Troubleshooting a System of Linear Equations

[Exulus]

I'm hoping someone can come along and give me a tap on the head for being so silly, but I've been trying this problem *all day* and i just can't seem to get a correct answer, even when checked with other people, we're all stumped! What we have:

3x - 2y + 5z = 0
x + y + 5z = 5
x - 2y - z = -4

I've then put this into an augmented matrix, and converted to reduce row echelon form which gives me these equations:

x + z = 2
y + z = 3

So I try a general solution set as (t, 1+t, 2-t) however it doesn't seem to work with the equations at the beginning. so i looked at what i'd written and went back a step before row echelon form which has these equations:

x + y + 5z = 5
y + z = 3

Giving a solution set of (4t-10, t, 3-t) but still this doesn't seem to be consistent. Any ideas where I am going wrong? Thanks :)

 

[quasar987]

You made a mistake reducing your matrix.

You should get x = 0 , y = 5/3 z = 2/3.

 

[Exulus]

Hmm, are you sure? Because the question asks for the "general solutions" rather than any specific one. According to the way i reduced my matrix, there are infinitely many solutions. I'll have a go at typing out what I've done:

\left(\begin{array}{ccc|c}3&-2&5&0\\1&1&5&5\\1&-2&-1&-4\end{array}\right)

R1 <-> R2

\left(\begin{array}{ l c c | r } 1 &amp; 1 &amp; 5 &amp; 5 \\ 3 &amp; -2 &amp; 5 &amp; 0 \\ 1 &amp; -2 &amp; -1 &amp; -4 \end{array}\right)

R2 -> R2 - 3R1
R3 -> R3 - R1

\left(\begin{array}{ccc|c}1&amp;1&amp;5&amp;5\\0&amp;-5&amp;-10&amp;-15\\0&amp;-3&amp;-6&amp;-9\end{array}\right)

R2 -> -1/5 x R2
R3 -> -1/3 x R3

\left(\begin{array}{ccc|c}1&amp;1&amp;5&amp;5\\0&amp;1&amp;2&amp;3\\0&amp;1&amp;2&amp;3\end{array}\right)

R3 -> R3 - R2
R1 -> R1 - R2

\left(\begin{array}{ccc|c}1&amp;0&amp;3&amp;2\\0&amp;1&amp;2&amp;3\\0&amp;0&amp;0&amp;0\end{array}\right)

Sorry for the dodgy formatting..it doesn't seem to like me..hope that made sense though? If anyone knows how to stop it cutting off the top of the numbers then that would be great :)

 

[snoble]

Well there's your problem. In your original post you dropped the coefficients for z.
So you have x+3z =2, y+2z=3, z=t. So x=2-3t, y=3-2t, z=t.

 

[Exulus]

I can't believe i didnt see that I knew i must've done something silly! Thanks for pointing that stupid mistake out...i think i'll go crawl in a hole now