Assuming that all matrices are nxn and invertible, solve for D in the equation.
$\displaystyle ABC^TDBA^T = AB^T$
How can I do it?
Please advise me the direction to solve this.
Thank you.
I am a bit rusty on these ones but i'll give it a shot.
We need to do some left multiplying on both sides
$\displaystyle ABC^TDBA^T = AB^T$
$\displaystyle A^{-1}ABC^TDBA^T = A^{-1}AB^T$
gives
$\displaystyle IBC^TDBA^T = IB^T$
$\displaystyle BC^TDBA^T = B^T$
try the same thing with B
$\displaystyle B^{-1}BC^TDBA^T = B^{-1}B^T$
gives
$\displaystyle IC^TDBA^T = B^{-1}B^T$
$\displaystyle C^TDBA^T = B^{-1}B^T$
this is where it gets tricky I would try
$\displaystyle C^{T^{-1}}C^TDBA^T = C^{T^{-1}}B^{-1}B^T$
which gives
$\displaystyle IDBA^T = C^{T^{-1}}B^{-1}B^T$
$\displaystyle DBA^T = C^{T^{-1}}B^{-1}B^T$
Now we need to do some right multiplying
$\displaystyle DBA^TA^{T^{-1}} = C^{T^{-1}}B^{-1}B^TA^{T^{-1}}$
$\displaystyle DBI = C^{T^{-1}}B^{-1}B^TA^{T^{-1}}$
$\displaystyle DB = C^{T^{-1}}B^{-1}B^TA^{T^{-1}}$
finally
$\displaystyle DBB^{-1} = C^{T^{-1}}B^{-1}B^TA^{T^{-1}}B^{-1}$
$\displaystyle DI = C^{T^{-1}}B^{-1}B^TA^{T^{-1}}B^{-1}$
$\displaystyle D = C^{T^{-1}}B^{-1}B^TA^{T^{-1}}B^{-1}$
I wouldn't take this as bible!