If all strict submatrices have full rank, does the matrix have full rank? Let $A$ be an $n \times n$ matrix. 


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

*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?
 A: 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).
A: I think that the conditions 1. and 2. are equivalent. Then I consider the condition 2.
Of course, the case $n=2$ is $very^{10}$ easy. That follows is a counter-example (easy to prove) for $n=3$.

Can someone find a counter example (and prove that it is one) for $n = 10$ ?
EDIT 1.
$\textbf{Proposition.}$ Let $S$ be the set of singular real or complex $n\times n$ matrices and $Z=\{A\in S; A \text{ satisfies condition 2}\}$. Then $Z$ is a dense Zariski-open subset of $S$.
$\textbf{Proof.}$ $B\in S\setminus Z$ iff $\det(B)=0$ and some strict submatrix of $B$ is singular. Thus $S\setminus Z$ is a finite union of Zariski-closed subsets of $S$, and therefore, is closed. It remains to prove that $Z\not= \emptyset$, that is, the user1551's work. $\square$
A matrix $W=[w_{i,j}]\in M_{p,q}(\mathbb{R})$ is said to be randomly chosen if the $w_{i,j}$'s follow independent standard normal laws. Now we randomly choose $U\in M_{n,n-1}(\mathbb{R}),V\in M_{n-1,n}(\mathbb{R})$ and we calculate $A=UV\in S$.
$\textbf{Corollary.}$ The above $A$ is in $Z$ with probability $1$.
EDIT 2.
$\textbf{Conjecture.}$ For every $n$, $Z$ contains, at least, one element in $M_n(\mathbb{Z})$.
$\textbf{Remark.}$ The conjecture works if $\mathbb{Q}^{n^2}\cap S$ is dense in $S$; is it true ?
EDIT 3. Answer to a OP's comment about the user1551's (pub for free) solution.
i) The first idea (standard and well-known) is to fill a large part of $A$ with real or complex elements $S=(a_i)_{i\leq k}$ s.t. the elements of $S$  do not satisfy any non trivial polynomial equation with coefficients in $\mathbb{Q}$. 
You can explicitly choose $S=\{1,e^1,e^{\sqrt{2}},e^{\sqrt{3}},e^{\sqrt{5}},\cdots\}$. Yet, the proof of that is not obvious; you can just report the existence of such a set as follows
The set $U_0$ of algebraic numbers over $\mathbb{Q}$ is countable (there are very few algebraic numbers but much more than human beings); then let $a_1\notin U_0$; the set $U_1$ of algebraic numbers over $\mathbb{Q}[a_1]$ is countable; then let $a_2\notin U_1$ and so on. 
ii) If $k=n^2$, then the subdeterminants of $A$ are $\not= 0$ but, unfortunately, $\det(A)$ too; then, necessarily $k<n^2$. The second idea (that I didn't have) is to take $k=n^2-n$ and to choose the last column as the sum of the previous ones. That's all folks ! 
A: ,In the same spirit of constructing a "totally invertible" matrix, we may construct a counterexample as follows. For every $n\ge2$, let $R\in\mathbb R^{n\times(n-1)}$ be an entrywise nonzero matrix $R$ such that each $r_{ij}$ is transcendental to the field of fractions generated by all entries preceding it in row-major lexicographic order. Let $\mathbf1\in\mathbb R^n$ be the vector of ones. Then $A=\pmatrix{R&R\mathbf1}\in\mathbb R^{n\times n}$ is singular, but each of its smaller square submatrices is invertible.
To illustrate, suppose $n=4$. Let
$$
A=\pmatrix{a&b&c&a+b+c\\ d&e&f&d+e+f\\ g&h&i&g+h+i\\ j&k&l&j+k+l}
$$
such that $a\ne0$, $b$ is transcendental to $\mathbb Q(a)$, $c$ is transcendental to $\mathbb Q(a,b)$, $d$ is transcendental to $\mathbb Q(a,b,c)$, etc.. Now consider a square submatrix of $A$, such as
$$
B=\pmatrix{c&a+b+c\\ f&d+e+f}.
$$
By subtracting the first column from the second one, we have $\det(B)=\det(C)$, where
$$
C=\pmatrix{c&a+b\\ f&d+e}.
$$
We have $\det(C)=p(f):=-(a+b)f+c(d+e)$. Since $p$ is a nonzero polynomial with coefficients in $\mathbb Q(a,b,c,d,e)$ but $f$ is transcendental to $\mathbb Q(a,b,c,d,e)$, the value of $p(f)$ cannot be zero. Hence $C$ and in turn $B$ are nonsingular.
