Let A and B be two matrices of order n `xx` n. Let A be non-singular and B be singular. Consider the following :
1. AB is singular
2. AB is non-singular
3. `A^(-1)` B is singular
4. `A^(-1)` B is non singular
Which of the above is/are correct ?

Options

A. 1 and 3
B. 2 and 4 only
C. 1 only
D. 3 only

Explanation

To solve the problem, we need to analyze the properties of the matrices A and B given that A is non-singular and B is singular.

### Step-by-step Solution:

1. **Understanding Non-Singular and Singular Matrices**:
- A matrix A is non-singular if its determinant is non-zero, i.e., \( \text{det}(A) \neq 0 \).
- A matrix B is singular if its determinant is zero, i.e., \( \text{det}(B) = 0 \).

2. **Analyzing the Product AB**:
- We need to determine whether the product \( AB \) is singular or non-singular.
- Since \( A \) is non-singular, we know that \( \text{det}(AB) = \text{det}(A) \cdot \text{det}(B) \).
- Given that \( \text{det}(B) = 0 \), we have:
\[ \text{det}(AB) = \text{det}(A) \cdot 0 = 0 \]
- Therefore, \( AB \) is singular.

3. **Analyzing the Matrix \( A^{-1}B \)**:
- Next, we need to check whether \( A^{-1}B \) is singular or non-singular.
- Since \( A \) is non-singular, \( A^{-1} \) exists and is also non-singular.
- The determinant of the product \( A^{-1}B \) can be calculated as:
\[ \text{det}(A^{-1}B) = \text{det}(A^{-1}) \cdot \text{det}(B) \]
- Since \( \text{det}(B) = 0 \), we have:
\[ \text{det}(A^{-1}B) = \text{det}(A^{-1}) \cdot 0 = 0 \]
- Thus, \( A^{-1}B \) is also singular.

4. **Conclusion**:
- From our analysis, we conclude:
- \( AB \) is singular (Statement 1 is correct).
- \( A^{-1}B \) is singular (Statement 3 is correct).
- Therefore, Statements 2 and 4 are incorrect.

### Summary of Correct Statements:
- **Correct Statements**: 1 and 3.
- **Incorrect Statements**: 2 and 4.

### Final Answer:
The correct options are 1 and 3.

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