Relationship between determinant of matrix and determinant of adjoint? We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. 
 A: If $A$ is of size $n\times n$, then
$$\det(adj(A))=\det(A)^{n-1}$$
also, you can verify that:
$$ A\cdot adj(A) = \det(A)\, I_{n\times n} $$
A: Suppose A is a square matrix of size $n \times n$. We will prove that $adj(A) A= A adj(A)=det(A)I.$
Denote the $(i,j)^{th}$ entry of A and adj(A) by $a_{ij}$ and $ã_{ij}$ respectively. Also let $A(i,j)$ be the submatrix of A obtained from eliminating the $i^{th}$ row and $j^{th}$ column of A.
For the $(i,i)^{th}$ entry, we have
$$\sum_{k=1}^{n} a_{ik}ã_{ki}=\sum_{k=1}^{n} (-1)^{i+k}a_{ik}det(A(i,k))$$
This is exactly what one gets if one expands det(A) along the $i^{th}$ row.
Now, consider the $(i,j)^{th}$ entry with $i \ne j$. By the same method as before, this equals 
$$\sum_{k=1}^{n} (-1)^{j+k}a_{ik}det(A(j,k))$$
which we'll call $D_{ij}$.
Notice that the ${i,j}^{th}$ entry of A, by still the same expansion performed along the $j^{th}$ row, is 
$$\sum_{k=1}^{n} (-1)^{j+k}a_{jk}det(A(j,k))$$
Hence, $D_{ij}$ is the determinant of a $n \times n$ matrix Ā that is almost identical to A. In fact, Ā only differs from A in the $j^{th}$ row, with the $(j,k)^{th}$ entry of Ā being the same as the $(i,k)^{th}$ entry of A for every $1 \le k \le n$.
Therefore, Ā has two identical rows, the $i^{th}$ row and the  $j^{th}$ row, and therefore det(Ā)=0 by the alternating property of determinants.
Hence $Aadj(A)$ is a diagonal matrix with all non-zero entries being $det(A)$, in other words,
$$Aadj(A)=det(A)I$$
Now, each row of $Aadj(A)$ can be regarded as the corresponding row of I scaled by $det(A)$, so by the multilinear (n-linear) property of determinants, 
$$det(Aadj(A))=det(A)^n$$
Using the well known property $det(MN)=det(M)det(N) \;\forall M,N \in \mathbb{M_n(R)}$, we have 
$$det(Aadj(A))=det(A)det(adj(A))=det(A)^n$$
If $det(A) \ne 0$, i.e. A is invertible, then 
$$det(adj(A))=det(A)^{n-1}$$ 
Q.E.D
