Matrix algebra linear transformation question Let $A = 4 \times 4$ matrix: 
$\begin{bmatrix}
 3 & 2 &  10 &  -6 \\ 
 1  & 0  &  2 & -4 \\ 
 0 &  1  & 2  & 3 \\ 
 1  &  4 &  10 &  8  
\end{bmatrix}$,
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. 
 A: 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 
$$A = \begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0
\end{pmatrix}.$$
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.
A: Another approach may use the concept of eigenvalues and eigenvectors.   
Main steps are: 


*

*Find transpose $A^T$.   

*Find eigenvalues of $A^T$    


*

*if there are only non zero eigenvalues then the vector $b$ lies in the
the span of $A$

*if there are zero eigenvalues then denote  as $v_1, v_2, \dots$ eigenvectors linked with obtained
$0$ eigenvalues - the matrix constructed from these vectors determines whether $b$ belongs the the span of $A$ or not, i.e. the vector belongs if $V=[v_1, v_2, \dots]$ generates equation $V^Tb=0$ (what is equivalent to: there is no non-zero components of the vector $b$ in $V$ ( $V$ is the orthogonal complement for $A$))



You can check that for the given data $V$ is the matrix of dimension $ 4 \times 2$ (two zero eigenvalues of $A^T  $ ) and for calculated $V$: $ \ $ $V^TA=0$ but $V^Tb \neq 0$ so the vector doesn't belong to the range of linear transformation $A$.
A: Hint: check to see if the columns of $A$ span $\mathbb{R}^4$.
