# Checking a solution for a linear system

Describe all solutions of the system:

$-x_1 + 2x_2 + x_3 + 4x_4 = 0$

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

I solved and got:

$\langle x_1, x_2, x_3, x_4\rangle = \langle x_3 +2x_4 + 2, x_3 +9x_4 - 1, x_3, x_4 \rangle$

$= x_3\langle 1, 1, 1, 0\rangle + x_4\langle 2, 9, 0, 1\rangle + \langle 2, -1, 0, 0\rangle$

How do I check my solution and what does my solution mean?

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$x_3\langle 1, 1, 1, 0\rangle + x_4\langle 2, 9, 0 1\rangle + \langle 2, -1, 0, 0\rangle$

First plug the values in to check, use first $\langle 2, -1, 0, 0\rangle$, which is for $x_3=0$ and $x_4=0$
Then also check with $x_3=1$ and $x_4=0$ or the input of $\langle 3, 0, 1, 0\rangle$
and similarly for $x_3=0$ and $x_4=1$ (though not strictly necessary)
This represents the entire solution space as $x_3$ and $x_4$ are the free variables. Every possible solution is within the two dimensional space defined by those parameters, start with them as any desired value, the result of the vector sum is a solution.

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This doesn't work? My solution is incorrect because I cannot add, subtract and multiply? – CodeKingPlusPlus Oct 4 '12 at 20:08
@Code: Well, clearly $\langle 2,-1,0,0\rangle$ is not a solution to the second equation (which then evaluates to $3=1$), so you must have done something wrong to arrive at that solution. – Henning Makholm Oct 4 '12 at 20:28
$\langle 2,-1,0,0\rangle$ does look incorrect... – adam W Oct 4 '12 at 20:29

Geometrically, your given equations are for two 3-dimensional spaces in $\mathbb{R}^4$. The intersection of these spaces is a plane you solved for.

Your answer,$\langle x_1, x_2, x_3, x_4\rangle = x_3\langle 1, 1, 1, 0\rangle + x_4\langle 2, 9, 0, 1\rangle + \langle 2, -1, 0, 0\rangle$, has three parts. The first two are vectors which span a plane. The last component is the vector by which the plane spanned by the first two vectors is offset from the origin.

To check your answers, plug $$\langle x_1, x_2, x_3, x_4\rangle = \langle x_3 +2x_4 + 2, x_3 +9x_4 - 1, x_3, x_4 \rangle$$ into your initial equations. The variables should drop out, and you'll be left with $0=0$ and $1=1$.

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Can I check my solution with the final form of vectors I have? – CodeKingPlusPlus Oct 4 '12 at 20:22
Also what does this mean? what is $x_3\langle 1, 1, 1, 0\rangle$ – CodeKingPlusPlus Oct 4 '12 at 20:23