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I am new to linear algebra and I am trying to solve a system of three equations with five unknowns. The system I have is the following:

$$ x_1+ x_2+ x_3+ x_4+ x_5=1 \\ -x_1+ x_2+ x_3 - 2x_4 - 2x_5=1 \\ 2x_1+ 2x_2 - x_3 -x_4+ x_5=1 $$

So what I did was set up the augmented matrix like this:

    1  1  1  1  1  1
   -1  1  1 -2 -2  1
    2  2 -1 -1  1  1

Then I try to obtain an identity matrix on the left side and end up with the following:

    1  0  0  3/2  3/2  0
    0  1  0 -3/2 -5/6  2/3
    0  0  1  1    1/3  1/3

So I think the answer is $x_1 = 0, x_2 = 2/3$ and $x_3 = 1/3$

But when I look in my answer sheet it reads:

$(x_1, x_2, x_3, x_4, x_5) = (0, 2/3, 1/3, 0, 0) + s(−3/2, 3/2, −1, 1, 0) + t(−3/2, 5/6, −1/3, 0, 1)$

I have no idea how to interpret this. My $x_1,x_2,x_3$ seems to match the first three in the first five-tuple but what are the other two five-tuples? Can someone explain what I am missing here? I would highly appreciate it.

  • 1
    is your profile name supposed to be funny? – 2017-02-07
  • 0
    More unknowns than equations almost never result in one unique solution, but a set of solutions. The $s$ and $t$ are **parameters** helping to describe these sets. The solution you found corresponds to coordinates $(0,0)$ in the parameter space $(s,t)$. You can try setting some other $s$ or $t$ and see if they work. – 2017-02-07

2 Answers 2

0

Your suggested solution assumes $x_4$ and $x_5$ are zero. The given solution acknowledges that there will be two remaining unknowns and set those to be $s=x_4$, $t=x_5$ (as you can see from the last two elements of the tuples they multiply) and solves for those.

[You can think of this as two more equations:
$0x_1+0x_2+0x_3+1x_4+0x_5=s$
$0x_1+0x_2+0x_3+0x_4+1x_5=t$
]

0

Rewrite the system as

 x1 +  x2 + x3 = 1 -  x4 -  x5
-x1 +  x2 + x3 = 1 + 2x4 + 2x5
2x1 + 2x2 - x3 = 1 +  x4 -  x5

Then for any choice of x4 and x5 (possibly renamed s and t) you can evaluate the RHSs and find a solution in x1, x2, x3.