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Let square matrix a satisfy a ^ 2-2a + 4E = O, prove that a + E and a-3e are invertible, and find their inverse matrix I know how to do it, but I don't understand why. Could you please explain how to do it? That is, list the process and explain each step. Thank you!
Let n-order square matrix a satisfy a * a = 10e, prove that a-3e is invertible, and find the inverse matrix of a-3e
(A-3E)(A+3E)=E
So a-3e is invertible, and the inverse matrix of a-3e is a + 3E
Find the n-order matrix a satisfying a + a-3e = 0, prove that a and a + 2E are invertible, and find their inverse matrix
Syndrome a is reversible
A²+A-3E=0
A（A+E）=3E
A（A+E）/3=E
So a is invertible and the inverse matrix of a is (a + e) / 3
Syndrome a + 2E reversible
A²+A-3E=0
（A+2E）（A-E）=E
So a + 2E is invertible and the inverse matrix of a + 2E is a-e
I wish you a happy study!
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