Is there any relationship between the value of a matrix and the determinant value of its adjoint matrix? The relationship between a * and a

Is there any relationship between the value of a matrix and the determinant value of its adjoint matrix? The relationship between a * and a

A^(-1)=A*/|A|
A*=A^(-1)|A|
|A*|
=|A^(-1)|A||
=|A|^n|A^(-1)|
=|A|^n|A|^(-1)
=|A|^(n-1)
Namely
│A*│=│A│^(n-1)

If the determinant of matrix A of order 3 is 2, then | 2A | = B?
Choose 2,4,12,16?

|2A|=(2^3)*|A|=8*2=16
2 ^ 3 is the cubic power of 2
*It's a multiple sign
It is obtained by the property of determinant

Matrix with rank 1: it can be decomposed into column matrix (vector) and row matrix (vector)
Matrix with rank 1: it can be decomposed into column matrix (vector) and row matrix (vector)
R (a) = 1, so let a = α β ^ t and then calculate a ^ n in this way is very convenient
Matrix with rank 1: it can be decomposed into column matrix (vector) and row matrix (vector)
Why?

Certification:
The rank of a is 1. Let the k th column of a be nonzero, denoted as α
Then all the other columns of a can be expressed linearly by α, that is, there are numbers
B1, B2, B3,..., BN make
a1＝b1α,a2＝b2α,...,an＝bnα,
Where a1, A2,..., an are the first, second,..., n columns of A
Let β = (B1, B2,..., BN) ^ t, then
A＝(a1,.,an)
＝(b1α,b2α,...,bnα)
＝α(b1,b2,...,bn)
＝αβ^T