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The image presents a problem with the following instructions:

- **Given:** \( A = S \Lambda S^{-1} \)
- **Task:** Take determinants to prove that \( \text{det}(A) = \text{det}(\Lambda) = \lambda_1 \lambda_2 \cdots \lambda_n \).
- **Additional Note:** This quick proof only works when \( A \) can be ______.

### Solution:

1. **Determinant of A:**
   Since \( A = S \Lambda S^{-1} \), take the determinant on both sides:
   \[
   \text{det}(A) = \text{det}(S \Lambda S^{-1})
   \]
   By the property of determinants, the determinant of a product of matrices equals the product of their determinants:
   \[
   \text{det}(A) = \text{det}(S) \cdot \text{det}(\Lambda) \cdot \text{det}(S^{-1})
   \]
   
2. **Determinant of the Inverse:**
   The determinant of the inverse of a matrix \( S \) is the reciprocal of the determinant of \( S \):
   \[
   \text{det}(S^{-1}) = \frac{1}{\text{det}(S)}
   \]
   Substituting this into the equation:
   \[
   \text{det}(A) = \text{det}(S) \cdot \text{det}(\Lambda) \cdot \frac{1}{\text{det}(S)}
   \]
   The \( \text{det}(S) \) terms cancel out:
   \[
   \text{det}(A) = \text{det}(\Lambda)
   \]

3. **Determinant of Lambda:**
   \( \Lambda \) is a diagonal matrix with eigenvalues \( \lambda_1, \lambda_2, \dots, \lambda_n \) on the diagonal. The determinant of a diagonal matrix is the product of its diagonal elements:
   \[
   \text{det}(\Lambda) = \lambda_1 \lambda_2 \cdots \lambda_n
   \]

4. **Conclusion:**
   Therefore:
   \[
   \text{det}(A) = \lambda_1 \lambda_2 \cdots \lambda_n
   \]

5. **Condition for Quick Proof:**
   The quick proof only works when \( A \) is **diagonalizable**. This is because \( A = S \Lambda S^{-1} \) assumes that \( A \) can be decomposed into a diagonalizable form.

### Final Answer:
The quick proof only works when \( A \) can be **diagonalized**.

---

Would you like more details on this solution or have any further questions?

### Related Questions:
1. What conditions must a matrix satisfy to be diagonalizable?
2. How is the determinant of a product of matrices related to the determinants of the individual matrices?
3. What is the significance of eigenvalues in determining the determinant of a matrix?
4. Can non-diagonalizable matrices be decomposed similarly? If so, how?
5. What are the implications of a matrix being non-diagonalizable?

### Tip:
Always verify if a matrix is diagonalizable by checking if it has a full set of linearly independent eigenvectors.

Understand how to prove that the determinant of matrix A equals the product of its eigenvalues using matrix diagonalization. This problem covers key concepts in linear algebra, including eigenvalues, determinants, and theorems related to matrix diagonalization. Suitable for advanced undergraduate students.

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