JEE MainMathematics

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Trace of a Matrix Power via Eigenvalues

Q: Why does the trace of I equal 3 and not 1?

I is a 3 × 3 identity matrix, so its diagonal elements are 1, 1, 1, and their sum is 1 + 1 + 1 = 3.

Q: Can we read the eigenvalues directly from the block diagonal form?

Yes, A is block diagonal with blocks [1] and [ 4 -1; 12 -3 ]. The first block gives eigenvalue 1. The 2 × 2 block has trace 1 and determinant 0, so its eigenvalues are 1 and 0.

Let A = [ 1 0 0; 0 4 -1; 0 12 -3 ]. Then the sum of the diagonal elements of the matrix (A + I)^11 is equal to

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A6144B2050C4097D4094

Step-by-step solutionView

Correct answer

The sum of the diagonal elements (trace) of $(A + I)^{11}$ is $2^{12} + 1 = 4097$ .

Option analysis

Why each option works or fails

A · 6144

Believing that all three eigenvalues of $A$ are $1$ , yielding eigenvalues of $A + I$ as $2, 2, 2$ , which gives a trace of $3 \times 2^{11} = 6144$ . Compute the eigenvalues of $A$ accurately from its characteristic equation: $\det (A - λ I) = (1 - λ) (λ^{2} - λ) = 0$ , giving eigenvalues $1, 1, 0$ rather than $1, 1, 1$ .

B · 2050

Using exponent 10 instead of 11, computing $2^{11} + 1 = 2049$ or miscalculating $2 \times 2^{10} + 2 = 2050$ . Keep the required power $n = 11$ consistent throughout: the trace is $2^{11} + 2^{11} + 1^{11} = 2048 + 2048 + 1 = 4097$ .

C · 4097

This is the correct option. Correctly identifying the eigenvalues of $A$ as $1, 1, 0$ , which gives eigenvalues for $(A + I)^{11}$ as $2^{11}, 2^{11}, 1^{11}$ , summing to $2048 + 2048 + 1 = 4097$ .

D · 4094

Subtracting rather than adding the third eigenvalue's contribution, or computing $2 \cdot 2^{11} - 2 = 4094$ . Ensure the eigenvalue $0$ of $A$ shifts to $0 + 1 = 1$ in $A + I$ , so its 11th power contributes $+ 1^{11} = + 1$ , not $- 2$ or $- 3$ .

Reviewed route

Solution

StepWorking

01Given

Given matrix $A = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{array})$ and identity matrix $I$ of size $3 \times 3$ .

02Goal

Find the sum of diagonal elements (trace) of the matrix $(A + I)^{11}$ .

03Approach

Compute $A^{2}$ to look for a pattern or idempotence. Then use the Binomial Theorem since $A$ and $I$ commute, and take the trace linearly.

04Execute

Compute $A^{2}$ : $A^{2} = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{array}) (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{array}) = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 16 - 12 & - 4 + 3 \\ 0 & 48 - 36 & - 12 + 9 \end{array}) = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{array}) = A$ . Thus $A^{k} = A$ for all integers $k \geq 1$ .

05Execute

Expand $(A + I)^{11}$ using the Binomial Theorem: $(A + I)^{11} = \sum_{k = 0}^{11} (\binom{11}{k}) A^{k} I^{11 - k} = I + \sum_{k = 1}^{11} (\binom{11}{k}) A^{k} = I + (\sum_{k = 1}^{11} (\binom{11}{k})) A$ . Since $\sum_{k = 0}^{11} (\binom{11}{k}) = 2^{11}$ , we have $\sum_{k = 1}^{11} (\binom{11}{k}) = 2^{11} - 1 = 2048 - 1 = 2047$ . So, $(A + I)^{11} = 2047 A + I$ .

06Execute

Find the trace: $tr (A) = 1 + 4 + (- 3) = 2$ . $tr (I) = 1 + 1 + 1 = 3$ . Therefore, $tr ((A + I)^{11}) = 2047 \cdot tr (A) + tr (I) = 2047 \times 2 + 3 = 4094 + 3 = 4097$ .

✓Verify

Check with eigenvalues: eigenvalues of $A$ are $1$ (from top-left block) and roots of $λ^{2} - tr (B) λ + \det (B) = 0$ for $B = (\begin{array}{cc} 4 & - 1 \\ 12 & - 3 \end{array})$ , where $tr (B) = 1$ , $\det (B) = 0$ , giving $λ = 1, 0$ . Thus eigenvalues of $A$ are $1, 1, 0$ . The eigenvalues of $A + I$ are $1 + 1, 1 + 1, 0 + 1 = 2, 2, 1$ . The sum of diagonal elements of $(A + I)^{11}$ is $2^{11} + 2^{11} + 1^{11} = 2048 + 2048 + 1 = 4097$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why can we use the binomial expansion on matrices

(A + I)^{11}

The binomial theorem applies to matrices if and only if the two matrices commute. Since $I$ commutes with any matrix ( $A I = I A = A$ ), expanding $(A + I)^{11}$ is completely valid.

Why does the trace of

I

equal 3 and not 1?

$I$ is a $3 \times 3$ identity matrix, so its diagonal elements are $1, 1, 1$ , and their sum is $1 + 1 + 1 = 3$ .

Can we read the eigenvalues directly from the block diagonal form?

Yes, $A$ is block diagonal with blocks $[1]$ and $(\begin{array}{cc} 4 & - 1 \\ 12 & - 3 \end{array})$ . The first block gives eigenvalue $1$ . The $2 \times 2$ block has trace $1$ and determinant $0$ , so its eigenvalues are $1$ and $0$ .

