Matrix raised to 14th power 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.
 A: 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.
A: $$A^n = \begin{bmatrix} 6^n & n \cdot 6^{n-1} & \dbinom{n}2 6^{n-2}\\ 0 & 6^n & n \cdot 6^{n-1}\\ 0 & 0 & 6^n\end{bmatrix}$$
Prove this by induction.
A: 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
$$
\begin{bmatrix}6&1&0\\0&6&1\\0&0&6\end{bmatrix}^n
=6^n\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}+6^{n-1}\binom{n}{1}\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}+6^{n-2}\binom{n}{2}\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}
$$
A: 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.
A: 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:
$$a_{14}=2^{14}*6^{91}$$
