> 
> 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.