# Thread: Solve for a matrix

1. ## Solve for a matrix

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?

Thank you.

2. 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!

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# Solve matrix(a b c)

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