Question

For a \( 3 \times 3 \) non singular matrix A, if \( \text{Adj}(\text{Adj}(\text{Adj}(\text{Adj}(A)))) = |A|^n A \), then \( n = \)

• A. \( 3 \)
• B. \( 4 \)
• C. \( 8 \)
• D. \( 5 \)

Hint:

The general formula for \(k\) applications of adjoint is \( \text{Adj}^{(k)}(A) = |A|^{(m-1)^k / m} \times \dots \) which simplifies significantly for \(m=3\). Remember: \( \text{Adj}(\text{Adj}(A)) = |A|^{m-2} A \).

Answer 1

The Correct Option is A

Step 1: Recall the key adjoint fact.
For any $m \times m$ non-singular matrix, $\operatorname{Adj}(A) = |A|\,A^{-1}$. Taking determinants, $|\operatorname{Adj}(A)| = |A|^{m-1}$. Here $m = 3$, so $|\operatorname{Adj}(A)| = |A|^{2}$.
Step 2: Find what one extra Adj does.
A standard result follows: $\operatorname{Adj}(\operatorname{Adj}(A)) = |A|^{m-2}A$. For $m=3$ this is $\operatorname{Adj}(\operatorname{Adj}(A)) = |A|^{1}A = |A|\,A$.
Step 3: Peel the four Adj's into two pairs.
Write the expression as $\operatorname{Adj}(\operatorname{Adj}(\,\operatorname{Adj}(\operatorname{Adj}(A))\,))$. The inner pair gives $\operatorname{Adj}(\operatorname{Adj}(A)) = |A|\,A$.
Step 4: Apply the outer pair.
So we need $\operatorname{Adj}(\operatorname{Adj}(|A|\,A))$. Using $\operatorname{Adj}(kM) = k^{m-1}\operatorname{Adj}(M)$ with $m=3$, the scalar $|A|$ comes out as $|A|^{2}$ across the two adjoints.
Step 5: Combine the scalars.
This gives $|A|^{2}\cdot \operatorname{Adj}(\operatorname{Adj}(A)) = |A|^{2}\cdot |A|\,A = |A|^{3}A$.
Step 6: Compare with $|A|^n A$.
Matching $|A|^{3}A = |A|^{n}A$ gives $n = 3$, which is option (A).
\[ \boxed{n = 3} \]