Let A = bmatrix 1 & 0 & 0 3 & 1 & 0 9 & 3 & 1 bmatrix and B = [b_...

Let $A = I + C$, where $I$ is the identity matrix and $C = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix}$.

Calculating the powers of $C$, we get:

$C^2 = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix}$

$C^3 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

Since $C^3 = 0$, all higher powers of $C$ are also zero matrices. Using the binomial expansion for $A^{99}$:

$A^{99} = (I + C)^{99} = I + 99C + ^{99}C_{2}C^2$

Given $B = A^{99} - I$, we have:

$B = 99C + 4851C^2$

Substituting the matrices $C$ and $C^2$:

$B = 99 \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} + 4851 \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix}$

From this, we can extract the required elements of $B$:

$b_{31} = 99 \times 9 + 4851 \times 9 = 9(99 + 4851) = 9 \times 4950 = 44550$

$b_{21} = 99 \times 3 = 297$

$b_{32} = 99 \times 3 = 297$

Now, substituting these values into the given expression:

$\dfrac{b_{31} - b_{21}}{b_{32}} = \dfrac{44550 - 297}{297} = \dfrac{44550}{297} - 1$

$\dfrac{b_{31} - b_{21}}{b_{32}} = 150 - 1 = 149$

Answer: $149$