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

By matrix multiplication from that M=[[cos theta,-sin theta],[sin theta,cos theta]] is orthogonal.

Simplify, costheta[[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]] , then what is A^(-1) ?

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

Find the matrix A , when A^(-1) = [[cos theta, sin theta],[-sin theta,cos theta]]

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

If A=[[cos theta, sin theta],[sin theta, cos theta]],B=[[cos phi, sin phi],[sin phi, cos phi]] show that AB=BA.

If A=[[cos theta, sin theta],[sin theta, cos theta]],B=[[cos phi, sin phi],[sin phi, cos phi]] show that AB=BA.

Prove the orthogonal matrices of order two are of the form [(cos theta,-sin theta),(sin theta,cos theta)] or [(cos theta,sin theta),(sin theta,-cos theta)]