Roboguru

Diketahui matriks . Tentukan nilai a dan b yang memenuhi !

Pertanyaan

Diketahui matriks A equals open square brackets table row 4 cell negative 1 end cell row cell negative 2 end cell 7 end table close square brackets comma space B equals open square brackets table row cell negative 4 end cell 1 row 2 cell negative 7 end cell end table close square brackets comma space C equals open square brackets table row cell negative 8 end cell a row b cell negative 14 end cell end table close square brackets. Tentukan nilai a dan b yang memenuhi A plus 3 B equals C!

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell A plus 3 B end cell equals C row cell open square brackets table row 4 cell negative 1 end cell row cell negative 2 end cell 7 end table close square brackets plus 3 open square brackets table row cell negative 4 end cell 1 row 2 cell negative 7 end cell end table close square brackets end cell equals cell open square brackets table row cell negative 8 end cell a row b cell negative 14 end cell end table close square brackets end cell row cell open square brackets table row 4 cell negative 1 end cell row cell negative 2 end cell 7 end table close square brackets plus open square brackets table row cell negative 12 end cell 3 row 6 cell negative 21 end cell end table close square brackets end cell equals cell open square brackets table row cell negative 8 end cell a row b cell negative 14 end cell end table close square brackets end cell row cell open square brackets table row cell negative 8 end cell 2 row 4 cell negative 14 end cell end table close square brackets end cell equals cell open square brackets table row cell negative 8 end cell a row b cell negative 14 end cell end table close square brackets end cell row a equals 2 row b equals 4 end table

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Herlanda

Mahasiswa/Alumni STKIP PGRI Jombang

Terakhir diupdate 07 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Jumlah umur kakak dan dua kali umur adik adalah 27 tahun. Selisih umur kakak dan umur adik adalah 3 tahun. Jika umur kakak x tahun dan umur adik y tahun, persamaan matriks yang sesuai dengan permasala...

Pembahasan Soal:

M i s a l k a n space u m u r space k a k a k space equals space x space s e d a n g k a n space u m u r space a d i k space equals space y.  x space plus 2 y space equals space 27  x space – space y space equals space 3  p e r s a m a a n space m a t r i k s space d a r i space i l u s t r a s i space t e r s e b u t space a d a l a h space  open parentheses table row 1 2 row 1 cell negative 1 end cell end table close parentheses open parentheses table row x row y end table close parentheses equals open parentheses table row 27 row 3 end table close parentheses  open parentheses table row x row y end table close parentheses equals open parentheses table row 1 2 row 1 cell negative 1 end cell end table close parentheses to the power of negative 1 end exponent open parentheses table row 27 row 3 end table close parentheses  open parentheses table row x row y end table close parentheses equals fraction numerator 1 over denominator negative 1 minus 2 end fraction open parentheses table row cell negative 1 end cell cell negative 2 end cell row cell negative 1 end cell 1 end table close parentheses open parentheses table row 27 row 3 end table close parentheses  open parentheses table row x row y end table close parentheses equals fraction numerator 1 over denominator negative 3 end fraction open parentheses table row cell negative 1 end cell cell negative 2 end cell row cell negative 1 end cell 1 end table close parentheses open parentheses table row 27 row 3 end table close parentheses  open parentheses table row x row y end table close parentheses equals open parentheses negative 1 close parentheses. open parentheses table row cell negative 1 end cell cell negative 2 end cell row cell negative 1 end cell 1 end table close parentheses.1 third open parentheses table row 27 row 3 end table close parentheses  open parentheses table row x row y end table close parentheses equals open parentheses table row 1 2 row 1 cell negative 1 end cell end table close parentheses open parentheses table row 9 row 1 end table close parentheses

0

Roboguru

Jika matriks  memenuhi persamaan:   maka determinan matriks  adalah ...

