Linear algebra problem: let a be a square matrix of order n, a * be the adjoint matrix of a, if / A / = a ≠ 0, then / A * / = () Let a be a square matrix of order n, and a * be the adjoint matrix of A. if / A / = a ≠ 0, then / A * / = () A、a B、an-1 C、1/a D、an In option B, n-1 is superscript, and in option D, n is superscript!

Linear algebra problem: let a be a square matrix of order n, a * be the adjoint matrix of a, if / A / = a ≠ 0, then / A * / = () Let a be a square matrix of order n, and a * be the adjoint matrix of A. if / A / = a ≠ 0, then / A * / = () A、a B、an-1 C、1/a D、an In option B, n-1 is superscript, and in option D, n is superscript!

|A|=a≠0
Then a is invertible, and a (- 1) is the inverse matrix of A
A(-1)= A*/|A|
A* = |A|A(-1)
AA * = |a|e (E is the identity matrix)
|A||A*| = ||A|E|=|A|^n
|A*|=|A|^(n-1)=a^(n-1)
Choose B

Let a be an n-dimensional sequence vector, a ^ t * a = 1, let H = e-2a * a ^ t, and prove that h is an orthogonal matrix

H^TH = (E-2aa^t)^T(E-2aa^t)
= (E-2aa^t)(E-2aa^t)
= E-2aa^t-2aa^t+4aa^taa^t
= E-4aa^t + 4 a(a^ta)a^t
= E - 4aa^t + 4aa^t
= E
So h is an orthogonal matrix

Let a be a matrix of order n, satisfying AA ^ t = e (E is the identity matrix of order n), a ^ t be the transpose matrix of a, and | a|

|A+E| = |A+AA^T|
= |A||E+A^T|
= |A||E+A|
So | a + e | = 0