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

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.

• you need to be careful on your usage of "adjoint" -- it is an overloaded term. I suspect you are referring to something akin to transpose but adjoint or classical adjoint also refers to adjugate matrix Feb 12, 2020 at 7:01
• @user8675309 You're right about it being overloaded. I will edit my main post with a description of what I mean. Feb 12, 2020 at 7:02

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.
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 $$\det(C)$$.
• Arthur, does this mean that the claim that $C^* = C^{-1}$ for a non-singular symmetric matrix $C^*$, where $C^*$ is the adjoint matrix of $C$, is false? After all, $\det(C) = 1$ is a specific case of non-singular matrix; if we had any other non-singular matrix, as you seem to claim, $C^* \not= C^{-1}$? Or, is it valid to say that $C^* = C^{-1}$ up to a scale? Feb 12, 2020 at 14:07
• @DomFomello A scaling (of the underlying coordinate system) won't really help, as the matrix would stay the same. But if you somehow scale the operator, sure. If the overall tone of the text is "we don't really care about scalings", then they are right. Otherwise they really ought to have specified that the determinant is $1$, yes. Feb 12, 2020 at 14:14
• I'm sorry, I should have elaborated: In the case that I am dealing with, $C$ is a conic, and it is claimed that $C^* = C^{-1}$ up to a constant. Does that change anything? If not, I'm going to send an email to the author, since this seems like a significant oversight. Feb 12, 2020 at 14:15
• @DomFomello "Up to a constant" is excactly the kind of scaling we are talking about here. That constant is $\det(C)$. So with that qualification, and knowing that $C$ is non-singular, I think that that statement is correct. Feb 12, 2020 at 14:25