Pembahasan Soal:

Diketahui B. open parentheses table row 3 2 row 1 cell negative 1 end cell end table close parentheses equals open parentheses table row 2 1 row 1 3 end table close parentheses comma maka:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell B. open parentheses table row 3 2 row 1 cell negative 1 end cell end table close parentheses end cell equals cell open parentheses table row 2 1 row 1 3 end table close parentheses end cell row cell B. open parentheses table row 3 2 row 1 cell negative 1 end cell end table close parentheses. open parentheses table row 3 2 row 1 cell negative 1 end cell end table close parentheses to the power of negative 1 end exponent end cell equals cell open parentheses table row 2 1 row 1 3 end table close parentheses. open parentheses table row 3 2 row 1 cell negative 1 end cell end table close parentheses to the power of negative 1 end exponent end cell end table end style

table attributes columnalign right center left columnspacing 0px end attributes row cell size 12px B size 12px. size 12px I end cell size 12px equals cell begin mathsize 12px style open parentheses table row 2 1 row 1 3 end table close parentheses end style size 12px. fraction numerator size 12px 1 over denominator size 12px minus size 12px 3 size 12px minus size 12px 2 end fraction begin mathsize 12px style open parentheses table row 3 2 row 1 cell negative 1 end cell end table close parentheses end style end cell row size 12px B size 12px equals cell size 12px minus size 12px 1 over size 12px 5 size 12px. begin mathsize 12px style open parentheses table row 2 1 row 1 3 end table close parentheses end style size 12px. begin mathsize 12px style open parentheses table row 3 2 row 1 cell negative 1 end cell end table close parentheses end style end cell row size 12px B size 12px equals cell size 12px minus size 12px 1 over size 12px 5 size 12px. begin mathsize 12px style open parentheses table row cell 2 cross times 3 plus 1 cross times 1 end cell cell 2 cross times 2 plus 1 cross times left parenthesis negative 1 right parenthesis end cell row cell 1 cross times 3 plus 3 cross times 1 end cell cell 1 cross times 2 plus 3 cross times left parenthesis negative 1 right parenthesis end cell end table close parentheses end style end cell row size 12px B size 12px equals cell size 12px minus size 12px 1 over size 12px 5 size 12px. begin mathsize 12px style open parentheses table row 7 3 row 6 cell negative 1 end cell end table close parentheses end style end cell row size 12px B size 12px equals cell begin mathsize 12px style open parentheses table row cell bevelled fraction numerator negative 7 over denominator 5 end fraction end cell cell bevelled fraction numerator negative 3 over denominator 5 end fraction end cell row cell bevelled fraction numerator negative 6 over denominator 5 end fraction end cell cell bevelled 1 fifth end cell end table close parentheses end style end cell end table  

Sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell det space B end cell equals cell negative 7 over 5 cross times 1 fifth minus open parentheses negative 6 over 5 cross times open parentheses negative 3 over 5 close parentheses close parentheses end cell row blank equals cell negative 7 over 25 minus 18 over 25 end cell row blank equals cell fraction numerator negative 25 over denominator 25 end fraction end cell row blank equals cell negative 1 end cell end table 

Jadi, jawaban yang tepat adalah A.

0

Roboguru

Tentukan jika

Pembahasan Soal:

Ingatlah rumus invers dan sifat invers matriks berikut:

1. A equals open parentheses table row a b row c d end table close parentheses rightwards arrow A to the power of negative 1 end exponent equals fraction numerator 1 over denominator a d minus b c end fraction open parentheses table row d cell negative b end cell row cell negative c end cell a end table close parentheses

2. A X equals B rightwards arrow X equals A to the power of negative 1 end exponent B

3. X A equals B rightwards arrow X equals B A to the power of negative 1 end exponent

Berdasarkan rumus dan sifat tersebut, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row 6 5 row a a end table close parentheses M end cell equals cell open parentheses table row cell negative 1 end cell cell negative 9 end cell row a cell 5 a end cell end table close parentheses end cell row cell open parentheses table row 6 5 row a a end table close parentheses end cell equals cell open parentheses table row cell negative 1 end cell cell negative 9 end cell row a cell 5 a end cell end table close parentheses M to the power of negative 1 end exponent end cell row cell open parentheses table row cell negative 1 end cell cell negative 9 end cell row a cell 5 a end cell end table close parentheses to the power of negative 1 end exponent open parentheses table row 6 5 row a a end table close parentheses end cell equals cell M to the power of negative 1 end exponent end cell row cell M to the power of negative 1 end exponent end cell equals cell fraction numerator 1 over denominator negative 1 times 5 a minus left parenthesis negative 9 right parenthesis times a end fraction open parentheses table row cell 5 a end cell 9 row cell negative a end cell cell negative 1 end cell end table close parentheses open parentheses table row 6 5 row a a end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator negative 5 a plus 9 a end fraction open parentheses table row cell 5 a end cell 9 row cell negative a end cell cell negative 1 end cell end table close parentheses open parentheses table row 6 5 row a a end table close parentheses end cell row blank blank blank end table

