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BITSAT 2019
Exam
BITSAT
Year
2019
Subject
Mathematics
Chapter
Matrices

BITSAT 2019 · Mathematics · Matrices

If A is an 3 × 3 non- singular matrix, such that AA^ ' = A^ ' A and...

If AA is an 3×33 \times 3 non- singular matrix, such that AA′=A′AAA^{'} = A^{'}A and B=A−1A′B = A^{- 1}A^{'}, then BB′BB^{'} equals to

Options

  1. A(B−1)′\left(B^{- 1}\right) '
  2. BI+BI + B
  3. CII
  4. DB−1B^{- 1}

Correct answer

C. II

Step-by-step solution

Given,

AA′=A′AAA^{'} = A^{'}A

and B=A−1A′B = A^{- 1}A^{'}

BB′=(A−1A′)(A−1A′)′BB^{'} = \left(A^{- 1}A^{'}\right) \left(A^{- 1}A^{'}\right) '

=(A−1A′)[(A′)′(A−1)′]= \left(A^{- 1}A^{'}\right)\left[ \left(A^{'}\right) '\left(A^{- 1}\right) '\right]

=(A−1A′)[A(A′)−1]= \left(A^{- 1}A^{'}\right) \left[A \left(A^{'}\right)- 1\right]

\left[∵ \left(A^{'}\right)' = A, \left(A^{- 1}\right)' = \left(A^{'}\right)- 1\right]</p><p></p><p>= A^{- 1} \left(A^{'}A\right) \left(A^{'}\right)−1- 1

=A−1(AA′(A′)−1)</p><p>= A^{- 1}\left(AA^{'}\left(A^{'}\right)- 1\right)</p><p>= \left(A^{- 1}A\right) \left(A^{- 1}A\right)′'

=I\cdotI=I= I \cdotI = I

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