If-A-2-1-1-2-then-A-2006-

Question Number 54938 by gunawan last updated on 14/Feb/19

If A = [\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix}], then A^{2006} = \dots

Commented by Abdo msup. last updated on 15/Feb/19

w e h a v e A = (\begin{matrix} 2 0 \\ 0 2 \end{matrix}) + (\begin{matrix} 0 1 \\ 1 0 \end{matrix}) = 2 I + J

J^{2} = (\begin{matrix} 0 1 \\ 1 0 \end{matrix}) . (\begin{matrix} 0 1 \\ 1 0 \end{matrix}) = (\begin{matrix} 1 0 \\ 0 1 \end{matrix}) = I \Rightarrow

J^{2 n} = I a n d J^{2 n + 1} = J a n d A^{n} = {(2 I + J)}^{n}

= \sum_{k = 0}^{n} C_{n}^{k} J^{k} {(2 I)}^{n - k}

= \sum_{p = 0}^{[\frac{n}{2}]} C_{n}^{2 p} j^{2 p} {(2 I)}^{n - 2 p} + \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} j^{2 p + 1} {(2 I)}^{n - 2 p - 1}

= \sum_{p = 0}^{[\frac{n}{2}]} C_{n}^{2 p} 2^{n - 2 p} I + \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} 2^{n - 2 p - 1} J \Rightarrow

A^{2006} = \sum_{p = 0}^{1003} C_{2006}^{2 p} 2^{2006 - 2 p} + \sum_{p = 0}^{2002} C_{2006}^{2 p + 1} 2^{2005 - 2 p}