Need some help with Row Echelon

[Flying Shogi]

So I've been told by my teacher that we get an infinite # of solutions when we deal with non square systems. However, in the hw, I've encountered a problem in which the # of variables is equal to the # of equations and the correct answer has infinite # of solutions. So how do I know when I have an infinite # of solutions?

If it helps, this is the question:

x+2y-7z=-4

2x+y+z=13

3x+9y-36z=-33

Ans this is the answer: -3a+10,5a-7,a

[Mikethfc]

Check your working, if you're getting infinite solutions you've gone wrong somewhere.

[spoiler=What I did]

Reduce it to

x+2y-z

-3y+11z=21

3y-15z=-21

Add the last two rows and you'll get -4z=0 and by extension z=0 input it in you get y=-7, z=10

Also what I've done isn't in Echelon form I was just too lazy to finish reducing it.

As for the infinite solutions it will be when some of the lines are collinear eg.

x-2y=5

2x-4y=10

so the second row is just two times the first equation, so they're effectively the same which means 2 unknowns and 1 equation, solution would be {5+2y,y}

[Flying Shogi]

This is straight from the textbook. Those answers are in the back of the book. I got the same answers but I just don't understand why.

[Mikethfc]

x+2y-7z=-4

2x+y+3z=13

3x+9y-36z=-33

Ans this is the answer: -3a+10,5a-7,a

Well if you put that into the second row you get

-6a+20+5a-7+3a=13

and the LHS simplifies to 13+2a, which is only possible if a=0

Also I double-checked on Wolfram Alpha and that seems to be agreeing with me.

http://www.wolframalpha.com/input/?i=x%2B2y-7z%3D-4%2C+2x%2By%2B3z%3D13%2C+3x%2B9y-36z%3D-33

So it would seem that the answers are needlessly complicating things

[Flying Shogi]

Sorry, it seems like I've miscopied the problem. However, the answer stays the same.

[Mikethfc]

Okay you cancel out the xs in rows 2 and 3 which leaves you with

x+2y-7z=-4

-3y+15z=21

3y-15z=-21

and you can see that they're co-linear because row 3 is equal to row 2 times minus 1 so you get infinite solutions

so let z=a

re-arranging to make y the subject gives

5a-7=y

inputtint these values into row 1 and re-arranging to make x the subject gives

x=-4-2y+7z=10-3a

So the solution is

{10-3a,5a-7,a}

[Flying Shogi]

Okay you cancel out the xs in rows 2 and 3 which leaves you with

x+2y-7z=-4

-3y+15z=21

3y-15z=-21

and you can see that they're co-linear because row 3 is equal to row 2 times minus 1 so you get infinite solutions

Thank you. Especially for this. My chances of failing tomorrow's test have decreased a little.