Solving for B matrix in terms of A. Can someone check my answer?

[thepatient]

Homework Statement

Suppose P is invertible and A = PBP^-1. Solve for B in terms of A.

Homework Equations

(AB)^-1 = B^-1*A^-1

The Attempt at a Solution

multiply from the left of each side of the equation by P^-1:
P^-1 *A = BP^-1

Take the inverse of both sides:

(P^-1*A)^-1 = (BP^-1)^-1
A^-1 * P = P*B^-1

Multiply from the left of each side of the equation by P^-1:
P^-1*A^-1*P = B^-1

Take the inverse of both sides:

[P^-1(A^-1*P)]^-1 = B

B = (A^-1*P)^-1*P
B = P^-1 *A * P


Is that correct?

 

[LCKurtz]

Yes, it looks OK but is much more complicated than it needs to be. At the first step where you have $P^{-1}A= BP^{-1}$ just multiply by $P$ on the right.

 

[thepatient]

Hahaaaa. That's right. Thanks. :]

 

[thepatient]

I wasn't sure if you can multiply from the right too.

 

[Ray Vickson]

I wasn't sure if you can multiply from the right too.

Why not? Two matrices can be multiplied as long as their row and column numbers match up properly. If all three of A, B and P (and P^(-1)) are nxn they can be multiplied in any order.

RGV