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NDA 2024 Maths · Mathematics · Matrices · easy

If A=beginpmatrix1&-1\\1&-1endpmatrix, then (I+A)^10 equals

Single correct option

If A=(1−11−1)A=\begin{pmatrix}1&-1\\1&-1\end{pmatrix}, then (I+A)10(I+A)^{10} equals
  1. A. (210000)\begin{pmatrix}2^{10}&0\\0&0\end{pmatrix}
  2. B. II
  3. C. (10−1010−10)\begin{pmatrix}10&-10\\10&-10\end{pmatrix}
  4. D. (11−1010−9)\begin{pmatrix}11&-10\\10&-9\end{pmatrix}
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Answer: (D) (11−1010−9)\begin{pmatrix}11&-10\\10&-9\end{pmatrix}

A2=0A^2=0 (check by multiplying), so the binomial expansion stops: (I+A)10=I+10A=(11−1010−9)(I+A)^{10}=I+10A=\begin{pmatrix}11&-10\\10&-9\end{pmatrix}. Trap: forgetting the identity term gives the last option.
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