Animated Solution for Mathematics - Matrices and Determinants: Let $S=\left\{\left[\begin{array}{cc}-1& a\\ 0& b\end{array}\right]:a,b\in \{1,2,3,\dots ,100\}\right\}$ 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

Imagine you are standing before a vast, structured landscape of matrices. You have a set $S$ defined by $A=\left[\begin{array}{cc}-1& a\\ 0& b\end{array}\right]$, where $a$ and $b$ are integers dancing between $1$ and $100$.

At first glance, this looks like a daunting combinatorial puzzle. How can we possibly find the intersection of one hundred different sets $T_n$?

Mathematics is not about brute force; it is about finding the hidden symmetry. Let us embark on this journey together.

The condition for a matrix to be in $T_n$ is $A^{n(n+1)}=I$. This looks intimidating, but let us start small by squaring the matrix.

We compute:

Performing the row-by-column multiplication, the top-left element is $(-1)(-1)+a(0)=1$. The top-right element becomes $(-1)(a)+a(b)=a(b-1)$.

The bottom-left is $0$, and the bottom-right is $b^2$. This is our first breakthrough; notice how the top-right element depends on $(b-1)$. If $b=1$, that term vanishes entirely!

Now, let us look at the diagonal elements. For any upper triangular matrix, raising it to a power $k$ simply raises its diagonal entries to that power.

Thus, the bottom-right element of $A^k$ is $b^k$. Since we require $A^{n(n+1)}=I$, the diagonal elements of $A^{n(n+1)}$ must be $1$.

This forces $b^{n(n+1)}=1$. Given that $b$ is a positive integer between $1$ and $100$, the only possible value is $b=1$. If $b$ were anything else, $b^{n(n+1)}$ would never be $1$ for all $n$.

Now, let us revisit the exponent $n(n+1)$. You might be worried about the complexity of this exponent, but remember your number theory: the product of any two consecutive integers is always even.

We can write $n(n+1)=2m$ for some integer $m$. This means:

We already know that if $b=1$, then $A^2=I$. And what happens when you raise the identity matrix to any power $m$? It remains the identity matrix, as $(I{)}^m=I$.

The condition is satisfied perfectly for every single $n$ from $1$ to $100$.

We have arrived at the finish line. We discovered that for a matrix to be in the intersection of all $T_n$, it must satisfy $b=1$.

The variable $a$, however, is completely free. It can be any integer from $1$ to $100$.

Since there is only $1$ choice for $b$ and $100$ choices for $a$, the total number of such matrices is $100\times 1=100$. You have navigated the complexity, simplified the matrix, and found the elegant solution hidden within.

The final answer is $100$.