Question

ANSWER ASAP :) ITS A MATHS CLASS 12 QUESTION

Answer 1

Answer:

              $\displaystyle A^{100}=\left(\begin{matrix}2^{99}&2^{99}\\2^{99}&2^{99}\end{matrix}\right)$

Step-by-step explanation:

Starting with the first few...

$\displaystyle\left(\begin{matrix}1&1\\1&1\end{matrix}\right)^2=\left(\begin{matrix}2&2\\2&2\end{matrix}\right)\\\\\left(\begin{matrix}1&1\\1&1\end{matrix}\right)^3=\left(\begin{matrix}4&4\\4&4\end{matrix}\right)$

it looks like it will be

$\displaystyle\left(\begin{matrix}1&1\\1&1\end{matrix}\right)^n=\left(\begin{matrix}2^{n-1}&2^{n-1}\\2^{n-1}&2^{n-1}\end{matrix}\right)$

Setting out to confirm this by induction, suppose this is true for some n.  Then..

$\displaystyle\left(\begin{matrix}1&1\\1&1\end{matrix}\right)^{n+1}=\left(\begin{matrix}1&1\\1&1\end{matrix}\right)\left(\begin{matrix}2^{n-1}&2^{n-1}\\2^{n-1}&2^{n-1}\end{matrix}\right)=\left(\begin{matrix}2^n&2^n\\2^n&2^n\end{matrix}\right)$

So yes, it follows by mathematical induction that the powers do indeed take the form suspected.

In particular,

$\displaystyle A^{100}=\left(\begin{matrix}2^{99}&2^{99}\\2^{99}&2^{99}\end{matrix}\right)$