Group Question

Question

If a, b are elements of a group G, then (ba)^-1 =

Comments

 (ba)^-1 = a^-1 *b^-1

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proof?

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Matrix multiplication is a group only if two matrices 'a' and 'b' are non singular.

Consider two non singular matrices 'a' and 'b' and perform ba(multiplication) and inverse of the matrix.

same as $b^{-1}*a^{-1}$.

If these two are same then it satisfies the above properety.

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we should know 2 facts before proving it:

1. Inverse is applicable only for square matrices.

2.Let X be any n*n square matrix, then a matrix Y if it exists such that XY=YX=In-------((Identity matrix of size n*n)

If this condition is true then we can say Both X and Y are inverses of each other.

X^-1 =Y  or Y^-1  = X

To prove :  (ba)^-1   = a^-1b^-1

Assume X =(ba)^-1

             Y=a^-1b^-1

My aim will be XY=YX=I_n

XY= (ba)(a^-1 b^-1 )

     =baa^-1b^-1 

     =I_n

YX= (a^-1 b^-1 )(ba)

      =a^-1b^-1 ba

      =I_n