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Let

$$A \sim \begin{pmatrix} 1 &2 &4&1\\ 0&0&1&2\\ 1&3&1&1\\ 0&0&0&0\\ \end{pmatrix}$$

such that the equivalence is achieved by elementary row transformations. Find all the solutions of the system $AX=B$ if $B$ is the difference between the first and the fourth column of $A$.

I hope I understood this: $$B=\begin{pmatrix} 0\\ -2\\ 0\\ 0 \end{pmatrix}$$

But I don't know what should I do here since I don't know what $A$ is exactly...I know $X$ should be a $4$x$1$ matrix and that's about it.

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    If you dont know $A$ then as information about $B$ is based on $A$ , then how do you know $B$?that is $B = $ difference of first and fourth column of $A$ and not that of row equivalent matrix – 2017-01-29
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    @BAYMAX Oops, you're right. Ok, I got it from Kanwaljit's answer. – 2017-01-29

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Hint -

You have A and B.

Solve AX = B.

$$\begin{pmatrix} 1 &2 &4&1\\ 0&0&1&2\\ 1&3&1&1\\ 0&0&0&0\\ \end{pmatrix} \cdot \begin{pmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{pmatrix} = \begin{pmatrix} 0\\ -2\\ 0\\ 0 \end{pmatrix}$$

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    See in my edit. – 2017-01-29
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    $X=\begin{pmatrix} 20+19x_4\\ -6-6x_4\\ -2-2x_4\\ x_4\end{pmatrix}$ Is this right? – 2017-01-29
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    Either show me complete steps. Or cross check with following equations to find values. $x_1+2x_2+4x_3+x_4=0, x_3+x_4=-2, x_1+3x_2+x_3+x_4=0$ – 2017-01-29
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    @Lewis Yes it is right. Another way to describe it would be the homogenous solutionspace, if that rings bells, here is how it'd be noted: $\begin{pmatrix} 20\\-6\\-2\\0\end{pmatrix}+ lin \begin{pmatrix}\begin{pmatrix} -19\\6\\22\\0\end{pmatrix} \end{pmatrix}$ – 2017-01-29
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    @WhatAMesh Thanks, never heard of homogenous solutionspace though. – 2017-01-29