Let a be a square matrix of order n, and satisfy AA ^ t = E and | a | = - 1, prove that the determinant | e + a | = 0 My question is why |A| |E+A'| = |A| |(E+A)'| = |A| |E+A|

Let a be a square matrix of order n, and satisfy AA ^ t = E and | a | = - 1, prove that the determinant | e + a | = 0 My question is why |A| |E+A'| = |A| |(E+A)'| = |A| |E+A|

Are you asking why the following three equations hold, or is it the title of your title?
If it's the following three equations
The first equation is because (E + a ') = e' + a '= (E + a)'
The second equation is because the determinant of a matrix is equal to its transposed determinant

Let a be a square matrix of order 2n + 1, and satisfy AA ^ t = e, | a | 0, prove the determinant | a-e|=

|A-E|
= |A-AA^T|
= |A(E-A^T)|
= |A||E-A^T|
= |A||E-A| --- (E-A^T)^T = E-A
= |A| (-1)^(2n+1) |A-E|
= -|A||A-E|
So | A-E | (1 + | a |) = 0
Because | a | > 0
So 1 + | a ≠ 0
So | A-E | = 0

Let a be a real square matrix of order n and AA ^ t = e, and prove that the determinant | a | = ± 1
5. Let a be a real square matrix of order n, and AA ^ t = e, prove that the determinant | a | = ± 1

Certification:
A A^T=E
|A| |A^T|=|E|
|A|^2=1
| A |= ±1.
Get proof
Property 1: | a | = | a ^ t|
Property 2: if the matrix AB = C has | a | B | = | C|