Will the value of $t$ affect the row, column, and solution spaces? 
Consider A = $\begin{bmatrix}4 & 2\\t & 1\\3&t\end{bmatrix}$.Is the column space of $A$ the same for all t$?$Is the row space of $A$ the same for all $t$? Is the solution space of $Ax = b$ for a given $b$ in the span of the column vectors the same for all $t$? 
  (Hint: Apply row reduction)

This is the question. I thought the column space of $A$ is not the same for all $t$. For instance, when $t = 0$, $\operatorname{span}{(4, t, 3), (2, 1, t)}$ will have only the vectors $(x, y, z)$ that satisfy the conditions $y = 1, z = 3$; when $t \ne 0$, $y$ does not have to be equal to $1$, $z$ does not have to be equal to $3$.
Knowing $b$ is in the span of the column vectors of $A$, which means the column space 
of A, I let $b = \begin{bmatrix}4c + 2k\\tc + k\\3c + tk\end{bmatrix}$. Then if $Ax = b$, $x = (c, k)$, it won't change with $t$'s changing. So, the solution space is the same for all $t$.
I couldn't say anything about the row space, I didn't apply row reduction and I'm not sure if the things I said constitute satisfactory answers to the questions.
 A: The rank of your matrix can never be $\lt 2$ (try to make the two columns multiples of each other -- you can't), so the row space always spans $\Bbb R^2$, regardless of $t$.
The column space does depend on $t$.  Just choose two different values of $t$ and you're almost guaranteed to get two different column spaces.
The solution space will be dependent on $t$.  Let $b=\begin{bmatrix} 1 \\ 2 \\ 3\end{bmatrix} = \begin{bmatrix} 4c+2k \\ tc+k\\ 3c+tk \end{bmatrix}$.  If $t$ changes, then the $c$'s and $k$'s have to change to keep the two sides of the equation equal.
A: $$\text{Row}A = \text{Col}A^T$$
So, row reduce $A^T$
$$\left[\begin{matrix}4 & t & 3\\2 & 1 & t\end{matrix} \right]\sim \left[\begin{matrix}4 & 2 & 2t\\0 & 2-t & 2t-3\end{matrix} \right]$$
If $t=2$ we can readily see the rowspace of $A$ is spanned by $(4,0)\sim (1,0)$ and $(4,1)$.
For $t\neq 2$ we can go farther:
$$\left[\begin{matrix}4 & 2 & 2t\\0 & 2-t & 2t-3\end{matrix} \right]\sim \left[\begin{matrix}4 & 0 & 2t-\dfrac{4t-6}{2-t}\\0 & 2-t & 2t-3\end{matrix} \right]$$
And you can see the rowspace is spanned by $(4,0)$ and $(0,2-t)$.
In either case, $\text{Row}A=\mathbb{R}^2$.
