Let a = (1,0,1) t, matrix A = AA, linear algebra Let a = (1,0,1) t, matrix A = AAT, find a ^ n and [2I + a]

Let a = (1,0,1) t, matrix A = AA, linear algebra Let a = (1,0,1) t, matrix A = AAT, find a ^ n and [2I + a]

Adjoint matrix: let a be a square matrix of order (n > = 2) and a * be the adjoint matrix of A. It is proved that the sufficient and necessary condition for R (a *) = n is R (a) = n-1
Is it necessary to combine the rank of matrix with the properties of Adjoint Matrix

Your conclusion is wrong. If R (a *) = n, then R (a) = n, that's right
I will prove a more difficult idea, that is, if R (a) = n-1, then R (a *) = 1
Since R (a) = n-1, there is a behavior 0 | a | = 0 in a, which has a non-zero subformula of order n-1, so r (a *) > = 1
Because AA * = |a|e = 0
r(A*)+r(A)

In linear algebra, let the eigenvalues of a matrix of order 3 be 1, - 1,2, and find | a * + 3a-2i |
In linear algebra, let the eigenvalues of a matrix of order 3 be 1, - 1,2, and find | a * + 3a-2i |. The answer is 9?

A*=|A|A^(-1)
|A|=1×（-1）×2=-2
Namely
A*+3A-2I=|A|A^(-1)+3A-2I
=-2A^(-1)+3A-2I
The characteristic equation is
-2/λ+3λ-2
therefore
The eigenvalue is: - 2 + 3-2 = - 1
-2/(-1)-3-2=-3
-2/2+6-2=3
thus
The original formula = - 1 × (- 3) × 3 = 9