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

A) Verified

B) Proved

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)=(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)=(1)/(a.d-b.c) \left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right]`

`B^(-1)=(1)/(e.h-f.g) \left[\begin{array}{ccc}h&-f\\-g&e\end{array}\right]`

`B^(-1)A^(-1)=(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)`