Soal-Soal Matematika/Matriks transformasi

Transformasi

Transformasi terdiri dari 2 jenis yaitu:

• Transformasi isometri

Transformasi isometri adalah transformasi yang dapat mengubah bentuknya. Contohnya translasi (penggeseran), refleksi (perpindahan) dan rotasi (perputaran).

• Transformasi nonisometri

Transformasi nonisometri adalah transformasi yang tidak dapat mengubah bentuknya. Contohnya dilatasi (perubahan), stretching (regangan) dan shearing (gusuran).

Translasi

Rumus translasi adalah: $\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\left(\begin{array}{c}a\\ b\end{array}\right)$ + $\left(\begin{array}{c}x\\ y\end{array}\right)$

Refleksi

Rumus refleksi adalah:

tanpa titik pusat

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\left(\begin{array}{cc}cos2\alpha & sin2\alpha \\ sin2\alpha & -cos2\alpha \end{array}\right)$ $\left(\begin{array}{c}x\\ y\end{array}\right)$

dengan titik pusat (a,b)

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\left(\begin{array}{cc}cos2\alpha & sin2\alpha \\ sin2\alpha & -cos2\alpha \end{array}\right)$ $\left(\begin{array}{c}x-a\\ y-b\end{array}\right)$ + $\left(\begin{array}{c}a\\ b\end{array}\right)$

Rotasi

Rumus rotasi adalah:

tanpa titik pusat

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\left(\begin{array}{cc}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{array}\right)$ $\left(\begin{array}{c}x\\ y\end{array}\right)$

dengan titik pusat (a,b)

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\left(\begin{array}{cc}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{array}\right)$ $\left(\begin{array}{c}x-a\\ y-b\end{array}\right)$ + $\left(\begin{array}{c}a\\ b\end{array}\right)$

Dilatasi

Rumus dilatasi adalah:

tanpa titik pusat

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\left(\begin{array}{cc}k& 0\\ 0& k\end{array}\right)$ $\left(\begin{array}{c}x\\ y\end{array}\right)$

dengan titik pusat (a,b)

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\left(\begin{array}{cc}k& 0\\ 0& k\end{array}\right)$ $\left(\begin{array}{c}x-a\\ y-b\end{array}\right)$ + $\left(\begin{array}{c}a\\ b\end{array}\right)$

Stretching

Rumus stretching adalah:

sumbu x

tanpa titik pusat

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\left(\begin{array}{cc}k& 0\\ 0& 1\end{array}\right)$ $\left(\begin{array}{c}x\\ y\end{array}\right)$

dengan titik pusat (a,b)

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\left(\begin{array}{cc}k& 0\\ 0& 1\end{array}\right)$ $\left(\begin{array}{c}x-a\\ y-b\end{array}\right)$ + $\left(\begin{array}{c}a\\ b\end{array}\right)$

sumbu y

tanpa titik pusat

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\left(\begin{array}{cc}1& 0\\ 0& k\end{array}\right)$ $\left(\begin{array}{c}x\\ y\end{array}\right)$

dengan titik pusat (a,b)

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\left(\begin{array}{cc}1& 0\\ 0& k\end{array}\right)$ $\left(\begin{array}{c}x-a\\ y-b\end{array}\right)$ + $\left(\begin{array}{c}a\\ b\end{array}\right)$

Shearing

Rumus shearing adalah:

sumbu x

tanpa titik pusat

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\left(\begin{array}{cc}1& k\\ 0& 1\end{array}\right)$ $\left(\begin{array}{c}x\\ y\end{array}\right)$

dengan titik pusat (a,b)

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\left(\begin{array}{cc}1& k\\ 0& 1\end{array}\right)$ $\left(\begin{array}{c}x-a\\ y-b\end{array}\right)$ + $\left(\begin{array}{c}a\\ b\end{array}\right)$

sumbu y

tanpa titik pusat

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\left(\begin{array}{cc}1& 0\\ k& 1\end{array}\right)$ $\left(\begin{array}{c}x\\ y\end{array}\right)$

dengan titik pusat (a,b)

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\left(\begin{array}{cc}1& 0\\ k& 1\end{array}\right)$ $\left(\begin{array}{c}x-a\\ y-b\end{array}\right)$ + $\left(\begin{array}{c}a\\ b\end{array}\right)$

