JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
If \(A =\frac{1}{5 ! 6 ! 7 !}\left[\begin{array}{lll}5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 !\end{array}\right]\), then \(|\operatorname{adj}(\operatorname{adj}(2 A ))|\) is equal to:
- A \(2^8\)
- B \(2^{12}\)
- C \(2^{20}\)
- D \(2^{16}\)
Answer & Solution
Correct Answer
(D) \(2^{16}\)
Step-by-step Solution
Detailed explanation
\(|\operatorname{adjadj}(2 A )|=|2 A |^{( n -1)^2}\) \(=|2 A |^4\) \(=\left(2^3| A |\right)^4\) \(=2^{12}| A |^4 \Rightarrow 2^{16}\) \(|A|=\frac{1}{5 ! 6 ! 7 !} 5 ! 6 !\left|\begin{array}{lll}1 & 6 & 42 \\ 1 & 7 & 56 \\ 1 & 8 & 72\end{array}\right|\)…
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