Jika bilangan real , , dan memenuhi persamaan: maka

Pertanyaan

Jika bilangan real $a$ , $b$ , dan $c$ memenuhi persamaan:

$a open parentheses table row 1 row 0 row 1 end table close parentheses minus 2 b open parentheses table row cell negative 1 end cell row 1 row 0 end table close parentheses plus c open parentheses table row 0 row cell negative 1 end cell row 1 end table close parentheses equals open parentheses table row 1 row 2 row 1 end table close parentheses$

maka $a plus b plus c equals...$

Pembahasan Soal:

Perkalian skalar dengan matriks.

$k open parentheses table row a row b end table close parentheses equals open parentheses table row cell k a end cell row cell k b end cell end table close parentheses$

dengan $k$ adalah konstanta.

Kesamaan matriks.

Jika terdapat persamaan $open parentheses table row a row b end table close parentheses equals open parentheses table row c row d end table close parentheses$ , maka:

$table attributes columnalign right center left columnspacing 0px end attributes row a equals c row b equals d end table$

Permasalahan di atas adalah kesamaan matriks, maka sederhanakan persamaan di ruas kiri.

$table attributes columnalign right center left columnspacing 0px end attributes row cell a open parentheses table row 1 row 0 row 1 end table close parentheses minus 2 b open parentheses table row cell negative 1 end cell row 1 row 0 end table close parentheses plus c open parentheses table row 0 row cell negative 1 end cell row 1 end table close parentheses end cell equals cell open parentheses table row 1 row 2 row 1 end table close parentheses end cell row cell open parentheses table row cell a times 1 end cell row cell a times 0 end cell row cell a times 1 end cell end table close parentheses minus open parentheses table row cell 2 b times left parenthesis negative 1 right parenthesis end cell row cell 2 b times 1 end cell row cell 2 b times 0 end cell end table close parentheses plus open parentheses table row cell c times 0 end cell row cell c times left parenthesis negative 1 right parenthesis end cell row cell c times 1 end cell end table close parentheses end cell equals cell open parentheses table row 1 row 2 row 1 end table close parentheses end cell row cell open parentheses table row cell a plus 2 b plus 0 end cell row cell 0 minus 2 b minus c end cell row cell a plus 0 plus c end cell end table close parentheses end cell equals cell open parentheses table row 1 row 2 row 1 end table close parentheses end cell row cell open parentheses table row cell a plus 2 b end cell row cell negative 2 b minus c end cell row cell a plus c end cell end table close parentheses end cell equals cell open parentheses table row 1 row 2 row 1 end table close parentheses end cell end table$

Sehingga, diperoleh:

$table attributes columnalign right center left columnspacing 0px end attributes row cell a plus 2 b end cell equals cell 1 space... left parenthesis straight i right parenthesis end cell row cell negative 2 b minus c end cell equals cell 2 space... left parenthesis ii right parenthesis end cell row cell a plus c end cell equals 1 row a equals cell 1 minus c space... left parenthesis iii right parenthesis end cell end table$

Substitusikan persamaan (iii) pada persamaan (i)

$table attributes columnalign right center left columnspacing 0px end attributes row cell a plus 2 b end cell equals 1 row cell 1 minus c plus 2 b end cell equals 1 row cell 2 b minus c end cell equals cell 0 space... open parentheses iv close parentheses end cell end table$

Eliminasi persamaan (ii) dengan persamaan (iv).

$negative 2 b minus c equals 2 bottom enclose space space space 2 b minus c equals 0 end enclose space plus space space space space space space minus 2 c equals 2 space space space space space space space space space space space c equals fraction numerator 2 over denominator negative 2 end fraction space space space space space space space space space space space c equals negative 1$

Substitusikan $c equals negative 1$ pada persamaan (iv) dan persamaan (iii).

$table attributes columnalign right center left columnspacing 0px end attributes row cell 2 b minus c end cell equals 0 row cell 2 b minus open parentheses negative 1 close parentheses end cell equals 0 row cell 2 b plus 1 end cell equals 0 row cell 2 b end cell equals cell negative 1 end cell row b equals cell negative 1 half end cell row a equals cell 1 minus c end cell row a equals cell 1 minus open parentheses negative 1 close parentheses end cell row a equals cell 1 plus 1 end cell row a equals 2 end table$

Tentukan nilai $a plus b plus c$ .

$table attributes columnalign right center left columnspacing 0px end attributes row cell a plus b plus c end cell equals cell 2 plus open parentheses negative 1 half close parentheses plus open parentheses negative 1 close parentheses end cell row blank equals cell 2 minus 1 minus 1 half end cell row blank equals cell 1 minus 1 half end cell row blank equals cell 1 half end cell end table$

Jadi, $table attributes columnalign right center left columnspacing 0px end attributes row blank blank a end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank b end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank c end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 half end cell end table$ .

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 09 Juni 2021

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