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

Rumus arah translasi T(a,b) atau $T\left(\begin{array}{c}a\\ b\end{array}\right)$

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)

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

terhadap persamaan y=mx+c

cari $\alpha$ bergradien $tan\alpha =m$ (untuk m bernilai hasil dari sudut istimewa)

kemudian rumus berikut:

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\left(\begin{array}{cc}cos2\alpha & sin2\alpha \\ sin2\alpha & -cos2\alpha \end{array}\right)$ $\left(\begin{array}{c}x\\ y-c\end{array}\right)$ + $\left(\begin{array}{c}0\\ c\end{array}\right)$

jika m merupakan bilangan real

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\frac{1}{1+m^2}\left(\begin{array}{cc}1-m^2& 2m\\ 2m& m^2-1\end{array}\right)$ $\left(\begin{array}{c}x\\ y-c\end{array}\right)$ + $\left(\begin{array}{c}0\\ c\end{array}\right)$

Pembuktian

$$\begin{array}{c}tan\alpha =m\text{ maka }sin\alpha =\frac{m}{\sqrt{1+m^2}},cos\alpha =\frac{1}{\sqrt{1+m^2}}\\ cos2\alpha =\frac{1-m^2}{1+m^2},sin2\alpha =\frac{2m}{1+m^2}\\ \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{cc}cos2\alpha & sin2\alpha \\ sin2\alpha & -cos2\alpha \end{array}\right]\left[\begin{array}{c}x\\ y-c\end{array}\right]+\left[\begin{array}{c}0\\ c\end{array}\right]& =\left[\begin{array}{cc}\frac{1-m^2}{1+m^2}& \frac{2m}{1+m^2}\\ \frac{2m}{1+m^2}& -(\frac{1-m^2}{1+m^2})\end{array}\right]\left[\begin{array}{c}x\\ y-c\end{array}\right]+\left[\begin{array}{c}0\\ c\end{array}\right]& =\frac{1}{1+m^2}\left[\begin{array}{cc}1-m^2& 2m\\ 2m& m^2-1\end{array}\right]\left[\begin{array}{c}x\\ y-c\end{array}\right]+\left[\begin{array}{c}0\\ c\end{array}\right]\end{array}$$

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)

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

terhadap persamaan y=mx+c

cari $\alpha$ bergradien $tan\alpha =m$

kemudian rumus berikut:

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\left(\begin{array}{cc}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{array}\right)$ $\left(\begin{array}{c}x\\ y-c\end{array}\right)$ + $\left(\begin{array}{c}0\\ c\end{array}\right)$

jika m merupakan bilangan real

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ = $\frac{1}{\sqrt{1+m^2}}\left(\begin{array}{cc}1& -m\\ m& 1\end{array}\right)$ $\left(\begin{array}{c}x\\ y-c\end{array}\right)$ + $\left(\begin{array}{c}0\\ c\end{array}\right)$

Pembuktian

$$\begin{array}{c}tan\alpha =m\text{ maka }sin\alpha =\frac{m}{\sqrt{1+m^2}},cos\alpha =\frac{1}{\sqrt{1+m^2}}\\ \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{cc}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{array}\right]\left[\begin{array}{c}x\\ y-c\end{array}\right]+\left[\begin{array}{c}0\\ c\end{array}\right]& =\left[\begin{array}{cc}\frac{1}{\sqrt{1+m^2}}& -(\frac{m}{\sqrt{1+m^2}})\\ \frac{m}{\sqrt{1+m^2}}& \frac{1}{\sqrt{1+m^2}})\end{array}\right]\left[\begin{array}{c}x\\ y-c\end{array}\right]+\left[\begin{array}{c}0\\ c\end{array}\right]& =\frac{1}{\sqrt{1+m^2}}\left[\begin{array}{cc}1& -m\\ m& 1\end{array}\right]\left[\begin{array}{c}x\\ y-c\end{array}\right]+\left[\begin{array}{c}0\\ c\end{array}\right]\end{array}$$

