Question

Prove that (AB)−1=B−1A−1

Answer 1

Answer with Step-by-step explanation:

Consider,

$(AB)^{-1}(AB)=I$ (Identity rule)

Multiplying by B⁻¹ on the both the sides, we get that

$(AB)^{-1}(AB)B^{-1}=IB^{-1}\\\\(AB)^{-1}A(BB^{-1})=B^{-1}$

And we know that BB⁻¹ = I

So, it becomes,

$(AB)^{-1}A=B^{-1}$

Now, multiplying by A⁻¹ on both the sides, we get that

$(AB)^{-1}AA^{-1}=B^{-1}A^{-1}\\\\(AB)^{-1}=B^{-1}A^{-1}$ (AA⁻¹=I)

Hence, proved.