Showing the inverse of the product of scalar and matrices

Question

Given that $k$ is a nonzero number, and $A$, $B$ and $C$ are non-singular matrices, how to show that

$(kABC)^{-1} = {1\over k}C^{-1}B^{-1}A^{-1}$

Answer 1

Direct calculation! $$(kABC)\left(\frac{1}{k}C^{-1}B^{-1}A{-1}\right) = \left(\frac{k}{k}\right)ABCC^{-1}B^{-1}A^{-1} = ABB^{-1}A^{-1} = AA^{-1} = I $$

Answer 2

Hint: First two steps I would do... Multiply both sides by $k$ and then multiply on the left of both sides by $C$. Giving you the following events so far: $\frac{1}{k}C^{-1}B^{-1}A^{-1}U=I$

$k \cdot \frac{1}{k}C^{-1}B^{-1}A^{-1}U=k \cdot I$

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

$IB^{-1}A^{-1}U=kC$

$B^{-1}A^{-1}U=kC$

In the one step above I wrote the constant first because it doesn't matter when that gets multiplied...

Answer 3

By using the fact that $(ST)^{-1}=T^{-1}S^{-1}$ and $(kS)^{-1}= \frac{1}{k}S^{-1}$ for any non-singular matrices $S,T$ and any non-zero number $k$.

$(kABC)^{-1}=(BC)^{-1}(kA)^{-1}=C^{-1}B^{-1}* \frac{1}{k}A^{-1}= \frac{1}{k}C^{-1}B^{-1}A^{-1}$