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.

• HINT Try multiplying the matrix out and finding a pattern by the 3rd or 4th power. – Don Larynx Dec 1 '13 at 17:11
• Closely related: for a $3 \times 3$ matrix A ,value of $A^{50}$ is – hardmath Dec 1 '13 at 17:23
• 14 is small enough that you can even do it by hand; doing only 5 matrix multiplications; namely: $A^2=A.A,A^4=A^2.A^2,A^6=A^2.A^4,A^8=A^4.A^4,A^{14}=A^6.A^8$! – Ali Dec 1 '13 at 19:04

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.

• You beat me to it and also remembered to mention that the two matrices commute. – Carsten S Dec 1 '13 at 17:15
• What's happened to your avatar, Harald? I remember a bee was on it!? :) – mrs Dec 1 '13 at 17:21
• @B.S. I still see a bee. – Git Gud Dec 1 '13 at 17:21
• @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. – Harald Hanche-Olsen Dec 1 '13 at 17:21

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.

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

• Does that mean you didn't induce it, but just found the pattern? – Thomas Ahle Dec 5 '13 at 15:46
• @ThomasAhle Yes. – user17762 Dec 5 '13 at 15:48

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 $$\begin{bmatrix}6&1&0\\0&6&1\\0&0&6\end{bmatrix}^n =6^n\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}+6^{n-1}\binom{n}{1}\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}+6^{n-2}\binom{n}{2}\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}$$

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: $$a_{14}=2^{14}*6^{91}$$