Suppose A and B are​ matrices, where Upper A Superscript negative 1​, Upper B Superscript negative 1​, and AB all exist. Show that left parenthesis AB right parenthesis Superscript negative 1 Baseline equals Upper B Superscript negative 1 Baseline Upper A Superscript negative 1.

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.

---

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.

Suppose A and B are matrices, where A^(-1), B^(-1), and AB all exist. Show that (AB)^(-1) = B^(-1)A^(-1).

This detailed solution explains how to prove that the inverse of a product of matrices (AB)^(-1) equals B^(-1)A^(-1). It uses the properties of matrix multiplication, the definition of a matrix inverse, and the identity matrix. Suitable for college-level students studying linear algebra.

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