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Pertanyaan

Jika matriks begin mathsize 14px style A equals open parentheses table row 2 1 row 3 5 end table close parentheses end style, maka matriks B yang memenuhi, A plus B to the power of t equals open parentheses A minus B close parentheses to the power of t adalah .... 

  1. begin mathsize 14px style open parentheses table row 2 3 row 1 5 end table close parentheses end style 

  2. begin mathsize 14px style open parentheses table row 0 2 row cell negative 2 end cell 0 end table close parentheses end style 

  3. begin mathsize 14px style open parentheses table row 0 cell negative 2 end cell row 2 0 end table close parentheses end style 

  4. open parentheses table row 0 1 row cell negative 1 end cell 0 end table close parentheses 

  5. open parentheses table row 0 cell negative 1 end cell row 1 0 end table close parentheses 

I. Ridha

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah E.space 

Pembahasan

Ingat kembali: Jika matriks A equals open parentheses table row a b row c d end table close parentheses dan B equals open parentheses table row m n row o p end table close parentheses, maka

  • A plus B equals open parentheses table row a b row c d end table close parentheses plus open parentheses table row m n row o p end table close parentheses equals open parentheses table row cell a plus m end cell cell b plus n end cell row cell c plus o end cell cell d plus p end cell end table close parentheses
  • A minus B equals open parentheses table row a b row c d end table close parentheses minus open parentheses table row m n row o p end table close parentheses equals open parentheses table row cell a minus m end cell cell b minus n end cell row cell c minus o end cell cell d minus p end cell end table close parentheses
  • matriks transpose dari A adalah A to the power of t equals open parentheses table row a c row b d end table close parentheses

Dengan demikian, misalkan matriks B adalah open parentheses table row a b row c d end table close parentheses, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell A plus B to the power of t end cell equals cell open parentheses A minus B close parentheses to the power of t end cell row cell open parentheses table row 2 1 row 3 5 end table close parentheses plus open parentheses table row a c row b d end table close parentheses end cell equals cell open parentheses open parentheses table row 2 1 row 3 5 end table close parentheses minus open parentheses table row a b row c d end table close parentheses close parentheses to the power of t end cell row cell open parentheses table row cell 2 plus a end cell cell 1 plus c end cell row cell 3 plus b end cell cell 5 plus d end cell end table close parentheses end cell equals cell open parentheses table row cell 2 minus a end cell cell 1 minus b end cell row cell 3 minus c end cell cell 5 minus d end cell end table close parentheses to the power of t end cell row cell open parentheses table row cell 2 plus a end cell cell 1 plus c end cell row cell 3 plus b end cell cell 5 plus d end cell end table close parentheses end cell equals cell open parentheses table row cell 2 minus a end cell cell 3 minus c end cell row cell 1 minus b end cell cell 5 minus d end cell end table close parentheses end cell end table

sehingga diperoleh

table row cell 2 plus a equals 2 minus a end cell blank cell midline horizontal ellipsis open parentheses 1 close parentheses end cell row cell 1 plus c equals 3 minus c end cell blank cell midline horizontal ellipsis open parentheses 2 close parentheses end cell row cell 3 plus b equals 1 minus b end cell blank cell midline horizontal ellipsis open parentheses 3 close parentheses end cell row cell 5 plus d equals 5 minus d end cell blank cell midline horizontal ellipsis open parentheses 4 close parentheses end cell end table

Dari open parentheses 1 close parentheses, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 plus a end cell equals cell 2 minus a end cell row cell 2 plus a plus a end cell equals 2 row cell 2 plus 2 a end cell equals 2 row cell 2 a end cell equals cell 2 minus 2 end cell row cell 2 a end cell equals 0 row a equals 0 end table

Dari open parentheses 2 close parentheses, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 plus c end cell equals cell 3 minus c end cell row cell 1 plus c plus c end cell equals 3 row cell 1 plus 2 c end cell equals 3 row cell 2 c end cell equals cell 3 minus 1 end cell row cell 2 c end cell equals 2 row c equals 1 end table

Dari open parentheses 3 close parentheses, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 plus b end cell equals cell 1 minus b end cell row cell 3 plus b plus b end cell equals 1 row cell 3 plus 2 b end cell equals 1 row cell 2 b end cell equals cell 1 minus 3 end cell row cell 2 b end cell equals cell negative 2 end cell row b equals cell negative 1 end cell end table

Dari open parentheses 4 close parentheses, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell 5 plus d end cell equals cell 5 minus d end cell row cell 5 plus d plus d end cell equals 5 row cell 5 plus 2 d end cell equals 5 row cell 2 d end cell equals cell 5 minus 5 end cell row cell 2 d end cell equals 0 row d equals 0 end table

Dengan menyubtitusikan a equals 0b equals negative 1c equals 1, dan d equals 0ke pemisalan B equals open parentheses table row a b row c d end table close parentheses, diperoleh B equals open parentheses table row 0 cell negative 1 end cell row 1 0 end table close parentheses.

Jadi, matriks B yang memenuhi, A plus B to the power of t equals open parentheses A minus B close parentheses to the power of t adalah open parentheses table row 0 cell negative 1 end cell row 1 0 end table close parentheses.

Oleh karena itu, jawaban yang benar adalah E.space 

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