Chapter 8, Problem 27RE

To determine

The values of x, y, zby Gauss Jordan elimination method.

Answer to Problem 27RE

  $x=1,y=2,z=2$

Explanation of Solution

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

  $\begin{array}{l}x+2y-z=3\\ x-y-z=-3\\ 2x+y+3z=10\end{array}$

Calculation:

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

  $\left[\begin{array}{ccc}1& 2& -1\\ 1& -1& -1\\ 2& 1& 3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}3\\ -3\\ 10\end{array}\right]$

Let A= $\left[\begin{array}{ccc}1& 2& -1\\ 1& -1& -1\\ 2& 1& 3\end{array}\right]$ and B= $\left[\begin{array}{c}3\\ -3\\ 10\end{array}\right]$

Its augmented matrix form is,

C= [A: B] = $\left[\begin{array}{ccccc}1& 2& -1& :& 3\\ 1& -1& -1& :& -3\\ 2& 1& 3& :& 10\end{array}\right]$

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.

  $C\sim \left[\begin{array}{ccccc}1& 2& -1& :& 3\\ 1& -1& -1& :& -3\\ 2& 1& 3& :& 10\end{array}\right]$

By applying operations $\begin{array}{l}{R}_2\to R_2-R_1\\ R_3\to R_3-2R_1\end{array}$ we get,

  $C\sim \left[\begin{array}{ccccc}1& 2& -1& :& 3\\ 0& -3& 0& :& -6\\ 0& -3& 5& :& 4\end{array}\right]$

By applying operation $R_3\to R_3-R_2$ we get,

  $C\sim \left[\begin{array}{ccccc}1& 2& -1& :& 3\\ 0& -3& 0& :& -6\\ 0& 0& 5& :& 10\end{array}\right]$

By applying operation $R_1\to R_1+\frac{2}{3}{R}_2+\frac{1}{3}{R}_3$ we get,

  $C\sim \left[\begin{array}{ccccc}1& 0& 0& :& 1\\ 0& -3& 0& :& -6\\ 0& 0& 5& :& 10\end{array}\right]$

Now given system of equations can be written as

  $\left[\begin{array}{ccc}1& 0& 0\\ 0& -3& 0\\ 0& 0& 5\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ -6\\ 10\end{array}\right]$

  $\begin{array}{l}\Rightarrow x=1\\ -3y=-6\\ 5z=10\end{array}$

  $\Rightarrow x=1,y=\frac{-6}{-3}=2,z=\frac{10}{5}=2$

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