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If A=[1tanx-tanx1], show that A^T A^(-1)...

If `A=[1tanx-tanx1],` show that `A^T A^(-1)=[cos2x-sin2xsin2xcos2x]`

A

`[(-"cos" 2x,"sin" 2x),(-sin 2x,cos 2x)]`

B

`[(cos 2x,-sin 2x),(sin 2x, cos 2x)]`

C

`[(cos 2x,cos 2x),(cos 2x, sin 2x)]`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B

`|A|=|(1,tan x),(-tan x,1)|=1+tan^(2) x ne 0`
So A is invertiable. Also,
adj `A=[(1,tan x),(-tan x,1)]^(T)=[(1,-tan x),(tan x,1)]`
Now, `A^(-1)=1/(|A|)` adj A
`= 1/((1+tan^(2) x)) [(1,-tan x),(tan x,1)]`
`=[(1/(1+tan^(2) x),(-tan x)/(1+tan^(2) x)),((tan x)/(1+tan^(2) x),1/(1+tan^(2)x))]`
`:. A^(T) A^(-1)=[(1,-tan x),(tan x,1)][(1/(1+tan^(2) x),(-tan x)/(1+tan^(2) x)),((tan x)/(1+tan^(2) x),1/(1+tan^(2)x))]`
`=[((1-tan^(2) x)/(1+tan^(2) x),(-2 tan x)/(1+tan^(2)x)),((2 tan x)/(1+tan^(2)x),(1-tan^(2) x)/(1+tan^(2) x))]`
`=[(cos 2x,-sin 2 x),(sin 2x,cos 2x)]`
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