Animated Solution for Mathematics - Matrices and Determinants: Let $A=\left(\begin{array}{ccc}1& -1& 0\\ 0& 1& -1\\ 0& 0& 1\end{array}\right)$ and $B=7A^{20}-20A^7+2I$, where $I$ is an identity matrix of order $3\times 3$. If $B=[b_{ij}]$, then $b_{13}$ is equal to

Imagine you are sitting in the examination hall, and you see $A^{20}$ staring back at you. Your first instinct might be to reach for your pen and start multiplying $A$ by itself, over and over. Stop!

In the JEE Advanced, if a problem looks like it requires brute force, it almost certainly requires a clever insight. The matrix

is not random. It is a structured, elegant object. The key to unlocking this problem lies in recognizing that $A$ is essentially the identity matrix $I$ with a small, manageable perturbation.

Let us decompose $A$. We can write $A=I+C$, where $I$ is the identity matrix and $C=A-I$.

If you perform this subtraction, you get

Look at $C$. It is a strictly upper triangular matrix with zeros on the diagonal. This is a special type of matrix.

Let us see what happens when we square it. Calculating $C^2=C\cdot C$, we find

And if we go one step further to $C^3$, we find that every single element becomes zero. We have found that $C^3=O$.

This is the magic key! Matrix $C$ is nilpotent of degree 3, meaning any power of $C$ greater than or equal to 3 vanishes into thin air.

Now, why does this matter? Because $I$ and $C$ commute, we can use the Binomial Theorem to expand $A^n=(I+C{)}^n$.

The expansion is

Because $C^3$ and all higher powers are the zero matrix, the infinite series collapses into a simple quadratic expression:

This is the moment where the difficulty of the problem evaporates. We no longer need to calculate massive powers; we just need to plug in the values of $n$.

For $n=20$, we get

Similarly, for $n=7$, we get

Now, we substitute these into our expression for $B=7A^{20}-20A^7+2I$.

Substituting our expansions, we get

Grouping the terms by their coefficients, we see the magic happen: the $I$ terms sum to $7-20+2=-11$, the $C$ terms sum to $140-140=0$, and the $C^2$ terms sum to $1330-420=910$.

Thus,

The question asks for $b_{13}$, the element in the first row and third column. Since $(I{)}_{13}=0$ and $(C^2{)}_{13}=1$, the final answer is simply 910.

It is a beautiful example of how pattern recognition and mathematical properties can turn a daunting calculation into a simple, elegant solution.