Pertanyaan

Tentukan nilai ( x , y , z ) dengan konsep OBE. ⎝ ⎛ − 3 2 − 1 1 0 3 0 1 − 2 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 3 − 4 2 ⎠ ⎞

Tentukan nilai $(x, y, z)$ dengan konsep OBE.

$⎝ ⎛ - 3 2 - 1 103 01 - 2 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 3 - 4 2 ⎠ ⎞$

A. Septianingsih

Master Teacher

Mahasiswa/Alumni Universitas Gadjah Mada

Jawaban terverifikasi

Pembahasan

Terlebih dahulu mengubah dalam bentuk matriks berikut :

$open parentheses table row cell negative 3 end cell 1 0 row 2 0 1 row cell negative 1 end cell 3 cell negative 2 end cell end table open vertical bar table row 3 row cell negative 4 end cell row 2 end table close vertical bar close parentheses Baris space 1 space colon left parenthesis negative 3 right parenthesis space left parenthesis agar space straight B subscript 11 equals 1 right parenthesis open parentheses table row 1 cell fraction numerator negative 1 over denominator 3 end fraction end cell 0 row 2 0 1 row cell negative 1 end cell 3 cell negative 2 end cell end table open vertical bar table row cell negative 1 end cell row cell negative 4 end cell row 2 end table close vertical bar close parentheses Baris space 2 space rightwards double arrow straight B subscript 2 minus 2 straight B subscript 1 space dan space Baris space 3 rightwards double arrow space space straight B subscript 3 plus straight B subscript 1 open parentheses table row 1 cell fraction numerator negative 1 over denominator 3 end fraction end cell 0 row 0 cell 2 over 3 end cell 1 row 0 cell 2 2 over 3 end cell cell negative 2 end cell end table open vertical bar table row cell negative 1 end cell row cell negative 2 end cell row 1 end table close vertical bar close parentheses Baris space 2 rightwards double arrow straight B subscript 2 x 3 over 2 open parentheses table row 1 cell fraction numerator negative 1 over denominator 3 end fraction end cell 0 row 0 1 cell 3 over 2 end cell row 0 cell 2 2 over 3 end cell cell negative 2 end cell end table open vertical bar table row cell negative 1 end cell row cell negative 3 end cell row 1 end table close vertical bar close parentheses Baris space 3 rightwards double arrow straight B subscript 3 minus 2 2 over 3 straight B subscript 2 open parentheses table row 1 cell fraction numerator negative 1 over denominator 3 end fraction end cell 0 row 0 1 cell 3 over 2 end cell row 0 0 cell negative 4 end cell end table open vertical bar table row cell negative 1 end cell row cell negative 3 end cell row 5 end table close vertical bar close parentheses Baris space 3 rightwards double arrow space straight B subscript 3 colon negative 4 open parentheses table row 1 cell fraction numerator negative 1 over denominator 3 end fraction end cell 0 row 0 1 cell 3 over 2 end cell row 0 0 1 end table open vertical bar table row cell negative 1 end cell row cell negative 3 end cell row cell fraction numerator negative 5 over denominator 4 end fraction end cell end table close vertical bar close parentheses Maka Baris space ke space 3 space diperoleh space straight z equals fraction numerator negative 5 over denominator 4 end fraction Baris space ke space 2 space diperoleh space straight y plus 3 over 2 straight z equals negative 3 straight y equals negative 3 minus open parentheses 3 over 2 close parentheses open parentheses fraction numerator negative 5 over denominator 4 end fraction close parentheses straight y equals fraction numerator negative 9 over denominator 8 end fraction Baris space ke space 1 space diperoleh straight x plus straight y plus 0. straight z equals negative 1 straight x plus fraction numerator negative 9 over denominator 8 end fraction equals 3 straight x equals 33 over 8$

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell Sehingga space diperoleh space nilai end cell row straight x equals cell 33 over 8 end cell row straight y equals cell fraction numerator negative 9 over denominator 8 end fraction end cell row straight z equals cell fraction numerator negative 5 over denominator 4 end fraction end cell end table$

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