Let $$S = \left\{\begin{pmatrix} -1 & a \\ 0 & b \end{pmatrix} ; a, b \in \{1, 2, 3, \ldots 100\}\right\}$$ and let $$T_n = \{A \in S : A^{n(n+1)} = I\}$$. Then the number of elements in $$\bigcap_{n=1}^{100} T_n$$ is ______.

We have the set $$S = \left\{\begin{pmatrix} -1 & a \\ 0 & b \end{pmatrix} : a, b \in \{1, 2, \ldots, 100\}\right\}$$.

Define $$T_n = \{A \in S : A^{n(n+1)} = I\}$$. We need to find $$\left|\bigcap_{n=1}^{100} T_n\right|$$.

For $$A = \begin{pmatrix} -1 & a \\ 0 & b \end{pmatrix}$$, we first compute $$A^2$$:

$$A^2 = \begin{pmatrix} -1 & a \\ 0 & b \end{pmatrix}\begin{pmatrix} -1 & a \\ 0 & b \end{pmatrix} = \begin{pmatrix} 1 & a(-1+b) \\ 0 & b^2 \end{pmatrix}$$

For $$A^m = I$$, the eigenvalues of $$A$$ must satisfy $$\lambda^m = 1$$. The eigenvalues of $$A$$ are $$-1$$ and $$b$$.

Condition on $$(-1)$$: $$(-1)^m = 1$$ requires $$m$$ to be even. Since $$n(n+1)$$ is always even (product of consecutive integers), this condition is automatically satisfied for all $$n$$.

Condition on $$b$$: $$b^{n(n+1)} = 1$$ for all $$n = 1, 2, \ldots, 100$$. Since $$b$$ is a positive integer, the only solution is $$b = 1$$.

To confirm: if $$b \geq 2$$, then $$b^{n(n+1)} \geq 2^2 = 4 \neq 1$$. So $$b = 1$$ is required.

Verifying $$A^2 = I$$ when $$b = 1$$:

$$A^2 = \begin{pmatrix} 1 & a(-1+1) \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$$

Since $$A^2 = I$$, for any even exponent $$m$$: $$A^m = (A^2)^{m/2} = I^{m/2} = I$$.

Since $$n(n+1)$$ is always even, $$A^{n(n+1)} = I$$ holds for all $$n$$.

Therefore, the condition is satisfied for $$b = 1$$ and any $$a \in \{1, 2, \ldots, 100\}$$.

The number of elements in $$\bigcap_{n=1}^{100} T_n$$ is $$100$$.

The correct answer is $$100$$.