Is the value of the determinant of the inverse matrix consistent with the original matrix?

Obviously not, they are reciprocal
|A^-1||A|=|E|=1

Let n-order non-zero real matrix a satisfy that the adjoint matrix of a is equal to the transpose of a, and prove that the determinant of a is equal to one, and a is an orthogonal matrix

Firstly, when n > 1, we have the following results about the rank of adjoint matrix
If R (a) = n, then R (a *) = n;
If R (a) = n-1, then R (a *) = 1;
If R (a) < n-1, then R (a *) = 0
It is proved that if R (a) = n, a is reversible, | a | ≠ 0
From a * a = | a | · e, a * = | a | · a ^ (- 1) is also reversible
If R (a) = n-1, a has non-zero subformula of order n-1, so a * ≠ 0, R (a *) ≥ 1
And a * a = | a | · e = 0, so r (a *) + R (a) ≤ R (a * a) + n = n, then R (a *) = 1
When R (a) < n-1, all the n-1 subformulas of a are 0, so a * = 0, R (a *) = n
Back to the original problem, we get R (a *) = R (a ') = R (a) from the condition a * = a'
When n > 2, according to the above conclusion, only R (a) = n, so | a | ≠ 0
For a * a = | a | · e, take the determinant to get | a * | ·| a | = | a | ^ n
So there is | a | ^ 2 = | a '| ·| a | = | a * | ·| a | = | a | ^ n, and the solution is | a | = 1 (| a | is a non-zero real number)
Then a'a = a * a = e is obtained, that is, a is an orthogonal matrix
There is a counterexample when n = 1,2, for example, a = 2E

The third power of matrix A is equal to 0 and the square of a is not equal to 0

A =
0 1 0
0 0 1
0 0 0
Meet the requirements: A ^ 3 = 0, a ^ 2 ≠ 0

N-order matrix A ^ 2 = a, R (a) = R, why is λ = 1 R multiple eigenvalue and 0 r multiple eigenvalue
How can we find out the multiplicity of 1 and 0?
It seems that there is no way to find the multiplicity of the eigenvalues of an abstract matrix. All of them are based on the power of the specific determinant value

This question 0 is n-r

Is the value of the determinant of the inverse matrix consistent with the original matrix?