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Diketahui: x + 3 y + z = 0 2 x − y + z = 5 3 x − 3 y + 2 z = 10 Maka nilai x ⋅ y ⋅ z = …

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nilai adalah

nilai x times y times z adalah negative 2

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Untuk sistem persamaan linier tiga variabel dapat ditulis dalam bentuk matriks dan penyelesaiannya dapat menggunakan aturan cramer berikut Bentuk matriks dari sistem persamaan tiga variabel tersebut dapat ditulis sebagai berikut Sehingga dengan aturan cramer diperoleh: Maka nilai Dengan demikian nilai adalah

Untuk sistem persamaan linier tiga variabel dapat ditulis dalam bentuk matriks 

open square brackets table row cell a subscript 1 end cell cell b subscript 1 end cell cell c subscript 1 end cell row cell a subscript 2 end cell cell b subscript 2 end cell cell c subscript 2 end cell row cell a subscript 3 end cell cell b subscript 3 end cell cell c subscript 3 end cell end table close square brackets open square brackets table row x row y row z end table close square brackets equals open square brackets table row cell d subscript 1 end cell row cell d subscript 2 end cell row cell d subscript 3 end cell end table close square brackets

dan penyelesaiannya dapat menggunakan aturan cramer berikut

D equals open vertical bar table row cell a subscript 1 end cell cell b subscript 1 end cell cell c subscript 1 end cell row cell a subscript 2 end cell cell b subscript 2 end cell cell c subscript 2 end cell row cell a subscript 3 end cell cell b subscript 3 end cell cell c subscript 3 end cell end table close vertical bar space D subscript x equals open vertical bar table row cell d subscript 1 end cell cell b subscript 1 end cell cell c subscript 1 end cell row cell d subscript 2 end cell cell b subscript 2 end cell cell c subscript 2 end cell row cell d subscript 3 end cell cell b subscript 3 end cell cell c subscript 3 end cell end table close vertical bar space space D subscript y equals open vertical bar table row cell a subscript 1 end cell cell d subscript 1 end cell cell c subscript 1 end cell row cell d a end cell cell d subscript 2 end cell cell c subscript 2 end cell row cell a subscript 3 end cell cell d subscript 3 end cell cell c subscript 3 end cell end table close vertical bar space space D subscript z equals open vertical bar table row cell a subscript 1 end cell cell b subscript 1 end cell cell d subscript 1 end cell row cell a subscript 2 end cell cell b subscript 2 end cell cell d subscript 2 end cell row cell a subscript 3 end cell cell b subscript 3 end cell cell d subscript 3 end cell end table close vertical bar

x equals D subscript x over D space space y equals D subscript y over D space space space z equals D subscript z over D

Bentuk matriks dari sistem persamaan tiga variabel tersebut dapat ditulis sebagai berikut

open square brackets table row 1 3 1 row 2 cell negative 1 end cell 1 row 3 cell negative 3 end cell 2 end table close square brackets open square brackets table row x row y row z end table close square brackets equals open square brackets table row 0 row 5 row 10 end table close square brackets

Sehingga dengan aturan cramer diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row D equals cell open vertical bar table row 1 3 1 row 2 cell negative 1 end cell 1 row 3 cell negative 3 end cell 2 end table close vertical bar table row 1 3 row 2 cell negative 1 end cell row 3 cell negative 3 end cell end table end cell row blank equals cell open parentheses negative 2 plus 9 minus 6 close parentheses minus open parentheses negative 3 minus 3 plus 12 close parentheses end cell row blank equals cell 1 minus 6 end cell row blank equals cell negative 5 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell D subscript x end cell equals cell open vertical bar table row 0 3 1 row 5 cell negative 1 end cell 1 row 10 cell negative 3 end cell 2 end table close vertical bar table row 0 3 row 5 cell negative 1 end cell row 10 cell negative 3 end cell end table end cell row blank equals cell open parentheses 0 plus 30 minus 15 close parentheses minus open parentheses negative 10 minus 0 plus 30 close parentheses end cell row blank equals cell 15 minus open parentheses 20 close parentheses end cell row blank equals cell 15 minus 20 end cell row blank equals cell negative 5 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell D subscript y end cell equals cell open vertical bar table row 1 0 1 row 2 5 1 row 3 10 2 end table close vertical bar table row 1 0 row 2 5 row 3 10 end table end cell row blank equals cell open parentheses 10 plus 0 plus 20 close parentheses minus open parentheses 15 plus 10 plus 0 close parentheses end cell row blank equals cell 30 minus 25 end cell row blank equals 5 end table

table attributes columnalign right center left columnspacing 0px end attributes row cell D subscript z end cell equals cell open vertical bar table row 1 3 0 row 2 cell negative 1 end cell 5 row 3 cell negative 3 end cell 10 end table close vertical bar table row 1 3 row 2 cell negative 1 end cell row 3 cell negative 3 end cell end table end cell row blank equals cell open parentheses negative 10 plus 45 plus 0 close parentheses minus open parentheses 0 minus 15 plus 60 close parentheses end cell row blank equals cell 35 minus 45 end cell row blank equals cell negative 10 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row x equals cell D subscript x over D space space space space space space y equals D subscript y over D space space space space space space space space z equals D subscript z over D end cell row blank equals cell fraction numerator negative 5 over denominator negative 5 end fraction space space space space space space space equals fraction numerator 5 over denominator negative 5 end fraction space space space space space space space space space space equals fraction numerator negative 10 over denominator negative 5 end fraction end cell row blank equals cell 1 space space space space space space space space space space space space equals negative 1 space space space space space space space space space space space equals 2 end cell end table

Maka nilai x times y times z equals 1 times negative 1 times 2 equals negative 2

Dengan demikian nilai x times y times z adalah negative 2

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Tentukan nilai x dan y dari persamaan linier 2 variabel dengan menggunakan: a. matriks b. aturan cramer c. eliminasi d. subsitusi e. gabungan { 5 x − 2 y = 21 − x + 2 y = − 9 ​

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