Answer

The sum of the diagonal elements (trace) of $(A + I)^{11}$ is $2^{12} + 1 = 4097$ .

Why each option works or fails

A: 6144 - Believing that all three eigenvalues of $A$ are $1$ , yielding eigenvalues of $A + I$ as $2, 2, 2$ , which gives a trace of $3 \times 2^{11} = 6144$ . Compute the eigenvalues of $A$ accurately from its characteristic equation: $\det (A - λ I) = (1 - λ) (λ^{2} - λ) = 0$ , giving eigenvalues $1, 1, 0$ rather than $1, 1, 1$ .
B: 2050 - Using exponent 10 instead of 11, computing $2^{11} + 1 = 2049$ or miscalculating $2 \times 2^{10} + 2 = 2050$ . Keep the required power $n = 11$ consistent throughout: the trace is $2^{11} + 2^{11} + 1^{11} = 2048 + 2048 + 1 = 4097$ .
C · correct: 4097 - This is the correct option. Correctly identifying the eigenvalues of $A$ as $1, 1, 0$ , which gives eigenvalues for $(A + I)^{11}$ as $2^{11}, 2^{11}, 1^{11}$ , summing to $2048 + 2048 + 1 = 4097$ .
D: 4094 - Subtracting rather than adding the third eigenvalue's contribution, or computing $2 \cdot 2^{11} - 2 = 4094$ . Ensure the eigenvalue $0$ of $A$ shifts to $0 + 1 = 1$ in $A + I$ , so its 11th power contributes $+ 1^{11} = + 1$ , not $- 2$ or $- 3$ .

Step-by-step solution

Given: Given matrix $A = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{array})$ and identity matrix $I$ of size $3 \times 3$ .
Goal: Find the sum of diagonal elements (trace) of the matrix $(A + I)^{11}$ .
Approach: Compute $A^{2}$ to look for a pattern or idempotence. Then use the Binomial Theorem since $A$ and $I$ commute, and take the trace linearly.
Execute: Compute $A^{2}$ : $A^{2} = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{array}) (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{array}) = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 16 - 12 & - 4 + 3 \\ 0 & 48 - 36 & - 12 + 9 \end{array}) = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{array}) = A$ . Thus $A^{k} = A$ for all integers $k \geq 1$ .
Execute: Expand $(A + I)^{11}$ using the Binomial Theorem: $(A + I)^{11} = \sum_{k = 0}^{11} (\binom{11}{k}) A^{k} I^{11 - k} = I + \sum_{k = 1}^{11} (\binom{11}{k}) A^{k} = I + (\sum_{k = 1}^{11} (\binom{11}{k})) A$ . Since $\sum_{k = 0}^{11} (\binom{11}{k}) = 2^{11}$ , we have $\sum_{k = 1}^{11} (\binom{11}{k}) = 2^{11} - 1 = 2048 - 1 = 2047$ . So, $(A + I)^{11} = 2047 A + I$ .
Execute: Find the trace: $tr (A) = 1 + 4 + (- 3) = 2$ . $tr (I) = 1 + 1 + 1 = 3$ . Therefore, $tr ((A + I)^{11}) = 2047 \cdot tr (A) + tr (I) = 2047 \times 2 + 3 = 4094 + 3 = 4097$ .
Verify: Check with eigenvalues: eigenvalues of $A$ are $1$ (from top-left block) and roots of $λ^{2} - tr (B) λ + \det (B) = 0$ for $B = (\begin{array}{cc} 4 & - 1 \\ 12 & - 3 \end{array})$ , where $tr (B) = 1$ , $\det (B) = 0$ , giving $λ = 1, 0$ . Thus eigenvalues of $A$ are $1, 1, 0$ . The eigenvalues of $A + I$ are $B = (\begin{array}{cc} 4 & - 1 \\ 12 & - 3 \end{array})$ . The sum of diagonal elements of $(A + I)^{11}$ is $2^{11} + 2^{11} + 1^{11} = 2048 + 2048 + 1 = 4097$ .

Shortcut:

When to use it: Fastest route when the trace of a high matrix power is required.

Given: $A = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{array})$ .

Goal: Compute $tr ((A + I)^{11})$ via eigenvalues of $A$ .

Approach: The eigenvalues of $(A + I)^{11}$ are $(λ_{i} + 1)^{11}$ , where $λ_{i}$ are the eigenvalues of $A$ . The trace is the sum of these eigenvalues.

Execute: The characteristic polynomial of $A$ is $(1 - λ) ((4 - λ) (- 3 - λ) + 12) = (1 - λ) (λ^{2} - λ) = - λ (λ - 1)^{2} = 0$ . Hence, eigenvalues of $A$ are $λ_{1} = 1, λ_{2} = 1, λ_{3} = 0$ .

Execute: Eigenvalues of $(A + I)^{11}$ are $(1 + 1)^{11}, (1 + 1)^{11}, (0 + 1)^{11}$ , which are $2^{11}, 2^{11}, 1^{11}$ . Sum of diagonal elements is $tr ((A + I)^{11}) = 2048 + 2048 + 1 = 4097$ .

Verify: $2048 + 2048 + 1 = 4097$ . Matches Option 2.

Trace of a Matrix Power via Eigenvalues

What feels right?

The sum of the diagonal elements (trace) of $(A + I)^{11}$ is $2^{12} + 1 = 4097$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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