You seem to already understand what you are asking.
The column space of a matrix is just the set of all linear combinations of the columns of $A$. It's the subspace spanned by those columns.
So, if $d^T$ is contained in this subspace, then the system is consistent for that specific $d^T$, otherwise it is not.
All of this is just another way of asking if row reducing the augmented matrix $[a^T b^T c^T d^T]$ is consistent. It sounds like maybe you have not got to subspaces yet, but they are not that difficult of a concept if you already understand spans.
Homogeneous linear systems of equation always have a (trivial) solution, $(x,y) = 0$. However I'm guessing you are referring to the case where the right side can be any vector $(b_1,b_2)$, so I will solve that case.
The matrix equation translates to
$$\begin{cases} Ax + By = b_1\\ Cx = b_2\end{cases}$$
Since $A$ is invertable, $x$ can be expressed from $A, B, y$ and $b_1$ the following way:
$$\begin{align*}Ax+By &= b_1 \\ Ax &= b_1-By \\ A^{-1}Ax &= A^{-1}(b_1-By) \\ x &= A^{-1}(b_1-By)\end{align*}$$
For any input $y$ vector, this will give a solution for $x$, since $A$ is always invertable and you can always multiply it with $b_1-By$. We can use the second condition to find such a $y$, by substituting this formula for $x$:
$$\begin{align*}Cx &= b_2 \\ CA^{-1}(b_1-By) &= b_2 \end{align*}$$
So if we can find a $y$ such that $CA^{-1}(b_1-By) = b_2$, then the original system of equations must have a solution, since if we know $y$, we can always find $x$ through the $x = A^{-1}(b_1-By)$ formula. After expanding the brackets, this is equivalent to the rowspace of $-CA^{-1}B$ containing $b_2+CA^{-1}b_1$, since
\begin{align*}CA^{-1}(b_1-By) &= b_2 \\ CA^{-1}b_1 - CA^{-1}By &= b_2 \\ -CA^{-1}By &= b_2 + CA^{-1}b_1\end{align*}
Answer:
So the original system does not have a solution exactly if the rowspace of $-CA^{-1}B$ does not contain $b_2+CA^{-1}By$, or equivalently if $CA^{-1}(b_1-By) = -b_2$ does not have a solution for $y$. If $b_2 = 0$, like in your case, then the system will always have the trivial $(x,y) = (0,0)$ solution.
Answer for the case $b_1 = 0$, $b_2 = 0$, with non-trivial solutions $(x,y) \ne (0,0)$: In this case, we can subsititute $b_1 = 0$ and $b_2 = 0$. The statement that "there is no non-trivial solution" is equivalent to the dimensionality of the rowspace of $−CA^{-1}B$ being $0$, which is further equivalent to the dimensionality of the rowspace of $CA^{-1}B$ being $0$. (I removed the negative sign.) This is true because in this case, the only solution the original system be $(x,y)=(0,0)$.
And, the dimensionality of the rowspace is the rank of the matrix, which yields a very simple equivalent condition:
If $\text{rank}(CA^{−1}B)=0$, then there are no non-trivial solutions to your original problem.
Best Answer
From an answer of mine:
Note that $\mathrm x_{\text{LN}}$ is in the column space of matrix $\mathrm V_1$, which is the orthogonal complement of the null space of matrix $\mathrm A$ (which is the column space of matrix $\mathrm V_2$).
For example, let $m = 1$ and $n=2$. In this case, the solution set of $\rm A x = b$ is a line in $\Bbb R^2$. The vector that spans this line is in the null space of matrix $\rm A$. The least-norm solution is the point on this line that is closest to the origin in the Euclidean norm. If one draws a line segment from the origin till the least-norm solution, this line segment will be orthogonal to the line. Thus, a vector orthogonal to the least-norm solution is in the null space. However, do note that I assumed that matrix $\rm A$ has full row rank. If matrix $\rm A$ does not have full row rank, then there is no unique least-norm solution.