Transformasi Matriks

TRANSFORMASI MATRIKS

Penggunaan matriks sangatlah membantu dalam menyelesaikan beberapa masalah geometri transformasi.

Cara mengubah penulisan transformasi dengan cara biasa ke dalam bentuk matriks :

Misalkan T$T$ adalah transformasi yang memetakan titik (x, y)$\left(x, y\right)$ ke titik (2x−y,x+3y)$\left(2x-y,x+3y\right)$

Maka dengan cara biasa penulisannya : (x,y) =⇒=======Transformasi T (2x−y, x+3y) $\left(x,y\right) \stackrel{Transformasi T}{\Rightarrow } \left(2x-y, x+3y\right)$

Dengan menggunakan matriks : (x'y')=(2x−yx+3y)$\left(\begin{array}{c}x\text{'}\\ y\text{'}\end{array}\right)=\left(\begin{array}{c}2x-y\\ x+3y\end{array}\right)$

(x'y')=(21−13)(xy)$\left(\begin{array}{c}x\text{'}\\ y\text{'}\end{array}\right)=\left(\begin{array}{cc}2& -1\\ 1& 3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)$

Contoh : Rotasi sebesar α$\alpha$ dengan pusat titik asal adalah Rα$R_{\alpha }$

(x, y)−→−−−−       Rα            (xcos α−ysinα, xsinα+ycos α)$\left(x, y\right)\stackrel{{ R}_{\alpha }}{\to }\left(x\cos \alpha -y\sin \alpha , x\sin \alpha +y\cos \alpha \right)$

Penulisan dengan matriksnya adalah

(x'y')=(cosαsinα−sinα cosα)(xy)$\left(\begin{array}{c}x\text{'}\\ y\text{'}\end{array}\right)=\left(\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)$

matriks transformasinya (cosαsinα−sinα cosα)$\left(\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right)$

Contoh : Mencari bayangan titik A(2, 3)$A\left(2, 3\right)$ , B(−1, 6)$B\left(-1, 6\right)$ , dan C(3, 4)$C\left(3, 4\right)$ oleh [(1, 9), −3]$\left[\left(1, 9\right), -3\right]$

Dengan cara biasa (tidak dengan matriks) prosesnya agak lama.

(x, y)−→−−−−−−−        [(a, b), k]      (k(x−a)+a, k(y−b)+b)$\left(x, y\right)\stackrel{ \left[\left(a, b\right), k\right] }{\to }\left(k\left(x-a\right)+a, k\left(y-b\right)+b\right)$

Dengan matriks (x'−ay'−b)=(k00k)(x−ay−b)$\left(\begin{array}{c}{x}^{\text{'}}-a\\ y^{\text{'}}-b\end{array}\right)=\left(\begin{array}{cc}k& 0\\ 0& k\end{array}\right)\left(\begin{array}{c}x-a\\ y-b\end{array}\right)$

Untuk soal di atas

(x'−1y'−9)=(−300−3)⎛⎝⎜2−13−9A−1−16−9B3−14−9C⎞⎠⎟$\left(\begin{array}{c}{x}^{\text{'}}-1\\ y^{\text{'}}-9\end{array}\right)=\left(\begin{array}{cc}-3& 0\\ 0& -3\end{array}\right)\left(\begin{array}{ccc}\underset{A}{\underbrace{\begin{array}{c}2-1\\ 3-9\end{array}}}& \underset{B}{\underbrace{\begin{array}{c}-1-1\\ 6-9\end{array}}}& \underset{C}{\underbrace{\begin{array}{c}3-1\\ 4-9\end{array}}}\end{array}\right)$

=(−300−3)(1−6−2−32−5)$=\left(\begin{array}{cc}-3& 0\\ 0& -3\end{array}\right)\left(\begin{array}{ccc}\begin{array}{c}1\\ -6\end{array}& \begin{array}{c}-2\\ -3\end{array}& \begin{array}{c}2\\ -5\end{array}\end{array}\right)$

=(−31869−615)$=\left(\begin{array}{ccc}-3& 6& -6\\ 18& 9& 15\end{array}\right)$

(x'y')=(−31869−615)+(191919)$\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\end{array}\right)=\left(\begin{array}{ccc}-3& 6& -6\\ 18& 9& 15\end{array}\right)+\left(\begin{array}{ccc}1& 1& 1\\ 9& 9& 9\end{array}\right)$

=(−227718−524)$=\left(\begin{array}{ccc}-2& 7& -5\\ 27& 18& 24\end{array}\right)$

Jadi bayangannya adalah A'(−2, 27)$A\text{'}\left(-2, 27\right)$ , B'(7, 18)$B\text{'}\left(7, 18\right)$ dan C'(−5, 24)