Q82 / 125 · BITSAT 2014 · Mathematics
If $\mathrm{A}=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$ then $\mathbf{A}^{100}$ :
- A$2^{100} \mathrm{~A}$
- Correct answerB$2^{99} \mathrm{~A}$
- C$2^{101} \mathrm{~A}$
- DNone of the above
Solution
Let $A=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$
$$
\begin{array}{l}
\mathrm{A}^{2}=\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]=2\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]=2 \mathrm{~A} \\
\mathrm{~A}^{3}=2^{2}\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right], \mathrm{A}^{4}=2^{3}\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right] \\
\mathrm{A}^{3}=2^{2} \mathrm{~A}, \quad \mathrm{~A}^{4}=2^{3} \mathrm{~A} \\
\therefore \mathrm{A}^{\mathrm{n}}=2^{\mathrm{n}-1}\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right] \\
\Rightarrow \mathrm{A}^{100}=2^{100-1} \mathrm{~A} \therefore \mathrm{A}^{100}=2^{99} \mathrm{~A}
\end{array}
$$
Hence, the correct answer is option: 'B'.