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Tuliskan sistem persamaan berikut ke sistem persamaan dalam bentuk matriks dan selesaikan dengan metode eliminasi-substitusi. b.

Pertanyaan

Tuliskan sistem persamaan berikut ke sistem persamaan dalam bentuk matriks dan selesaikan dengan metode eliminasi-substitusi.

b.  begin mathsize 14px style open curly brackets table attributes columnalign left end attributes row cell fraction numerator x y over denominator x plus y end fraction equals 2 end cell row cell fraction numerator y z over denominator y plus z end fraction equals 3 end cell row cell fraction numerator z x over denominator z plus x end fraction equals 4 end cell end table close end style 

Pembahasan Soal:

Berdasarkan soal di atas, diberikan sistem persamaan yang bentuknya dapat diubah sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open curly brackets table attributes columnalign left end attributes row cell fraction numerator x y over denominator x plus y end fraction equals 2 left right double arrow 2 x plus 2 y equals x y end cell row cell fraction numerator y z over denominator y plus z end fraction equals 3 left right double arrow 3 y plus 3 z equals y z end cell row cell fraction numerator z x over denominator z plus x end fraction equals 4 left right double arrow 4 x plus 4 z equals z x end cell end table close end cell end table 

Sehingga, jika persamaan tersebut ditulis dalam bentuk matriks akan menjadi sebagai berikut:

open parentheses table row 2 2 0 row 0 3 3 row 4 0 4 end table close parentheses open parentheses table row x row y row z end table close parentheses equals open parentheses table row cell x y end cell row cell y z end cell row cell z x end cell end table close parentheses  

Penyelesaiaan sistem persamaan di atas, dapat dengan mengubah persamaan pertama sebagai berikut:

fraction numerator x y over denominator x plus y end fraction equals 2 left right double arrow fraction numerator x y over denominator 2 end fraction equals x plus y left right double arrow fraction numerator x y over denominator 2 end fraction minus y equals x rightwards double arrow fraction numerator x y minus 2 y over denominator 2 end fraction equals x rightwards double arrow fraction numerator open parentheses x minus 2 close parentheses y over denominator 2 end fraction equals x space space space space space space space space space space space space rightwards double arrow y equals fraction numerator 2 x over denominator x minus 2 end fraction  

Kemudian substitusi ke persamaan kedua sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator y z over denominator y plus z end fraction end cell equals 3 row blank left right double arrow cell fraction numerator y z over denominator 3 end fraction equals y plus z end cell row blank left right double arrow cell fraction numerator y z over denominator 3 end fraction minus z equals y end cell row blank rightwards double arrow cell fraction numerator y z minus 3 z over denominator 3 end fraction equals y end cell row blank left right double arrow cell fraction numerator open parentheses y minus 3 close parentheses z over denominator 3 end fraction equals y end cell row blank rightwards double arrow cell z equals fraction numerator 3 open parentheses begin display style fraction numerator 2 x over denominator x minus 2 end fraction end style close parentheses over denominator y minus 3 end fraction end cell row blank rightwards double arrow cell z equals fraction numerator 3 open parentheses begin display style fraction numerator 2 x over denominator x minus 2 end fraction end style close parentheses over denominator open parentheses begin display style fraction numerator 2 x over denominator x minus 2 end fraction end style close parentheses minus 3 end fraction end cell row blank rightwards double arrow cell z equals fraction numerator begin display style fraction numerator 6 x over denominator x minus 2 end fraction end style over denominator begin display style fraction numerator 2 x minus 3 open parentheses x minus 2 close parentheses over denominator x minus 2 end fraction end style end fraction end cell row blank rightwards double arrow cell z equals fraction numerator 6 x over denominator 2 x minus 3 x plus 6 end fraction end cell row blank rightwards double arrow cell z equals fraction numerator 6 x over denominator negative x plus 6 end fraction end cell row blank blank blank end table

Kemudian substitusikan ke persamaan ketiga sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator z x over denominator z plus x end fraction end cell equals 4 row blank left right double arrow cell z x equals 4 open parentheses z plus x close parentheses end cell row blank rightwards double arrow cell open parentheses fraction numerator 6 x over denominator negative x plus 6 end fraction close parentheses x equals 4 open parentheses fraction numerator 6 x over denominator negative x plus 6 end fraction plus x close parentheses end cell row blank rightwards double arrow cell fraction numerator 6 x squared over denominator negative x plus 6 end fraction equals 4 open parentheses fraction numerator 6 x plus x open parentheses negative x plus 6 close parentheses over denominator negative x plus 6 end fraction close parentheses end cell row blank rightwards double arrow cell fraction numerator 6 x squared over denominator negative x plus 6 end fraction equals 4 open parentheses fraction numerator 6 x minus x squared plus 6 x over denominator negative x plus 6 end fraction close parentheses end cell row blank rightwards double arrow cell fraction numerator 6 x squared over denominator negative x plus 6 end fraction equals fraction numerator 24 x minus 4 x squared plus 24 x over denominator negative x plus 6 end fraction end cell row blank rightwards double arrow cell 6 x squared equals 48 x minus 4 x squared end cell row blank rightwards double arrow cell 2 x squared minus 48 x equals 0 end cell row blank rightwards double arrow cell x squared minus 24 x equals 0 end cell row blank rightwards double arrow cell x open parentheses x minus 24 close parentheses equals 0 end cell end table 

table row cell x equals 0 end cell logical or cell x minus 24 equals 0 end cell row blank blank cell x equals 24 end cell end table 

Diperoleh nilai x equals 0 atau x equals 24.

Jika nilai x equals 0, maka nilai y dan z dapat ditenttukan sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row y equals cell fraction numerator 2 x over denominator x minus 2 end fraction end cell row blank rightwards double arrow cell y equals fraction numerator 2 open parentheses 0 close parentheses over denominator 0 minus 2 end fraction end cell row blank rightwards double arrow cell y equals 0 end cell row blank blank blank row z equals cell fraction numerator 6 x over denominator negative x plus 6 end fraction end cell row blank rightwards double arrow cell z equals fraction numerator 6 open parentheses 0 close parentheses over denominator negative 0 plus 6 end fraction end cell row blank rightwards double arrow cell z equals 0 end cell end table 

Jika nilai x equals 24, maka nilai y dan z dapat ditenttukan sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row y equals cell fraction numerator 2 x over denominator x minus 2 end fraction end cell row blank rightwards double arrow cell y equals fraction numerator 2 open parentheses 24 close parentheses over denominator 24 minus 2 end fraction end cell row blank rightwards double arrow cell y equals 48 over 22 end cell row blank rightwards double arrow cell y equals 24 over 11 end cell row z equals cell fraction numerator 6 x over denominator negative x plus 6 end fraction end cell row blank rightwards double arrow cell z equals fraction numerator 6 open parentheses 24 close parentheses over denominator negative 24 plus 6 end fraction end cell row blank rightwards double arrow cell z equals fraction numerator 144 over denominator negative 18 end fraction end cell end table rightwards double arrow z equals negative 8 

Jadi, penyelesaian sistem persamaan di atas adalah open parentheses x comma y comma z close parentheses equals open parentheses 0 comma 0 comma 0 close parentheses atau open parentheses x comma y comma z close parentheses equals open parentheses 24 comma 24 over 11 comma negative 8 close parentheses.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 19 Juli 2021

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