Pertanyaan

Jika , nilai ( z − x ) adalah ....

Jika $A equals open parentheses table row cell negative 1 end cell cell negative 1 end cell 0 row cell negative 1 end cell 1 2 end table close parentheses comma space B equals open parentheses table row cell negative 1 end cell x row 1 y row 0 z end table close parentheses. space text dan end text space left parenthesis A B right parenthesis to the power of negative 1 end exponent equals open parentheses table row cell negative 1 end cell cell fraction numerator 1 over denominator 2 end fraction end cell row cell fraction numerator 1 over denominator 2 end fraction end cell cell 0 end cell end table close parentheses space space$ , nilai $(z - x)$ adalah $....$

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Klaim

I. Sutiawan

Master Teacher

Mahasiswa/Alumni Universitas Pasundan

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B

Pembahasan

Diketahui . Ingat! Invers matriks ordo dari adalah . Dari kesamaan di atas, didapat dan , maka: Eliminasi (1) dan (2), maka: Jadi, jawaban yang tepat adalah B

Diketahui $A equals open parentheses table row cell negative 1 end cell cell negative 1 end cell 0 row 1 1 2 end table close parentheses comma space B equals open parentheses table row cell negative 1 end cell x row 1 y row 0 z end table close parentheses. space text dan end text space left parenthesis A B right parenthesis to the power of negative 1 end exponent equals open parentheses table row cell negative 1 end cell cell fraction numerator 1 over denominator 2 end fraction end cell row cell fraction numerator 1 over denominator 2 end fraction end cell cell 0 end cell end table close parentheses space space$ .

Ingat!

Invers matriks ordo 2 cross times 2 dari A equals open parentheses table row a b row c d end table close parentheses adalah $A to the power of negative 1 end exponent equals fraction numerator 1 over denominator a d minus b c end fraction open parentheses table row d cell negative b end cell row cell negative c end cell a end table close parentheses$ .

$table attributes columnalign right center left columnspacing 0px end attributes row cell A B end cell equals cell open parentheses table row cell negative 1 end cell cell negative 1 end cell 0 row cell negative 1 end cell 1 2 end table close parentheses open parentheses table row cell negative 1 end cell x row 1 y row 0 z end table close parentheses end cell row blank equals cell open parentheses table row 0 cell negative x minus y end cell row 2 cell space minus x plus y plus 2 z end cell end table close parentheses end cell row blank blank blank row cell open parentheses A B close parentheses to the power of negative 1 end exponent end cell equals cell fraction numerator 1 over denominator 0 minus 2 left parenthesis negative x minus y right parenthesis end fraction open parentheses table row cell negative x plus y plus 2 z end cell cell x plus y end cell row cell negative 2 end cell cell space 0 end cell end table close parentheses end cell row cell open parentheses table row cell negative 1 end cell cell 1 half end cell row cell 1 half end cell 0 end table close parentheses end cell equals cell fraction numerator 1 over denominator 2 x plus 2 y end fraction open parentheses table row cell negative x plus y plus 2 z end cell cell x plus y end cell row cell negative 2 end cell cell space 0 end cell end table close parentheses end cell row cell open parentheses table row cell negative 1 end cell cell 1 half end cell row cell 1 half end cell 0 end table close parentheses end cell equals cell open parentheses table row cell fraction numerator negative x plus y plus 2 z over denominator 2 x plus 2 y end fraction end cell cell 1 half end cell row cell negative fraction numerator 1 over denominator x plus y end fraction end cell cell space 0 end cell end table close parentheses end cell end table$

Dari kesamaan di atas, didapat $1 half equals negative fraction numerator 1 over denominator x plus y end fraction$ dan $table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator negative x plus y plus 2 z over denominator 2 x plus 2 y end fraction end cell end table equals negative 1$ , maka:

$table attributes columnalign right center left columnspacing 0px end attributes row cell 1 half end cell equals cell negative fraction numerator 1 over denominator x plus y end fraction end cell row cell x plus y end cell equals cell negative 2 space.......... space left parenthesis 1 right parenthesis end cell row blank blank blank row cell fraction numerator negative x plus y plus 2 z over denominator 2 x plus 2 y end fraction end cell equals cell negative 1 end cell row cell negative x plus y plus 2 z end cell equals cell negative 1 left parenthesis 2 x plus 2 y right parenthesis end cell row cell negative x plus y plus 2 z end cell equals cell negative 2 x minus 2 y end cell row cell x plus 3 y plus 2 z end cell equals cell 0 space........... space left parenthesis 2 right parenthesis end cell end table$

Eliminasi (1) dan (2), maka:

Jadi, jawaban yang tepat adalah B