Matrices & Determinants
General
Grade 12
Question:
If $A$ is square matrix of order $n$ then $\text{adj}(\text{adj } A) =$
|A|^(n-1) A
|A|^(n-2) A
|A|^(n-2)
|A|^n A
Step-by-Step Solution
Key Concept: General
<div>We know that $A \cdot \text{adj } A = |A| I$ and $|\text{adj } A| = |A|^{n-1}$<br/>$\therefore \text{adj } A \cdot (\text{adj}(\text{adj } A)) = |\text{adj } A| I$<br/>Pre-multiplying by $A$:<br/>$\Rightarrow A \cdot \text{adj } A \cdot (\text{adj}(\text{adj } A)) = A |A|^{n-1} I$<br/>$\Rightarrow |A| I (\text{adj}(\text{adj } A)) = A |A|^{n-1}$<br/>$\Rightarrow \text{adj}(\text{adj } A) = A |A|^{n-2}$</div>
Correct Answer: B