How to check if a given vector is in the image of a linear transformation? 
Determine whether or not $(1,3,1)$ or $(-1,-1,-2)$ is in the image of the linear transformation $T: \mathbb R^4 \to \mathbb R^3$ defined by $T(x) = A x$, where
$$A = \left(\begin{array}{crc}
 1 &  2 &  -1 & -1\\
1 & 0 & 1 &  1\\
 2 &  -4 &  6 & 2\\
 \end{array}\right)$$

I don't know how to do this problem and would greatly appreciate some help!
 A: One way or another, you're going to have to do some form of row-reduction.  If we think of the "usual" case, then four dimensions should easily be able to be mapped to three dimensions meaning that any $3$-vector would be possible.  So the best thing to do is to row-reduce your matrix and find the rank of your matrix (The maximum number of linearly independent vectors in a matrix is equal to the number of non-zero rows in its row echelon matrix).
$$
\begin{pmatrix}
1&2&-1&-1\\
1&0&1&1\\
2&-4&6&2
\end{pmatrix}
$$
$$
\begin{pmatrix}
1&2&-1&-1\\
0&2&-2&-2\\
0&8&-8&-4
\end{pmatrix}
\sim
\begin{pmatrix}
1&2&-1&-1\\
0&1&-1&-1\\
0&2&-2&-1
\end{pmatrix}
$$
$$
\begin{pmatrix}
1&2&-1&-1\\
0&1&-1&-1\\
0&0&0&-1
\end{pmatrix}
$$
You can see from this that this matrix spans $\mathbb{R}^3$ (for it not to, it would have to have at least one row that is all zeros).  Therefore you can pick any vector in $\mathbb{R}^3$ and this matrix could produce it from a $4$-vector.
A: The correct procedure computationally is to bring thee matrix to echelon form by row operations (gaussian elimination), and mark the columns corresponding to leading 1's. Now again repeat the same procedure by appending the $b$ vector as extra column.  If no new columns are introduced by way of leading 1's then the said vector is in the image. Otherwise not.
A: Using the Gaussian elimination we have $T(2,0,\frac{-5}{4},\frac{9}{4})=(1,3,1)$ and $T(-1,0,0,0)=(-1,-1,-2)$.
A: The comment made by florence has the most obvious approach. 

$b$ will be in the image of $T$ if and only if there exists an $x$ such that $T(x)=b$. This last equation simplifies to just be $Ax=b$, which corresponds to a system of linear equations for the components of $x$. If we can solve this system then we know $b$ is in the image of $T$. If the equations are inconsistent then $b$ is not in the image of $T$. 
As an exaple consider the following case, 
$$A= \left(\begin{array} \ 1 & 1 \\ 1 & 1 \end{array}\right) \qquad b=\left(\begin{array} \ 3 \\ 3 \end{array}\right),$$
then $Ax=b$ becomes, 
$$ \left(\begin{array} \ 1 & 1 \\ 1 & 1 \end{array}\right) \left(\begin{array}\  x_1 \\ x_2 \end{array}\right)  =\left(\begin{array} \ 3 \\ 3 \end{array}\right),$$
which leads to the equation, 
$$ x_1 + x_2 = 3,$$
since there is a solution $x_1=x_2=1.5$ we have that $b$ is in the range of $T$. 
Now consider the case where $b=(1,-1)^T$. The equation $Ax=b$ is now, 
$$ \left(\begin{array} \ 1 & 1 \\ 1 & 1 \end{array}\right) \left(\begin{array}\  x_1 \\ x_2 \end{array}\right)  =\left(\begin{array} \ 1 \\ -1 \end{array}\right),$$
which corresponds to the following two equations, 
$$ x_1+x_2= 1,$$
$$x_1+x_2=-1,$$
in other words $x_1+x_2 \neq x_1+x_2$ which is contradictory. The equations are not consistent and there is no pair of real numbers $x_1$ and $x_2$ can simultaneously solve both of them. For this reason $b$ is not in the range of $T$. 
