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    Hi, i'm currently a bit confused about proving that two matricies e.g A and B are non singular. It says in my textbook that (AB)-1=B-1A-1 if both are non singular, however i'm a bit confused to how they've got this, any help would be appreciated
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    (Original post by Season One)
    Hi, i'm currently a bit confused about proving that two matricies e.g A and B are non singular. It says in my textbook that (AB)-1=B-1A-1 if both are non singular, however i'm a bit confused to how they've got this, any help would be appreciated
    Well if (AB)-1=B-1A-1, then (AB)(B-1 A-1) = I

    We can the write this as A(BB-1)A-1 [ Since matrices are associative i.e. A(BC)=(AB)C ]

    This means that (AB)(B-1 A-1) = AIA-1 = AA-1 = I, which follows from (AB)-1=B-1A-1
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    More simply than above:
    Without loss of generality, a matrix is non-singular if it has a determinant that is non zero.

    the inverse of a singular matrix is not defined, so calculating the determinants of A & B and showing them to be zero is sufficient.
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    (Original post by dknt)
    Well if (AB)-1=B-1A-1, then (AB)(B-1 A-1) = I

    We can the write this as A(BB-1)A-1 [ Since matrices are associative i.e. A(BC)=(AB)C ]

    This means that (AB)(B-1 A-1) = AIA-1 = AA-1 = I, which follows from (AB)-1=B-1A-1
    Here 'Well if (AB)-1=B-1A-1, then (AB)(B-1 A-1) = I' have you just multiplied both sides by AB to get the identity on the right? If so then i think i get it now :P
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    (Original post by c471)
    More simply than above:
    Without loss of generality, a matrix is non-singular if it has a determinant that is non zero.

    the inverse of a singular matrix is not defined, so calculating the determinants of A & B and showing them to be zero is sufficient.

    Ahh okay so say you had a matrix m where a=x b=2x-7 c=-1 and d=x+4 would you be proving its singular by finding the deteriment, equating them to 0 then solving the quadratic to see if it had real roots?
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    (Original post by Season One)
    Ahh okay so say you had a matrix m where a=x b=2x-7 c=-1 and d=x+4 would you be proving its singular by finding the deteriment, equating them to 0 then solving the quadratic to see if it had real roots?
    well no, if the matrix is singular, the determinant is equal to 0, but that method is not valid. try it on any 2x2 matrix and see.

    if you want to see if a matrix of that form is invertable, you have to invert them both and then evaluate if the property explained above is valid.

    If you are dealing with numbers as elements of your matrix, and are asked to determine whether it is invertable, then calculating the determinate is as far as you need (and possibly can) go.
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    (Original post by c471)
    well no, if the matrix is singular, the determinant is equal to 0, but that method is not valid. try it on any 2x2 matrix and see.

    if you want to see if a matrix of that form is invertable, you have to invert them both and then evaluate if the property explained above is valid.

    If you are dealing with numbers as elements of your matrix, and are asked to determine whether it is invertable, then calculating the determinate is as far as you need (and possibly can) go.

    okay cheers, thanks for your help
 
 
 
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