# The Solution Sets for linear equations in three variables

1. Aug 14, 2008

### takercena

1. The problem statement, all variables and given/known data
This is from Further Pure Mathematics by L. Bonstock S. Chandler and C. Rourke (question 11 pg 113). The question is find the set of points which are mapped to the point (1, -1, -1) by the transformation
$$\left( \begin{array}{ccc} 1 & -2 & 4 \\ 3 & 4 & 6 \\ 1 & 3 & 1 \\ \end{array} \right) \right. \left( \begin{array}{c} x \\ y \\ z \\ \end{array} \right) = \right. \left( \begin{array}{c} X \\ Y \\ Z \\ \end{array} \right)$$

2. The attempt at a solution
I found that the equation have infinite solution since the matrix will give 0 in column 1 by using reduction method. And that's the problem because I still don't understand how to solve the infinite solution problem. I try to compare from the example from this book,

Still, no success.

2. Aug 14, 2008

### tiny-tim

Hi takercena!

Nicely set-out problem …

I assume you've got to 5y = 3z - 2.

Now eliminate z from (say) the second line.

That gives you another equation for y.

3. Aug 14, 2008

### Defennder

Well you should post the reduced row echelon form of the matrix then we can help you more explicitly. As it is, all I can tell you is to first reduce it to the RREF form, then let one of the variables x,y,z be a parameter and try to express the rest in terms of that.

4. Aug 14, 2008

### takercena

$$\left( \begin{array}{ccc} 0 & -5 & 3 \\ 3 & 14 & 0 \\ 0 & 5 & -3 \\ \end{array} \left| \begin{array}{c} 2 & -5 & -2 \\ \end{array} \right)$$

By the way, I already get it. The answer is y/3 = (z-2)/3 = (x+(5/3))/-14. Thanks both of you

Last edited: Aug 14, 2008