Question about invertible matrices

[Cyborg31]

Homework Statement

A, B, I-Y^-1, X, Y are n x n invertible matrices

(A(I-Y^-1))^-1 = YB

Solve for Y

Homework Equations

The Attempt at a Solution

(I - Y^-1)^-1A^-1 = YB
(I - Y^-1)^-1A^-1B^-1 = Y
A^-1B^-1 = (I - Y^-1)Y
A^-1B^-1 = Y - I
A^-1B^-1*(A^-1B^-1)^-1 = Y - I(A^-1B^-1)^-1
I = Y - (A^-1B^-1)^-1
Y^-1*I = Y^-1*Y - (A^-1B^-1)^-1
Y^-1 = I - (A^-1B^-1)^-1
Inverse both sides
Y = (I - (A^-1B^-1)^-1)^-1
or is it
Y = I^-1 - A^-1B^-1
Y = I - A^-1B^-1

Is this correct?

 

[HallsofIvy]

Homework Statement

A, B, I-Y^-1, X, Y are n x n invertible matrices

(A(I-Y^-1))^-1 = YB

Solve for Y

Homework Equations

The Attempt at a Solution

(I - Y^-1)^-1A^-1 = YB
(I - Y^-1)^-1A^-1B^-1 = Y
A^-1B^-1 = (I - Y^-1)Y
A^-1B^-1 = Y - I

You can solve for Y from here immediately:
A^-1B^-2+ I= Y

A^-1B^-1*(A^-1B^-1)^-1 = Y - I(A^-1B^-1)^-1

Don't you mean = (Y- I)(A^-1B^-1)^-1

I = Y - (A^-1B^-1)^-1

This is wrong now.

Y^-1*I = Y^-1*Y - (A^-1B^-1)^-1
Y^-1 = I - (A^-1B^-1)^-1
Inverse both sides
Y = (I - (A^-1B^-1)^-1)^-1
or is it
Y = I^-1 - A^-1B^-1
Y = I - A^-1B^-1

Is this correct?

 

[Cyborg31]

Didn't know you could move -I to the other side like normal algebra.

Is the B^-2 a typo? Cause I don't know what that's supposed to represent.