A is a matrix of order 3, a * is the adjoint matrix of a, | a | = 1, find | (2a) ^ - 1 + 3a*|

A is a matrix of order 3, a * is the adjoint matrix of a, | a | = 1, find | (2a) ^ - 1 + 3a*|

|（2A）^-1 +3A*|
= | 1/2 A^-1 + 3|A|A^-1 |
= | -5/2A^-1 |
= (-5/2)^3 |A|^-1
= - 125/8

Let n-order matrix a satisfy a ＾ 2-2a + 2I = 0, prove that matrix a-3i is invertible, and find (a-3i) ＾ - 1

Reduce 5I on both sides at the same time
A ＾ 2-2a-3i = - 5I was obtained
(a-3i)(a+i)=-5i
(-1/5(a+i))(a-3i)=i
So the inverse matrix of a-3i is - 1 / 5 (a + I)
Because there is an inverse matrix, it is invertible

Let n-order matrix a satisfy a ＾ 2 + a-3i = 0, prove that matrix a-2i is invertible, and find (a-2i) ＾ - 1

Note: I should be written as capital I, but it looks like 1. It can also be written as E
Because a ^ 2 + a-3e = 0
So a (a-2e) + 3 (a-2e) + 3E = 0
That is, (a + 3e) (a-2e) = - 3E
So a-2e is reversible and (a-2e) ^ - 1 = (- 1 / 3) (a + 3e)