Luas bayangan persegi panjang PQRS dengan P(-1,2), Q(3,2),R(3,-1), S(-1,-1) karena dilatasi [0,3] dilanjutkan rotasi pusat 0 dang bersudut Π/2 ( phi /2) adalah.

jawab

Luas PQRS = PQ x PS
PQ = 3 - (-1) = 4
PS = 2 -(-1) = 3
L1 = 4(3) = 12

T2 o T1 =

[tex] \left[\begin{array}{ccc}cos\,90&-sin\,90\\sin\,90&cos\,90\end{array}\right] \left[\begin{array}{ccc}3&0\\0&3\end{array}\right]\,= \left[\begin{array}{ccc}0&-1\\1&0\end{array}\right] \,\left[\begin{array}{ccc}3&0\\0&3\end{array}\right]\,=\left[\begin{array}{ccc}0&-3\\3&0\end{array}\right][/tex]

det T2 o T1 = 0 -3(-3) = 9

Luas hasil transformasi = L2
L2 = det T2oT1  ( L1)
L2 = 9(12)
L2 = 108