Trick to finding general solution of Ax=b?
In my online linear algebra homework, we had a question to find the general solution (a solution as the span of 3 linearly independent vectors) of the equation $Ax=b$, where $$A=\begin{bmatrix} 1 & -5& 4&-4 &-1&-5\\ 0 & 0& 1&4&0&-2\\ 0 & 0& 0&0&1&-1\\ 0&0&0&0&0&0 \end{bmatrix} $$ and $$b=\begin{bmatrix} 19\\ -5\\ 3\\ 0 \end{bmatrix}.$$ To solve this, I wrote an augmented matrix and reduced it to row echelon form, and then wrote out the pivot variables in terms of the others to create a span of 3 vectors. My answer was correct, however to do the question my instructor added a "hint" saying : let $a_1, a_2, a_3, a_4, a_5, a_6$ be the column vectors of matrix $A$. Notice that $b=-3a_1+5a_2+a_3-4a_4-2a_5-5a_6$. Use this to find the general solution. I have no idea how I could use this, and I was wondering if this could be used as a shortcut?
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$\begingroup$In general, if $a_1,\ldots,a_n$ are the column vectors of a matrix $A$ and $$ x = \begin{bmatrix}x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}, $$ then $Ax = x_1a_1+\cdots+x_na_n$. This is essentially the definition of matrix multiplication.
Thus if $b=-3a_1+5a_2+a_3-4a_4-2a_5-5a_6$, then we have $Ax=b$ where $$ x = \begin{bmatrix} -3 \\ 5 \\ 1 \\ -4 \\ -2 \\ -5 \end{bmatrix}. $$
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