Question

Let $A$ is a matrix of order 3 such that $|A| = -4$, then the value of $|\text{adj}(\text{adj}(2\text{adj} A)^{-1})|$ is

• A. $\frac{1}{2^{26}}$
• B. $\frac{1}{2^{27}}$
• C. $\frac{1}{2^{28}}$
• D. $\frac{1}{2^{29}}$

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We use determinant properties:
1. $|\text{adj}(M)| = |M|^{n-1}$
2. $|\text{adj}(\text{adj} M)| = |M|^{(n-1)^2}$
3. $|kM| = k^n |M|$ for order $n$.
4. $|M^{-1}| = \frac{1}{|M|}$
Step 2: Key Formula or Approach:
Let $B = 2\text{adj} A$. We need to find $| \text{adj}(\text{adj}(B^{-1})) |$.
For $n=3$, the formula $| \text{adj}(\text{adj} M) | = |M|^{(3-1)^2} = |M|^4$.
Substituting $M = B^{-1}$:
\[ |\text{adj}(\text{adj}(B^{-1}))| = |B^{-1}|^4 = \frac{1}{|B|^4} \]
Step 3: Detailed Explanation:
1. Find $|B| = |2\text{adj} A|$:
Since $A$ is $3 \times 3$, $|2\text{adj} A| = 2^3 |\text{adj} A|$.
We know $|\text{adj} A| = |A|^{n-1} = |A|^2 = (-4)^2 = 16$.
So, $|B| = 8 \times 16 = 128 = 2^7$.
2. Now calculate the final expression:
\[ \frac{1}{|B|^4} = \frac{1}{(2^7)^4} = \frac{1}{2^{28}} \]
Step 4: Final Answer:
The result is $\frac{1}{2^{28}}$.