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NDA 2023 Maths · Mathematics · Matrices · hard

If A=beginpmatrix0&1\\-1&0endpmatrix, then A^2023 equals

Single correct option

If A=(01−10)A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}, then A2023A^{2023} equals
  1. A. (01−10)\begin{pmatrix}0&1\\-1&0\end{pmatrix}
  2. B. (1001)\begin{pmatrix}1&0\\0&1\end{pmatrix}
  3. C. (−100−1)\begin{pmatrix}-1&0\\0&-1\end{pmatrix}
  4. D. (0−110)\begin{pmatrix}0&-1\\1&0\end{pmatrix}
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Answer: (D) (0−110)\begin{pmatrix}0&-1\\1&0\end{pmatrix}

A2=(−100−1)=−IA^2=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}=-I, so A4=IA^4=I and the powers repeat with period 44. Since 2023=4⋅505+32023=4\cdot505+3, A2023=A3=−A=(0−110)A^{2023}=A^3=-A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}. Choosing AA itself comes from wrongly reducing the exponent to 11.
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Original question written in the NDA 2023 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake