Let A=pmatrix1&0&03&0&09&3&0pmatrix and B=I.

Question

Let $A=\begin{pmatrix}1&0&0\3&0&0\9&3&0\end{pmatrix}$ and $B=I$. If $B_n=[b_{ij}]$, $1\le i,j\le3$, is given by $B=A^{99}-I$, then the value of $\dfrac{b_{31}-b_{21}}{b_{32}}$ is:

Accepted Answer

Write $A=P+I$ where $P=\begin{pmatrix}0&0&0\3&0&0\9&3&0\end{pmatrix}$ is nilpotent with $P^3=O$.

By the binomial theorem, $A^{99}=(P+I)^{99}=I+99P+\binom{99}{2}P^2$ (higher terms vanish since $P^3=0$).

$P^2=\begin{pmatrix}0&0&0\0&0&0\9&0&0\end{pmatrix}$, so
$$A^{99}-I = 99P+\binom{99}{2}P^2=\begin{pmatrix}0&0&0\297&0&0\44550&297&0\end{pmatrix}.$$

So $b_{31}=44550,\ b_{21}=297,\ b_{32}=297$.

$$\frac{b_{31}-b_{21}}{b_{32}}=\frac{44550-297}{297}=\frac{44253}{297}=149.$$

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