Cofactors and determinant

Question

Anyone can explain to me why $\det(C)=\det(A)^{n-1}$ where $A$ is $n$-by-$n$ matrix and $C$ is the matrix of cofactors of $A$. I have been thinking, anyone can help? Thanks!

Your identity is equivalent to $\det(C^T)=\det(A)^{n-1}$, in which case others have asked about it before. For example, see: http://math.stackexchange.com/q/972471/136641. — Kim Jong Un, Oct 15 '14 at 13:14
I know how to prove the identity u stated. But I am not sure my ques~~ — UnusualSkill, Oct 15 '14 at 13:16
If you can show $\det(C^T)=\det(A)^{n-1}$ then the original question follows immediately because $\det(C)=\det(C^T)$. Note also that people typically write $C^T=\text{adj}(A)$. — Kim Jong Un, Oct 15 '14 at 13:17
Anyway, do u know any online resources(like e-book) have these types of proving questions and solutions? — UnusualSkill, Oct 15 '14 at 13:23

score 0 · Answer 1 · answered Oct 15 '14 at 13:13

For $A \in \mathcal{M}_{n}(\mathbb{R})$ : $$ A \times \mathrm{Com}(A)^{\top} = \mathrm{Com}(A)^{\top} \times A = \det(A) \mathrm{I}_{n} \tag{$\star$} $$ where $\mathrm{Com}(A)$ is the matrix of cofactors of $A$. As a consequence, if $A$ is invertible, take the det in $(\star)$ and you get :

$$ \det(C) = \det(A)^{n-1}. $$

Alfred Chern · Answer 2 · 2014-10-15T14:53:13.680

As $C$ is the matrix of cofactors of $A$ , so $CA=\text{det}(A)I_{n}$ and note that for arbitrary two matrices $P,Q\in M_{n}(R)$, $\text{rank}(P)+\text{rank}(Q)\leq\text{rank}(PQ)+n$.

Case 1: If $A$ is invertible, i.e. $\text{det}(A)\neq0$, then take det in both sides of $CA=\text{det}(A)I_{n}$, we can get that $\text{det}(C)\text{det}(A)=(\text{det}(A))^{n}$, i.e. $\text{det}(C)=(\text{det}(A))^{n-1}$.

Case 2: If $A$ is singular and $A\neq0$, then $\text{det}(A)=0$ and $\text{rank}(A)>0$, $CA=0$ and $\text{rank}(C)+\text{rank}(A)\leq n$, $\text{rank}(C)<n$, $C$ is singular and $\text{det}(C)=0=(\text{det}(A))^{n-1}$.

Case 3: If $A=0$, then $C=0$, then the result is obvious.

Cofactors and determinant

2 Answers2