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(C) Let $A = \begin{bmatrix} -1 & a \ 0 & b \end{bmatrix}$.First,calculate $A^2$:$A^2 = \begin{bmatrix} -1 & a \ 0 & b \end{bmatrix} \begin{bmatrix}

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$100$

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(C) Let $A = \begin{bmatrix} -1 & a \ 0 & b \end{bmatrix}$.First,calculate $A^2$:$A^2 = \begin{bmatrix} -1 & a \ 0 & b \end{bmatrix} \begin{bmatrix} -1 & a \ 0 & b \end{bmatrix} = \begin{bmatrix} 1 & -a + ab \ 0 & b^2 \end{bmatrix}$.For $A^{n(n+1)} = I$ to hold for all $n \in \{1, 2, \ldots, 100\}$,we examine the condition $A^{n(n+1)} = I$.If $b = 1$,then $A^2 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I$.Since $n(n+1)$ is always even for any integer $n \ge 1$,$A^{n(n+1)} = (A^2)^{\frac{n(n+1)}{2}} = I^{\frac{n(n+1)}{2}} = I$.Thus,if $b = 1$,$A \in T_n$ for all $n$.If $b 
eq 1$,then $A^2 = \begin{bmatrix} 1 & a(b-1) \ 0 & b^2 \end{bmatrix}$.For $A^{n(n+1)} = I$,we require $b^{n(n+1)} = 1$ and the top-right entry to be $0$.Since $b \in \{1, 2, \ldots, 100\}$,$b^{n(n+1)} = 1$ implies $b = 1$ (as $b > 0$).Therefore,only matrices with $b = 1$ satisfy the condition for all $n$.With $b = 1$,$a$ can be any value in $\{1, 2, \ldots, 100\}$.There are $100$ such elements.

Let $S = \left\{ \begin{bmatrix} -1 & a \\ 0 & b \end{bmatrix} : a, b \in \{1, 2, 3, \ldots, 100\} \right\}$ 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

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