The problem of line generation adjoint matrix If a is an invertible matrix of order n and a * is an adjoint matrix of a, then (a *) * = () A ㄧAㄧ^n-1 A B ㄧAㄧ^n-2 A C ㄧAㄧ^n+1 A D ㄧAㄧ^n+2 A Don't just choose the answer of ABCD, explain the detailed process, answer the detailed and clear additional bonus points, thank you!

The problem of line generation adjoint matrix If a is an invertible matrix of order n and a * is an adjoint matrix of a, then (a *) * = () A ㄧAㄧ^n-1 A B ㄧAㄧ^n-2 A C ㄧAㄧ^n+1 A D ㄧAㄧ^n+2 A Don't just choose the answer of ABCD, explain the detailed process, answer the detailed and clear additional bonus points, thank you!

B ㄧAㄧ^n-2 A
There is an important equation in Linear Algebra: AA * = ㄧㄧ a ㄧ e, whether it is invertible or not. When a is invertible, we can derive the inverse matrix a '= a * / ㄧ a ㄧ, which is a very impractical method to find the inverse matrix before we learn to transform
Now use a * instead of a in the equation AA * = ㄧ a ㄧ E
A * (a *) * = ㄧ a * ㄧ e is obtained
So (a *) * = ㄧ a * ㄧ (a *) '
Bring in ㄧ a * ㄧ a = ㄧ a ㄧ n-1 and a * = ㄧ a ㄧ a '
The results show that (a * * = ㄧ a ㄧ ^ n-1 × (ㄧ a ㄧ a ')' = ㄧ a ㄧ ^ n-1 × (A / ㄧ a ㄧ) = ㄧ a ㄧ ^ n-2 a
Supplement 1: the formula a * = ㄧㄧㄧㄧㄧㄧㄧㄧ a 'is obtained directly from AA * = ㄧㄧㄧㄧ E
Supplement 2: the formula ㄧ a * ㄧ = ㄧ a ㄧ ^ n-1 is obtained by taking determinants on both sides of AA * = ㄧ a ㄧ E
ㄧAㄧㄧA*ㄧ=ㄧㄧAㄧE ㄧ= ㄧAㄧ^n
So ㄧ a * ㄧ = ㄧ a ㄧ ^ n-1
Supplement 3: (ㄧ a ㄧ a ')' = A / ㄧ a ㄧ is derived from the formula (KA) '= (1 / k) a and (a') '= a
Supplement 4: this has nothing to do with the topic, but it's very useful. This kind of abstract matrix with no given number is used most in random derivation. Invertible matrix A, inversion, transpose and adjoint have nothing to do with the order, and can be arbitrarily reversed. For example, (a *) '= (a') *, the formula shows that the three operation symbols in the upper right corner can be exchanged

The solution of adjoint matrix
I can't get the answer

Let d be a determinant of order n, and AIJ (I, J are the lower corner marks) be the element on the i-th row and j-th column of D. after the i-th row and j-th column of AIJ are crossed out in D, the remaining determinant of order n-1 is called the "remainder" of element AIJ, denoted as mij

How to calculate matrix A ^ 3? Do you calculate a * a first and then (a * a) * a?
For example, a = 0 1 1
0 0 1
0 0 0

Yes
A^3=(A*A)*A
A^3=A*(A*A）