Question

Match List-I (Matrix expressions) with List-II (Properties).

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

Hint:

Memorize 4 core identities of adjoint: product, transpose, determinant, and adjoint of adjoint — they cover almost every exam question.

Answer 1

The Correct Option is A

Concept: We use standard adjoint and determinant identities:
• $\text{adj}(AB)=\text{adj}(B)\,\text{adj}(A)$
• $\text{adj}(\text{adj}A)=|A|^{n-2}A$
• $\text{adj}(A^T)=(\text{adj}A)^T$
• $|\text{adj}A|=|A|^{n-1}$ ---

Step 1: {Match (A): adj(AB)}
Using identity: \[ \text{adj}(AB)=\text{adj}(B)\text{adj}(A) \] So: \[ (A) \rightarrow (II) \] ---

Step 2: {Match (B): adj(adj A)}
Standard identity: \[ \text{adj}(\text{adj}A)=|A|^{n-2}A \] So: \[ (B) \rightarrow (IV) \] ---

Step 3: {Match (C): adj($A^T$)}
Property: \[ \text{adj}(A^T)=(\text{adj}A)^T \] So: \[ (C) \rightarrow (III) \] ---

Step 4: {Match (D): $|\text{adj}A|$}
Determinant identity: \[ |\text{adj}A|=|A|^{n-1} \] So: \[ (D) \rightarrow (I) \] --- Final Matching: \[ (A)-(II),\ (B)-(IV),\ (C)-(III),\ (D)-(I) \] ---