The Solution Sets for linear equations in three variables

[takercena]

Homework Statement

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
<br /> \left(<br /> \begin{array}{ccc} <br /> 1 &amp; -2 &amp; 4 \\<br /> 3 &amp; 4 &amp; 6 \\<br /> 1 &amp; 3 &amp; 1 \\<br /> \end{array} <br /> \right) <br /> <br /> \right.<br /> <br /> \left(<br /> \begin{array}{c} <br /> x \\<br /> y \\<br /> z \\<br /> \end{array} <br /> \right) <br /> <br /> =<br /> <br /> \right.<br /> <br /> \left(<br /> \begin{array}{c} <br /> X \\<br /> Y \\<br /> Z \\<br /> \end{array} <br /> \right) <br /> <br />

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,

x - y + z = 4
2x + y - 2z = 1
5x - 2y + z = 13

eq 1 and eq 2 give z = 3x - 5
eq 2 and eq 3 also give z = 3x - 5

and by substitution into the first equation
z = (1/4)(3y + 7)
Hence 3x - 5 = (3y + 7)/4 = z

Still, no success.

 

[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.

 

[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.

 

[takercena]

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.

<br /> \left( <br /> \begin{array}{ccc} <br /> 0 &amp; -5 &amp; 3 \\<br /> 3 &amp; 14 &amp; 0 \\<br /> 0 &amp; 5 &amp; -3 \\ <br /> \end{array} <br /> \left|<br /> \begin{array}{c}<br /> 2 &amp; -5 &amp; -2 \\<br /> \end{array}<br /> \right)<br />

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