Which vectors $b$ in $\mathbb{R}^3$ does the matrix equation have a solution?

Question

Let A = [{1,2,-4},{-3,-3,3},{1,3,-7}]

For which vectors $b$ in $\mathbb{R}^3$ does the matrix equation $Ax=b$ have a solution?

What is the span of the columns $A$?

Can someone refresh my memory on what we are looking for here? I have row reduced the matrix to :

[{1,0,2},{0,1,-3},{0,0,0}]

But I am unclear as to what the question is asking. I think the answer is all vectors b but I am not quite sure why. I at first thought it was b = {2,-3,0}

Thanks to anyone looking at this!

Answer 1

Row-reduce the augmented matrix $$\pmatrix{1&2&-4&|&b_1\cr -3&-3&3&|&b_2\cr 1&3&-7&|&b_3\cr}\ .$$ I'm trusting your working for the LHS and have not actually done the RHS, but say for example it reduces to $$\pmatrix{1&0&2&|&b_1+2b_2+3b_3\cr 0&1&-3&|&4b_1+5b_2+6b_3\cr 0&0&0&|&7b_1+8b_2+9b_3\cr}\ .$$ Then $A{\bf x}={\bf b}$ has a solution if and only if $7b_1+8b_2+9b_3=0$, and the span of the columns is $$\{{\bf b}\in{\Bbb R}^3\mid 7b_1+8b_2+9b_3=0\}\ .$$ Now see if you can do it for yourself with the correct calculations.

Answer 2

The second line states that you have to find all the vectors $b$ for which $Ax=b$ has a solution. Why he is asking you to find the solution is a philosophical discussion, and I'll leave it for another time.

The hint is the third line, which is asking you to find the span of the columns of $A$, which gives you an idea of what you can "reach" using $Ax$. For example, if

$$ A_{1} = \begin{bmatrix} 1&0\\ 0&0 \end{bmatrix}, $$ then the span of $A$ is $$ \begin{bmatrix} 1\\0 \end{bmatrix} x_{1} + \begin{bmatrix} 0\\0 \end{bmatrix} x_{2} = \begin{bmatrix} \mathbb{R}\\0 \end{bmatrix}, $$ which means you can reach any value of in the first dimension, but the only value you can get for the second dimension is 0. So if someone asked me for which values of $b$ does $A_{1}x=b$ have a solution, I would look at the span, and conclude accordingly.