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JEE Advanced 2024 Paper 2 · Mathematics · Matrices and Determinants · easy

Then det (A^100-100A) equals

Single correct option

Let A=(2−110)A=\begin{pmatrix}2&-1\\1&0\end{pmatrix}. Then det⁡(A100−100A)\det\left(A^{100}-100A\right) equals
  1. A. 1000010000
  2. B. 96049604
  3. C. 98019801
  4. D. 99009900
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Answer: (C) 98019801

Write A=I+NA=I+N with N=(1−11−1)N=\begin{pmatrix}1&-1\\1&-1\end{pmatrix}. Then N2=0N^2=0, so by the binomial theorem (valid as II commutes with NN) An=I+nNA^{n}=I+nN. Therefore A100−100A=I+100N−100I−100N=−99IA^{100}-100A=I+100N-100I-100N=-99I, and det⁡(−99I)=(−99)2=9801\det(-99I)=(-99)^2=9801. The distractors come from off-by-one slips in the scalar: 10000=det⁡(−100I)10000=\det(-100I), 9604=det⁡(−98I)9604=\det(-98I), and 9900=99⋅1009900=99\cdot100 mixes the two.
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