Matrix with determinant 0 If $A \in M_3(\mathbb{R})$ is a $3 \times 3$ matrix with $\det(A)=0$ and the square of each element equals its cofactor, do we necessarily have $A=0_3$?
$a_{ij}^2=A_{ij}$, where $A_{ij}=(-1)^{i+j}M_{ij}$ and $M_{ij}$ is the minor of $a_{ij}$.
 A: The answer is yes, $A$ must be zero. Observe that

(a) If the adjugate matrix of a $3\times3$ real matrix $B$ is zero, the rank of $B$ is at most 1.
(b) If $B$ is an entrywise nonnegative matrix of rank at most $1$, then $B=pq^T$ for some entrywise nonnegative vectors $p$ and $q$.

Let $C=\operatorname{adj}(A)^T$ be the cofactor matrix of $A$. By the given assumption, $C$ is the entrywise square of $A$.
Using the identity $\operatorname{adj}\left(\operatorname{adj}(A)\right)=\det(A)^{n-2}A$ for a generic square matrix $A$ of size $n$ (see q92837 or q162966), we get, in our case, $\operatorname{adj}(C^T)=0$. By observation (a), the rank of $C$ is at most $1$. As $C$ is the entrywise square of $A$, it is also entrywise nonnegative. So, by (b), $C=pq^T$ for some nonnegative vectors $p,q$. Taking entrywise square root on both sides, we get $A=\sqrt{p}\sqrt{q}^T$, so that $\operatorname{rank}(A)\le1$. Yet, the cofactor matrix of every $3\times3$ rank $1$ matrix is zero. Therefore $C=0$ and its entrywise square root, $A$, is also zero.
By the way, that $n=3$ is important here. When $n=2$, the cofactor matrix of $A=\pmatrix{1&-1\\ -1&1}\ne0$ is $C=\pmatrix{1&1\\ 1&1}$, which is the entrywise square of $A$.
A: From the given assumptions one can easily show for row 1 that
a11^3 + a12^3 + a13^3 = 0
and similarly for the other rows and columns. Such equations will be denoted below by (A), while the given assumption involving the cofactor will be denoted by (B).
Now consider different possible cases for row 1.
(1) All elemements are nonzero. Consider their signs. We need evidently only to treat the cases +++ and ++-. +++ is clearly impossible for (A). ++- is impossible for (A) due to FLT.
(2) Two elements are nonzero: These must have the same magnitude but of opposite signs due to (A).
(3) One element is nonzero: Impossble for (A).
(4) All elements are zero: Yet possible.
Thus we are left with (2) and (4).
Let's assume (4) holds for row 1. If (4) also holds for row 2, then it must also hold for row 3 upon consideraton of the columns (since (2) can't apply) and hence all elemments of the matrix are zero. In the other case row 2 must have two nonzeoro elements since (2) is now the only alternative. Let's first consider 0 C -C for row 2. Considering column 2
and column 3, we have 0 -C C for row 3. But then (B) is violated for a22. Similarly -C C 0 is also not possible for row 2.
Similarly we can deal with the cases where (4) holds for row 2 or row 3. Hence from now on we need only to consider (2). 
Let row 1 be 0 C -C. Considering column 2, a22 can only be 0 or -C. We apply (2) to determine the other elements akin to above, In the first case we have -C 0 C for row 2 and C -C 0 for row 3 and (B) is violated for a33. In the second case row 2 can't be 0 -C C, since otherwise, considering column 2 and column 3 we would have a32 = a33 = 0 which violates (2) for row 3. Thus we have C -C 0 as row 2 and consequently row 3 is -C 0 C. But then (B) is violated for a23.
Similarly one can show that C 0 -C etc. are also impossible for row 1. QED
