Help with matrices property of inverse.

Question

I recently learnt to prove $(AB)^{-1} = B^{-1} A^{-1} $ where $A$ and $B$ are non-singular matrices.

I'm curious, is this property applicable for more than $2$ non singular matrices?

Eg: Will this property hold for, $$(A_1 \cdot A_2 \cdot A_3 \cdot \ ... \ \cdot A_n)^{-1} = A_n^{-1} \cdot A_{n-1}^{-1} \cdot \ ... \ \cdot A_3^{-1} \cdot A_2^{-1} \cdot A_1^{-1} $$

Thoughts?

Answer 1

Yes, and the proof is pretty straight forward: for any n, $(A_n^{-1}A_{n-1}^{-1}\cdot\cdot\cdot A_2^{-1}A_1^{-1})(A_1A_2\cdot\cdot\cdot A_{n-1}A_n)= (A_n^{-1}A_{n-1}^{-1}\cdot\cdot\cdot A_2^{-1})(A_1^{-1}A_1)(A_2\cdot\cdot\cdot A_{n-1}A_n)= (A_n^{-1}A_{n-1}^{-1}\cdot\cdot\cdot A_2^{-1})(A_2\cdot\cdot\cdot A_{n-1}A_n)= (A_n^{-1}A_{n-1}^{-1}\cdot\cdot\cdot A_3^{-1})(A_2^{-1}A_2)(A_3\cdot\cdot\cdot A_{n-1}A_n)$, etc.

Answer 2

Yes. For example, for 3 matrices: $$(ABC)^{-1}=(BC)^{-1}A^{-1}=C^{-1}B^{-1}A^{-1}$$ And you can do the same for as many matrices as you want.