IF A and B are squre matrices of order 3, then the true statement is/are (where I is unit matrix).

Options

A. det(-A)=-detA
B. If AB is singular then atleast one of A or B is singular
C. `det(A+I)=1+detA`
D. `det(2A)=2^(3)"detA"`

Answer

To solve the problem, we need to analyze the statements given regarding square matrices A and B of order 3. We will evaluate each statement one by one.

### Step-by-Step Solution:

1. **Statement 1**: The determinant of -A is equal to -det(A).
- The determinant of a matrix has the property that for any scalar \( k \), \( \text{det}(kA) = k^n \cdot \text{det}(A) \) where \( n \) is the order of the matrix. For \( k = -1 \) and \( n = 3 \), we have:
\[ \text{det}(-A) = (-1)^3 \cdot \text{det}(A) = -\text{det}(A). \]
- **Conclusion**: This statement is **true**.

2. **Statement 2**: If AB is singular, then at least one of A or B is singular.
- A matrix is singular if its determinant is zero. If \( AB \) is singular, then:
\[ \text{det}(AB) = \text{det}(A) \cdot \text{det}(B) = 0. \]
- This implies that at least one of \( \text{det}(A) = 0 \) or \( \text{det}(B) = 0 \) must hold true.
- **Conclusion**: This statement is **true**.

3. **Statement 3**: \( \text{det}(A + I) = \text{det}(A) + \text{det}(I) \).
- The determinant of the sum of two matrices is not equal to the sum of their determinants. This is a known property in linear algebra. Therefore, we cannot say:
\[ \text{det}(A + I) \neq \text{det}(A) + \text{det}(I). \]
- **Conclusion**: This statement is **false**.

4. **Statement 4**: \( \text{det}(2A) = 2^3 \cdot \text{det}(A) \).
- Using the property of determinants again, we have:
\[ \text{det}(kA) = k^n \cdot \text{det}(A) \quad \text{for } n = 3. \]
- Thus, \( \text{det}(2A) = 2^3 \cdot \text{det}(A) = 8 \cdot \text{det}(A) \).
- **Conclusion**: This statement is **true**.

### Final Conclusion:
The true statements are 1, 2, and 4. Statement 3 is false.