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) atau [O,$\alpha$]
0°	$(x,y)$	$(x,y)$
90°	$(x,y)$	$(-y,x)$
180°	$(x,y)$	$(-x,-y)$
270° atau -90°	$(x,y)$	$(y,-x)$
berpusat (0,0) atau [O,k]
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	$(x,y)$	$(x,x+k\cdot y)$
Rotasi
berpusat (a,b) atau [(a,b),$\alpha$]
0°	$(x,y)$	$(x,y)$
90°	$(x,y)$	$(-y+a+b,x-a+b)$
180°	$(x,y)$	$(-x+2a,-y+2b)$
270° atau -90°	$(x,y)$	$(y-a+b,-x+a+b)$
berpusat (a,b) atau [(a,b),k]
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	$(x,y)$	$(x,y+k\cdot (x-a))$

• keterangan:

1. berlawanan arah dengan jarum jam adalah sudut positif sedangkan searah jarum jam adalah sudut negatif

Luas

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

cara 1

titik awal diubah menjadi titik bayangan.

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

cara 2

Luas = | det M | x luas awal

contoh

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

cara 1

Jawaban

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

cara 2

Jawaban

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

1. Tentukan persamaan bayangan dari persamaan parabola y=x²-2x-3 jika ditransformasikan oleh translasi berpusat (0,0) yaitu $\left[\begin{array}{c}2\\ 5\end{array}\right]$ lalu oleh refleksi berpusat (0,0) yaitu y=x?

cara 1

Jawaban

$$\begin{array}{cc}\left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{c}2\\ 5\end{array}\right]+\left[\begin{array}{c}x\\ y\end{array}\right]\\ & =\left[\begin{array}{c}2+x\\ 5+y\end{array}\right]\\ \text{refleksi berpusat (0,0) dan y=x berarti sudut 45 derajat. }\\ \left[\begin{array}{c}{x}^{″}\\ y^{″}\end{array}\right]& =\left[\begin{array}{cc}cos90^{\circ }& sin90^{\circ }\\ sin90^{\circ }& -cos90^{\circ }\end{array}\right]\cdot \left[\begin{array}{c}2+x\\ 5+y\end{array}\right]\\ & =\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\cdot \left[\begin{array}{c}2+x\\ 5+y\end{array}\right]\\ \left[\begin{array}{c}{x}^{″}\\ y^{″}\end{array}\right]& =\left[\begin{array}{c}5+y\\ 2+x\end{array}\right]\\ \text{jadi x'' = 5+y dan y'' = 2+x }\\ \text{substitusi y = x''-5 dan x = y''-2 }\\ \text{masukkan x dan y ke dalam persamaan }\\ y& =x^2-2x-3\\ x^{″}-5& =(y^{″}-2{)}^2-2(y^{″}-2)-3\\ x^{″}-5& =y^{\prime }{\text{'}}^2-4y^{″}+4-2y^{″}+4-3\\ x^{″}& =y^{\prime }{\text{'}}^2-6y^{″}+10\\ x& =y^2-6y+10\\ \text{jadi persamaan bayangan adalah }x=y^2-6y+10\end{array}$$

1. Tentukan persamaan bayangan dari persamaan parabola y=x²-5x+6 jika ditransformasikan oleh refleksi persamaan y-x=3?

cara 1

Jawaban

$$\begin{array}{cc}y-x& =3\text{ diubah menjadi }y=x+3\\ tan\alpha & =1\\ \alpha & =45^{\circ }\\ \text{refleksi berpusat (0,0) dan y=x berarti sudut 45 derajat. }\\ \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{cc}cos90^{\circ }& sin90^{\circ }\\ sin90^{\circ }& -cos90^{\circ }\end{array}\right]\cdot \left[\begin{array}{c}x\\ y-3\end{array}\right]+\left[\begin{array}{c}0\\ 3\end{array}\right]\\ & =\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\cdot \left[\begin{array}{c}x\\ y-3\end{array}\right]+\left[\begin{array}{c}0\\ 3\end{array}\right]\\ \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{c}y-3\\ x\end{array}\right]+\left[\begin{array}{c}0\\ 3\end{array}\right]\\ \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{c}y-3\\ x+3\end{array}\right]\\ \text{jadi x' = y-3 dan y' = x+3 }\\ \text{substitusi y = x'+3 dan x = y'-3 }\\ \text{masukkan x dan y ke dalam persamaan }\\ y& =x^2-5x+6\\ x^{\prime }+3& =(y^{\prime }-3{)}^2-5(y^{\prime }-3)+6\\ x^{\prime }+3& =y{\text{'}}^2-6y^{\prime }+9-5y^{\prime }+15+6\\ x^{\prime }& =y{\text{'}}^2-11y^{″}+27\\ x& =y^2-11y+27\\ \text{jadi persamaan bayangan adalah }x=y^2-11y+27\end{array}$$

