Linear transformation 
If $\mathbf{A} = \begin{pmatrix} 2&-1&5&8 \\ 2&4&0&5 \\ 1&3&-1&4
 \end{pmatrix}$, let $\mathbf{x}$ denote the column vector $(x_1, x_2,
 x_3, x_4)^T$, and let the linear map $T$ be given by the equation
   $T(\mathbf{x}) = \mathbf{Ax}$ then (a) find and describe the kernel
   (null space), (b) find $T(3e_1-2e_2-e_3)$; (c) solve
   $T(\mathbf{x}) = 3e_1-2e_2-2e_3.$

I row reduced the matrix to $\begin{pmatrix} 1&0&2&0 \\ 0&1&-1&0 \\ 0&0&0&1 \end{pmatrix} $ then found the kernel to be $\mathbf{x} = x_{3}(-2,1,1,0)^{T}$. It's a line through the origin but I don't know anything else about it. Also, I need help with the rest. 
Also, I thought $T(3e_1-2e_2-e_3)=\begin{pmatrix} 1&0&2&0 \\ 0&1&-1&0 \\ 0&0&0&1 \end{pmatrix}(3,-2,-1)$ but the answer is a vector. 
 A: The row reduction looks good, and that vector is indeed a basis for the kernel. I'm not sure I'd write ${\bf x} = x_3(-2, 1, 1, 0)$, it's kind of analogous to writing $x = \Bbb R$ when you mean to say that $x$ can be anything in $\Bbb R$. So perhaps write $\ker T = \{x_3(-2, 1, 1, 0): x_3 \in \Bbb R\}$. 
I honestly don't think there's a better way to describe the kernel than you have, just saying it's a one-dimensional subspace, hence a line through the origin in $\Bbb R^4$.

As mentioned in the comment, since the transformation $T$ is a map $\Bbb R^4 \to \Bbb R^3$, we need to rewrite $3e_1-2e_2-2e_3$ as $3e_1-2e_2-2e_3+0e_4$, adding in the "missing" basis vector.
But you can't use the row-reduced ${\bf A}$ to compute $T(3e_1-2e_2-2e_3)$, you need to use the original matrix (you should get $(-2, -2, -1)$). To see why this is, consider the dilation matrix 
${\bf D} = \begin{pmatrix}2&0\\0&2\end{pmatrix}$. It should be pretty easy to see that ${\bf D}$ sends a vector $(x, y)$ to $(2x, 2y)$, but if you row-reduce $D$ to the identity, then you'll incorrectly think $D$ sends $(x, y)$ to itself (in fact, you'll think all invertible matrices do the same thing as the identity matrix)!

But for part (c), you can row-reduce the augmented matrix $\left(\begin{array}{cccc|c} 2&-1&5&8&3 \\ 2&4&0&5&-2 \\ 1&3&-1&4&-2
 \end{array}\right)$.
