Let A = [[1, 0, 0], [0, 4, -1], [0, 12, -3]]. Then the sum of the diagonal elements of the matrix (A

Q: Wait, why can't we just multiply the matrix 11 times? It's just arithmetic, right?

In the heat of a JEE Advanced exam, time is your most precious currency. While you could technically multiply it, you would likely make a calculation error by the third or fourth iteration. JEE problems are designed to test your ability to spot patterns, not your endurance for tedious arithmetic. Always look for the 'shortcut'—the mathematical structure hidden beneath the numbers.

Q: Why does the Binomial Theorem work here? I thought matrix multiplication wasn't commutative?

You are absolutely right to be cautious! Matrix multiplication is generally non-commutative ($AB \neq BA$). However, the identity matrix $I$ commutes with every square matrix ($AI = IA = A$). Because $A$ and $I$ commute, the Binomial Theorem holds perfectly. This is a crucial distinction to remember.

Q: How do I know if a matrix is idempotent without checking every power?

You don't need to check every power! The definition of an idempotent matrix is simply $A^2 = A$. If you find that $A^2 = A$, then by induction, $A^3 = A^2 \cdot A = A \cdot A = A$, and so on. It is a beautiful, self-replicating property. Once you prove $A^2 = A$, you have unlocked the key to all higher powers.

Question

Let $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 & 12 & -3 \end{bmatrix}$. Then the sum of the diagonal elements of the matrix $(A+I)^{11}$ is equal to:

Options:

(A) 6144

(B) 4094

(C) 4097

(D) 2050

Sigma Percentile · Accepted Answer

### The Illusion of Complexity: A Journey into Matrix Patterns

Welcome, future engineer. Today, we stand before a problem that, at first glance, looks like a nightmare of arithmetic. You see a matrix

$$A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 & 12 & -3 \end{bmatrix}$$

raised to the power of 11, and your instinct might be to panic. You might think, "Do I really have to multiply this matrix eleven times?" The answer, as it often is in the JEE Advanced, is a resounding "No." Let us peel back the layers of this problem together.

### Phase 1: The Investigation

When you encounter a matrix raised to a high power, your first move should never be brute force. It should be investigation. We need to see how this matrix behaves when it interacts with itself.

Let us calculate $A^2$. We perform the row-by-column multiplication:

$$A^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 & 12 & -3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 & 12 & -3 \end{bmatrix}$$

As we compute the elements, watch closely. The first row remains unchanged. The second row: $4(4) + (-1)(12) = 16 - 12 = 4$, and $4(-1) + (-1)(-3) = -4 + 3 = -1$.

The third row: $12(4) + (-3)(12) = 48 - 36 = 12$, and $12(-1) + (-3)(-3) = -12 + 9 = -3$. Look at the result!

$$A^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 & 12 & -3 \end{bmatrix}$$

It is identical to $A$. This is the "Aha!" moment. We have discovered that $A^2 = A$. This is known as an **idempotent matrix**. It is a rare and beautiful property where the matrix, when squared, remains unchanged. This implies that for any integer $k \ge 1$, $A^k = A$. The complexity of the power 11 has just evaporated.

### Phase 2: The Binomial Weapon

Now that we know $A^k = A$, we can tackle the expression $(A+I)^{11}$. Because the identity matrix $I$ commutes with $A$, we are permitted to use the Binomial Theorem:

$$(A+I)^{11} = \sum_{k=0}^{11} \binom{11}{k} A^k I^{11-k}$$

Let us expand this carefully. We separate the $k=0$ term, because $A^0 = I$. So,

$$(A+I)^{11} = \binom{11}{0} I + \sum_{k=1}^{11} \binom{11}{k} A^k$$

Since $A^k = A$ for all $k \ge 1$, we can substitute $A$ into the summation:

$$(A+I)^{11} = I + A \left( \sum_{k=1}^{11} \binom{11}{k} ight)$$

The sum of binomial coefficients $\sum_{k=0}^{n} \binom{n}{k} = 2^n$. Therefore,

$$\sum_{k=1}^{11} \binom{11}{k} = 2^{11} - \binom{11}{0} = 2048 - 1 = 2047$$

Our expression simplifies beautifully to $(A+I)^{11} = I + 2047A$.

### Phase 3: The Final Trace

The problem asks for the sum of the diagonal elements, which is the trace, denoted as $	ext{tr}(M)$. We need $	ext{tr}(I + 2047A)$.

Thanks to the linearity of the trace operator, we can write this as $	ext{tr}(I) + 2047 \cdot 	ext{tr}(A)$. The trace of the identity matrix $I$ is simply the sum of its diagonal elements: $1 + 1 + 1 = 3$.

The trace of $A$ is the sum of its diagonal elements: $1 + 4 + (-3) = 2$. Now, we perform the final calculation:

$$3 + 2047(2) = 3 + 4094 = 4097$$

We have arrived at the solution. The journey from a terrifying matrix power to a simple arithmetic sum was made possible only by our willingness to look for the pattern. Remember, in JEE Advanced, the math is rarely about brute force; it is about finding the elegance hidden in the structure.

**Final Answer: 4097**

Animated Solution for Mathematics - Matrices and Determinants: Let $A = 1000412 0 - 1 - 3$ . Then the sum of the diagonal elements of the matrix $(A + I)^{11}$ is equal to:

Select Answer:

Visualized Solution

Understanding the Matrix $A$

Testing for Idempotency: Calculating $A^{2}$

Matrix Multiplication

Verifying $A^{2} = A$

Applying Binomial Expansion

Separating the $k = 0$ Term

Simplifying with Idempotency

Calculating the Binomial Sum

Applying the Trace Property

Calculating Individual Traces

Final Substitution

Conclusion & Key Takeaway

The Sigma Insight: Algebraic Operations on Matrices

The Illusion of Complexity

Phase 1

Phase 2

Phase 3

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