If $A=\left(\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right)$, then $A^{10}$ is equal to

• A

  $2^8 A$
• Correct answer
  B

  $2^9 \mathrm{~A}$
• C

  $2^{10} A$
• D

  $2^{11} A$

Solution

$\because \quad A=\left(\begin{array}{ll}1 & 1 \\ & 1\end{array}\right)$, $\therefore \quad A^2=\left(\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right)\left(\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right)=\left(\begin{array}{ll}1+1 & 1+1 \\ 1+1 & 1+1\end{array}\right)$ $=\left(\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right)=2\left(\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right)=2 A$ $\therefore \quad A^3=\left(\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right)\left(\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right)$ $=\left(\begin{array}{ll}2+2 & 2+2 \\ 2+2 & 2+2\end{array}\right)=\left(\begin{array}{ll}4 & 4 \\ 4 & 4\end{array}\right)$ $=4\left(\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right)=4 A$

Hence, the correct answer is option: 'B'.