Rumus sederhana

Keterangan	Posisi	Hasil
Translasi
penggeseran (a,b)	$(x,y)$	$(x+a,y+b)$
Refleksi
sumbu x [0°]	$(x,y)$	$(x,-y)$
sumbu y [90°]	$(x,y)$	$(-x,y)$
y=x [45°]	$(x,y)$	$(y,x)$
y=-x [135°]	$(x,y)$	$(-y,-x)$
pusat (0,0) [0° dan 90°]	$(x,y)$	$(-x,-y)$
pusat (a,b) [0° dan 90°]	$(x,y)$	$(2a-x,2b-y)$
pusat (a,0) [0° dan 90°]	$(x,y)$	$(2a-x,y)$
pusat (0,b) [0° dan 90°]	$(x,y)$	$(x,2b-y)$
Rotasi
berpusat (0,0)
90°	$(x,y)$	$(-y,x)$
-90°	$(x,y)$	$(y,-x)$
180°	$(x,y)$	$(-x,-y)$
berpusat (a,b)
90°	$(x,y)$	$(-y+a+b,x-a+b)$
-90°	$(x,y)$	$(y-a+b,-x+a+b)$
180°	$(x,y)$	$(-x+2a,-y+2b)$
berpusat (0,0)
Dilatasi
skala k	$(x,y)$	$(k\cdot x,k\cdot y)$
Stretching
sumbu x dan skala k	$(x,y)$	$(k\cdot x,y)$
sumbu y dan skala k	$(x,y)$	$(x,k\cdot y)$
Shearing
sumbu x dan skala k	$(x,y)$	$(k\cdot x+y,y)$
sumbu y dan skala k	$(x,y)$	$(x,x+k\cdot y)$
berpusat (a,b)
Dilatasi
skala k	$(x,y)$	$(k\cdot x+(1-k)a,k\cdot y+(1-k)b)$
Stretching
sumbu x dan skala k	$(x,y)$	$(k\cdot x+(1-k)a,y)$
sumbu y dan skala k	$(x,y)$	$(x,k\cdot y+(1-k)b)$
Shearing
sumbu x dan skala k	$(x,y)$	$(x+k\cdot (y-b)),y)$
sumbu y dan skala k	$(x,y)$	$(x,y+k\cdot (x-a))$

Luas

misalkan A (x1,y1), B (x2,y2) dan C (x3,y3)

cara 1

titik awal diubah menjadi titik bayangan.

$\begin{array}{c}L=\frac{1}{2}\cdot \begin{array}{cccc}{x_1}^{\prime }& {x_2}^{\prime }& {x_3}^{\prime }& {x_1}^{\prime }\\ {y_1}^{\prime }& {y_2}^{\prime }& {y_3}^{\prime }& {y_1}^{\prime }\\ -& -+& -+& +\end{array}\\ =\frac{1}{2}\cdot ({x_1}^{\prime }{y_2}^{\prime }+{x_2}^{\prime }{y_3}^{\prime }+{x_3}^{\prime }{y_1}^{\prime }-({x_1}^{\prime }{y_3}^{\prime }+{x_3}^{\prime }{y_2}^{\prime }+{x_2}^{\prime }{y_1}^{\prime }))\\ \end{array}$

cara 2

Luas = | det M | x luas awal

contoh

1. Tentukan persamaan bayangan dari persamaan garis 2x−3y=5 jika ditransformasikan oleh matriks $\left[\begin{array}{cc}2& -1\\ 5& -3\end{array}\right]$?

