Let `A` be an nth-order square matrix and `B` be its adjoint, then `|A B+K I_n|` is (where `K` is a scalar quantity) `(|A|+K)^(n-2)` b. `(|A|+K)^n` c. `(|A|+K)^(n-1)` d. none of these

Options

A. `(|A|+K)^(n-2)`
B. `(|A|+K)^(n)`
C. `(|A|+K)^(n-1)`
D. none of these

Explanation

To solve the problem, we need to find the value of the determinant \( |A B + K I_n| \), where \( A \) is an \( n \)-th order square matrix, \( B \) is its adjoint, and \( K \) is a scalar quantity.

### Step-by-Step Solution:

1. **Understanding the Adjoint**:
The adjoint \( B \) of a matrix \( A \) is defined such that:
\[ A B = |A| I_n \]
where \( |A| \) is the determinant of \( A \) and \( I_n \) is the identity matrix of order \( n \).

2. **Substituting for \( A B \)**:
We can substitute \( A B \) in our expression:
\[ |A B + K I_n| = | |A| I_n + K I_n | \]
This simplifies to:
\[ |( |A| + K ) I_n| \]

3. **Using the Determinant Property**:
The determinant of a scalar multiple of the identity matrix is given by:
\[ |c I_n| = c^n \]
where \( c \) is a scalar. In our case, \( c = |A| + K \).

4. **Calculating the Determinant**:
Therefore, we have:
\[ |( |A| + K ) I_n| = (|A| + K)^n \]

5. **Final Result**:
Thus, the value of \( |A B + K I_n| \) is:
\[ (|A| + K)^n \]

### Conclusion:
The correct answer is:
**b. \( (|A| + K)^n \)**