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Hitunglah ( − 8 x + y ) − ( 4 z ) x = [ − 5 0 ​ 6 − 2 ​ ] , y = [ 1 − 2 ​ − 4 3 ​ ] , z = [ 0 − 4 ​ − 10 2 ​ ]

Hitunglah  

 

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R. Novianto

Master Teacher

Mahasiswa/Alumni Universitas Tanjungpura Pontianak

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 begin mathsize 14px style open parentheses negative 8 straight x plus straight y close parentheses minus open parentheses 4 straight z close parentheses equals open square brackets table row 41 cell negative 12 end cell row 14 11 end table close square brackets end style  

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Pembahasan

Diketahui Maka; Jadi

Diketahui begin mathsize 14px style straight x equals open square brackets table row cell negative 5 end cell 6 row 0 cell negative 2 end cell end table close square brackets comma space straight y equals open square brackets table row 1 cell negative 4 end cell row cell negative 2 end cell 3 end table close square brackets comma space straight z equals open square brackets table row 0 cell negative 10 end cell row cell negative 4 end cell 2 end table close square brackets end style

Maka; 

begin mathsize 14px style left parenthesis negative 8 straight x plus straight y right parenthesis minus left parenthesis 4 straight z right parenthesis equals space open parentheses negative 8 times open square brackets table row cell negative 5 end cell 6 row 0 cell negative 2 end cell end table close square brackets plus open square brackets table row 1 cell negative 4 end cell row cell negative 2 end cell 3 end table close square brackets close parentheses minus open parentheses 4 times open square brackets table row 0 cell negative 10 end cell row cell negative 4 end cell 2 end table close square brackets close parentheses equals open parentheses open square brackets table row 40 cell negative 48 end cell row 0 16 end table close square brackets plus open square brackets table row 1 cell negative 4 end cell row cell negative 2 end cell 3 end table close square brackets close parentheses minus open square brackets table row 0 cell negative 40 end cell row cell negative 16 end cell 8 end table close square brackets equals open square brackets table row cell 40 plus 1 end cell cell negative 48 plus open parentheses negative 4 close parentheses end cell row cell 0 plus negative 2 end cell cell 16 plus 3 end cell end table close square brackets minus open square brackets table row 0 cell negative 40 end cell row cell negative 16 end cell 8 end table close square brackets equals open square brackets table row 41 cell negative 52 end cell row cell negative 2 end cell 19 end table close square brackets minus open square brackets table row 0 cell negative 40 end cell row cell negative 16 end cell 8 end table close square brackets equals open square brackets table row cell 41 minus 0 end cell cell negative 52 minus open parentheses negative 40 close parentheses end cell row cell negative 2 minus open parentheses negative 16 close parentheses end cell cell 19 minus 8 end cell end table close square brackets equals space open square brackets table row 41 cell negative 12 end cell row 14 11 end table close square brackets end style  
 

Jadi begin mathsize 14px style open parentheses negative 8 straight x plus straight y close parentheses minus open parentheses 4 straight z close parentheses equals open square brackets table row 41 cell negative 12 end cell row 14 11 end table close square brackets end style  

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Jika matriks A dan B berordo 2 × 2 yang memenuhi sistem persamaan matriks berikut. ⎩ ⎨ ⎧ ​ A + 2 B = ( 1 3 ​ 2 − 1 ​ ) B − 3 A = ( 0 1 ​ − 1 0 ​ ) ​ Tentukan matriks dan .

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