If there is $A B C = D$. and $A$, $B$ and $D$ are given.

How can we solve for $C$?

I'm able to solve when there's two matrices, but not sure when there are three ($A$, $B$, and $C$)

Thank you.

  • 2
    $\begingroup$ Note that $(AB)^{-1}=B^{-1}A^{-1}$... $\endgroup$ – J. M. isn't a mathematician Aug 10 '11 at 8:33
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    $\begingroup$ Give the matrix AB the name E. Then you have EC=D. Does that help? $\endgroup$ – Jonas Meyer Aug 10 '11 at 8:33

Assuming $A$ and $B$ are invertible (square and columns are linearly independent, or one of the many thousands (exaggeration) of possibilities for the Invertible Matrix Theorem) which implies $C$ and $D$ are square, then just break it down into steps:

$A \cdot B \cdot C = D$

Now multiply each side by $A^{-1}$

$A^{-1}\cdot A \cdot B \cdot C = A^{-1} \cdot D $

$I$ is the n x n identity matrix.

$I \cdot B \cdot C = A^{-1} \cdot D $

$B \cdot C = A^{-1} \cdot D$

Now, rinse and repeat.

$C = B^{-1} \cdot A^{-1} \cdot D$

And if you look at it as the others mentioned, if you let $E = AB$, then $E^{-1} = B^{-1} A^{-1}$, so:

$E C = D \implies C = E^{-1} D \implies C = B^{-1} A^{-1} D$

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    $\begingroup$ I would recommend using \cdot or nothing, instead of \times. $\endgroup$ – Mikael Öhman Aug 10 '11 at 11:43
  • $\begingroup$ @Mikael Noted and used. $\endgroup$ – norcalli Aug 10 '11 at 11:45

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