Let A = [[3, -4], [1, -1]] and B be two matrices such that A¹⁰⁰ = 100B + I. Then the sum of all the

Q: Wait, why can't we just multiply $A$ by itself 100 times?

Calculating $A^{100}$ by brute force is a trap designed to waste your time. In JEE Advanced, high powers of matrices almost always involve a pattern, like nilpotency or periodicity, which simplifies the expression significantly.

Q: How do I know when to use $A = I + M$?

Whenever you see a matrix $A$ and a high power $A^n$, check if $A - I$ or $A + I$ has a simple structure. If $A - I$ is a nilpotent matrix (like $M^2 = O$), the binomial expansion becomes trivial.

Q: Why does $M^2 = O$ make all higher powers zero?

If $M^2 = O$, then $M^3 = M^2 \cdot M = O \cdot M = O$. By induction, $M^k = O$ for all $k \geq 2$. This is the definition of a nilpotent matrix, and it's a powerful tool for simplifying matrix expressions.

Question

Let $A = \begin{bmatrix} 3 & -4 \ 1 & -1 \end{bmatrix}$ and $B$ be two matrices such that $A^{100} = 100B + I$. Then the sum of all the elements of $B^{100}$ is .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to the battlefield of JEE Advanced. Today, we face a classic adversary: the 'Matrix Monster.'

You see $A^{100}$ and your first instinct might be to panic, to reach for a calculator, or to start multiplying $A$ by itself until your hand cramps. But stop!

In the world of competitive mathematics, brute force is rarely the intended path. There is always a secret, a hidden geometric or algebraic reality waiting to be uncovered. Let us peel back the layers of this problem together.

### The Decomposition Strategy

The secret to handling high powers of a matrix is to break it down into something manageable. We look at $A = \begin{bmatrix} 3 & -4 \ 1 & -1 \end{bmatrix}$ and we ask: can we write this as a sum of a simple matrix and the identity matrix $I$?

Let $A = I + M$. If we define $M = A - I$, we get:

$$M = \begin{bmatrix} 3 & -4 \ 1 & -1 \end{bmatrix} - \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -4 \ 1 & -2 \end{bmatrix}$$

Why do we do this? Because the identity matrix $I$ is the '1' of the matrix world; it commutes with everything and makes powers trivial.

### The Nilpotent Revelation

Now, let us test the behavior of our new matrix $M$. We calculate $M^2$:

$$M^2 = \begin{bmatrix} 2 & -4 \ 1 & -2 \end{bmatrix} \begin{bmatrix} 2 & -4 \ 1 & -2 \end{bmatrix} = \begin{bmatrix} (2)(2) + (-4)(1) & (2)(-4) + (-4)(-2) \ (1)(2) + (-2)(1) & (1)(-4) + (-2)(-2) \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$$

This is the 'Aha!' moment! $M^2$ is the null matrix $O$.

This means $M$ is **nilpotent**. Any power of $M$ greater than or equal to 2 will also be the null matrix. This is the key that unlocks the entire problem.

### The Binomial Shortcut

With $M^2 = O$, we can use the Binomial Theorem to expand $(I + M)^{100}$. Since $I$ and $M$ commute, we have:

$$(I + M)^{100} = \binom{100}{0}I + \binom{100}{1}M + \binom{100}{2}M^2 + \dots + \binom{100}{100}M^{100}$$

Because $M^2 = O$, every term from $\binom{100}{2}M^2$ onwards vanishes into thin air. We are left with only the first two terms:

$$A^{100} = I + 100M$$

### The Final Comparison

The problem states $A^{100} = 100B + I$. We have just derived $A^{100} = I + 100M$.

Comparing these two expressions, it is immediate that $100B = 100M$, which implies $B = M$. The question asks for the sum of all elements of $B^{100}$.

Since $B = M$, $B^{100} = M^{100}$. And since $M^2 = O$, $M^{100}$ is also the null matrix $O$.

The sum of all elements of a null matrix is, quite elegantly, **0**. You see, my friend, the JEE is not testing your ability to multiply matrices for an hour; it is testing your ability to spot the structure, to find the nilpotency, and to simplify the complex into the beautiful.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = [31 - 4 - 1]$ and $B$ be two matrices such that $A^{100} = 100 B + I$ . Then the sum of all the elements of $B^{100}$ is .........

Enter Numerical Value:

Visualized Solution

Analyzing Matrix $A$

Matrix Decomposition

Defining Matrix $M$

Calculating $M^{2}$

Nilpotency of $M$

Binomial Expansion

Simplifying $A^{100}$

Comparing with Given Equation

Finding $B^{100}$

Final Sum of Elements

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Decomposition Strategy

The Nilpotent Revelation

The Binomial Shortcut

The Final Comparison

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Visualized Solution

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

Matrix Decomposition

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

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M2

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

Binomial Expansion

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A100