Question

Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	If A is a square matrix of order 3 such that \(|A| = 3\), then \(|adj (adj A)| =\)	I.	16
B.	If A is a non-singular matrix and \(|A| = \frac{1}{16}\), then \(|A^{-1}| =\)	II.	512
C.	If A is a square matrix of order 4 such that \(|A| = 2\), then \(|adj (adj A)| =\)	III.	9
D.	If A is a square matrix of order 2 such that \(|A| = 9\), then \(|adj (adj A)| =\)	IV.	81

Choose the correct answer from the options given below:

• A. A - IV, B - I, C - III, D - II
• B. A - IV, B - I, C - II, D - III
• C. A - I, B - IV, C - III, D - II
• D. A - III, B - I, C - IV, D - II

Hint:

Use \(|adj(adj A)| = |A|^{(n-1)^2}\) and \(|A^{-1}| = 1/|A|\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We use the identity \(|adj A| = |A|^{n-1}\) for a matrix of order n. Applying it twice gives \(|adj(adj A)| = |A|^{(n-1)^2}\). For the inverse we use \(|A^{-1}| = 1/|A|\).

Step 2: Match A.
Here \(n = 3\) and \(|A| = 3\). So \(|adj(adj A)| = 3^{(3-1)^2} = 3^4 = 81\). This is IV.

Step 3: Match B.
\(|A^{-1}| = \frac{1}{|A|} = \frac{1}{1/16} = 16\). This is I.

Step 4: Match C.
Here \(n = 4\) and \(|A| = 2\). So \(|adj(adj A)| = 2^{(4-1)^2} = 2^9 = 512\). This is II.

Step 5: Match D.
Here \(n = 2\) and \(|A| = 9\). So \(|adj(adj A)| = 9^{(2-1)^2} = 9^1 = 9\). This is III.

Step 6: Compare the options.
We got A - IV, B - I, C - II, D - III. That is option 2. Option 1 swaps C and D. Option 3 gives A - I, which is wrong because 81 is not 16. Option 4 gives A - III, which is wrong too.

Final Answer:
The correct matching is A - IV, B - I, C - II, D - III. \[ \boxed{\text{Option 2}} \]