Chapter 7.3, Problem 8E

Solution

To identify: The system of equations by using Gaussian elimination.

  $\begin{array}{c}\frac{1}{2}x-y+z-w=1\\ -x+y+z+2w=-3\\ x-z=2\\ y+w=0\end{array}$

The value of $x,y,z,w$ is $2,1,0,-1$ respectively.

Given information:

The given is.

  $\begin{array}{c}\frac{1}{2}x-y+z-w=1\\ -x+y+z+2w=-3\\ x-z=2\\ y+w=0\end{array}$

Explanation:

Consider the given information.

Add the first equation to the second equation, replacing the second equation with result:

  $\begin{array}{l}\underset{¯}{\begin{array}{l}\frac{1}{2}x-y+z-w=1\\ -x+y+z+2w=-3\end{array}}\\ -\frac{1}{2}x+2z+w=-2\end{array}$

Add the first equation to the fourth equation, replacing the fourth equation with result:

  $\begin{array}{l}\underset{¯}{\begin{array}{l}\frac{1}{2}x-y+z-w=1\\ y+w=0\end{array}}\\ \frac{1}{2}x+z=1\end{array}$

Add the third equation to the fourth equation, replacing the fourth equation with result:

  $\begin{array}{l}\underset{¯}{\begin{array}{l}x-z=2\\ \frac{1}{2}x+z=1\end{array}}\\ \frac{3}{2}x=3\end{array}$

So, $x=2$

Multiply the fourth equation by $-1$ and add the result to the third equation, replacing the third equation.

  $\begin{array}{l}\underset{¯}{\begin{array}{l}x-z=2\\ -x=-2\end{array}}\\ z=0\end{array}$

So, $z=0$

Substitute the value for $x\text{and}z$ into the second equation to solve for $w$

  $\begin{array}{c}-\frac{1}{2}\left(2\right)+2\left(0\right)+w=-2\\ -1+w=-2\\ w=-1\end{array}$

Substitute the value for $x,z\text{and}w$ into the first equation to solve for $y$ :

  $\begin{array}{c}-\frac{1}{2}\left(2\right)-y\left(0\right)-\left(-1\right)=1\\ 1-y+1=1\\ y=1\end{array}$

So, the value of $x,y,z,w$ is $2,1,0,-1$ respectively.