MCQ
If $A = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$and $B = \left[ {\begin{array}{*{20}{c}}{\cos \beta }&{ - \sin \beta }\\{\sin \beta }&{\cos \beta }\end{array}} \right]$, then the correct relation is
  • A
    ${A^2} = {B^2}$
  • B
    $A + B = B - A$
  • $AB = BA$
  • D
    None of these

Answer

Correct option: C.
$AB = BA$
c
(c) Clearly, $AB = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\cos \beta }&{ - \sin \beta }\\{\sin \beta }&{\cos \beta }\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}{\cos (\alpha + \beta )}&{ - \sin (\alpha + \beta )}\\{\sin (\alpha + \beta )}&{\cos (\alpha + \beta )}\end{array}} \right] = BA$ (verify).

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