Question

Let \( A \) be a matrix of order 3 such that \( |A| = -4 \). Then, the value of \( |adj(adj(2adj(A)))^{-1}| \) is:

• A. \( \dfrac{1}{2^{16}} \)
• B. \( \dfrac{1}{2^{28}} \)
• C. \( \dfrac{1}{2^{30}} \)
• D. \( \dfrac{1}{2^{20}} \)

Hint:

For a matrix of order 3, the determinant of \( adj(adj(A)) \) is always \( |A|^4 \). Be extremely careful with the scalar \( k \); it must be raised to the power of \( n \) (the order) before being multiplied by the determinant.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We use the properties of determinants and adjoints for a matrix of order \( n \): 1. \( |adj(A)| = |A|^{n-1} \) 2. \( |adj(adj(A))| = |A|^{(n-1)^2} \) 3. \( |kA| = k^n |A| \) 4. \( |M^{-1}| = 1/|M| \)
Step 2: Key Formula or Approach:
For \( n=3 \), the power in the adjoint of adjoint is \( (3-1)^2 = 4 \). Let \( B = 2adj(A) \). We need to find \( \frac{1}{|adj(adj(B))|} = \frac{1}{|B|^4} \).
Step 3: Detailed Explanation:
First, find \( |B| \): \[ |B| = |2adj(A)| = 2^3 |adj(A)| = 8 |A|^{3-1} = 8 |A|^2 \] Since \( |A| = -4 \), then \( |A|^2 = 16 \). \[ |B| = 8 \times 16 = 128 = 2^7 \] Now, find the determinant of the double adjoint of \( B \): \[ |adj(adj(B))| = |B|^{(3-1)^2} = |B|^4 \] \[ |B|^4 = (2^7)^4 = 2^{28} \] The required value is the determinant of the inverse: \[ |(adj(adj(B)))^{-1}| = \frac{1}{|adj(adj(B))|} = \frac{1}{2^{28}} \]
Step 4: Final Answer:
The value is \( \frac{1}{2^{28}} \).