Animated Solution for Mathematics - Matrices and Determinants: Let $A=[a_{ij}]$ be a square matrix of order 3 such that $a_{ij}=2^{j-i}$, for all $i,j=1,2,3$. Then, the matrix $A^2+A^3+\cdots +A^{10}$ is equal to

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of brute-force calculation. You see a matrix $A$ of order 3, and you are asked to find the sum $A^2+A^3+\cdots +A^{10}$.

Your instinct might be to panic—how on earth are we supposed to calculate $A^{10}$? But here is the secret of the JEE Advanced: whenever you see a high power of a matrix, there is almost always a hidden, elegant pattern waiting to be discovered. Let us embark on this journey together.

Constructing the Matrix

First, let us demystify the matrix $A$. We are given the rule $a_{ij}=2^{j-i}$. This is not random; it is a structured definition.

Let us write it out explicitly: For $i=1$: $a_{11}=2^{1-1}=1$, $a_{12}=2^{2-1}=2$, $a_{13}=2^{3-1}=4$. For $i=2$: $a_{21}=2^{1-2}=\frac{1}{2}$, $a_{22}=2^{2-2}=1$, $a_{23}=2^{3-2}=2$. For $i=3$: $a_{31}=2^{1-3}=\frac{1}{4}$, $a_{32}=2^{2-3}=\frac{1}{2}$, $a_{33}=2^{3-3}=1$.

So, our matrix $A$ is:

Look at this matrix. It has a beautiful, rhythmic structure. Each row is a geometric progression with a common ratio of 2. This structure is the key to everything that follows.

The Discovery of the Recursive Relation

Now, instead of blindly calculating $A^{10}$, let us calculate $A^2$. This is where the magic happens. When we perform the matrix multiplication $A\times A$, we are essentially taking the dot product of the rows of the first matrix with the columns of the second.

Let us calculate the first element, $(A^2{)}_{11}$:

Now, let us look at the next element, $(A^2{)}_{12}$:

And the third, $(A^2{)}_{13}$:

Do you see it? The first row of $A^2$ is $[3,6,12]$. Compare this to the first row of $A$, which is $[1,2,4]$. It is exactly 3 times the original!

If you continue this for the other rows, you will find that every single element in $A^2$ is exactly 3 times the corresponding element in $A$. Thus, we have discovered the recursive relation:

This is the "Aha!" moment. This single equation collapses the complexity of the problem.

The Power of Generalization

If $A^2=3A$, what happens when we go higher? Let us find $A^3$:

And $A^4$:

By induction, we can confidently state that for any integer $k\ge 1$:

This is the power of mathematical induction. We have tamed the beast. We no longer fear $A^{10}$; we know it is simply $3^{9}A$.

The Final Summation

We are asked to find the sum $S=A^2+A^3+\cdots +A^{10}$. Substituting our generalized formula, we get:

We can factor out the matrix $A$:

The expression inside the parentheses is a classic Geometric Progression (GP). The first term $a=3$, the common ratio $r=3$, and the number of terms $n=9$. The sum of a GP is given by $\frac{a(r^n-1)}{r-1}$.

Substituting our values:

Therefore, our final result is:

Look at what we have achieved. We started with a daunting matrix power problem and reduced it to a simple geometric series through observation and recursive logic. This is the essence of JEE Advanced mathematics—it is not about brute force; it is about finding the elegant path. Keep this mindset, and you will conquer any problem they throw at you.