If `I=[(1,0),(0,1)]=J[(0,1),(-1,0)]` and `B=[(cos theta, sin theta),(-sin theta, cos theta)]` then `B=`

If A=[(cos theta, sin theta, 0),(-sin theta, cos theta, 0),(0,0,1)] then

If I= [[1,0],[0,1]] and J= [[0,1],[-1,0]] and B = [[cos theta, sin theta],[-sin theta, cos theta]] , then B=

If I=[[1,0] , [0,1]] and B=[[0,1] , [-1,0]] and C=[[cos theta, sin theta] , [-sin theta, cos theta]] then C=

If I=det[[1,00,1]],,B=det[[0,1-1,0]] and C=det[[cos theta,sin theta-sin theta,cos theta]] the C=

If the matrix [(cos theta, sin theta, 0),(sin theta, cos theta, 0),(0,0,1)] is singular then theta=

The inverse of [(cos theta, -sin theta, 0),(sin theta, cos theta, 0),(0,0,1)] is

cos^(2)theta-sin theta.cos theta-(1)/(2)=0

cos^(2)theta-sin theta*cos theta-(1)/(2)=0