Let A=pmatrix1&a&a0&1&b0&0&1pmatrix, a,bC.

Question

Let $A=\begin{pmatrix}1&a&a\0&1&b\0&0&1\end{pmatrix}$, $a,b\in\mathbb{C}$. If for some $n\in\mathbb{N}$, $A^n=\begin{pmatrix}1&48&2160\0&1&96\0&0&1\end{pmatrix}$, then $n+a+b$ is equal to

Answer: 24

Accepted Answer

Write $A=I+B$ where $B=\begin{pmatrix}0&a&a\0&0&b\0&0&0\end{pmatrix}$, so $B^3=O$ and
$$A^n=(I+B)^n=I+nB+\binom n2B^2=\begin{pmatrix}1&na&na+\binom n2ab\0&1&nb\0&0&1\end{pmatrix}.$$

Comparing: $na=48$, $nb=96$, $na+\binom n2ab=2160$.

From these, $b=2a$, so $na=48$ and $\dfrac{n(n-1)}2\cdot2a^2=2160-48=2112 \Rightarrow n(n-1)a^2=2112$.

Testing $n=12$: $a=4$, $n(n-1)a^2=12\times11\times16=2112$ , and $b=8$.

So $n=12,\ a=4,\ b=8$, giving $n+a+b=24$.

Solutions · 1