Let A = [[1, 0, 0], [0, 1, 1], [1, 0, 0]]. Then A²⁰²⁵ - A²⁰²⁰ is equal to :

Q: Why can't I just use the Cayley-Hamilton theorem to find $A^{2025}$?

You absolutely could! The characteristic equation would give you a relation between $A^2$ and $A$. However, for this specific matrix, the pattern recognition method is significantly faster and less prone to calculation errors. In an exam, speed is as important as accuracy, and pattern recognition is a vital skill for the JEE.

Q: How do I know for sure that the pattern $n-1$ holds for all $n$?

Great question! While observation is enough for the exam, you can rigorously prove it using the Principle of Mathematical Induction. Assume it holds for $k$, then multiply $A^k$ by $A$ to show it holds for $k+1$. It is a beautiful way to solidify your logic and ensure your intuition is mathematically sound.

Q: Is it possible that I made a mistake in the multiplication? How can I be sure?

Always check the identity elements. In this matrix, the first and third rows are very stable. If you find that those rows are changing during your multiplication, stop and re-check your arithmetic. The stability of those rows is your 'sanity check'—if they change, you know you've taken a wrong turn.

Question

Let $A = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 1 \ 1 & 0 & 0 \end{pmatrix}$. Then $A^{2025} - A^{2020}$ is equal to :

Options:

(A) $A^6 - A$

(B) $A^5$

(C) $A^5 - A$

(D) $A^6$

Sigma Percentile · Accepted Answer

### The Illusion of Complexity: Cracking the Matrix Code

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare. You see $A^{2025} - A^{2020}$ and your instinct might be to panic.

You might think, "Do I really have to multiply this matrix two thousand times?" The answer, of course, is a resounding "No."

In the world of JEE Advanced, whenever you encounter a matrix raised to a massive power, you are not looking at a calculation problem; you are looking at a pattern recognition puzzle. Let us embark on this journey together.

### Phase 1: The Anatomy of the Matrix

Let us look at our given matrix $A$:

$$A = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 1 \ 1 & 0 & 0 \end{pmatrix}$$

Before we touch a pen to paper, observe the structure. It is sparse, containing zeros and ones, which is a huge hint.

Sparse matrices are designed to behave predictably under multiplication. Our goal is to find a general form for $A^n$ by calculating the first few powers to identify the "DNA" of the matrix.

### Phase 2: The Detective Work

Let us calculate $A^2$ using standard row-by-column multiplication:

$$A^2 = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 1 \ 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 1 \ 1 & 0 & 0 \end{pmatrix}$$

When you multiply the first row of the first matrix by the columns of the second, the first row remains $(1, 0, 0)$. The third row also remains $(1, 0, 0)$.

However, the element at position $(2, 1)$, denoted as $a_{21}$, changes. Calculating it: $(0 \cdot 1) + (1 \cdot 0) + (1 \cdot 1) = 1$. Thus, we have:

$$A^2 = \begin{pmatrix} 1 & 0 & 0 \ 1 & 1 & 1 \ 1 & 0 & 0 \end{pmatrix}$$

Now, let us push further to $A^3$ by multiplying $A^2$ by $A$:

$$A^3 = \begin{pmatrix} 1 & 0 & 0 \ 1 & 1 & 1 \ 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 1 \ 1 & 0 & 0 \end{pmatrix}$$

Again, the first and third rows remain unchanged. For the second row, the element $a_{21}$ becomes $(1 \cdot 1) + (1 \cdot 0) + (1 \cdot 1) = 2$.

### Phase 3: The Generalization

Look at the progression of $a_{21}$:
- For $A^1$, $a_{21} = 0$
- For $A^2$, $a_{21} = 1$
- For $A^3$, $a_{21} = 2$

The value of the element is always $n-1$. This is the beauty of mathematics; we have reduced a problem involving the year 2025 to a simple linear relationship.

We can confidently state that for any power $n$:

$$A^n = \begin{pmatrix} 1 & 0 & 0 \ n-1 & 1 & 1 \ 1 & 0 & 0 \end{pmatrix}$$

### Phase 4: The Final Calculation

Now, the "scary" part of the question becomes trivial. We need $A^{2025} - A^{2020}$. Using our formula:

$$A^{2025} = \begin{pmatrix} 1 & 0 & 0 \ 2024 & 1 & 1 \ 1 & 0 & 0 \end{pmatrix}, \quad A^{2020} = \begin{pmatrix} 1 & 0 & 0 \ 2019 & 1 & 1 \ 1 & 0 & 0 \end{pmatrix}$$

Subtracting these two matrices is a matter of simple arithmetic. The ones and zeros in the other positions cancel out perfectly:

$$A^{2025} - A^{2020} = \begin{pmatrix} 0 & 0 & 0 \ 2024 - 2019 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \ 5 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix}$$

### Phase 5: The Victory Lap

We have our result. If we check the options, we find that $A^6 - A$ yields the same result.

Using our formula for $n=6$, $A^6$ has $6-1 = 5$ in the $(2, 1)$ position, while matrix $A$ has $0$. Subtracting them gives $5 - 0 = 5$, which is a perfect match.

The problem was never about the number 2025; it was about your ability to stay calm, observe the structure, and find the underlying rule. You have mastered the matrix.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = 101010010$ . Then $A^{2025} - A^{2020}$ is equal to :

Select Answer:

Visualized Solution

Given Matrix $A$

The Strategy: Finding a Pattern

Calculating $A^{2}$

Result of $A^{2}$

Calculating $A^{3}$

General Formula for $A^{n}$

The Changing Element

Applying the Formula

Calculating the Difference

Final Difference Matrix

Checking Option 1: $A^{6} - A$

Verifying Option 1

The Sigma Insight: Algebraic Operations on Matrices

The Illusion of Complexity

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A=​101​010​010​​. Then A2025−A2020 is equal to :

Select Answer:

Visualized Solution

Given Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

The Strategy: Finding a Pattern

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2

Result of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A3

General Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } An