Matrix Flipping

For any pair of invertible matrices $A$ and $B$ which can be multiplied, the inverse of the matrix $AB$ is $\left( AB \right)^{-1} = B^{-1}A^{-1}$ .

Now, we can proceed by induction on $n$ for the inverse of the product of $n$ invertible $m \times m$ matrices: $\left( A_1 \cdot A_2 \cdots A_{n-1} \cdot A_n \right)^{-1} = \left( \left( A_1 \cdot A_2 \cdots A_{n-1} \right) \cdot A_n \right)^{-1} \\ = A_n^{-1} \cdot \left( A_1 \cdot A_2 \cdots A_{n-2} \cdot A_{n-1} \right)^{-1} = \cdots = A_n^{-1} \cdot A_{n-1}^{-1} \cdots A_2^{-1} \cdot A_1^{-1}.$

Hence, the statement is true for all $n$ .