" 29."[[cos theta,-sin theta],[sin theta,cos theta]]" का व्युत्कम है : "

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

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 IAI=

Simplify, costheta[[cos theta, sin theta],[-sin theta, cos theta]] + sin theta [[sin theta, -cos theta],[cos theta, sin theta]]

Verify that [(cos theta, sin theta),(-sin theta, cos theta)] and [(cos theta, - sin theta),(sin theta, cos theta)] are inverse of each other.

Simplify: cos theta[[cos theta , sin theta],[-sin theta , cos theta]]+sin theta[[sin theta, -cos theta],[cos theta ,sin theta]] .

If A = [(cos theta, sin theta),(-sin theta, cos theta)] and B = [(sin theta, - cos theta), (cos theta, sin theta)] , evaluate A cos theta + B sin theta .

Evaluate the following determinants: (b) |(cos theta, -sin theta),(sin theta, cos theta)| = cos theta (cos theta) - sin theta(-sin theta) = cos^(2) theta + sin^(2) theta = 1