How about $A$ if $\text{adj}(A)$ is symmetric?

Question

If $\text{adj}(A)$ is a symmetric matrix with $A$ is $n, n$ real matrix, then is $A$ symmetric also?

(The inverse is true, which is well-known, i.e $A$ is symmetric then $\text{adj}(A)$ is symmetric also.)

Answer 1

$\text{adj}(\text{adj}(A))=\text{det}(A)^{n-2}A$

This is easy to prove for invertible matrices. It then follows from the converse you stated that A is also symmetric if it is invertible.

It is not true for non-invertible matrices, as other answers have indicated

Answer 2

No, this is not true. Take for example the matrix $$A = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$ which has the property that ever minor's determinant (and thus every co-factor) is zero. (To see why this is, note that ever minor must have a zero row, since there are two zero rows in $A$ and you can't eliminate both when creating a minor).

Answer 3

$A$ is non-symmetric and $\operatorname{adj}(A)$ is symmetric if and only if $A$ is orthogonally similar to $\pmatrix{0&0\\ 0&B}$ for some non-symmetric matrix $B\in M_{n-1}(\mathbb R)$. (It follows that the answer to your question is negative and $A=0\oplus B$ is a counterexample when $B$ is non-symmetric.)

Suppose $\operatorname{adj}(A)$ is symmetric but $A$ is non-symmetric. Then $A$ is singular (otherwise $A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A)$ is symmetric and so is $A$) but nonzero (otherwise $A$ is symmetric). Hence the rank of $\operatorname{adj}(A)$ is at most $1$ and we may write $\operatorname{adj}(A)=auu^T$ for some nonzero scalar $a$ and some unit vector $u$. Since $a(Au)u^T=A\operatorname{adj}(A)=0=\operatorname{adj}(A)A=au(u^TA)$, we have $Au=0$ and $u^TA=0$. Pick any orthogonal matrix $Q$ whose first column is $u$. Then $Q^TAQ=\pmatrix{0&0\\ 0&B}$ for some matrix $B$. Since $A$ is non-symmetric, so must be $B$.

Conversely, suppose $Q^TAQ=\pmatrix{0&0\\ 0&B}$ for some orthogonal matrix $Q$ and non-symmetric matrix $B$. Then \begin{aligned} \pmatrix{\det(B)&0\\ 0&0} &=\operatorname{adj}\pmatrix{0&0\\ 0&B}\\ &=\operatorname{adj}(Q^TAQ)\\ &=\operatorname{adj}(Q)\operatorname{adj}(A)\operatorname{adj}(Q^T)\\ &=\left(\frac{1}{\det(Q)}Q^T\right)\operatorname{adj}(A)\left(\frac{1}{\det(Q^T)}Q\right)\\ &=Q^T\operatorname{adj}(A)Q. \end{aligned} Therefore $\operatorname{adj}(A)=Q\pmatrix{\det(B)&0\\ 0&0}Q^T$ is symmetric.