Chapter 8, Problem 25RE

To determine

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

Answer to Problem 25RE

  $x=5,y=2,z=-6$

Explanation of Solution

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

  $\begin{array}{l}4x+3y+z=10\\ 2x-3y-3z=22\\ 4x-2y+3z=-2\end{array}$

Calculation:

The system of given equation in matrix form is

  $\left(\begin{array}{ccc}4& 3& 1\\ 2& -3& -3\\ 4& -2& 3\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}10\\ 22\\ -2\end{array}\right)$

Its augmented matrix form is

  $C=\left[A:B\right]=\left(\begin{array}{ccccc}4& 3& 1& :& 10\\ 2& -3& -3& :& 22\\ 4& -2& 3& :& -2\end{array}\right)$

Applying operations, $R_3\to R_{{}_3}+R_2$ we get

  $C\sim \left(\begin{array}{ccccc}2& 3& 1& :& 10\\ 2& -3& -3& :& 22\\ 6& -5& 0& :& 20\end{array}\right)$

Applying operations, $R_2\to R_2-R_1$ we get

  $C\sim \left(\begin{array}{ccccc}2& 3& 1& :& 10\\ 0& -6& -4& :& 12\\ 6& -5& 0& :& 20\end{array}\right)$

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

  $C\sim \left(\begin{array}{ccccc}2& 0& -1& :& 16\\ 0& -6& -4& :& 12\\ 6& -5& 0& :& 20\end{array}\right)$

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

  $C\sim \left(\begin{array}{ccccc}2& 0& -1& :& 16\\ 0& -6& -4& :& 12\\ 0& -5& 3& :& -28\end{array}\right)$

Applying operation, $R_3\to R_3+\frac{3}{4}{R}_2$ we get

  $C\sim \left(\begin{array}{ccccc}2& 0& -1& :& 16\\ 0& -6& -4& :& 12\\ 0& \frac{-19}{2}& 0& :& -19\end{array}\right)$

The given equation can now be expressed as

  $\begin{array}{l}2x-z=16\mathrm{............}\text{(1)}\\ -6y-4z=12\mathrm{.............}\text{(2)}\\ \frac{-19}{2}y=-19\mathrm{..............}\text{(3)}\end{array}$

Solving equation (3) we get,

  $\text{ y=}\frac{-19\times 2}{-19}=2$

Putting y=2 in equation (2) we get,

  $\begin{array}{l}-6y-4z=12\\ \Rightarrow -6\times 2-4z=12\\ \Rightarrow -12-4z=12\\ \Rightarrow -4z=12+12=24\\ \Rightarrow z=\frac{24}{-4}=-6\end{array}$

Now putting z=-6 in equation (1) we get,

  $\begin{array}{l}2x-z=16\\ \Rightarrow 2x-(-6)=16\\ \Rightarrow 2x+6=16\\ \Rightarrow 2x=16-6=10\\ \Rightarrow x=\frac{10}{2}=5\end{array}$

And hence, $x=5,y=2,z=-6$