If A is a square matrix of order `n xx n` and k is a scalar, then `adj (kA)` is equal to (1) `k adj A` (2) `k^n adj A` (3) `k^(n-1) adj A` (4) `k^(n+1) adj A`

Options

A. k adj A
B. `k^n adj A`
C. `k^(n-1)adjA`
D. `k^(n+1)adjA`

Answer

To solve the problem, we need to find the relationship between the adjoint of a scalar multiplied by a matrix and the adjoint of the matrix itself. Let's denote the matrix \( A \) as an \( n \times n \) square matrix and \( k \) as a scalar.

### Step-by-Step Solution:

1. **Understanding the Adjoint**:
The adjoint of a matrix \( A \), denoted as \( \text{adj}(A) \), is related to the inverse of the matrix. The formula for the inverse of a matrix can be expressed as:
\[ A^{-1} = \frac{\text{adj}(A)}{\det(A)} \]

2. **Finding the Adjoint of \( kA \)**:
We need to find \( \text{adj}(kA) \). We can use the property of determinants which states that for a scalar \( k \) and an \( n \times n \) matrix \( A \):
\[ \det(kA) = k^n \det(A) \]
This means that when we multiply a matrix by a scalar, the determinant of the resulting matrix is the scalar raised to the power of the order of the matrix multiplied by the determinant of the original matrix.

3. **Using the Inverse Formula**:
We can express \( (kA)^{-1} \) as follows:
\[ (kA)^{-1} = \frac{1}{k} A^{-1} \]
This is because the scalar \( k \) can be factored out when taking the inverse.

4. **Combining the Results**:
Now substituting back into the adjoint formula:
\[ \text{adj}(kA) = \det(kA) \cdot (kA)^{-1} \]
Substituting the expressions we derived:
\[ \text{adj}(kA) = k^n \det(A) \cdot \left(\frac{1}{k} A^{-1}\right) \]

5. **Simplifying the Expression**:
This simplifies to:
\[ \text{adj}(kA) = k^{n-1} \cdot \det(A) \cdot A^{-1} \]
Since \( \det(A) \cdot A^{-1} = \text{adj}(A) \), we can write:
\[ \text{adj}(kA) = k^{n-1} \cdot \text{adj}(A) \]

6. **Final Result**:
Therefore, the final result is:
\[ \text{adj}(kA) = k^{n-1} \cdot \text{adj}(A) \]
This means the correct answer is option (3) \( k^{n-1} \text{adj}(A) \).