Question

Suppose that A and B are nonsingular matrices. Then AB is also nonsingular. Furthermore, a theorem from linear algebra then states that (AB)-1= B-1A-1 . (a) Verify this theorem for 2× 2 matrices. (b) Prove the theorem for any n × n matrix

Answer 1

Answer:

A) Verified

B) Proved

Step-by-step explanation:

a) Let's verify it for 2 x 2 matrix,

$A=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$ and $B=\left[\begin{array}{ccc}e&f\\g&h\end{array}\right]$

$AB=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]\left[\begin{array}{ccc}e&f\\g&h\end{array}\right]=\left[\begin{array}{ccc}a.e+b.g&a.f+b.h\\c.e+d.g&c.f+d.h\end{array}\right]$

$(AB)^{-1}=\frac{1}{(a.e+b.g)(c.f+d.h)-(a.f+b.h)(c.e+d.g)}\left[\begin{array}{ccc}c.f+d.h&-(a.f+b.h)\\-(c.e+d.g)&a.e+b.g\end{array}\right]$

$A^{-1}=\frac{1}{a.d-b.c} \left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right]$

$B^{-1}=\frac{1}{e.h-f.g} \left[\begin{array}{ccc}h&-f\\-g&e\end{array}\right]$

$B^{-1}A^{-1}=\frac{1}{(a.e+b.g)(c.f+d.h)-(a.f+b.h)(c.e+d.g)}\left[\begin{array}{ccc}c.f+d.h&-(a.f+b.h)\\-(c.e+d.g)&a.e+b.g\end{array}\right]$

So it is proved that the results are same.

b) Now, let's prove it for any n x n matrix.

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