  $C\sim \left[\begin{array}{ccccc}1& -3& 1& :& 2\\ 0& 8& -4& :& -12\\ 0& -2& -2& :& 0\end{array}\right]$

By applying operation $R_3\to R_3+\frac{1}{4}{R}_2$ we get,

  $C\sim \left[\begin{array}{ccccc}1& -3& 1& :& 2\\ 0& 8& -4& :& -12\\ 0& 0& -3& :& -3\end{array}\right]$

By applying operations $\begin{array}{l}{R}_1\to R_1+\frac{1}{3}{R}_3\\ R_2\to R_2-\frac{4}{3}{R}_3\end{array}$ we get,

  $C\sim \left[\begin{array}{ccccc}1& -3& 0& :& 1\\ 0& 8& 0& :& -8\\ 0& 0& -3& :& -3\end{array}\right]$

By applying operation $R_1\to R_1+\frac{3}{8}{R}_2$ we get,

  $C\sim \left[\begin{array}{ccccc}1& 0& 0& :& -2\\ 0& 8& 0& :& -8\\ 0& 0& -3& :& -3\end{array}\right]$

Now given system of equations can be written as

  $\left[\begin{array}{ccc}1& 0& 0\\ 0& 8& 0\\ 0& 0& -3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-2\\ -8\\ -3\end{array}\right]$

  $\begin{array}{l}\Rightarrow x=-2\\ 8y=-8\\ -3z=-3\end{array}$

  $\Rightarrow x=-2,y=\frac{-8}{8}=-1,z=\frac{-3}{-3}=1$

And hence, $x=-2,y=-1,z=1$