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Calculate $\left(\begin{matrix} 6&1&0\\0&6&1\\0&0&6\end{matrix}\right)^{14}$

Whould I do it one by one, and then find a pattern? I sense $6^{14}$ on the diagonal, and zeroes in the "lower triangle", but the "upper triangle" I'm not sure. Was thinking $14 \cdot 6^{13} $ but that's not correct.

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    $\begingroup$ HINT Try multiplying the matrix out and finding a pattern by the 3rd or 4th power. $\endgroup$
    – Don Larynx
    Dec 1, 2013 at 17:11
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    $\begingroup$ Closely related: for a $3 \times 3$ matrix A ,value of $ A^{50} $ is $\endgroup$
    – hardmath
    Dec 1, 2013 at 17:23
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    $\begingroup$ 14 is small enough that you can even do it by hand; doing only 5 matrix multiplications; namely: $A^2=A.A,A^4=A^2.A^2,A^6=A^2.A^4,A^8=A^4.A^4,A^{14}=A^6.A^8$! $\endgroup$
    – Ali
    Dec 1, 2013 at 19:04

5 Answers 5

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Write the matrix as $6I+N$ where $$N=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}$$ and expand $(6I+N)^{14}$ using the binomial formula (which is valid here because $I$ and $N$ commute). Note that since $N^3=0$, you don't need to compute most of the coefficients.

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    $\begingroup$ You beat me to it and also remembered to mention that the two matrices commute. $\endgroup$
    – Carsten S
    Dec 1, 2013 at 17:15
  • $\begingroup$ What's happened to your avatar, Harald? I remember a bee was on it!? :) $\endgroup$
    – Mikasa
    Dec 1, 2013 at 17:21
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    $\begingroup$ @B.S. I still see a bee. $\endgroup$
    – Git Gud
    Dec 1, 2013 at 17:21
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    $\begingroup$ @B.S. It still is. No, it's a fly actually. If you don't see it, it's a problem with the gravatar site. Most likely temporary. $\endgroup$ Dec 1, 2013 at 17:21
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Write it as $\left(\begin{matrix} 6&1&0\\0&6&1\\0&0&6\end{matrix}\right)=\left(\begin{matrix} 6&0&0\\0&6&0\\0&0&6\end{matrix}\right)+\left(\begin{matrix} 0&1&0\\0&0&1\\0&0&0\end{matrix}\right)$, use the binomial formula and see what you can say about powers of these two matrices.

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$$A^n = \begin{bmatrix} 6^n & n \cdot 6^{n-1} & \dbinom{n}2 6^{n-2}\\ 0 & 6^n & n \cdot 6^{n-1}\\ 0 & 0 & 6^n\end{bmatrix}$$ Prove this by induction.

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  • $\begingroup$ Does that mean you didn't induce it, but just found the pattern? $\endgroup$ Dec 5, 2013 at 15:46
  • $\begingroup$ @ThomasAhle Yes. $\endgroup$
    – user17762
    Dec 5, 2013 at 15:48
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I didn't notice that this was suggested by Harald Hanche-Olsen until just now. Consider this an expansion on his answer.

Since the identity matrix commutes with any matrix, we can use the binomial theorem with $$ \left(6\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}+\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}\right)^n $$ while noting that $$ \begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}^2=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix} \quad\text{and}\quad \begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}^3=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix} $$ To get $$ \begin{bmatrix}6&1&0\\0&6&1\\0&0&6\end{bmatrix}^n =6^n\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}+6^{n-1}\binom{n}{1}\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}+6^{n-2}\binom{n}{2}\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix} $$

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Denote the element in the top center (now 1) in $A^n$ by $a_n$. From the matrice multiple and the diagonal values we can see that $$a_{n+1}=6^na_n+6^na_n=2*6^na_n$$ from this we can make the general nth element $$a_n=2*6^{n-1}*2*6^{n-2}*...*2*1=2^n*6^{(n-1)+(n-2)+...+1}=2^n*6^{\frac{n(n-1)}{2}}$$ Putting n=14 gives: $$a_{14}=2^{14}*6^{91}$$

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