The answer for $n \leq 2$ is yes, and for $n \geq 3$, it's no. (The $n=1$ case is left to the reader)

**The case $n=2$:** If $v_i$ are the rows of the matrix, then each equation $v_1 \cdot x = 0$ and $v_2 \cdot x = 0$ describes a co-dimension $1$ space. And the four corners of the square $\{0,1\}^2$ cannot be covered by two lines passing through the origin.

**The case $n \geq 3$:** For $n \geq 3$, we can cover all the vertices of the cube with three independent hyperplanes through the origin. Namely...
$$x_1 = 0$$
$$x_2 -x_3 = 0$$
$$x_1 - x_2 - x_3=0$$
It's easy to see these three conditions describe a co-dimension three space [i.e., the conditions are independent]. To see they cover the vertices of the cube, note that failure of the first and second equations would imply $x_1 =1$ and $x_2 \neq x_3$. Since each variable is $0,1$, this would force $x_2 + x_3 = 1$, so the third equation will be satisfied.

**The $n\geq 3$ case phrased as a matrix:**
Make the first row $[1, 0, 0, 0, 0, \ldots]$. Then the second row $[0, 1, -1, 0, 0, \ldots]$. Then make the third row $[1, -1, -1, 0, 0, 0, \ldots]$. These three rows are independent (yay!), so just fill up the rest of the matrix with more linearly independent rows until you have a basis (i.e., an invertible matrix). Then see above argument why $A \vec{x}$ will have at least one of the first three coordinates equal to $0$.

**Literature:**
Also, there are some really cool results about covering the cube with hyperplanes. See for instance this article about covering all but one of the vertices and also this article about covering the vertices in a way where each equation is needed, and every variable actually shows up.

**Edit 1:** I changed the equations to handle the $n=3$ as well.

**Edit 2 (this part added to address orthogonal case):** As per comment, perhaps you’d like the matrix to be orthogonal as well. That’s also fine (at least for $n \geq 4$. I haven’t thought about $n=3$ in this case).

Take the first three rows to be $[1, -1, 0, 0, 0, 0, \ldots]$ and $[0,0,1,-1,0,0,\ldots]$ and $[1,1,-1,-1,0,0,\ldots]$ (and renormalize these so that they each have length 1). Then extend this to an orthonormal basis and fill the rest of the matrix with those rows. This works in the same ways as the above example. (This was actually the first construction I gave, but then I swapped it out for the above, which only needed $n=3$.) ((I notice that I don’t enjoy thinking about the orthogonal $n=3$ case. It might be that you can’t make any orthogonal matrix to do it. On the other hand, it could be that I’m bad at visualizing rotations of the cube that put all its corners onto three orthogonal planes...))

**Edit 3:** You seem to want this to be true for really special matrices. I don’t think that’s really going to happen. The cube fits inside a pretty darn small space (union of three planes), so this seems to give us a lot of wiggle room for all sorts of stuff. :-/

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