Animated Solution for Mathematics - Matrices and Determinants: If $A=\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right]$ and $M=A+A^2+A^3+\cdots +A^{20}$, then the sum of all the elements of the matrix $M$ is equal to

Imagine you are staring at the matrix:

At first glance, it looks like a simple upper triangular matrix. However, in the context of JEE Advanced, we must look for a more elegant path than manual multiplication.

The secret lies in decomposition. Notice that $A$ is the sum of the identity matrix $I$ and a strictly upper triangular matrix $C$:

Let us explore the properties of $C$. If you calculate $C^2$, you obtain:

Now, multiply by $C$ again to find $C^3$. Every element becomes zero, meaning $C$ is a nilpotent matrix. Any power of $C$ greater than or equal to $3$ is the zero matrix.

Because $I$ and $C$ commute, we can use the Binomial Theorem:

Since all terms from $C^3$ onwards vanish, we are left with only three terms:

Now, let us substitute our matrices back into this expansion:

Adding these together, we get a beautiful, generalized formula for $A^n$:

The sum of all elements in this matrix, $S(n)$, is $1+n+\frac{n^2+n}{2}+0+1+n+0+0+1$. This simplifies to:

We need to calculate the sum of $S(n)$ for $n$ from $1$ to $20$:

Using standard summation formulas:

1. The constant term: $3\times 20=60$. 2. The linear term: $\frac{5}{2}\times \frac{20\times 21}{2}=525$. 3. The quadratic term: $\frac{1}{2}\times \frac{20\times 21\times 41}{6}=1435$.

Adding these together, we get:

The elegance of matrix algebra has turned a daunting calculation into a series of simple, satisfying steps.