Question

Show that cumulative law holds good for the product of the two matrices.
$A=\left[\begin{array}{cc}a& b\\ -b& a\end{array}\right]andB=\left[\begin{array}{cc}x& y\\ -y& x\end{array}\right]$
If a, b, x, y are all different from zero, find the inverse of A, B and verify that
$(AB{)}^{-1}=B^{-1}{A}^{-1}$.

Answer 1

It is easy to verify that

AB = BA = $\left[\begin{array}{cc}ax-by& ay+bx\\ -(ay+bx)& ax-by\end{array}\right]$

$\therefore Adj.A=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right],Adj.B=\left[\begin{array}{cc}x& -y\\ y& x\end{array}\right]$

Adj. AB = $\left[\begin{array}{cc}ax-by& -(ay+bx)\\ (ay+bx)& ax-by\end{array}\right]$

$A^{-1}=\frac{1}{|A|}Adj.A=\frac{1}{(a^2+b^2)}\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$ .......(1)

$B^{-1}=\frac{1}{|B|}Adj.B=\frac{1}{(x^2+y^2)}\left[\begin{array}{cc}x& -y\\ y& x\end{array}\right]$ .......(2)

$A{B}^{-1}=\frac{1}{|AB|}Adj.AB=\frac{1}{(ax-by{)}^2+(ay+bx{)}^2)}\left[\begin{array}{cc}ax-by& -(ay+bx)\\ ay+bxax-by& \end{array}\right]$

But $(ax-by{)}^2+(ay+bx{)}^2=a^2(x^2+y^2)+b^2(x^2+y^2)=(a^2+b^2)(x^2+y^2)$

$\therefore (AB{)}^{-1}=\frac{1}{(a^2+b^2)(x^2+y^2)}\left[\begin{array}{cc}ax-by& -(ay+bx)\\ ay+bx& x-by\end{array}\right]$ .......(3)

Also $B^{-1}{A}^{-1}=\frac{1}{(a^2+b^2)(x^2+y^2)}\left[\begin{array}{cc}x& -y\\ y& x\end{array}\right]\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$

= $\frac{1}{(a^2+b^2)(x^2+y^2)}\left[\begin{array}{cc}ax-by& -(ay+bx)\\ ay+bx& ax-by\end{array}\right]$ ..............(4)

From (3) and (4) we verify that $A{B}^{-1}=B^{-1}{A}^{-1}$.