Let n-order square matrix a satisfy a * A-A + e = 0, and prove that a is an invertible matrix

Let n-order square matrix a satisfy a * A-A + e = 0, and prove that a is an invertible matrix

Sweat, it's Square I thought it was accompanied by
A²-A+E=0
E=A-A²=A(E-A)
(E-A)A=A-A²=E
So a is invertible and the inverse matrix is e-a

Let the square matrix a satisfy a * a-a-2e = 0, prove that the matrix A + e is invertible and find it

A * a-a-2e should be written as: A ^ 2-a-2e,
A^2-A-2E=（A+E)（A-2E）?
It's impossible to have a + e reversibility,

Let a and B be square matrices of order n, e be identity matrices of order n, and ab = a-b. it is proved that a + B is invertible

Most problematic, there can be counter examples, such as let a = b = 0
Let AB = A-B = 0
But a + B = 0, irreversible