Answer:
A) Verified
B) Proved
Step-by-step explanation:
a) Let's verify it for 2 x 2 matrix,
[tex]A=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right][/tex] and [tex]B=\left[\begin{array}{ccc}e&f\\g&h\end{array}\right][/tex]
[tex]AB=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]\left[\begin{array}{ccc}e&f\\g&h\end{array}\right]=\left[\begin{array}{ccc}a.e+b.g&a.f+b.h\\c.e+d.g&c.f+d.h\end{array}\right][/tex]
[tex](AB)^{-1}=\frac{1}{(a.e+b.g)(c.f+d.h)-(a.f+b.h)(c.e+d.g)}\left[\begin{array}{ccc}c.f+d.h&-(a.f+b.h)\\-(c.e+d.g)&a.e+b.g\end{array}\right][/tex]
[tex]A^{-1}=\frac{1}{a.d-b.c} \left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right][/tex]
[tex]B^{-1}=\frac{1}{e.h-f.g} \left[\begin{array}{ccc}h&-f\\-g&e\end{array}\right][/tex]
[tex]B^{-1}A^{-1}=\frac{1}{(a.e+b.g)(c.f+d.h)-(a.f+b.h)(c.e+d.g)}\left[\begin{array}{ccc}c.f+d.h&-(a.f+b.h)\\-(c.e+d.g)&a.e+b.g\end{array}\right][/tex]
So it is proved that the results are same.
b) Now, let's prove it for any n x n matrix.
[tex](AB)(AB)^{-1}=I\\\\A^{-1}(AB)(AB)^{-1}=A^{-1}I\\\\IB(AB)^{=1}=A^{-1}I\\\\B(AB)^{=1}=A^{-1}\\\\B^{-1}B(AB)^{=1}=B^{-1}A^{-1}\\\\I(AB)^{=1}=B^{-1}A^{-1}\\\\(AB)^{=1}=B^{-1}A^{-1}[/tex]
