12. Use the properties of the determinant function to show that if A and B are any n × n matrices, then: A •B is invertible if and only if A and B are both invertible.

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### Matrix Invertibility and Determinants

**Exercise 12:**
Use the properties of the determinant function to show that if \( A \) and \( B \) are any \( n \times n \) matrices, then \( A \cdot B \) is invertible **if and only if** \( A \) and \( B \) are both invertible.

**Explanation:**

To tackle this exercise, you will need to delve into the fundamental properties of determinants in linear algebra. Recall that for any square matrices \( A \) and \( B \), the following relationship holds:

\[ \det(A \cdot B) = \det(A) \cdot \det(B) \]

Using this property, you can ascertain the invertibility condition.

- **Invertibility Condition:**
  - A matrix \( A \) is invertible if and only if \( \det(A) \neq 0 \).
  - Similarly, \( B \) is invertible if and only if \( \det(B) \neq 0 \).

Hence, for the product \( A \cdot B \) to be invertible, both \( \det(A) \neq 0 \) and \( \det(B) \neq 0 \) must hold true.

Putting it all together, \( A \cdot B \) is invertible **if and only if** both \( A \) and \( B \) are invertible.

By following this path, you'll be able to demonstrate the invertibility of the matrix product using determinant properties effectively.
Transcribed Image Text:### Matrix Invertibility and Determinants **Exercise 12:** Use the properties of the determinant function to show that if \( A \) and \( B \) are any \( n \times n \) matrices, then \( A \cdot B \) is invertible **if and only if** \( A \) and \( B \) are both invertible. **Explanation:** To tackle this exercise, you will need to delve into the fundamental properties of determinants in linear algebra. Recall that for any square matrices \( A \) and \( B \), the following relationship holds: \[ \det(A \cdot B) = \det(A) \cdot \det(B) \] Using this property, you can ascertain the invertibility condition. - **Invertibility Condition:** - A matrix \( A \) is invertible if and only if \( \det(A) \neq 0 \). - Similarly, \( B \) is invertible if and only if \( \det(B) \neq 0 \). Hence, for the product \( A \cdot B \) to be invertible, both \( \det(A) \neq 0 \) and \( \det(B) \neq 0 \) must hold true. Putting it all together, \( A \cdot B \) is invertible **if and only if** both \( A \) and \( B \) are invertible. By following this path, you'll be able to demonstrate the invertibility of the matrix product using determinant properties effectively.
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