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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Mathematics
Chapter
Matrices

BITSAT 2014 · Mathematics · Matrices

If A=[ array ll 1 & 1 1 & 1 array ] then ^ 100 :

If \mathrm{A}=\left[1111\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right] then\mathbfA^{100}$ :

Options

  1. A2100A2^{100} \mathrm{ A}
  2. B299A2^{99} \mathrm{ A}
  3. C2101A2^{101} \mathrm{ A}
  4. DNone of the above

Correct answer

B. 299A2^{99} \mathrm{ A}

Step-by-step solution

Let A=\left[1111\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]\begin{array}{l} \mathrm{A}^{2}=\left[\begin{array}{ll}1 & 1 \\ 1 & 1 \end{array}\right]\left[1111\begin{array}{ll}1 & 1 \\ 1 & 1 \end{array}\right]=2\left[1111\begin{array}{ll}1 & 1 \\ 1 & 1 \end{array}\right]=2 \mathrm{ A}\\ \mathrm{ A}^{3}=2^{2}\left[1111\begin{array}{ll}1 & 1 \\ 1 & 1 \end{array}\right], \mathrm{A}^{4}=2^{3}\left[1111\begin{array}{ll}1 & 1 \\ 1 & 1 \end{array}\right] \\ \mathrm{A}^{3}=2^{2} \mathrm{ A}, \quad \mathrm{ A}^{4}=2^{3} \mathrm{ A}\\ \therefore \mathrm{A}^{\mathrm{n}}=2^{\mathrm{n}-1}\left[1111\begin{array}{ll}1 & 1 \\ 1 & 1 \end{array}\right] \\ \Rightarrow \mathrm{A}^{100}=2^{100-1} \mathrm{ A} \therefore A100=299A\mathrm{A}^{100}=2^{99} \mathrm{ A} \end{array}
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