How to solve a linear system when there isn't an exact solution?

Question

I've been trying to solve this linear system using Gaussian elimination, but I can't seem to finish it.

$ x +0y - z + w = 0 $

$ 0x + 2y + 2z + 2w = 2 $

$ x + y + 0z + 2w = 1 $

$$ \left[ \begin{array}{cccc|c} 1 & 0 & -1 & 1 & 0 \\ 0 & 2 & 2 & 2 & 2 \\ 1 & 1 & 0 & 2 & 1 \end{array} \right] \Rightarrow \left[ \begin{array}{cccc|c} 1 & 0 & -1 & 1 & 2 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 \end{array} \right] $$

$$ \\ -R_1\rightarrow R_3 \\ \frac{1}{2} R_2 \\ $$

And what then? I couldn't come up with a conventional solution where I just start with finding the rightmost variable and all other variables show themselves easily then. $ -R_2 \rightarrow R_3 $ doesn't end up in anything sensible. Could you please give me a hint on how to move on from here?

p.s. First time using the formatting, pardon my mistakes.

Answer 1

From

$\left[ \begin{array}{cccc|c} 1 & 0 & -1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right]$

add the free variables $z=s$, $w=t$ to get

$\left[ \begin{array}{cccc|c} 1 & 0 & -1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & s\\ 0 & 0 & 0 & 1 & t \end{array} \right]\to \left[\begin{array}{cccc|c} 1 & 0 & 0 & 1 & s \\ 0 & 1 & 0 & 1 & 1-s \\ 0 & 0 & 1 & 0 & s\\ 0 & 0 & 0 & 1 & t \end{array} \right]\to \left[\begin{array}{cccc|c} 1 & 0 & 0 & 0 & s-t \\ 0 & 1 & 0 & 0 & 1-s-t \\ 0 & 0 & 1 & 0 & s\\ 0 & 0 & 0 & 1 & t \end{array} \right]$

obtaining $$\left( \begin{array}{c} x\\y\\z\\w \end{array} \right)= \left( \begin{array}{c} 0\\1\\0\\0 \end{array} \right)+ \left(\begin{array}{r} 1\\-1\\1\\0 \end{array} \right)s+ \left(\begin{array}{r} -1\\-1\\0\\1 \end{array} \right)t$$

Answer 2

If you continue the elimination process, you get this reduced row echelon form: $$\left[ \begin{array}{cccc|c} 1 & 0 & -1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right]$$ Since there are no pivots in the $z$ or $w$ columns, this means you can choose $z$ and $w$ freely when constructing a solution. The remaining equations, e.g. $$ y + 1z+1w = 1 $$ will then allow you to find the corresponding values for $x$ and $y$ that satisfy all the equations.

Answer 3

Next you subtract row 2 from row 3 to get a row of all $0$'s. Thus you have two free variables ($z$ and $w$), whose values are arbitrary; the top two rows tell you what $x$ and $y$ are in terms of $z$ and $w$.