Chapter 8, Problem 26RE

To determine

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

Answer to Problem 26RE

  $x=\frac{1}{2},y=\frac{-1}{3},z=1$

Explanation of Solution

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

  $\begin{array}{l}2x+3y+3z=3\\ 6x+6y+12z=13\\ 12x+9y-z=2\end{array}$

Calculation:

A system of given equation in matrix form is

  $\left(\begin{array}{ccc}2& 3& 3\\ 6& 6& 12\\ 12& 9& -1\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}3\\ 13\\ 2\end{array}\right)$

Its augmented matrix form is

  $C=(A:B)=\left(\begin{array}{ccccc}2& 3& 3& :& 3\\ 6& 6& 12& :& 13\\ 12& 9& -1& :& 2\end{array}\right)$

By applying operations on the above matrix our main motive is to bring two zeros in any row and one zero in the any other row.

Applying operation, $R_3\to R_3-3R_1$ we get,

  $C\sim \left(\begin{array}{ccccc}2& 3& 3& :& 3\\ 6& 6& 12& :& 13\\ 6& 0& -10& :& -7\end{array}\right)$

Applying operation, $R_2\to R_2-R_3$ we get,

  $C\sim \left(\begin{array}{ccccc}2& 3& 3& :& 3\\ 0& 6& 22& :& 20\\ 6& 0& -10& :& -7\end{array}\right)$

Applying operation, $R_1\to R_1-\frac{1}{2}{R}_2$ we get,

  $C\sim \left(\begin{array}{ccccc}2& 0& -8& :& -7\\ 0& 6& 22& :& 20\\ 6& 0& -10& :& -7\end{array}\right)$

Applying operation, $R_3\to R_3-3R_1$ we get,

  $C\sim \left(\begin{array}{ccccc}2& 0& -8& :& -7\\ 0& 6& 22& :& 20\\ 0& 0& 14& :& 14\end{array}\right)$

Now the given equations can be written as,

  $\begin{array}{l}2x-8z=-7\mathrm{...............}\text{(1)}\\ 6y+22z=20\mathrm{................}\text{(2)}\\ 14z=14\mathrm{................}\text{(3)}\end{array}$

Solving equation (3) we get,

  $\begin{array}{l}14z=14\\ \Rightarrow z=\frac{14}{14}=1\end{array}$

Putting z=1 in equation (1) we get,

  $\begin{array}{l}2x-8z=-7\\ \Rightarrow 2x-8\times 1=-7\\ \Rightarrow 2x=-7+8=1\\ \Rightarrow x=\frac{1}{2}\end{array}$

Again putting z=1 in equation (2) we get,

  $\begin{array}{l}6y+22z=20\\ \Rightarrow 6y+22\times 1=20\\ \Rightarrow 6y=20-22=-2\\ \Rightarrow y=\frac{-2}{6}=\frac{-1}{3}\end{array}$

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