Let X and y be n-dimensional column vectors, and x ^ t * y = 1, matrix A = e + X * y ^ t, then a is an invertible matrix, and find a ^ - 1

Let X and y be n-dimensional column vectors, and x ^ t * y = 1, matrix A = e + X * y ^ t, then a is an invertible matrix, and find a ^ - 1

A^2=E+2X*Y^T+X*Y^TX*Y^T=E+3X*Y^T=3A-2E
A^2-3A=-2E
A(A-3E)=-2E
A^-1=0.5(3E-A)

Let n-dimensional vector α (a, 0,0.0, a), a

Because a and B are reversible
So e = AB = (E - α ^ t) [e - (1 / a) α ^ t] = e - (1 / a) α ^ t - (1 / a) (α ^ t) ^ 2
So o = α ^ t + (α ^ t) ^ 2
So a ^ 2 + a = 0
So a = - 1

Let n-dimensional determinant a = (1 / 2,0,..., 0,1 / 2), matrix A = I-A ^ t, B = I + 2A ^ TA, where I is the unit vector of order n, then AB =?
Hope that the master guidance, write a detailed process, thank you

I'll change the identity matrix I to e so that I can see more clearly
AB=(E-a^Ta)(E+2a^Ta)=E-a^Ta+2a^Ta-2(a^Ta)(a^Ta)=E+a^Ta-2a^T(aa^T)a
=E+a^Ta-2[(1/2)(1/2)+(1/2)(1/2)]a^Ta=E

Let a = e-2aat, where e is the n-order identity matrix and a is the n-dimensional identity column vector. It is proved that any n-dimensional vector B has / / AB / / = / / b//

There are three steps
1. Because a is an n-dimensional unit column vector, it has
A'a = 1 (a '= at)
2.A'A = (E-2aa')(E-2aa') = E - 4aa' + 4aa'aa' = E-4aa'+4aa' = E
3.||AB|| = √(AB)'(AB) = √B'A'AB = √B'B = ||B||.