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Anku posted an Question

April 25, 2021 • 03:15 am

30 points

IIT JAM
Mathematics (MA)

Please once tell

please once tell . . .. . .. . . . . . . . . .. . . . . . .. . . . . .

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Gattu uday

See dude here we are applying inverse property of group not finding a Inverse matrix for the given matrix ,both are different ... And u know set of all matrices forms a group under multiplication that means it includes all matrices invertible,non invertible,singular etc all
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Deepak singh 1 Best Answer

Refer attached proof
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sir only tell me isnt it violating property that a singular matrix is non invertible...? rest I understood
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yes , you are right , your this property holds in collection of all n×n matrix over R (in linear algebra ) . But Here G is collection of special set , ( all elements of G have determinant 0 and special type of element ) . We check group axioms on a given set .
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Alka gupta

also see this...
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Alka gupta

see ....
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mam only tell me isnt it violating property that a singular matrix is non invertible...?
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this is a set....and it is a element of this set here not any type of matrix this is
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now we have to prove that this set is a group....and given matrix multiplication is operation....so we have to check all 4 properties of group using operation....
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so here we can not say it is singular ....aur any other type matrix...it just a element of a set
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when you will find inverse of A by adjoint method then you have to check that it is singular or not.....
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