# Strange results when using Gaussian Elimination on systems that have an infinite number of solutions.

I am learning Gaussian Elimination in linear algebra class and just worked through the problems at the end of the chapter. For each elimination problem with systems that have infinite solutions, I ended up with an answer that was wrong but sort of correct. For each problem, I arrived at my answer by a sequence of steps that hinged on reducing one of the equations to $ax_i = 0$ and then solving the other two equations from which the $x_i$ term had just been eliminated.

For example

$$\begin{matrix} 2x +y +4z = 4 \\ 4x +0 +2z = 10\\ -2x +3y -13z = -8\\ \end{matrix}$$

$$\begin{matrix} 2x +y +4z = 4\\ 4x +0 +2z = 10\\ 0 +4y -9z = -4\\ \end{matrix}$$

$$\begin{matrix} 2x +y +4z = 4\\ 0 -2y -6z = 2\\ 0 +4y -9z = -4\\ \end{matrix}$$

$$\begin{matrix} 2x +y +4z = 4\\ 0 -2y -6z = 2\\ 0 +0 -21z = 0\\ \end{matrix}$$

$$\begin{matrix} 2x -1 = 4\\ y = -1\\ z = 0\\ \end{matrix}$$

$$x =5/2, y =-1, z=0$$

The correct answer is

$$x = 5/2 - t/2, y = 4t -1, z=t$$ which would have occurred in my attempt had I found that $x=t$. This same error comes up every time I do these problems, so it occurs to me that, whenever I find $x_i = 0$, maybe I can just change that to $x_i = t$ and then obtain the correct answer.

So, where am I going wrong and why am I getting close to the correct answer in spite of that. Also, could I really just modify this method and reliably get the correct answer?

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## 1 Answer

Assuming you have copied the problem correctly, your answer is correct. Try substituting in the so-called correct answer and see if it really works.

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AGh! you're right. But the others weren't mis-copied, so I still have to figure out where I went wrong with them. thanks. –  Matt Munson Sep 9 '12 at 0:54