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JEE Advanced 2024 Paper 1 · Mathematics · Matrices and Determinants · hard

Let A=beginpmatrix2&1&0\\0&2&1\\0&0&2endpmatrix. Which of the following statements is/are correct?

One or more options correct

Let A=(210021002)A=\begin{pmatrix}2&1&0\\0&2&1\\0&0&2\end{pmatrix}. Which of the following statements is/are correct?
  1. A. The (1,3)(1,3) entry of A6A^6 is 120120
  2. B. tr⁡(A5)=80\operatorname{tr}(A^5)=80
  3. C. det⁡(adj⁡(adj⁡A))=4096\det\big(\operatorname{adj}(\operatorname{adj}A)\big)=4096
  4. D. The (1,3)(1,3) entry of A−1A^{-1} is 116\dfrac1{16}
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Answer: C

Write A=2I+NA=2I+N with N=(010001000)N=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}, N3=0N^3=0, and 2I2I commuting with NN. Then An=2nI+n2n−1N+(n2)2n−2N2A^n=2^nI+n2^{n-1}N+\binom n2 2^{n-2}N^2, and N2N^2 has a single 11 in position (1,3)(1,3). (A) For n=6n=6 the (1,3)(1,3) entry is (62)24=15⋅16=240\binom62 2^4=15\cdot16=240, not 120120, so false. (B) tr⁡(A5)=3⋅25=96\operatorname{tr}(A^5)=3\cdot2^5=96 because NN and N2N^2 have zero diagonals, so false. (C) For a 3×33\times3 matrix adj⁡(adj⁡A)=(det⁡A)3−2A=8A\operatorname{adj}(\operatorname{adj}A)=(\det A)^{3-2}A=8A, and det⁡(8A)=83det⁡A=512⋅8=4096\det(8A)=8^3\det A=512\cdot8=4096, so true. (D) A−1=12(I+N/2)−1=12(I−N2+N24)A^{-1}=\tfrac12(I+N/2)^{-1}=\tfrac12\left(I-\tfrac N2+\tfrac{N^2}{4}\right), whose (1,3)(1,3) entry is 18\tfrac18, not 116\tfrac1{16}, so false. Only (C) is correct.
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Original question written in the JEE Advanced 2024 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake