"If "A=[[cos theta,sin theta],[-sin theta,cos theta]]" and "A(adj A)=kI" then "k" is "

If A=[[cos theta,sin theta],[-sin theta,cos theta]] and A(adj A)=[[k,0],[0,k]] , then k =

(i) |(cos theta,-sin theta),(sin theta,cos theta)|

cos theta/(1+sin theta)=(1-sin theta)/cos theta

If A=[(-cos theta, sin theta),(sin theta, cos theta)] then IAI=

Evaluate |[cos theta, -sin theta], [sin theta, cos theta]|

cos theta+sin theta=cos2 theta+sin2 theta

The inverse of A=[[cos theta, sin theta],[-sin theta,cos theta]] is :

If A= [[cos theta,-sin theta],[sin theta,cos theta]] ,then Adj A is (a) [[cos theta,-sin theta],[cos theta,sin theta]] (b) [[1,0],[0,1]] (c) [[cos theta,sin theta],[-sin theta,cos theta]] (d) [[-1,0],[0,-1]]

If A=[{:(cos theta, sin theta),(-sin theta, cos theta):}] then "A A"^(T) is :