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Diketahui P = ⎝ ⎛ ​ 1 − 2 − 9 ​ 5 3 2 x + y ​ x − 2 y 4 − 2 ​ ⎠ ⎞ ​ dan Q = ⎝ ⎛ ​ 1 3 5 ​ − 1 0 x − y ​ 3 x + 2 y − 3 2 ​ ⎠ ⎞ ​ . Jika 2 P − 4 Q = ⎝ ⎛ ​ − 2 − 16 − 38 ​ 14 6 − 6 ​ 2 20 − 12 ​ ⎠ ⎞ ​ ,maka nilai dari 3 x + 2 y adalah ....

Diketahui  dan  . Jika  , maka nilai dari   adalah ....

  1. begin mathsize 14px style negative 5 end style

  2. begin mathsize 14px style negative 1 end style

  3. begin mathsize 14px style 1 end style

  4. begin mathsize 14px style 2 end style

  5. begin mathsize 14px style 5 end style

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Jawaban terverifikasi

Jawaban

jawabannya adalah C.

jawabannya adalah C.

Pembahasan

Perhatikan perhitungan berikut ini! Dari kesamaan matriks di atas, kita peroleh dan Sehingga didapat Jadi, jawabannya adalah C.

Perhatikan perhitungan berikut ini!

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 P minus 4 Q end cell equals cell open parentheses table row cell negative 2 end cell 14 2 row cell negative 16 end cell 6 20 row cell negative 38 end cell cell negative 6 end cell cell negative 12 end cell end table close parentheses end cell row cell 2 open parentheses table row 1 5 cell x minus 2 y end cell row cell negative 2 end cell 3 4 row cell negative 9 end cell cell 2 x plus y end cell cell negative 2 end cell end table close parentheses minus 4 open parentheses table row 1 cell negative 1 end cell cell 3 x plus 2 y end cell row 3 0 cell negative 3 end cell row 5 cell x minus y end cell 2 end table close parentheses end cell equals cell open parentheses table row cell negative 2 end cell 14 2 row cell negative 16 end cell 6 20 row cell negative 38 end cell cell negative 6 end cell cell negative 12 end cell end table close parentheses end cell row cell open parentheses table row 2 10 cell 2 x minus 4 y end cell row cell negative 4 end cell 6 8 row cell negative 18 end cell cell 4 x plus 2 y end cell cell negative 4 end cell end table close parentheses minus open parentheses table row 4 cell negative 4 end cell cell 12 x plus 8 y end cell row 12 0 cell negative 12 end cell row 20 cell 4 x minus 4 y end cell 8 end table close parentheses end cell equals cell open parentheses table row cell negative 2 end cell 14 2 row cell negative 16 end cell 6 20 row cell negative 38 end cell cell negative 6 end cell cell negative 12 end cell end table close parentheses end cell row cell open parentheses table row cell 2 minus 4 end cell cell 10 minus open parentheses negative 4 close parentheses end cell cell 2 x minus 4 y minus open parentheses 12 x plus 8 y close parentheses end cell row cell negative 4 minus 12 end cell cell 6 minus 0 end cell cell 8 minus open parentheses negative 12 close parentheses end cell row cell negative 18 minus 20 end cell cell 4 x plus 2 y minus open parentheses 4 x minus 4 y close parentheses end cell cell negative 4 minus 8 end cell end table close parentheses end cell equals cell open parentheses table row cell negative 2 end cell 14 2 row cell negative 16 end cell 6 20 row cell negative 38 end cell cell negative 6 end cell cell negative 12 end cell end table close parentheses end cell row cell open parentheses table row cell negative 2 end cell 14 cell negative 10 x minus 12 y end cell row cell negative 16 end cell 6 20 row cell negative 38 end cell cell 6 y end cell cell negative 12 end cell end table close parentheses end cell equals cell open parentheses table row cell negative 2 end cell 14 2 row cell negative 16 end cell 6 20 row cell negative 38 end cell cell negative 6 end cell cell negative 12 end cell end table close parentheses end cell end table end style 

Dari kesamaan matriks di atas, kita peroleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 6 y end cell equals cell negative 6 end cell row y equals cell fraction numerator negative 6 over denominator 6 end fraction end cell row y equals cell negative 1 end cell end table end style 

dan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative 10 x minus 12 y end cell equals 2 row cell negative 10 x minus 12 open parentheses negative 1 close parentheses end cell equals 2 row cell negative 10 x plus 12 end cell equals 2 row cell negative 10 x end cell equals cell 2 minus 12 end cell row cell negative 10 x end cell equals cell negative 10 end cell row x equals cell fraction numerator negative 10 over denominator negative 10 end fraction end cell row x equals 1 end table end style 

Sehingga didapat 

table attributes columnalign right center left columnspacing 0px end attributes row cell size 14px 3 size 14px x size 14px plus size 14px 2 size 14px y end cell size 14px equals cell size 14px 3 begin mathsize 14px style left parenthesis 1 right parenthesis end style size 14px plus size 14px 2 begin mathsize 14px style left parenthesis negative 1 right parenthesis end style end cell row blank size 14px equals cell size 14px 3 size 14px minus size 14px 2 end cell row blank size 14px equals size 14px 1 end table 

Jadi, jawabannya adalah C.

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