Answer with Step-by-step explanation:
Consider,
[tex](AB)^{-1}(AB)=I[/tex] (Identity rule)
Multiplying by B⁻¹ on the both the sides, we get that
[tex](AB)^{-1}(AB)B^{-1}=IB^{-1}\\\\(AB)^{-1}A(BB^{-1})=B^{-1}[/tex]
And we know that BB⁻¹ = I
So, it becomes,
[tex](AB)^{-1}A=B^{-1}[/tex]
Now, multiplying by A⁻¹ on both the sides, we get that
[tex](AB)^{-1}AA^{-1}=B^{-1}A^{-1}\\\\(AB)^{-1}=B^{-1}A^{-1}[/tex] (AA⁻¹=I)
Hence, proved.
