If Aα=cos αsin α-sin αcos α then Aα Aβ=

Aα=cos αsin α-sin αcos αA(α)A(β)=cos αsin α-sin αcos αcos βsin β-sin βcos β =cos(α+β)sin (α+β)-sin (α+β)cos (α+β) =A((α+β))

If Aα=cos αsin α-sin αcos α then Aα Aβ=

Aα=cos αsin α-sin αcos αA(α)A(β)=cos αsin α-sin αcos αcos βsin β-sin βcos β =cos(α+β)sin (α+β)-sin (α+β)cos (α+β) =A((α+β))

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If $A (α) = [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}]$ then $A (α) A (β) =$

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$A (α + β)$

$A (α - β)$

$A (α) + A (β)$

$A (α) - A (β)$

answer is D.

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Detailed Solution

$\begin{array}{rcl} A (α) & = & [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}] \\ A (α) A (β) & = & [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}] [\begin{matrix} \cos β & \sin β \\ - \sin β & \cos β \end{matrix}] \\ = & [\begin{matrix} \cos (α + β) & \sin (α + β) \\ - \sin (α + β) & \cos (α + β) \end{matrix}] \\ = & A ((α + β)) \end{array}$