Let $n$ an integer strictly larger than 1. Let two matrices $A,B \in \mathbb{R}^{n \times n}$ and linearly independent vectors $x,y \in \mathbb{R}^n$. Is it true that: If $Ax+By = 0$, then $\begin{bmatrix} A & B \end{bmatrix}$ is not full rank.
Edit: by "not full rank", I mean "rank strictly smaller than $n$".
The matrix $\begin{bmatrix}A&B\end{bmatrix}$ is not full rank if and only if the rank is less than $n$.
Take $A$ to be the identity and $x=e_1$. We'd like to find $B$ such that $e_1+Be_2=0$ (where $\{e_1,e_2,\dots,e_n\}$ is the canonical basis for $\mathbb{R}^n$).
Clearly $\begin{bmatrix}I&B\end{bmatrix}$ has rank $n$, for every matrix $B$; take $$ B=\begin{bmatrix} 0 & -e_1 & 0 & \dots & 0\end{bmatrix} $$ and you have your counterexample.