If `A=[(1,tanx),(-tanx, 1)], " show that " A' A^(-1)=[(cos 2x,-sin 2x),(sin 2x, cos 2x)]`.

Explanation

We have `|A|=[(1,tanx),(-tan x,1)]= (1+tan^(2)x)=sec^(2)x !=0`.
So, A is invertible.
The cofactors of the elements of |A| are
`A_(11)=1, A_(12)=tan x, A_(21)= -tan x, A_(22)=1`.
` :." "(adj A)=[(1,tanx),(-tanx,1)]^(')=[(1,-tanx),(tan x,1)]`
`rArr" "A^(-1)=1/|A|(adj A)=1/sec^(2)x.[(1,-tan x),(tan x,1)]`
`=cos^(2)x.[(1,-tan x),(tan x,1)]=[(cos^(2)x,-tan x cos^(2)x),(tan x cos^(2)x, cos^(2)x)]`
`[(cos^(2)x,-sin x cos x),(sin x cos x, cos^(2)x)]`
`rArr" "A'A^(-1)=[(1,-tan x),(tan x,1)][(cos^(2)x, -sin x cos x),(sin x cos x ,cos^(2)x)]`
`=[(cos^(2)x-sin^(2)x,-2sin x cos x),(2sin 2x,cos 2x)]`
`=[(cos 2x ,-sin 2x),(sin 2x,cos 2x)]`
Hence, `A'A^(-1)=[(cos 2x ,-sin x),(sin 2x, cos 2x)]`.