At the very beginning, I'm going to refer to posts dealing with a similar problem:
Writing a matrix as a product of elementary matrices.
If two linear systems have the same solutions then the corresponding matrices are row equivalent
Task:
Let$A\sim\begin{bmatrix}1&2&4&1\\0&0&1&3\\1&3&1&1\\0&0&0&0\end{bmatrix}$ where the given equivalence relation is realised by row-transformations only.
Find all the solutions of $AX=B$ if:
$(i)$ $B$ is the difference between $1^{\text{st}}\;\&\;4^{\text{th}}\;\text{column}$
$(ii)$$B$ is the difference between $1^{\text{st}}\;\&\;2^{\text{nd}}\;\text{column}$
My attempts:
Although I'm trying to solve this in advance, I would like to return to the beginning and see my non-sequuntur.
'Two matrices of the same size are row equivalent iff the corresponding homogeneous systems have the same set of solutions, or equivalently the matrices have the same $\text{null space/kernel}$. Because elementary row operations are reversible, row equivalence is an equivalence relation.'
$$\text{and}$$
Every matrix is row equivalent to a unique matrix with reduced row echelon form.
I tried to reduce the given matrix to row echelon form, but the given matrix isn't regular, so I didn't even get an upper triangular matrix (which might be a synonym for the echelon form when it comes to regular matrices.)
$A\sim\begin{bmatrix}1&2&4&1\\ 0&0&1&3\\ 1&3&1&1\\ 0&0&0&0\end{bmatrix}\sim\begin{bmatrix}1&2&4&1\\1&3&1&1\\0&0&1&3\\0&0&0&0\end{bmatrix}\sim\begin{bmatrix}\;1&\;2&\;4&\;1\\\;0&-1&-3&\;0\\\;0&\;0&\;1&\;3\\\;0&\;0&\;0&\;0\end{bmatrix}$
It seems there are infinitely many solutions because $A$ is $\text{singular}$ and there are more unknowns than equations in the system. Then the equation looks either like this (but the problem is that the systems aren't homogeneous): $$A\cdot X=\begin{bmatrix}x_1+2x_2+4x_3+x_4\\-x_2-3x_1\\x_3+3x_4&\\0 \end{bmatrix}=\begin{bmatrix}0\\0\\3\\0\end{bmatrix}$$ or this: $$A\cdot X=\begin{bmatrix}x_1+2x_2+4x_3+x_4\\-x_2-3x_1\\x_3+3x_4&\\0 \end{bmatrix}=\begin{bmatrix}\;1\\-1\\\;0\\\;0\end{bmatrix}$$ Are these $\;\begin{cases}x_2=-3x_1,\;x_3=3-3x_4\implies\;-5x_1-11x_4=-12\\ x_2=1-3x_1~;x_3=-3x_4\implies\;-5x_1-11x_4=-1\end{cases}\;$ cases for the solutions sets?
To begin with, the matrix that you’ve derived from $A$ is in fact upper-triangular. The only requirement is that all of the elements below the main diagonal be zero. Nothing says that the elements along the main diagonal can’t be zero. I’m not sure how not being regular is relevant here. Every matrix can be made upper-triangular via a series of elementary row operations.
Since one of your quote blocks mentions reduced row echelon form, though, you’re not there yet. Continuing, you would multiply the second row by $-1$ and clear elements above the pivots to obtain $$R=\begin{bmatrix}1&0&0&-29\\0&1&0&9\\0&0&1&3\\0&0&0&0\end{bmatrix}.$$ Since $R$ is a rank-three matrix, its null space is one dimensional and, referring to the only non-pivot column of $R$, is spanned by $(29,-9,-3,1)^T$. To put it another way, the system represented by $A\mathbf x=0$ reduces to $x_1=29x_4$, $x_2=-9x_4$, $x_3=-3x_4$. This clearly has one free variable. Choosing $x_4$, we see immediately that every solution has the form $x_4(29,-9,-3,1)^T$.
Moving to the inhomogeneous equation $A\mathbf x=\mathbf b$, if $\mathbf x_0$ is a solution and $\mathbf v\in\mathcal N(A)$, then $A(\mathbf x_0+\mathbf v)=A\mathbf x_0+A\mathbf v=\mathbf b+0\mathbf b$, so $\mathbf x_0+\mathbf v$ is also a solution. In fact, as you’ve no doubt learned, every solution to the inhomogeneous equation can be written in this way. So, for your problem, the solutions to both parts will be of the form $x_0+\lambda(29,-9,-3,1)^T$. The challenge is to find a particular solution $\mathbf x_0$ for each of the equations.
That’s the easy part, though. One way to interpret the product $A\mathbf x$ is that it is a linear combination of the columns of $A$ with coefficients given by the corresponding entries of $\mathbf x$. So, for instance, the difference between the first and fourth columns of $A$ can be expressed as the product $A(1,0,0,-1)^T$ (or $A(-1,0,0,1)^T$, depending on how you interpret “difference”). The equation for part (i) can thus be written as $A\mathbf x=A(1,0,0,-1)^T$, with an I hope obvious solution.
Before closing, I’ll correct one other thing that you wrote. $A\cdot X$ is not equal to the expression that follows near the end of your question. You don’t even know what $A$ is in the first place, so how would you know? Using an equal sign there is inappropriate. The system of equations represented by $A\cdot X=B$ is, however, equivalent to the stuff that follows $A\cdot X$ in the last part of your question.