# If A = BCD show that $C^{-1}$ = $DA^{-1}B$

I came across this question in a past paper, If A = BCD show that $C^{-1}$ = $DA^{-1}B$. All these matrices are sqaure and have inverses.

I attempted a solution but I am not sure if-
1. The solution is correct and;
2. Is this the best solution

Here is my attempt: *

$A = BCD$
$A^{-1}A = I = A^{-1}BCD$ This implies that $D^{-1} = A^{-1}BC$ and now multiply both sides by $D$ and we have:
$DD^{-1} = I = DA^{-1}BC$ from which the result follows.

• Right-multiply both sides with the inverse of the rightmost matrix on the side where $C$ is until $C$ has moved (as $C^{-1}$) to the other side. Then left-multiply with the inverse of the leftmost matrix on that side until $C^{-1}$ stands alone to get the desired result.
– user139000
Oct 29 '14 at 11:16
• Thanks, but is my solution correct or no?
– dan
Oct 29 '14 at 11:18
• Yours works fine! Oct 29 '14 at 11:24
• Asking if your (correct) solution is "best" is mostly a matter of opinion. It is fairly terse, but some might object to a lack of "motivation". Certainly if there were a number of such formulas to verify, one might want to develop a more systematic approach. Oct 29 '14 at 12:49
• I agree my use of 'best' was slightly misleading, I was merely asking if my solution (the method) was appropriate :)
– dan
Oct 29 '14 at 17:58

$$A=BCD$$ $$B^{-1}A=CD$$ $$C^{-1}B^{-1}A=D$$ $$C^{-1}B^{-1}=DA^{-1}$$ $$C^{-1}=DA^{-1}B$$
Note that we do not consider the size of the matrices as they are all square, say if $A$ is $n \times n$ and B is not, the question itself would be not well-defined.
\begin{align} A&=BCD\\ B^{-1}A&=CD\\ C^{-1}B^{-1}A&=D\\ C^{-1}B^{-1}&=DA^{-1}\\ C^{-1}&=DA^{-1}B\\ \end{align}