Matrix algebra linear transformation question Let $A = 4 \times 4$ matrix: 
$\begin{bmatrix}
 3 &amp; 2 &amp;  10 &amp;  -6 \\ 
 1  &amp; 0  &amp;  2 &amp; -4 \\ 
 0 &amp;  1  &amp; 2  &amp; 3 \\ 
 1  &amp;  4 &amp;  10 &amp;  8  
\end{bmatrix}$,
let $b = 4 \times 1$  matrix: 
$\begin{bmatrix}
 -1  \\
 3  \\
 -1  \\
 4  \\
\end{bmatrix}$

Is $b$ in the range of linear transformation $x \rightarrow Ax$?
Why so or why not?
I'm not really sure what the question is asking. Any help would be greatly appreciated.   

Sorry I don't know how to properly format the mathematical equations on this website yet, I did my best to make it legible.
This is a counterexample to the answer Hugh Mungus gave: consider the linear transformation $\mathbb{R}^3 \to \mathbb{R}^3: x \mapsto Ax$, where 
$$A = \begin{pmatrix}
1 &amp; 0 &amp; 0\\
0 &amp; 1 &amp; 0\\
0 &amp; 0 &amp; 0
\end{pmatrix}.$$
Clearly the columns of $A$ do not span $\mathbb{R}^3$. What the map actually does is it sends $(x_1, x_2, x_3)^T$ to $(x_1, x_2, 0)^T$. Therefore, it is not difficult to see that the vector $(2,2,0)^T$ is an element of the range of the linear map. 

The way to proceed is to check if the vector $b$ is a linear combination of the columns of $A$, which result in performing Gaussian Elimination on the augmented matrix (for example) and see if you can find an inconsistency (which you did, seeing your comment on Hugh Mungus' answer). 

I just wanted to point out that it is not necessary for the columns to span the whole space in order to see if $b$ is in the range of not. Note however that if you would have found that the columns of $A$ do span the whole space, you are obviously done, but in the case it does not, you are still not sure if your vector is or isn't part of the range.
Hint: check to see if the columns of $A$ span $\mathbb{R}^4$.