Question

Assuming that all matrices are n x n and invertible, solve for D ?? A B (A^T) D B (C^T) C = A B^T.

Answer 1

I'm assuming \(A^T\) is the transpose of \(A\)?

Answer 2

If so, just write out what you have. \[A \cdot B \cdot A^T \cdot D \cdot B \cdot C^T \cdot C=A B^T\]

Answer 3

Then start multiplying by inverses on either side until you only have \(D\) on the left side. \[(A B A^T)^{-1} \cdot (ABA^T) \cdot D \cdot (BC^TC)(BC^TC)^{-1}=D=\]\[(ABA^T)^{-1} AB^T (BC^TC)^{-1}\]

Answer 4

Did this make sense?

Answer 5

\[ D = B^{-1} A^{-T} B^{T} B^{-1} C^{-T} C^{-1} This is true ?? =(\]

Answer 6

Close. Rather, \[D=(A^T)^{-1} B^{-1} B^T C^{-1} (C^T)^{-1}B^{-1}\]Remember, if you have three invertible matrices \(X, Y, Z\) then\[(XYZ)^{-1} = Z^{-1}Y^{-1}X^{-1}\]

Answer 7

\[OK ,, D = ( B A^{T} B C^{T} C )^{-1} B^{T} \]

Answer 8

No... \[(BA^TBC^TC)^{−1}=C^{-1} (C^T)^{-1} B^{-1}(A^T)^{-1} B^{-1}\]This is not the same result as what we obtained above.

Answer 9

why? =""
this is same =$

Answer 10

King has it correct
\[ D=(A^T)^{-1} B^{-1} B^T C^{-1} (C^T)^{-1}B^{-1} \]
The easiest way to get this is to premultiply on the left:

multiply both sides on the left by A^-1, then B^-1 then A^-T (inverse and transpose)
post multiply on the right in the order
C^-1
the C^-T
then B^-1

Answer 11

Thx king and phi
I understand what you are saying
but the issue from the outset so
A B (A^T) D B (C^T) C = A B^T.
First : (A A^-1) B (A^T) D B (C^T) C = (AA^-1) B^T
: B (A^{T}) D B (C^{T}) C = B^{T}
: (BB^{-1}) (A^{T} A^{-T}) D B C^{T} C = B^{T} B^{-1} A^{-T}
: D B (C^{T}) C = B^{T} B^{-1} A^{-T}
: D (C^{T} C^{-T}) (C C^{-1}) (B B^{-1}) = B^{T} B^{-1} A^{-T} C^{-T} C^{-1} B^{-1}
: D = B^{T} B^{-1} A^{-T} C^{-T} C^{-1} B^{-1}
: D = ( B C C^{T} A^{T} B )^{-1} B^{T}
This is reached =$

Answer 12

As you know matrix multiplication is (in general) not commutative. That is
AB≠BA
So you must be careful when multiplying, to always multiply on the same side:

So it is more correct to write
( A^-1A) B (A^T) D B (C^T) C = (A^-1A) B^T (A^-1 goes on the left)
than what you wrote:(A A^-1) B (A^T) D B (C^T) C = (AA^-1) B^T

Going from
: B (A^{T}) D B (C^{T}) C = B^{T}
to
: (BB^{-1}) (A^{T} A^{-T}) D B C^{T} C = B^{T} B^{-1} A^{-T}
is WRONG. (You cannot switch the order of B^T and B^-1 on the right-hand side.)

Answer 13

أهاا ,,لم انتبه لذلك اشكرك :(

Answer 14

A B (A^T) D B (C^T) C = A B^T.
(A^{-1} A) B (A^{T}) D B (C^{T}) C = (A^{-1}A) B^{T}
(B^{-1} B) A^{T} D B (C^{T}) C = B^{T} B^{-1}
(A^-1)^{T} A^{T} D B (C^{T}) C = (A^-1)^{T} B^{T} B^{-1}
D (B^{-1}B) (C^{T}) C = (A^-1)^{T} B^{T} B^{-1} B^{-1}
D (C^-1)^{T} (C^{T}) C = (A^-1)^{T} B^{T} B^{-1} B^{-1} (C^T)^{-1}
D C^{-1} C = (A^-1)^{T} B^{T} B^{-1} B^{-1} (C^T)^{-1} C^{-1}
D = (A^-1)^{T} B^{T} B^{-1} B^{-1} (C^T)^{-1} C^{-1}
<<I hope that to be true phi

Answer 15

On your 2nd line you start with
B (A^{T}) D B (C^{T}) C = B^{T}
(where I have removed A^-1 A because that is just I, the identity matrix)

you then wrote
(B^{-1} B) A^{T} D B (C^{T}) C = B^{T} B^{-1}

BUT this is not correct. You multiply B^-1 on the left side (you pre-multiplied) on the LHS (left hand side of the equal sign). But on the RHS of the equal you post-multiplied by B^-1. You should have written
(B^{-1} B) A^{T} D B (C^{T}) C = B^{-1}B^{T}

Remember order matters. If you start with the equation
A=A
and then multiply both sides by B:
BA= BA is correct, but
BA=AB is not correct.