If a is a square matrix of order n, e is a unit matrix of order n, and a ^ 3 = O, it is proved that A-E is an invertible matrix!

If a is a square matrix of order n, e is a unit matrix of order n, and a ^ 3 = O, it is proved that A-E is an invertible matrix!

A ^ 3 = 0 deduces a ^ 3-E = - E. then (A-E) (a ^ 2 + A + e) = - E (this covariance formula holds because the multiplication of identity matrix E and a has commutative law), that is, (A-E) (- A ^ 2-a-e) = E
From the definition of matrix invertibility, we know that A-E is invertible, and its inverse matrix is a ^ 2-a-e

Let n-order square matrix a satisfy a ^ 2-A + e = 0, prove that a is an invertible matrix, and find the expression of a ^ - 1?
Why is a reversible if a (e-A) = e

Prove: because a ^ 2-A + e = 0, so a (e-A) = e, so a is invertible, and a ^ - 1 = e-a. supplement: This is a theorem, there should be in the textbook: if AB = e, then a, B are invertible, and a ^ - 1 = b, B ^ - 1 = a. the proof is very simple. Because AB = e both sides of the determinant | a | B | = | e | = 1, so | a | ≠ 0, | B | ≠ 0, so a

Let B be an invertible matrix, a be a square matrix of the same order as B, and satisfy A2 + AB + B2 = 0 {a square b square}. It is proved that both a and B are invertible matrices
How to prove that B is an inverse matrix?