JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Symmetry of Matrix Commutators Under Powers of Symmetric and Skew-Symmetric Matrices

Q: Why did (-B)^26 become +B^26?

Because (-1)^26 = +1 since the power 26 is even.

Let A, B, C be 3 × 3 matrices such that A is symmetric and B and C are skew-symmetric. Consider the statements (S1) A^13 B^26 - B^26 A^13 is symmetric (S2) A^26 C^13 - C^13 A^26 is symmetric Then,

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What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

AOnly S2 is trueBBoth S1 and S2 are falseCOnly S1 is trueDBoth S1 and S2 are true

Step-by-step solutionView

Correct answer

Only statement S2 is true because the transpose of the matrix expression in S1 equals its negative (making it skew-symmetric), whereas in S2 the transpose equals the matrix itself.

Option analysis

Why each option works or fails

A · Only S2 is true

This option correctly evaluates the transposes of both expressions. Correctly applying $(X Y)^{T} = Y^{T} X^{T}$ and accounting for the sign change from the odd power of the skew-symmetric matrix confirms S1 is skew-symmetric and S2 is symmetric.

B · Both S1 and S2 are false

Believing that the commutator $X Y - Y X$ of two symmetric matrices or a symmetric and skew-symmetric matrix is always skew-symmetric regardless of signs. Trace the signs carefully: because $C^{13}$ is skew-symmetric, $(A^{26} C^{13} - C^{13} A^{26})^{T} = - (C^{13} A^{26} - A^{26} C^{13}) = A^{26} C^{13} - C^{13} A^{26}$ , which is symmetric.

C · Only S1 is true

Confusing the behavior of even versus odd powers of skew-symmetric matrices, treating $B^{26}$ as skew-symmetric instead of symmetric. Remember that an even power of a skew-symmetric matrix is symmetric: $(B^{26})^{T} = (B^{T})^{26} = (- B)^{26} = B^{26}$ . The commutator of two symmetric matrices is skew-symmetric, not symmetric.

D · Both S1 and S2 are true

Assuming that any expression of the form $X Y - Y X$ preserves symmetry if the components are symmetric or skew-symmetric. Apply the transpose operation directly: $(X Y - Y X)^{T} = Y^{T} X^{T} - X^{T} Y^{T}$ . When $X$ and $Y$ are symmetric, this evaluates to $Y X - X Y = - (X Y - Y X)$ , which is skew-symmetric.

Reviewed route

Solution

StepWorking

01Given

$A, B, C$ are $3 \times 3$ matrices with $A^{T} = A$ , $B^{T} = - B$ , and $C^{T} = - C$ .

02Goal

Determine the truth values of (S1): $M_{1} = A^{13} B^{26} - B^{26} A^{13}$ is symmetric, and (S2): $M_{2} = A^{26} C^{13} - C^{13} A^{26}$ is symmetric.

03Approach

Apply the transpose operation using $(X - Y)^{T} = X^{T} - Y^{T}$ , $(X Y)^{T} = Y^{T} X^{T}$ , and $(X^{k})^{T} = (X^{T})^{k}$ . If $M^{T} = M$ , it is symmetric; if $M^{T} = - M$ , it is skew-symmetric.

04Execute

Evaluate $M_{1}^{T}$ : $M_{1}^{T} = (A^{13} B^{26} - B^{26} A^{13})^{T} = (B^{26})^{T} (A^{13})^{T} - (A^{13})^{T} (B^{26})^{T}$ . Since $(B^{T})^{26} = (- B)^{26} = B^{26}$ and $(A^{T})^{13} = A^{13}$ , we have: $M_{1}^{T} = B^{26} A^{13} - A^{13} B^{26} = - (A^{13} B^{26} - B^{26} A^{13}) = - M_{1}$ . Thus, $M_{1}$ is skew-symmetric, so S1 is false.

05Execute

Evaluate $M_{2}^{T}$ : $M_{2}^{T} = (A^{26} C^{13} - C^{13} A^{26})^{T} = (C^{13})^{T} (A^{26})^{T} - (A^{26})^{T} (C^{13})^{T}$ . Since $(C^{T})^{13} = (- C)^{13} = - C^{13}$ and $(A^{T})^{26} = A^{26}$ , we have: $M_{2}^{T} = (- C^{13}) A^{26} - A^{26} (- C^{13}) = - C^{13} A^{26} + A^{26} C^{13} = A^{26} C^{13} - C^{13} A^{26} = M_{2}$ . Thus, $M_{2}$ is symmetric, so S2 is true.

✓Verify

Observe the structure: for any symmetric $X$ and skew-symmetric $Y$ , $[X, Y] = X Y - Y X$ satisfies $[X, Y]^{T} = Y^{T} X^{T} - X^{T} Y^{T} = (- Y) X - X (- Y) = - (Y X - X Y) = X Y - Y X = [X, Y]$ (symmetric). Here, $B^{26}$ is symmetric, so $[A^{13}, B^{26}]$ is the commutator of two symmetric matrices, which is always skew-symmetric. In contrast, $C^{13}$ is skew-symmetric while $A^{26}$ is symmetric, so $[A^{26}, C^{13}]$ is the commutator of a symmetric and a skew-symmetric matrix, which is always symmetric. This confirms S1 is false and S2 is true.

