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# Question below?

2. Prove that if A, B elements of Mn(R) and nonsingular, then (kA)^-1=(k^-1A^-1)

where k element of R*, where R* is the set of nonzero real number.

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- Anonymous1 decade agoFavorite Answer
you need to multiply kA by (k^-1A^-1)

kA(k^-1A^-1)=(kk^{-1}) AA^{-1} = 1 I = I

so (k^-1A^-1) has to be the inverse of kA,

i.e. (kA)^{-1} = (k^-1A^-1) .

- 1 decade ago
multiply kA by (k^-1A^-1), if you get the identity it means that (k^-1A^-1) is the inverse of kA:

kA(k^-1A^-1)=(kk^{-1}) AA^{-1} = 1 I = I

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