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BITSAT 2024Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Mathematics
Chapter
Matrices

BITSAT 2024 · Paper 2 · Mathematics · Matrices

If A and B are symmetric matrices of same order such that A B+B A=X...

If AA and BB are symmetric matrices of same order such that AB+BA=XA B+B A=X and AB−BA=YA B-B A=Y, then (XY)T=(X Y)^{\mathrm{T}}=

Options

  1. AXYX Y
  2. BX⊤Y⊤X^{\top} Y^{\top}
  3. C-YX
  4. D−Y⊤X⊤-Y^{\top} X^{\top}

Correct answer

C. -YX

Step-by-step solution

Since XY=(AB+BA)(AB−BA)=(AB)AB+(BA)(AB)−(AB)(BA)−(BA)(BA)X Y=(A B+B A)(A B-B A)=(\mathrm{AB}) \mathrm{AB}+(\mathrm{BA})(\mathrm{AB})-(\mathrm{AB})(\mathrm{BA})-(\mathrm{BA})(\mathrm{BA}) Now (XY)T=((AB)⋅(AB))T+(BA⋅AB)T−(AB⋅BA)T(X Y)^{\mathrm{T}}=((\mathrm{AB}) \cdot(\mathrm{AB}))^{\mathrm{T}}+(\mathrm{BA} \cdot \mathrm{AB})^{\mathrm{T}}-(\mathrm{AB} \cdot \mathrm{BA})^{\mathrm{T}}=(AB)T⋅(AB)T+(AB)T⋅(BA)T−(BA)T(AB)T−(BA)T(BA)T\begin{array}{r}=(\mathrm{AB})^{\mathrm{T}} \cdot(\mathrm{AB})^{\mathrm{T}}+(\mathrm{AB})^{\mathrm{T}} \cdot(\mathrm{BA})^{\mathrm{T}}-(\mathrm{BA})^{\mathrm{T}}(\mathrm{AB})^{\mathrm{T}} \\ -(\mathrm{BA})^{\mathrm{T}}(\mathrm{BA})^{\mathrm{T}}\end{array}=\left(B⊤B^{\top}\cdotA^{\top}\right)\left(B^{\top} ⋅\cdotA^{\top}\right)+\left(B⊤B^{\top}\cdotA^{\top}\right)\cdot\left(A^{\top} B^{\top}\right)-\left(\mathrm{A}^{\top} \mathrm{B}^{\top}\right)\left(\mathrm{B}^{\top} \mathrm{A}^{\top}\right)- \left(\mathrm{A}^{\top} \mathrm{B}^{\top}\right)\left(\mathrm{A}^{\top} \mathrm{B}^{\top}\right) Since, A & B are symmetric matrix.=(BA)(BA)+(BA)(AB)−(AB)(BA)−(AB)(AB)=(BA−AB)(BA+AB)=−YX\mathrm{BA})(\mathrm{BA})+(\mathrm{BA})(\mathrm{AB})-(\mathrm{AB})(\mathrm{BA})-(\mathrm{AB})(\mathrm{AB})=(\mathrm{BA}-\mathrm{AB})(\mathrm{BA}+\mathrm{AB})=-\mathrm{YX}
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