If all strict submatrices have full rank, does the matrix have full rank

Question

Let $A$ be an $n \times n$ matrix.

1. If all submatrices $B$ (for $B \neq A$) have full rank, does $A$ have full rank?

2. If all square submatrices $B$ (for $B \neq A$) have full rank, does $A$ have full rank?

Also, what if $A$ isn't square? Do the above hold?

Edit: As pointed out in the comments, it isn't true in the case of $2 \times 2$ matrices. What about $n > 2$? Can we always find a counterexample?

Answer 1

No, given $A$ square matrix, being all strict submatrices $B$ ($B\neq A$) of full rank does not imply that $A$ has full rank too.

It's easy to give an example that does not verify it: $$A=\pmatrix{1&1\\1&1}$$ As you see, $rank(A)=1$ (not full rank), but it's square strict submatrices are $(1)$, which obviously has rank $1$ (full rank).