If `A` is a singular matrix, then adj `A` is a. singular b. non singular c. symmetric d. not defined

Options

A. singular
B. non-singular
C. symmetic
D. not defined

Answer

To solve the problem, we need to determine the nature of the adjoint of a singular matrix \( A \).

### Step-by-Step Solution:

1. **Understanding Singular Matrix**:
A matrix \( A \) is said to be singular if its determinant is zero. This means that \( \text{det}(A) = 0 \).

**Hint**: Recall that a matrix is singular if it does not have an inverse, which occurs when its determinant is zero.

2. **Using the Property of Adjoint**:
We know the property that relates a matrix and its adjoint:
\[ A \cdot \text{adj}(A) = \text{det}(A) \cdot I \]
where \( I \) is the identity matrix of the same order as \( A \).

**Hint**: Remember that the adjoint of a matrix is used to find the inverse, and this property is fundamental in matrix algebra.

3. **Taking Determinants**:
Taking the determinant of both sides of the equation:
\[ \text{det}(A \cdot \text{adj}(A)) = \text{det}(\text{det}(A) \cdot I) \]
This simplifies to:
\[ \text{det}(A) \cdot \text{det}(\text{adj}(A)) = \text{det}(A)^n \]
where \( n \) is the order of the matrix \( A \).

**Hint**: Use the property that the determinant of a product of matrices is the product of their determinants.

4. **Substituting Determinant of A**:
Since \( A \) is singular, we have \( \text{det}(A) = 0 \). Substituting this into the equation gives:
\[ 0 \cdot \text{det}(\text{adj}(A)) = 0^n \]
This simplifies to:
\[ 0 = 0 \]
which is always true.

**Hint**: This means we cannot conclude anything directly about \( \text{det}(\text{adj}(A)) \) from this equation alone.

5. **Finding the Determinant of the Adjoint**:
From the property of the determinant of the adjoint, we know:
\[ \text{det}(\text{adj}(A)) = \text{det}(A)^{n-1} \]
Since \( \text{det}(A) = 0 \), we have:
\[ \text{det}(\text{adj}(A)) = 0^{n-1} = 0 \]

**Hint**: The determinant of the adjoint is related to the determinant of the original matrix raised to the power of \( n-1 \).

6. **Conclusion**:
Since \( \text{det}(\text{adj}(A)) = 0 \), the adjoint of a singular matrix \( A \) is also singular.

**Final Answer**: The correct option is **a. singular**.