# System of linear equations...

• Jun 4th 2008, 07:40 PM
amor_vincit_omnia
System of linear equations...
Hi, I tried to solve these by way of the reduced row echelon, and posted my answers. If anyone could check em', I'd appreciate it.

x+ y + 2z = -1
x- 2y + z = -5
3x+ y+ z = 3

(For this one, I got x= 1, y=2, z= -2)
Anyone agree/disagree?
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Next,

x+y+z=0
x+z=0
2x+y-2z=o
x+5y+5z=0

looks to me like x=y=z=0? But when I reduced the rows, it was showing that 1=0, so could the answer be no solution?

Thanks a lot
• Jun 4th 2008, 07:58 PM
o_O
1. You can always plug in your values of x, y, and z to see if they satisfy the three equations.

2. Every homogeneous system of equations has only the trivial solution (x,y,z) = (0,0,0) or infinitely many non-zero solutions in which the trivial solution is one of them (specifically when there are more unknowns than equations which isn't the case here). There are no other alternatives.

When you say you got 1 = 0, did you mean that in your augmented matrix you had a row that was similar to: $\displaystyle \left[ \begin{array}{ccccc} 0 & 0 & 1 & | & 0 \end{array} \right]$. This would mean z = 0, giving your trivial solution.
• Jun 4th 2008, 11:53 PM
amor_vincit_omnia
Yes, exactly. I thought it was saying 1=0. Thanks.