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If A is an 3 x 3 non - singular matrix such that AA' = A'A   and  B = A-1  A' , then BB' equals :

  • Option 1)

    B-1

  • Option 2)

    (B-1)'

  • Option 3)

    I + B

  • Option 4)

    I

 

Answers (2)

As we learnt in 

Property of Transpose -

\left ( AB \right ){}'=B{}'A{}'

- wherein

A ' is the conjugate matrix of A

 

 \Rightarrow AA^{1}= A^{1}A

and \: \beta = A^{-1}A^{1}\:\:\:(given)

So that    BA= A^{-1}A^{1}A\:\:\:\:\left [ multiply\: by \:A \right ]

                         = A^{-1}AA^{1}

                         = IA^{1}

                         = A^{1}

Now, \left ( BA \right )^{1}= \left ( A ^{1}\right )^{1}= A

          A^{1}B^{1}= A\\ A^{-1}A^{1}B^{1}= A^{-1}A= I\\\beta B^{1}= I            [since A-1A = I and multiply by A-1]


Option 1)

B-1

This option is incorrect.

Option 2)

(B-1)'

This option is incorrect.

Option 3)

I + B

This option is incorrect.

Option 4)

I

This option is correct.

Posted by

Sabhrant Ambastha

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