Jawaban

$$\begin{array}{cc}\left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{cc}2& -1\\ 5& -3\end{array}\right]\cdot \left[\begin{array}{c}x\\ y\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]& ={\left[\begin{array}{cc}2& -1\\ 5& -3\end{array}\right]}^{-1}\cdot \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]\\ & =\frac{1}{2\cdot (-3)-((-1)\cdot 5)}\cdot \left[\begin{array}{cc}-3& 1\\ -5& 2\end{array}\right]\cdot \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]\\ & =\frac{1}{-6-(-5)}\cdot \left[\begin{array}{cc}-3& 1\\ -5& 2\end{array}\right]\cdot \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]\\ & =-1\cdot \left[\begin{array}{cc}-3& 1\\ -5& 2\end{array}\right]\cdot \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]\\ & =\left[\begin{array}{cc}3& -1\\ 5& -2\end{array}\right]\cdot \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]\\ & =\left[\begin{array}{c}3x^{\prime }-y^{\prime }\\ 5x^{\prime }-2y^{\prime }\end{array}\right]\\ x& =3x^{\prime }-y^{\prime }\\ y& =5x^{\prime }-2y^{\prime }\\ 2x-3y& =5\\ 2(3x^{\prime }-y^{\prime })-3(5x^{\prime }-2y^{\prime })& =5\\ 6x^{\prime }-2y^{\prime }-15x^{\prime }+6y^{\prime }& =5\\ -9x^{\prime }+4y^{\prime }& =5\\ \end{array}$$

1. Diketahui segitiga ABC dengan titik koordinat sudut-sudutnya yaitu A(1,3), B(-2,4), dan C(-1,-1). Jika segitiga ABC ditransformasikan oleh matriks yang bersesuaian dengan matriks $\left[\begin{array}{cc}2& -3\\ 1& 4\end{array}\right]$, maka tentukan luas bayangan segitiga ABC tersebut?

cara 1

Jawaban

$$\begin{array}{c};\text{menentukan titik bayangan A}\\ \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{cc}2& -3\\ 1& 4\end{array}\right]\cdot \left[\begin{array}{c}x\\ y\end{array}\right]\\ \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{cc}2& -3\\ 1& 4\end{array}\right]\cdot \left[\begin{array}{c}1\\ 3\end{array}\right]\\ \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{c}-7\\ 13\end{array}\right]\\ ;\text{menentukan titik bayangan B}\\ \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{cc}2& -3\\ 1& 4\end{array}\right]\cdot \left[\begin{array}{c}x\\ y\end{array}\right]\\ \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{cc}2& -3\\ 1& 4\end{array}\right]\cdot \left[\begin{array}{c}-2\\ 4\end{array}\right]\\ \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{c}-16\\ 14\end{array}\right]\\ ;\text{menentukan titik bayangan C}\\ \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{cc}2& -3\\ 1& 4\end{array}\right]\cdot \left[\begin{array}{c}x\\ y\end{array}\right]\\ \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{cc}2& -3\\ 1& 4\end{array}\right]\cdot \left[\begin{array}{c}-1\\ -1\end{array}\right]\\ \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{c}1\\ -5\end{array}\right]\\ L& =\frac{1}{2}\cdot \begin{array}{cccc}-7& -16& 1& -7\\ 13& 14& -5& 13\\ -& -+& -+& +\end{array}\\ & =\frac{1}{2}\cdot ((-7)\cdot 14+(-16)\cdot (-5)+1\cdot 13-((-7)\cdot (-5)+1\cdot 14+(-16)\cdot 13))\\ & =\frac{1}{2}\cdot (-98+80-13-(35+14-208))\\ & =\frac{1}{2}\cdot (-5-(-159))\\ & =\frac{1}{2}\cdot 154\\ & =77\\ \end{array}$$

cara 2

Jawaban

$$\begin{array}{cc}{L}_a& =\frac{1}{2}\cdot \begin{array}{cccc}1& -2& -1& 1\\ 3& 4& -1& 3\\ -& -+& -+& +\end{array}\\ & =\frac{1}{2}\cdot (1\cdot 4+(-2)\cdot (-1)+(-1)\cdot 3-(1\cdot (-1)+(-1)\cdot 4+(-2)\cdot 3))\\ & =\frac{1}{2}\cdot (4+2-3-(-1-4-6))\\ & =\frac{1}{2}\cdot (3-(-11))\\ & =\frac{1}{2}\cdot 14\\ & =7\\ L_b& =|\text{det M}|\cdot L_a\\ & =\left[\begin{array}{cc}2& -3\\ 1& 4\end{array}\right]\cdot 7\\ & =(2\cdot 4-(-3)\cdot 1)\cdot 7\\ & =(8+3)\cdot 7\\ & =11\cdot 7\\ & =77\\ \end{array}$$