LINEAR algebra 231

Question

Assuming all matrices are nxn and invertible solve for D

BCA^T DCB^T A= BC^T 

 

Answer 1

The useful fact here is that the inverse of the transpose is the transpose of the inverse, i.e. (A^T)^{-1} = (A^{-1})^T. I'll use A' to mean A^T (not derivative!) in order to keep the notation simpler.

We're given that

BCA'DCB'A = BC'

We can cancel B on the left by multiplying with B^{-1} so that we get

CA'DCB'A = C'

Now multiply both on the left by C^{-1}

A'DCB'A = C^{-1} C'

Multiply by the inverse of A' on the left to get

DCB'A = (A')^{-1}C^{-1}C'

(You can't cancel the terms C^{-1} and C'.)

Now start canceling on the right, beginning by multiplying by A^{-1}

DCB' = (A')^{-1}C^{-1}C' A^{-1}

Now multiply by (B')^{-1}

DC = (A')^{-1}C^{-1}C' A^{-1} (B')^{-1}

Finally cancel the C by multiplying both sides by C^{-1}

D = (A')^{-1}C^{-1}C' A^{-1} (B')^{-1} C^{-1}