If `A=[[costheta, sintheta], [-sintheta, costheta]]`, then

Show that AB=BA if A=[[costheta,sintheta],[-sintheta,costheta]],B=[[costheta,-sintheta],[sintheta,costheta]]

If A=[[costheta,sintheta],[-sintheta,costheta]] and beta=A^4+A then determinant of beta=

Show that AB=BA where, A=[[costheta,sintheta],[-sintheta,costheta]],B=[[costheta,-sintheta],[sintheta,costheta]]

costheta[[costheta, sintheta], [-sintheta, costheta]]+sintheta[[sintheta, -costheta], [costheta, sintheta]]=

Let A=[{:(costheta,sintheta),(-sintheta,costheta):}] , then |2A| is equal to

Let A=[{:(costheta,sintheta),(-sintheta,costheta):}] , then |2A| is equal to

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

Let A={:[(costheta,-sintheta),(-sintheta,-costheta)]:} then inverse of A is

If A={:[(costheta,-sintheta),(-sintheta,costheta)]:}, and AB=BA=I" then "B=

If A=((costheta, sintheta), (-sintheta, costheta)) prove that AA^(T)=I