JEE Challenger
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MathMedMarks to AllAdv 2011 • Paper 1

Simplifying Expression of Commuting Skew-Symmetric Matrices

Let MM and NN be two 3×33\times 3 non-singular skew-symmetric matrices such that MN=NMMN = NM . If PTP^T denotes the transpose of PP, then M2N2(MTN)−1(MN−1)TM^2N^2 (M^T N)^{-1} (MN^{-1})^T is equal to

Official Notice: Marks Awarded to All

This question was dropped / full marks were awarded to all candidates in the official answer key by the exam conducting body due to an ambiguity or error in the question or options.

Options

A

M2M^2

B

−N2-N^2

C

−M2-M^2

D

MNMN

Topics & Concepts

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Step-by-Step Solution

To find the value of the given matrix expression, we first analyze the properties given in the problem:

  1. Existence of 3×33 \times 3 Non-Singular Skew-Symmetric Matrices:
    For any n×nn \times n skew-symmetric matrix AA, by definition:
    AT=−AA^T = -A
    Taking the determinant on both sides:
    det⁡(AT)=det⁡(−A)=(−1)ndet⁡(A)\det(A^T) = \det(-A) = (-1)^n \det(A)
    Since det⁡(AT)=det⁡(A)\det(A^T) = \det(A), for an odd order n=3n = 3:
    det⁡(A)=−det⁡(A)  ⟹  2det⁡(A)=0  ⟹  det⁡(A)=0\det(A) = -\det(A) \implies 2\det(A) = 0 \implies \det(A) = 0
    Therefore, any 3×33 \times 3 skew-symmetric matrix is necessarily singular (det⁡(A)=0\det(A) = 0), which implies its inverse does not exist.

    Because no non-singular 3×33 \times 3 skew-symmetric matrix can exist, the premise of the question is mathematically inconsistent. For this reason, the official answer key for this question was declared as MARKS TO ALL.

  1. Formal Algebraic Simplification (Intended Solution):
    Assuming formally that MM and NN are invertible, commuting skew-symmetric matrices (MT=−MM^T = -M, NT=−NN^T = -N, and MN=NMMN = NM):

    Since MN=NMMN = NM, their inverses and transposes also commute with each other:
    MN−1=N−1M,M−1N=NM−1,M−1N−1=N−1M−1MN^{-1} = N^{-1}M, \quad M^{-1}N = NM^{-1}, \quad M^{-1}N^{-1} = N^{-1}M^{-1}

    Now, we simplify the terms in the given expression:
    E=M2N2(MTN)−1(MN−1)TE = M^2N^2 (M^T N)^{-1} (MN^{-1})^T

    • First term:
      (MTN)−1=(−MN)−1=−(MN)−1=−N−1M−1(M^T N)^{-1} = (-MN)^{-1} = -(MN)^{-1} = -N^{-1}M^{-1}

    • Second term:
      (MN−1)T=(N−1)TMT=(NT)−1MT=(−N)−1(−M)=(−N−1)(−M)=N−1M(MN^{-1})^T = (N^{-1})^T M^T = (N^T)^{-1} M^T = (-N)^{-1}(-M) = (-N^{-1})(-M) = N^{-1}M

    Substituting these into the expression:
    E=M2N2(−N−1M−1)(N−1M)E = M^2 N^2 \left( -N^{-1}M^{-1} \right) \left( N^{-1}M \right)

    Since all matrices in the product commute with one another, we can rearrange the factors:
    E=−(M2⋅M−1⋅M)(N2⋅N−1⋅N−1)E = - \left( M^2 \cdot M^{-1} \cdot M \right) \left( N^2 \cdot N^{-1} \cdot N^{-1} \right)
    E=−(M2−1+1)(N2−1−1)E = - \left( M^{2 - 1 + 1} \right) \left( N^{2 - 1 - 1} \right)
    E=−M2N0=−M2E = - M^2 N^0 = -M^2

Thus, the algebraically intended answer is −M2-M^2 (Option C), but due to the non-existence of non-singular 3×33 \times 3 skew-symmetric matrices, this question was awarded MARKS TO ALL.

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