Rearranging matrix

Question

I have a question in one of my assignments that is as follows:

Find the matrix A (in terms of B,C, and D) if

$(2B^{-1}A^{-1}C)^{-1} = D$

Possible options for answer are as follows:

A.) $2BDC$

B.) $\frac{1}{2}B^{-1}DC $

C.) $\frac{1}{2}BDC$

D.) $2CDB^{-1}$

E.) $\frac{1}{2}CDB $

F.) $2CDB$

G.) $\frac{1}{2}CDB^{-1} $

H.) $2B^{-1}DC $

So, my solution was:

$(2B^{-1}A^{-1}C)^{-1} = D$

$I = (2B^{-1} A^{-1}C) D $

$ IA = (2B^{-1} C) D$

$A = 2B^{-1}CD$

My answer is obviously not listed...any ideas? I mean mine is close in to D, but obviously multiplication order matters with matrices so its wrong.

Answer 1

Your mistake is here $IA = (2B^{-1} C) D$. Why did $A$ go to the other side ?

The correct way to do it:

Use $(X_1X_2X_3)^{-1} = X_3^{-1}X_2^{-1}X_1^{-1}$ if the three matrices are square of course (and invertible). \begin{equation} (2B^{-1}A^{-1}C)^{-1} = D \end{equation}

\begin{equation} \frac{1}{2} C^{-1}AB = D \end{equation} So \begin{equation} C^{-1}AB = 2D \end{equation} So \begin{equation} AB = 2CD \end{equation} So \begin{equation} A = 2CDB^{-1} \end{equation} So $D$ is one choice.

Answer 2

Use that

$$(2B^{-1}A^{-1}C)^{-1} =\frac12C^{-1}AB= D \implies C^{-1}AB=2 D \implies CC^{-1}ABB^{-1}=2 CDB^{-1} $$