Question
Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{bmatrix}$ and $B = \left\lbrack b_{ij} \right\rbrack,1 \leq i,j \leq 3$.
If $B = A^{99} - I$, then the value of $\frac{b_{31} - b_{21}}{b_{32}}$ is:
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Solution
Sol. [Key Concept: $A = P + I$ with $P$ nilpotent ($P^{3} = 0$), so $A^{99} = I + 99P + {}^{99}C_{2}P^{2}$; read off the entries $b_{31},b_{21},b_{32}$ of $B = A^{99} - I$.]
$${A = P + I }{A = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}}$$
$$ A = P + I$$ where $P^{3} = 0$
$$A^{99} = I + 99P + \ ^{99}C_{2}P^{2}$$
$$ {A^{99} - I = 99\begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} + \ ^{99}C_{2}\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix}}$$
$$ {= \begin{bmatrix} 0 & 0 & 0 \\ 99 \times 3 & 0 & 0 \\ 44550 & 99 \times 3 & 0 \end{bmatrix}}$$
$$ {b_{31} = 44550,{\text{ }b}_{21} = 99 \times 3,{\text{ }b}_{32} = 99 \times 3}$$
$$ {\frac{44550 - 297}{297} = 149 }$$
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