To show that $(AB)^{-1} = B^{-1}A^{-1}$, let's use the following properties of matrix multiplication and inverses:

1. Definition of a Matrix Inverse: For any invertible matrix $M$, $M^{-1}$ is the matrix such that: $M M^{-1} = M^{-1} M = I,$ where $I$ is the identity matrix.

2. Associativity of Matrix Multiplication: Matrix multiplication is associative, meaning that for matrices $A$, $B$, and $C$: $A(BC) = (AB)C.$

Now, given that $A^{-1}$, $B^{-1}$, and $AB$ exist, we want to show that $(AB)^{-1} = B^{-1}A^{-1}$.

Proof:

We want to prove that $(AB)(B^{-1}A^{-1}) = I$, where $I$ is the identity matrix.

1. Start with the product $(AB)(B^{-1}A^{-1})$: $(AB)(B^{-1}A^{-1}).$

2. By associativity of matrix multiplication, group the terms as follows: $= A(BB^{-1})A^{-1}.$

3. Since $BB^{-1} = I$, the identity matrix, we have: $= AIA^{-1}.$

4. The identity matrix $I$ acts as a neutral element in matrix multiplication, so $AI = A$. Thus, we get: $= AA^{-1}.$

5. Since $AA^{-1} = I$, we have: $= I.$

Therefore, $(AB)(B^{-1}A^{-1}) = I$, which shows that $B^{-1}A^{-1}$ is a right inverse of $AB$.

To check if $B^{-1}A^{-1}$ is also a left inverse, we compute:

$(B^{-1}A^{-1})(AB).$

Again, using the associativity of matrix multiplication:

$= B^{-1}(A^{-1}A)B = B^{-1}IB = B^{-1}B = I.$

Thus, $B^{-1}A^{-1}$ is both a left and right inverse of $AB$, meaning:

$(AB)^{-1} = B^{-1}A^{-1}.$

This completes the proof.

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Would you like further details on this, or do you have any questions?

Here are some related questions to further expand your understanding:

1. What happens when only $A$ or $B$ is invertible, but not both?
2. How does this result extend to the product of more than two matrices?
3. Can we apply this proof to non-square matrices?
4. How does the proof change if $A$ and $B$ are not invertible?
5. What are the consequences of this result for block matrices?

Tip: Always check whether the inverse of a product exists before applying this property in any computation.