1. Given set $S$

We have

$$S=\left\{\begin{pmatrix}-1 & a\\ 0 & b\end{pmatrix}: a,b\in\{1,2,3,\dots,100\}\right\}.$$

For each $n$,

$$T_n=\{A\in S: A^{n(n+1)}=I\}.$$

We need the number of matrices in

$$\bigcap_{n=1}^{100} T_n.$$

That means we want all matrices $A\in S$ such that

$$A^{n(n+1)}=I \quad \text{for every } n=1,2,\dots,100.$$

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2. Use the condition for $n=1$

Since the intersection includes $T_1$, any matrix in the intersection must satisfy

$$A^{1(1+1)}=A^2=I.$$

So it is enough to find matrices in $S$ with

$$A^2=I.$$

Because if $A^2=I$, then for every $n$,

$$A^{n(n+1)}=(A^2)^{\frac{n(n+1)}{2}}=I,$$

noting that $n(n+1)$ is always even.

So the problem reduces to counting matrices in $S$ such that $A^2=I$.

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3. Compute $A^2$

Let

$$A=\begin{pmatrix}-1 & a\\ 0 & b\end{pmatrix}.$$

Then

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

Thus,

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

For this to equal the identity matrix

$$I=\begin{pmatrix}1&0\\0&1\end{pmatrix},$$

we need

$$a(b-1)=0$$

and

$$b^2=1.$$

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4. Use the allowed values of $a,b$

Since

$$a,b\in\{1,2,3,\dots,100\},$$

we have $a\ge 1$, so $a\ne 0$.

Also $b$ is a positive integer, and $b^2=1$ implies

$$b=1.$$

Then automatically

$$a(b-1)=a(1-1)=0.$$

So for every choice of $a\in\{1,2,\dots,100\}$ with $b=1$, we get $A^2=I$.

Hence all such matrices are

$$\begin{pmatrix}-1 & a\\0 & 1\end{pmatrix}, \quad a=1,2,\dots,100.$$

There are exactly

$$100$$

such matrices.

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5. Check they belong to all $T_n$

If $A^2=I$, then for every $n$,

$$A^{n(n+1)}=I$$

because $n(n+1)$ is even. Thus these 100 matrices lie in every $T_n$.

So,

$$\left|\bigcap_{n=1}^{100} T_n\right|=100.$$

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6. Compare with stored answer

Derived answer = $100$.

Stored correct answer = $100$.

They agree.