Consider the following matrix:
\begin{bmatrix} 1 & 0 & -1 & -1 & t-6\\ -t & 0 & 3 & t & 9\\ -1 & 0 & t-6 & 1 & 3\\ 0 & 0 & 0 & t-3 & 0\end{bmatrix}
What are bases for the column and row space and how can I determine the column and row ranks in dependence of $t$?
I've tried Gauss-Jordan but it kind of leads me nowhere.
Note that one column is zero, so we can delete that column, and look at the matrix $$B:=\begin{bmatrix}1&-1&-1&t-6\\-t&3&t&9\\-1&t-6&1&3\\0&0&t-3&0\end{bmatrix},$$ which is square.
Now we can compute the determinant, $$\det B = (t-3)\det \begin{bmatrix} 1&-1&t-6 \\ -t&3&9\\-1&t-6&3 \end{bmatrix}$$ $$ =(t-3)(9+9-t(t-6)^2+3(t-6)-9(t-6)-3t) $$ $$ =(t-3)(-t(t-6)^2-9t+54) =(t-3)(t-6)(-t(t-6)-9) $$ $$ =-(t-3)(t-6)(t^2-6t+9) =-(t-3)^3(t-6). $$
Thus when $t\ne 3,6$, $B$ (and thus $A$) has full rank, i.e. rank 4, and the nonzero rows and columns of $A$ are bases for the column/row space.
Now we just need to look at what happens when $t=3$ or $t=6$.
First if $t=3$, we have $$B=\begin{bmatrix}1&-1&-1&-3\\-3&3&3&9\\-1&-3&1&3\\0&0&0&0\end{bmatrix}.$$ Noting that the second row is $-3$ times the first row, and the last row is 0, the rank is at most two. Since the third row is not a multiple of the first row, they are not linearly dependent. Hence the rank of $B$ is exactly two.
The first and third rows of $B$ and hence $A$ form a basis for the row space in this case. The first and second columns of $B$ and hence the corresponding columns in $A$ are linearly independent and hence form a basis for the column space.
Now if $t=6$, $$B=\begin{bmatrix}1&-1&-1&0\\-6&3&6&9\\-1&0&1&3\\0&0&3&0\end{bmatrix}.$$ Note that the first second and third columns are linearly independent, since they are nonzero, and the first and second are linearly independent being separated by projection on the third coordinate, and the first and second are linearly independent from the third column, since they're separated by projection on the fourth coordinate. Thus since the rank is at most 3, the rank must be at least 3.
Thus the columns corresponding to the first, second, and third columns of $B$ form a basis for the column space of $A$. On the other hand for the rows, by a similar argument as for the columns, the first, second and fourth rows are linearly independent, so the corresponding rows in $A$ form a basis for the row space of $A$.