Hints that build this answer step by step

What are the transposes of $A^{13}$ , $B^{26}$ , $A^{26}$ , and $C^{13}$ given that $A^{T} = A$ , $B^{T} = - B$ , and $C^{T} = - C$ ?
$(A^{13})^{T} = A^{13}$ , $(B^{26})^{T} = B^{26}$ , $(A^{26})^{T} = A^{26}$ , and $(C^{13})^{T} = - C^{13}$
Let $M = A^{13} B^{26} - B^{26} A^{13}$ . Using $(X Y)^{T} = Y^{T} X^{T}$ and the results from step 1, what is $M^{T}$ ?
$M^{T} = - M$ , so $M$ is skew-symmetric
Let $N = A^{26} C^{13} - C^{13} A^{26}$ . Using $(C^{13})^{T} = - C^{13}$ and $(A^{26})^{T} = A^{26}$ , what is $N^{T}$ ?
$N^{T} = N$ , so $N$ is symmetric

✓ 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 did

(- B)^{26}

become

+ B^{26}

Because $(- 1)^{26} = + 1$ since the power 26 is even.

Answer

Only statement S2 is true because the transpose of the matrix expression in S1 equals its negative (making it skew-symmetric), whereas in S2 the transpose equals the matrix itself.

Why each option works or fails

A · correct: Only S2 is true - This option correctly evaluates the transposes of both expressions. Correctly applying $(X Y)^{T} = Y^{T} X^{T}$ and accounting for the sign change from the odd power of the skew-symmetric matrix confirms S1 is skew-symmetric and S2 is symmetric.
B: Both S1 and S2 are false - Believing that the commutator $X Y - Y X$ of two symmetric matrices or a symmetric and skew-symmetric matrix is always skew-symmetric regardless of signs. Trace the signs carefully: because $C^{13}$ is skew-symmetric, $(A^{26} C^{13} - C^{13} A^{26})^{T} = - (C^{13} A^{26} - A^{26} C^{13}) = A^{26} C^{13} - C^{13} A^{26}$ , which is symmetric.
C: Only S1 is true - Confusing the behavior of even versus odd powers of skew-symmetric matrices, treating $B^{26}$ as skew-symmetric instead of symmetric. Remember that an even power of a skew-symmetric matrix is symmetric: $(B^{26})^{T} = (B^{T})^{26} = (- B)^{26} = B^{26}$ . The commutator of two symmetric matrices is skew-symmetric, not symmetric.
D: Both S1 and S2 are true - Assuming that any expression of the form $X Y - Y X$ preserves symmetry if the components are symmetric or skew-symmetric. Apply the transpose operation directly: $(X Y - Y X)^{T} = Y^{T} X^{T} - X^{T} Y^{T}$ . When $X$ and $Y$ are symmetric, this evaluates to $Y X - X Y = - (X Y - Y X)$ , which is skew-symmetric.

Step-by-step solution

Given: $A, B, C$ are $3 \times 3$ matrices with $A^{T} = A$ , $B^{T} = - B$ , and $C^{T} = - C$ .
Goal: Determine the truth values of (S1): $M_{1} = A^{13} B^{26} - B^{26} A^{13}$ is symmetric, and (S2): $M_{2} = A^{26} C^{13} - C^{13} A^{26}$ is symmetric.
Approach: Apply the transpose operation using $(X - Y)^{T} = X^{T} - Y^{T}$ , $(X Y)^{T} = Y^{T} X^{T}$ , and $(X^{k})^{T} = (X^{T})^{k}$ . If $M^{T} = M$ , it is symmetric; if $M^{T} = - M$ , it is skew-symmetric.
Execute: Evaluate $M_{1}^{T}$ : $M_{1}^{T} = (A^{13} B^{26} - B^{26} A^{13})^{T} = (B^{26})^{T} (A^{13})^{T} - (A^{13})^{T} (B^{26})^{T}$ . Since $(B^{T})^{26} = (- B)^{26} = B^{26}$ and $(A^{T})^{13} = A^{13}$ , we have: $M_{1} = A^{13} B^{26} - B^{26} A^{13}$ . Thus, $M_{1}$ is skew-symmetric, so S1 is false.
Execute: Evaluate $M_{2}^{T}$ : $M_{2}^{T} = (A^{26} C^{13} - C^{13} A^{26})^{T} = (C^{13})^{T} (A^{26})^{T} - (A^{26})^{T} (C^{13})^{T}$ . Since $(C^{T})^{13} = (- C)^{13} = - C^{13}$ and $(A^{T})^{26} = A^{26}$ , we have: $M_{2}^{T} = (- C^{13}) A^{26} - A^{26} (- C^{13}) = - C^{13} A^{26} + A^{26} C^{13} = A^{26} C^{13} - C^{13} A^{26} = M_{2}$ . Thus, $M_{2}$ is symmetric, so S2 is true.
Verify: Observe the structure: for any symmetric $X$ and skew-symmetric $Y$ , $[X, Y] = X Y - Y X$ satisfies $[X, Y]^{T} = Y^{T} X^{T} - X^{T} Y^{T} = (- Y) X - X (- Y) = - (Y X - X Y) = X Y - Y X = [X, Y]$ (symmetric). Here, $B^{26}$ is symmetric, so $[A^{13}, B^{26}]$ is the commutator of two symmetric matrices, which is always skew-symmetric. In contrast, $C^{13}$ is skew-symmetric while $A^{26}$ is symmetric, so $[A^{26}, C^{13}]$ is the commutator of a symmetric and a skew-symmetric matrix, which is always symmetric. This confirms S1 is false and S2 is true.

Symmetry of Matrix Commutators Under Powers of Symmetric and Skew-Symmetric Matrices

What feels right?

Only statement S2 is true because the transpose of the matrix expression in S1 equals its negative (making it skew-symmetric), whereas in S2 the transpose equals the matrix itself.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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