1. Tentukan persamaan bayangan dari persamaan parabola x=y²+3y-4 jika ditransformasikan oleh rotasi berpusat (0,0) dan sumbu y lalu oleh dilatasi yaitu berpusat (0,0) dan skala 4?

cara 1

Jawaban

$$\begin{array}{c}\text{rotasi sumbu y berarti sudut 90 derajat. }\\ \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{cc}cos90^{\circ }& -sin90^{\circ }\\ sin90^{\circ }& cos90^{\circ }\end{array}\right]\cdot \left[\begin{array}{c}x\\ y\end{array}\right]\\ & =\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\cdot \left[\begin{array}{c}x\\ y\end{array}\right]\\ & =\left[\begin{array}{c}-y\\ x\end{array}\right]\\ \text{dilatasi skala 4. }\\ & =\left[\begin{array}{cc}4& 0\\ 0& 4\end{array}\right]\cdot \left[\begin{array}{c}-y\\ x\end{array}\right]\\ & =\left[\begin{array}{c}-4y\\ 4x\end{array}\right]\\ \text{jadi x'' = -4y dan y'' = 4x }\\ \text{diubah jadi }y=\frac{-x^{″}}{4}\text{dan }x=\frac{y^{″}}{4}\\ \text{masukkan x dan y ke dalam persamaan }\\ x& =y^2+3y-4\\ \frac{y^{″}}{4}& =(\frac{-x^{″}}{4}{)}^2+3(\frac{-x^{″}}{4})-4\\ \frac{y^{″}}{4}& =\frac{x^{\prime }{\text{'}}^2}{16}+3(\frac{-x^{″}}{4})-4\\ 4y^{″}& =16x^{\prime }{\text{'}}^2-12x^{″}-64\\ 4y& =16x^2-12x-64\\ \text{jadi persamaan bayangan adalah }4y& =16x^2-12x-64\end{array}$$

cara 2

Jawaban

$$\begin{array}{cc}\left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{cc}4& 0\\ 0& 4\end{array}\right]\cdot \left[\begin{array}{cc}cos90^{\circ }& -sin90^{\circ }\\ sin90^{\circ }& cos90^{\circ }\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]\\ & =\left[\begin{array}{cc}4& 0\\ 0& 4\end{array}\right]\cdot \left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\cdot \left[\begin{array}{c}x\\ y\end{array}\right]\\ & =\left[\begin{array}{cc}0& -4\\ 4& 0\end{array}\right]\cdot \left[\begin{array}{c}-y\\ x\end{array}\right]\\ & ={\left[\begin{array}{cc}0& -4\\ 4& 0\end{array}\right]}^{-1}\cdot \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]\\ & =\frac{1}{0\cdot 0-((-4)\cdot 4)}\cdot \left[\begin{array}{cc}0& 4\\ -4& 0\end{array}\right]\cdot \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]\\ & =\frac{1}{0-(-16)}\cdot \left[\begin{array}{cc}0& 4\\ -4& 0\end{array}\right]\cdot \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]\\ & =\frac{1}{16}\cdot \left[\begin{array}{cc}0& 4\\ -4& 0\end{array}\right]\cdot \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]\\ & =\left[\begin{array}{cc}0& \frac{1}{4}\\ -\frac{1}{4}& 0\end{array}\right]\cdot \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]\\ & =\left[\begin{array}{c}\frac{y^{\prime }}{4}\\ -\frac{x^{\prime }}{4}\end{array}\right]\\ x& =\frac{y^{\prime }}{4}\\ y& =-\frac{x^{\prime }}{4}\\ x& =y^2+3y-4\\ \frac{y^{\prime }}{4}& =(\frac{-x^{″}}{4}{)}^2+3(\frac{-x^{″}}{4})-4\\ \frac{y^{″}}{4}& =\frac{x^{\prime }{\text{'}}^2}{16}+3(\frac{-x^{″}}{4})-4\\ 4y^{″}& =16x^{\prime }{\text{'}}^2-12x^{″}-64\\ 4y& =16x^2-12x-64\\ \text{jadi persamaan bayangan adalah }4y& =16x^2-12x-64\end{array}$$

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

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

cara 2

Jawaban

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