Is it true that all non-singular, symmetric adjoint matrices are equal to the inverse of their base matrix?

Question

I have a non-singular symmetric matrix $C^*$, where $C^*$ is the adjoint matrix of $C$. I am told that $C^* = C^{-1}$. I'm wondering if it is true that all non-singular, symmetric adjoint matrices are equal to the inverse of their base matrix (the base matrix here being $C$)? Because, as I understand it, this is not necessarily true for all symmetric or adjoint matrices alone, right?

What do I mean here by "adjoint"? Let $M$ be a square matrix. By $M^*$ is meant the matrix of cofactors of $M$. That is, the $(i, j)$-th entry of the matrix $M^*$ is equal to $(-1)^{i + j} \det(\hat{M}_{i, j})$, where $\hat{M}_{i, j}$ is the matrix obtained from $M$ by striking out the $i$-th row and $j$-th column. The transpose of the cofactor matrix $M^∗$ is known as the adjoint of $M$, and denoted $\text{adj}(M)$. This definition is a bit confusing, since it is said that $M^*$ is the matrix of cofactors in this definition, but $C^*$ is the adjoint in $C^* = C^{-1}$, but that's what I have.

Thank you.

Answer 1

Note that $$\operatorname{adj}(C)\cdot C = \det(C)\cdot I$$ So if $\det(C) = 1$, then we do indeed have $\operatorname{adj}(C) = C^{-1}$, but not otherwise.

Edit: Knowing that the original statement included the qualification "up to a constant", I am inclined to say that it is true. The adjoint of $C$ and the inverse of $C$ are indeed equal up to a constant. That constant is $\det(C)$.

Answer 2

Your notation for " * " is confusing. It sometimes seems to mean the cofactor matrix and sometimes the transpose of the cofactor matrix.For any square matrix $M$, we define $$adj(M)=(cof(M))^T$$. Then it is true that $$M(adj(M))=(adj(M))M=(det(M))I $$ so $adj(M)$ is the inverse of $M$ iff $det(M)=1$. It doesn't matter whether or not $M$ or $adj(M)$ is symmetric.