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Pertanyaan

Diberikan open parentheses table row 3 5 row 1 2 end table close parentheses times open parentheses table row a 0 row cell a plus b end cell cell c plus 2 end cell end table close parentheses equals open parentheses table row 1 cell negative 5 end cell row 0 cell negative 2 end cell end table close parentheses. Cari nilai dari open parentheses a plus b minus c close parentheses.space 

I. Ridha

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

nilai dari open parentheses a plus b minus c close parentheses adalah 2.space 

Pembahasan

Ingat kembali:

  • Perkalian matriks open parentheses table row a b row c d end table close parentheses open parentheses table row m n row o p end table close parentheses equals open parentheses table row cell a m plus b o end cell cell a n plus b p end cell row cell c m plus d o end cell cell c n plus d p end cell end table close parentheses
  • Jika open parentheses table row a b row c d end table close parentheses equals open parentheses table row m n row o p end table close parentheses, maka a equals mb equals nc equals o, dan d equals p

Dengan demikian, 

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row 3 5 row 1 2 end table close parentheses times open parentheses table row a 0 row cell a plus b end cell cell c plus 2 end cell end table close parentheses end cell equals cell open parentheses table row 1 cell negative 5 end cell row 0 cell negative 2 end cell end table close parentheses end cell row cell open parentheses table row cell 3 open parentheses a close parentheses plus 5 open parentheses a plus b close parentheses end cell cell 3 open parentheses 0 close parentheses plus 5 open parentheses c plus 2 close parentheses end cell row cell 1 open parentheses a close parentheses plus 2 open parentheses a plus b close parentheses end cell cell 1 open parentheses 0 close parentheses plus 2 open parentheses c plus 2 close parentheses end cell end table close parentheses end cell equals cell open parentheses table row 1 cell negative 5 end cell row 0 cell negative 2 end cell end table close parentheses end cell row cell open parentheses table row cell 3 a plus 5 a plus 5 b end cell cell 0 plus 5 c plus 10 end cell row cell a plus 2 a plus 2 b end cell cell 0 plus 2 c plus 4 end cell end table close parentheses end cell equals cell open parentheses table row 1 cell negative 5 end cell row 0 cell negative 2 end cell end table close parentheses end cell row cell open parentheses table row cell 8 a plus 5 b end cell cell 5 c plus 10 end cell row cell 3 a plus 2 b end cell cell 2 c plus 4 end cell end table close parentheses end cell equals cell open parentheses table row 1 cell negative 5 end cell row 0 cell negative 2 end cell end table close parentheses end cell end table

sehingga diperoleh

table row cell 8 a plus 5 b equals 1 end cell blank cell midline horizontal ellipsis open parentheses 1 close parentheses end cell row cell 5 c plus 10 equals negative 5 end cell blank cell midline horizontal ellipsis open parentheses 2 close parentheses end cell row cell 3 a plus 2 b equals 0 end cell blank cell midline horizontal ellipsis open parentheses 3 close parentheses end cell row cell 2 c plus 4 equals negative 2 end cell blank cell midline horizontal ellipsis open parentheses 4 close parentheses end cell end table

Dari open parentheses 2 close parentheses, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell 5 c plus 10 end cell equals cell negative 5 end cell row cell 5 c end cell equals cell negative 5 minus 10 end cell row cell 5 c end cell equals cell negative 15 end cell row c equals cell fraction numerator negative 15 over denominator 5 end fraction end cell row c equals cell negative 3 end cell end table

Hasil yang sama diperoleh dari open parentheses 4 close parentheses, yaitu c equals negative 3.

Dengan mengalikan open parentheses 1 close parentheses dengan 2, mengalikan open parentheses 3 close parentheses dengan 5, dan kemudian mengeliminasi variabel b, diperoleh

table row cell table row cell 8 a plus 5 b equals 1 end cell row cell 3 a plus 2 b equals 0 end cell row blank end table end cell blank cell table row cell open vertical bar table row cell cross times 2 end cell row cell cross times 5 end cell end table close vertical bar end cell row blank end table end cell cell table row cell 16 a plus 10 b equals 2 end cell row cell 15 a plus 10 b equals 0 end cell row cell a equals 2 end cell end table end cell cell table row blank row minus end table end cell end table

Dengan menyubtitusi a equals 2 ke open parentheses 3 close parentheses, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 open parentheses 2 close parentheses plus 2 b end cell equals 0 row cell 6 plus 2 b end cell equals 0 row cell 2 b end cell equals cell 0 minus 6 end cell row cell 2 b end cell equals cell negative 6 end cell row b equals cell fraction numerator negative 6 over denominator 2 end fraction end cell row b equals cell negative 3 end cell end table

Oleh karena diperoleh a equals 2b equals negative 3, dan c equals negative 3, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a plus b minus c close parentheses end cell equals cell 2 plus open parentheses negative 3 close parentheses minus open parentheses negative 3 close parentheses end cell row blank equals cell 2 minus 3 plus 3 end cell row blank equals 2 end table

Jadi, nilai dari open parentheses a plus b minus c close parentheses adalah 2.space 

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