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I need to prove that if A and B are invertible matrices, then this statement is valid:

$(A*B)^{-1}=B^{-1}*A^{-1}$

Is it enough if I take some random matrices with variables and prove that left side and the right side give the same result? How else could I go about doing this?

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    Just compute: $B^{-1}A^{-1}AB$ – 2017-02-01
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    Well your approach would work if you would know the general form of an invertible matrix, but that's an extreme amount of work for something you can prove in on line. – 2017-02-01

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Consider the definition of an inverse. If $A$ is an $n$ x $n$ matrix, then $B$ is the inverse of A if and only if $AB = BA = I_n$. So as Zach is saying, we can see that $(AB) \cdot (B^{-1}A^{-1}) = ABB^{-1}A^{-1} = AI_nA^{-1} =AA^{-1} =I_n $ and therefore $(AB)^{-1} = (B^{-1}A^{-1})$.

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    You also use (in your last equation) that the inverse (if it exists) is unique. – 2017-02-01
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    The uniqueness comes from the definition of an inverse. If two matrices $AB = BA = I_n$, then $B$ is the only inverse for $A$ (because it is an if and only if statement). – 2017-02-01
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    Yes, i only said that you find two possible inverses, so they have to be the same (which might not have been clear to the OP). – 2017-02-01