Question

If \(A=\begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}\), then
Match the LIST-I with LIST-II

LIST-I	LIST-II
A. \(|\text{adj }A|\)	I. 81
B. \(|A|\)	II. \(-27\)
C. \(|A\ \text{adj }A|\)	III. 9
D. \(|\text{adj (adj }A)|\)	IV. \(-3\)

Choose the correct answer from the options given below:

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

Hint:

Find \(|A|=-3\) first. Then use \(|\text{adj }A|=|A|^{2}\) and \(|\text{adj adj }A|=|A|^{4}\) for order 3.

Answer 1

The Correct Option is D

Step 1: Find |A|.
The matrix is lower triangular, so its determinant is the product of the diagonal entries.
\[ |A| = 1 \times 3 \times (-1) = -3 \] So B matches IV.

Step 2: Recall the adjoint formulas.
For a square matrix of order \(n\): \(|\text{adj }A| = |A|^{n-1}\), and \(A\ (\text{adj }A) = |A| I\). Here \(n=3\).

Step 3: Find |adj A|.
\[ |\text{adj }A| = |A|^{2} = (-3)^2 = 9 \] So A matches III.

Step 4: Find |A adj A|.
\(A\ \text{adj }A = |A| I = -3I\). The determinant of \(-3I\) (order 3) is \((-3)^3 = -27\). So C matches II.

Step 5: Find |adj(adj A)|.
Use the formula \(|\text{adj}(\text{adj }A)| = |A|^{(n-1)^2}\). With \(n=3\) the power is 4.
\[ (-3)^4 = 81 \] So D matches I.

Step 6: Compare with the options.
We have A-III, B-IV, C-II, D-I. This is option 4. Option 1 puts A with II, which is \(-27\) and wrong. Option 2 puts A with I (81), wrong. Option 3 puts A with IV, wrong.

Final Answer:
The matching is A-III, B-IV, C-II, D-I, option 4. \[ \boxed{\text{A-III, B-IV, C-II, D-I}} \]