As Andre says this is not true. On the other hand if your question is why is the reason for which a system (of $n$ equation and $n$ unknowns) with complete rank has always a unique solution. In my opinion the better way to see it is by linear maps. Let $n$ be fixed and select a collection of $a_{i,j}\in \mathbb{F}$ for $(i,j) \in \{\ 1, \ldots,n\} \times \{\ 1, \ldots,n\}$. Define the linear map $f$
$$(x_1 \ldots,x_n)\longmapsto \bigg(\sum_{1\le j\le n} a_{1,j}x_j \ldots\sum_{1\le j\le n} a_{n,j}x_j \bigg)$$
When we say that the rank is complete, we say that for all $(c_1 \ldots,c_n)$ always exists a $(x_1 \ldots,x_n)$ such that $f(x_1 \ldots,x_n)=(c_1 \ldots,c_n)$, i.e.,
$$\bigg(\sum_{1\le j\le n} a_{1,j}x_j \ldots\sum_{1\le j\le n} a_{n,j}x_j \bigg)=(c_1 \ldots,c_n)$$
With really means that $\sum_{1\le j\le n} a_{i,j}x_j =c_j$ for our system.
Since the $\text{rank }f=n$. Then by the dimention formula we can conclude that its kernel is trivial, i.e., $\ker f = \{0\}$. Then $f$ is an injective map. So we know that always exists a solution because is surjective (the rank has the same dimension as the target space) and also this solution is unique since is one-to-one.
Using our above notation. What happens when the rank is less than $n$. Then there exists some values in the target space which are "outside" of the image and in these cases there is no solution. Since the rank is less than $n$ so the linear map cannot be surjective and exists vectors in $\mathbb{F}^n \backslash f(\mathbb{F}^n)$.
For the question: when does it have an infinite number of solution? Assuming again that the rank is less than $n$. Let $(c_1, \ldots ,c_n) \in f(\mathbb{F^n})$ (a vector in the image), because is in the image there is a vector in the domain that under $f$ is mapped in $(c_1, \ldots ,c_n)$, let call it $x=(x_1, \ldots, x_n)$. But also we know that the kernel is not trivial, again using the dimension formula we conclude $\dim(\ker f)= n-\text{rank} f>0$. Then there is some vector $(z_1, \ldots, z_n) \not= 0$ such that $f(z_1, \ldots, z_n)= (0, \ldots, 0)$.
Now consider $(x_1,\ldots, x_n)+k(z_1, \ldots, z_n)$, where $k\in \mathbb{F}$ and see what happens under $f$.
\begin{align}f((x_1,\ldots, x_n)+k(z_1, \ldots, z_n))=f(x_1,\ldots, x_n)+kf(z_1, \ldots, z_n)\\
=f(x_1,\ldots, x_n)+0= (c_1 \ldots, c_n) \end{align}
This occurs because $f$ is linear and the vector $(z_1, \ldots, z_n)$ is in the kernel (is zero under $f$). Thus $(x_1,\ldots, x_n)+k(z_1, \ldots, z_n)$ is also a solution. And indeed the set
$$x+\ker f: = \{x+k: x=(x_1,\ldots,x_n)+k, \text{ and } k\in \ker f\}$$
by the same argument as above contain all the solutions. In that case there is a infinite number of solutions.
Best Answer
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.