Why does the trace of equal 3 and not 1?

is a identity matrix, so its diagonal elements are , and their sum is .

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Matrices and Determinants: Mathematics | JEE Main

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

Yes, is block diagonal with blocks and . The first block gives eigenvalue . The block has trace and determinant , so its eigenvalues are and .

Let

\text{A} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix}

. Then the sum of the diagonal elements of the matrix

(\text{A} + \text{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: $\det(A - \lambda I) = (1 - \lambda)(\lambda^2 - \lambda) = 0$ , 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

None. 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 $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix}$ 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 $A^2$ : $A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 16 - 12 & -4 + 3 \\ 0 & 48 - 36 & -12 + 9 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix} = A$ . Thus $A^k = A$ for all integers $k \ge 1$ .

05execute

Expand $(A + I)^{11}$ using the Binomial Theorem: $(A + I)^{11} = \sum_{k=0}^{11} \binom{11}{k} A^k I^{11-k} = I + \sum_{k=1}^{11} \binom{11}{k} A^k = I + \left(\sum_{k=1}^{11} \binom{11}{k}\right) A$ . 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: $\text{tr}(A) = 1 + 4 + (-3) = 2$ . $\text{tr}(I) = 1 + 1 + 1 = 3$ . Therefore, $\text{tr}((A+I)^{11}) = 2047 \cdot \text{tr}(A) + \text{tr}(I) = 2047 \times 2 + 3 = 4094 + 3 = 4097$ .

✓verify

Check with eigenvalues: eigenvalues of $A$ are $1$ (from top-left block) and roots of $\lambda^2 - \text{tr}(B)\lambda + \det(B) = 0$ for $B = \begin{pmatrix} 4 & -1 \\ 12 & -3 \end{pmatrix}$ , where $\text{tr}(B) = 1$ , $\det(B) = 0$ , giving $\lambda = 1, 0$ . 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
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 ( $AI = IA = A$ ), 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 $\begin{pmatrix} 4 & -1 \\ 12 & -3 \end{pmatrix}$ . 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 $\text{A} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix}$ as $2, 2, 2$ , which gives a trace of $3 \times 2^{11} = 6144$ . Compute the eigenvalues of $A$ accurately from its characteristic equation: $\det(A - \lambda I) = (1 - \lambda)(\lambda^2 - \lambda) = 0$ , 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 $A$ consistent throughout: the trace is $2^{11} + 2^{11} + 1^{11} = 2048 + 2048 + 1 = 4097$ .
C · correct: 4097 - None. 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 $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix}$ 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 $A^2$ : $A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 16 - 12 & -4 + 3 \\ 0 & 48 - 36 & -12 + 9 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix} = A$ . Thus $A^k = A$ for all integers $k \ge 1$ .
execute: Expand $(A + I)^{11}$ using the Binomial Theorem: $(A + I)^{11} = \sum_{k=0}^{11} \binom{11}{k} A^k I^{11-k} = I + \sum_{k=1}^{11} \binom{11}{k} A^k = I + \left(\sum_{k=1}^{11} \binom{11}{k}\right) A$ . 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: $\text{tr}(A) = 1 + 4 + (-3) = 2$ . $\text{tr}(I) = 1 + 1 + 1 = 3$ . Therefore, $\text{tr}((A+I)^{11}) = 2047 \cdot \text{tr}(A) + \text{tr}(I) = 2047 \times 2 + 3 = 4094 + 3 = 4097$ .
verify: Check with eigenvalues: eigenvalues of $A$ are $1$ (from top-left block) and roots of $\lambda^2 - \text{tr}(B)\lambda + \det(B) = 0$ for $B = \begin{pmatrix} 4 & -1 \\ 12 & -3 \end{pmatrix}$ , where $\text{tr}(B) = 1$ , $\det(B) = 0$ , giving $\lambda = 1, 0$ . 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 $2^{11} + 2^{11} + 1^{11} = 2048 + 2048 + 1 = 4097$ .

Shortcut:

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

given: $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix}$ .

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

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

execute: The characteristic polynomial of $A$ is $(1 - \lambda)((4 - \lambda)(-3 - \lambda) + 12) = (1 - \lambda)(\lambda^2 - \lambda) = -\lambda(\lambda - 1)^2 = 0$ . Hence, eigenvalues of $A$ are $\lambda_1 = 1, \lambda_2 = 1, \lambda_3 = 0$ .

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 $\text{tr}((A+I)^{11}) = 2048 + 2048 + 1 = 4097$ .

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

Matrices and Determinants: Mathematics | JEE Main

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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