Linear algebra proof?

Question

If A is invertible and 'a' cannot equal zero, then aA is invertible and
(aA)^-1 = 1/a(A^-1)

I'm confused with this question. It doesn't say what small 'a' is supposed to be, is it any number? We are still in the beginning of linear algebra, so I can't use anything too advanced to prove it. I'm not sure where to start.

Answer 1

If A is invertible and 'a' cannot equal zero, then aA is invertible and
(aA)^-1 = 1/a(A^-1)

a is ANY number, as long as a is different from 0 .

here all you need to do is to multiply aA by (1/a)A^{-1}. If you get the identity, then (aA)^-1 = 1/a(A^-1), if not, then the equality is NOT true.
so let us check:

aA ( (1/a) A^{-1} ) = a(1/a) A A^{-1} = 1 I = I
.

Answer 2

that is fake counterexamples are available to locate operating party, take V=R^2 with time-honored starting up up e1=(0,a million), e2=(a million,0) Any line that passes by using way of 0 inspite of the actuality that isn't an axis is a vectorial section. obviously those do now no longer look generated by using way of e1 or e2 In known, evaluate V had starting up up u,v The subspace generated by using way of u+v has length a million inspite of the actuality that isn't generated by using way of u or v edit: i change into typing on the similar time you printed. stop commenting. i do now no longer care about factors and BA as a lot as you

Answer 3

(aA)^1 = (a^-1)(A^-1)
But (a^-1) is the same as 1/a
So (aA)^-1= (a^-1)(A^-1) = 1/a(A^-1)