If A is a square matrix such that `{:A(adjA)=[(4,0,0),(0,4,0),(0,0,4)]:}` then `abs(adjA)`=

Options

A. 4
B. 16
C. 64
D. 256

Answer

To solve the problem, we need to find the absolute value of the adjugate of matrix \( A \) given that \( A \cdot \text{adj}(A) = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix} \).

### Step-by-Step Solution:

1. **Understanding the Given Equation**:
We know that for any square matrix \( A \), the relationship between \( A \) and its adjugate \( \text{adj}(A) \) is given by:
\[ A \cdot \text{adj}(A) = \det(A) \cdot I_n \]
where \( I_n \) is the identity matrix of order \( n \).

2. **Identifying the Matrix Order**:
The given matrix \( \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix} \) is a \( 3 \times 3 \) matrix, indicating that \( n = 3 \).

3. **Setting Up the Equation**:
From the relationship, we can equate:
\[ A \cdot \text{adj}(A) = \det(A) \cdot I_3 \]
This implies:
\[ \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix} = \det(A) \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

4. **Finding the Determinant of \( A \)**:
From the above equation, we can see that:
\[ \det(A) \cdot I_3 = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix} \]
This means:
\[ \det(A) = 4 \]

5. **Using the Formula for the Determinant of the Adjugate**:
The determinant of the adjugate of \( A \) is given by the formula:
\[ \det(\text{adj}(A)) = \det(A)^{n-1} \]
where \( n \) is the order of the matrix. Since \( n = 3 \):
\[ \det(\text{adj}(A)) = \det(A)^{3-1} = \det(A)^2 = 4^2 = 16 \]

6. **Conclusion**:
Therefore, the absolute value of the determinant of the adjugate of \( A \) is:
\[ |\det(\text{adj}(A))| = 16 \]

### Final Answer:
\[ \text{abs(adjA)} = 16 \]