Roboguru

Misalkan  dan  adalah hasil bayangan titik  dan  oleh transformasi matriks  berordo . Jika  adalah bayangan titik  oleh transformasi tersebut, titik  adalah ...

Pertanyaan

Misalkan straight A ’ left parenthesis negative 1 comma space minus 2 right parenthesis dan straight B ’ left parenthesis 3 comma space 7 right parenthesis adalah hasil bayangan titik straight A left parenthesis negative 1 comma space 0 right parenthesis dan straight B left parenthesis 2 comma space 1 right parenthesis oleh transformasi matriks straight X berordo 2 cross times 2. Jika straight C ’ left parenthesis 0 comma space 1 right parenthesis adalah bayangan titik straight C oleh transformasi tersebut, titik straight C adalah ... 

Pembahasan Soal:

Misalkan matriks transformasi tersebut adalah X equals open square brackets table row straight a straight b row straight c straight d end table close square brackets, bayangan titik straight A left parenthesis negative 1 comma space 0 right parenthesis adalah straight A ’ left parenthesis negative 1 comma space minus 2 right parenthesis, maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell straight A apostrophe end cell equals cell straight X times straight A end cell row cell open square brackets table row cell negative 1 end cell row cell negative 2 end cell end table close square brackets end cell equals cell open square brackets table row straight a straight b row straight c straight d end table close square brackets times open square brackets table row cell negative 1 end cell row 0 end table close square brackets end cell row cell open square brackets table row cell negative 1 end cell row cell negative 2 end cell end table close square brackets end cell equals cell open square brackets table row cell negative straight a end cell row cell negative straight c end cell end table close square brackets end cell end table 

Maka straight a equals 1 dan straight c equals 2 

Bayangan titik straight B left parenthesis 2 comma space 1 right parenthesis adalah straight B ’ left parenthesis 3 comma space 7 right parenthesis, maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell straight B apostrophe end cell equals cell straight X times straight B end cell row cell open square brackets table row 3 row 7 end table close square brackets end cell equals cell open square brackets table row straight a straight b row straight c straight d end table close square brackets times open square brackets table row 2 row 1 end table close square brackets end cell row cell open square brackets table row 3 row 7 end table close square brackets end cell equals cell open square brackets table row 1 straight b row 2 straight d end table close square brackets times open square brackets table row 2 row 1 end table close square brackets end cell row cell open square brackets table row 3 row 7 end table close square brackets end cell equals cell open square brackets table row cell 2 plus straight b end cell row cell 4 plus straight d end cell end table close square brackets end cell end table 

Maka 

table attributes columnalign right center left columnspacing 0px end attributes row 3 equals cell 2 plus straight b space rightwards arrow space straight b equals 1 end cell row 7 equals cell 4 plus straight d space rightwards arrow space straight d equals 3 end cell end table 

Jadi matriks transformasi tersebut adalah 

X equals open square brackets table row 1 1 row 2 3 end table close square brackets 

Misalkan koordinati titik straight C adalah straight C open parentheses x comma space y close parentheses, maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell straight C apostrophe end cell equals cell straight X times straight C end cell row cell open square brackets table row 0 row 1 end table close square brackets end cell equals cell open square brackets table row 1 1 row 2 3 end table close square brackets times open square brackets table row x row y end table close square brackets end cell row cell open square brackets table row 0 row 1 end table close square brackets end cell equals cell open square brackets table row cell x plus y end cell row cell 2 x plus 3 y end cell end table close square brackets end cell end table 

jadi 

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus y end cell equals cell 0 space space.... open parentheses 1 close parentheses end cell row cell 2 x plus 3 y end cell equals cell 1 space..... open parentheses 2 close parentheses end cell end table 

Persamaan open parentheses 1 close parentheses kalikan 2 diperoleh 

2 x plus 2 y equals 0 space space... open parentheses 3 close parentheses 

Eliminasi x dari persamaan open parentheses 2 close parentheses space dan space open parentheses 3 close parentheses diperoleh 

table row cell 2 x end cell plus cell 3 y end cell equals 1 space row cell 2 x end cell plus cell 2 y end cell equals 0 minus row space space y equals 1 space end table 

Substitusi nilai y equals 1 ke persamaan open parentheses 3 close parentheses, diperoleh 

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x plus 2 y end cell equals 0 row cell 2 x plus 2 open parentheses 1 close parentheses end cell equals 0 row cell 2 x plus 2 end cell equals 0 row cell 2 x end cell equals cell negative 2 end cell row x equals cell negative 1 end cell end table 

Dengan demikian, koordinat titik straight C adalah straight C open parentheses negative 1 comma space 1 close parentheses

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 04 Mei 2021

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