RMO District Round, Bucharest 2008, 11th Grade

[thaithuan_GC]

For any positive integer $n $ compute $A^n $ , where :
$A=\left(\begin{array}{ccc}1 & 1 & 3 \\ 1 & 3 & 1 \\ 3 & 1 & 1 \end{array} \right) \in \mathcal{M}_3(\mathbb{R}) $
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[mhar_teens]

@admin::why did you ban me??

$A = J + 2B, $ where $B = \begin{bmatrix}0 & 0 & 1 \\0 & 1 & 0 \\1 & 0 & 0\end{bmatrix}. $
We have that $B^2 = I $ and $JB = BJ = JI = IJ = J. $ We also have $J^2 = 3J. $

$A^2 = 3J + 2J + 2J + 4I = 7J + 4I $

$A^3 = (J + 2B)(7J + 4I) = 21J + 4J + 14J + 8B = 39J + 8B $

$A^4 = (J + 2B)(39J + 8B) = 117J + 8J + 78J + 16I = 203J + 16I $

The pattern is a multiple of $J$ plus a multiple of $I $ (for even powers) or $B $ (for odd powers). What remains is determining the coefficients.

Notice the row sums. If $x = [1\ 1\ 1]^T, $ then $Ax = 5x, $ so $A^nx = 5^nx. $ Since we also have the coefficient of $B $ or $I, $ we can see how to finish:

For $n $ odd, $A^n = \left(\frac {5^n - 2^n}{3}\right)J + 2^nB. $

For $n $ even, $A^n = \left(\frac {5^n - 2^n}{3}\right)J + 2^nI $
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