Matrices & Determinants
Matrices & Determinants
star_batch_jee_advanced_2025
Grade 12
Question:
Let $A = [a_{rs}]$ be a $n \times n$ matrix such that $a_{rs} = (r-s)2^{(r-s)}$ where $i = \sqrt{-1}$, then $A = (\bar{A} \text{ denotes } [\bar{a}_{rs}] \text{ complex conjugate})$
\bar{A}
-\bar{A}
(\bar{A})^T
-(\bar{A})^T
Step-by-Step Solution
Key Concept: The adjugate of a matrix relates to the inverse through $A \cdot \text{adj}(A) = |A|I$.
For $a_{rs} = (r-s)2^{(r-s)}$, the matrix $A$ has entries that depend on differences $r-s$. The adjugate matrix satisfies $\widetilde{A} = [(r-s)2^{-(r-s)}]$ and $A = -(\widetilde{A})^T$. This leads to the relation $K = |A|^{(n-1)^n}$ giving $|A| = K^{1/(n-1)^n}$.
Correct Answer: 4