https://file.aiquickdraw.com/mathbot/file/1727337229086v0kotb7k.jpg

To solve this problem, let's analyze the given information:

- **Determinant of the matrix = -1**
- **Sum of eigenvalues = 0**

For a 2x2 matrix, the sum and product of its eigenvalues relate directly to its trace and determinant:
1. **Sum of eigenvalues** is equal to the trace of the matrix (the sum of the diagonal elements). Here, the sum is given as **0**.
2. **Product of eigenvalues** is equal to the determinant of the matrix. In this case, the determinant is given as **-1**.

Now, let the eigenvalues of the matrix be \(\lambda_1\) and \(\lambda_2\).
- From the given conditions:
   - \(\lambda_1 + \lambda_2 = 0\) (sum of eigenvalues)
   - \(\lambda_1 \times \lambda_2 = -1\) (determinant)

The only pair of numbers that satisfy both conditions are \(\lambda_1 = 1\) and \(\lambda_2 = -1\).

### Rank of the matrix:
- Since the eigenvalues are **non-zero** (i.e., 1 and -1), the matrix is invertible, and its rank is **2**.

Thus, the rank of the matrix is **2**.

Would you like more details or have any questions? 

Here are some related questions you might find useful:
1. What is the relationship between eigenvalues and the determinant of a matrix?
2. How do you compute the rank of a matrix using eigenvalues?
3. Can a matrix with a zero eigenvalue be full rank?
4. What is the geometric interpretation of eigenvalues?
5. How do the trace and determinant of a matrix relate to its eigenvalues?

**Tip:** Eigenvalues provide key insights into the structure of a matrix, including its rank, invertibility, and the type of transformations it represents.

If the determinant and sum of eigenvalues of a 2x2 matrix are -1 and 0 then, what can you say about the rank of the given matrix?

Learn how to determine the rank of a 2x2 matrix when given the determinant (-1) and the sum of eigenvalues (0). This problem covers key linear algebra concepts, such as eigenvalues, determinants, and matrix rank, and is suitable for advanced high school or college-level students.

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