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Diketahui P = ( − x − 9 ​ 2 − 2 ​ ) , Q = ( 2 y 7 ​ − 5 3 x ​ ) dan R = ( 10 7 ​ 2 − y ​ ) .Jika 3 P + 2 Q − R = ( − 16 − 20 ​ − 6 − 3 ​ ) , maka nilai 3 x - 2 y = ....

Diketahui  dan  . Jika  , maka nilai 32=  ....

  1. -5

  2. -4

  3. 0

  4. 2

  5. 4

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M. Robo

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

jawaban yang tepat adalah E.

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Pembahasan

Dari kesamaan matriks di atas, kita peroleh dan Subtitusikan y = 3 - 6 x ke - 3 x + 4 y = -6 sehingga Subtitusikan sehingga Jadi, Dengan demikian, jawaban yang tepat adalah E.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 3 P plus 2 Q minus R end cell equals cell open parentheses table row cell negative 16 end cell cell negative 6 end cell row cell negative 20 end cell cell negative 3 end cell end table close parentheses end cell row cell 3 open parentheses table row cell negative x end cell 2 row cell negative 9 end cell cell negative 2 end cell end table close parentheses plus 2 open parentheses table row cell 2 y end cell cell negative 5 end cell row 7 cell 3 x end cell end table close parentheses minus open parentheses table row 10 2 row 7 cell negative y end cell end table close parentheses end cell equals cell open parentheses table row cell negative 16 end cell cell negative 6 end cell row cell negative 20 end cell cell negative 3 end cell end table close parentheses end cell row cell open parentheses table row cell negative 3 x end cell 6 row cell negative 27 end cell cell negative 6 end cell end table close parentheses plus open parentheses table row cell 4 y end cell cell negative 10 end cell row 14 cell 6 x end cell end table close parentheses minus open parentheses table row 10 2 row 7 cell negative y end cell end table close parentheses end cell equals cell open parentheses table row cell negative 16 end cell cell negative 6 end cell row cell negative 20 end cell cell negative 3 end cell end table close parentheses end cell row cell open parentheses table row cell negative 3 x plus 4 y minus 10 end cell cell 6 plus open parentheses negative 10 close parentheses minus 2 end cell row cell negative 27 plus 14 minus 7 end cell cell negative 6 plus 6 x minus open parentheses negative y close parentheses end cell end table close parentheses end cell equals cell open parentheses table row cell negative 16 end cell cell negative 6 end cell row cell negative 20 end cell cell negative 3 end cell end table close parentheses end cell row cell open parentheses table row cell negative 3 x plus 4 y minus 10 end cell cell negative 6 end cell row cell negative 20 end cell cell negative 6 plus 6 x plus y end cell end table close parentheses end cell equals cell open parentheses table row cell negative 16 end cell cell negative 6 end cell row cell negative 20 end cell cell negative 3 end cell end table close parentheses end cell row blank blank blank 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 negative 3 x plus 4 y minus 10 end cell equals cell negative 16 end cell row cell negative 3 x plus 4 y end cell equals cell negative 6 end cell end table end style 

dan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative 6 plus 6 x plus y end cell equals cell negative 3 end cell row cell 6 x plus y end cell equals 3 row y equals cell 3 minus 6 x end cell end table end style 

Subtitusikan = 3 - 6x  ke -3+ 4= -6  sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative 3 x plus 4 open parentheses 3 minus 6 x close parentheses end cell equals cell negative 6 end cell row cell negative 3 x plus 12 minus 24 x end cell equals cell negative 6 end cell row cell negative 27 x end cell equals cell negative 18 end cell row x equals cell fraction numerator negative 18 over denominator negative 27 end fraction end cell row x equals cell 2 over 3 end cell end table end style 

Subtitusikan begin mathsize 14px style x equals 2 over 3 space k e space y equals 3 minus 6 x end style sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell 3 minus 6 open parentheses 2 over 3 close parentheses end cell row blank equals cell 3 minus 4 end cell row blank equals cell negative 1 end cell end table end style 

Jadi, begin mathsize 14px style 3 x minus 2 y equals 3 open parentheses 2 over 3 close parentheses minus 2 open parentheses negative 1 close parentheses equals 2 plus 2 equals 4 end style 

Dengan demikian, jawaban yang tepat adalah E.

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begin mathsize 14px style 2 over 3 open parentheses table row cell negative 3 over 2 end cell 6 row cell negative 3 end cell cell negative 3 over 2 end cell end table close parentheses minus 2 over 5 open parentheses table row 0 cell negative 5 end cell row 0 10 end table close parentheses plus open parentheses table row 0 cell negative 1 end cell row cell negative 1 end cell 0 end table close parentheses equals... end style 

39

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