If `A=[[cosalpha,sinalpha],[-sinalpha,cosalpha]]` then prove that `A A^'=I` Hence find `A^(-1)`.

If A=[[cosalpha, sinalpha], [-sinalpha, cosalpha]] , then A^(10)=

if A=[[cosalpha,sinalpha],[-sinalpha,cosalpha]] be such that A+A'=I then alpha

If A=[{:(cosalpha,sinalpha),(-sinalpha,cosalpha):}] , show that A'A=I.

If A_(alpha)=[(cosalpha,-sinalpha),(sinalpha,cosalpha)] , then

If A=[(sinalpha,cosalpha),(-cosalpha,sinalpha)] , the prove that A'A=I .

If (i) A=[[cosalpha,sinalpha],[-sinalpha,cosalpha]] , then verify that AprimeA" "=" "I . (ii) A=[[sinalpha,cosalpha],[-cosalpha,sinalpha]] , then verify that AprimeA" "=" "I .

A = [ [ cosalpha , sinalpha ], [ sinalpha , cosalpha ] ] ,then find | A |

If A_(alpha)=[(cosalpha,sinalpha),(-sinalpha,cosalpha)] then (A_(alpha))^2=?

If A=[[cosalpha, sinalpha],[-sinalpha, cosalpha]] , prove by mathematicasl induction that, A^n=[[cosnalpha, sin nalpha],[-sin nalpha,cos nalpha]] for every natural number n