Selanjutnya ingat operasi perkalian matriks berikut:

open parentheses table row a b row c d end table close parentheses open parentheses table row e f row g h end table close parentheses equals open parentheses table row cell a e plus b g end cell cell a f plus b h end cell row cell c e plus d g end cell cell c f plus d h end cell end table close parentheses

Sehingga,

table attributes columnalign right center left columnspacing 0px end attributes row cell M to the power of negative 1 end exponent end cell equals cell fraction numerator 1 over denominator negative 5 a plus 9 a end fraction open parentheses table row cell 5 a end cell 9 row cell negative a end cell cell negative 1 end cell end table close parentheses open parentheses table row 6 5 row a a end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator 4 a end fraction open parentheses table row cell 30 a plus 9 a end cell cell 25 a plus 9 a end cell row cell negative 6 a minus a end cell cell negative 5 a minus a end cell end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator 4 a end fraction open parentheses table row cell 39 a end cell cell 34 a end cell row cell negative 7 a end cell cell negative 6 a end cell end table close parentheses end cell row blank equals cell open parentheses table row cell fraction numerator 39 a over denominator 4 a end fraction end cell cell fraction numerator 34 a over denominator 4 a end fraction end cell row cell fraction numerator negative 7 a over denominator 4 a end fraction end cell cell fraction numerator negative 6 a over denominator 4 a end fraction end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 39 over 4 end cell cell 34 over 4 end cell row cell negative 7 over 4 end cell cell negative 3 over 2 end cell end table close parentheses end cell end table

Jadi, diperoleh M to the power of negative 1 end exponent equals table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses table row cell 39 over 4 end cell cell 34 over 4 end cell row cell negative 7 over 4 end cell cell negative 3 over 2 end cell end table close parentheses end cell end table.

0

Roboguru

Diketahui  adalah hasil kali matriks  dengan , yaitu  dan . Jika , matriks  ...

Pembahasan Soal:

Persamaan matriks dengan A adalah matriks persegi yang mempunyai invers atau A to the power of negative 1 end exponent, berlaku:

  • A X equals B

table attributes columnalign right center left columnspacing 0px end attributes row cell A X end cell equals B row cell A to the power of negative 1 end exponent A X end cell equals cell A to the power of negative 1 end exponent B end cell row cell I X end cell equals cell A to the power of negative 1 end exponent B end cell row X equals cell A to the power of negative 1 end exponent B end cell end table

Rumus invers matriks 2 cross times 2 yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row A equals cell open parentheses table row a b row c d end table close parentheses end cell row cell A to the power of negative 1 end exponent end cell equals cell fraction numerator 1 over denominator a times d minus b times c end fraction open parentheses table row d cell negative b end cell row cell negative c end cell a end table close parentheses end cell end table

Diperoleh penyelesaiannya yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row C equals cell A B end cell row cell open parentheses table row 6 7 row 19 18 end table close parentheses end cell equals cell A times open parentheses table row 4 3 row 1 2 end table close parentheses end cell row cell open parentheses table row 6 7 row 19 18 end table close parentheses times open parentheses table row 4 3 row 1 2 end table close parentheses to the power of negative 1 end exponent end cell equals A row cell open parentheses table row 6 7 row 19 18 end table close parentheses times fraction numerator 1 over denominator 4 open parentheses 2 close parentheses minus 3 open parentheses 1 close parentheses end fraction open parentheses table row 2 cell negative 3 end cell row cell negative 1 end cell 4 end table close parentheses end cell equals A row cell open parentheses table row 6 7 row 19 18 end table close parentheses times fraction numerator 1 over denominator 8 minus 3 end fraction open parentheses table row 2 cell negative 3 end cell row cell negative 1 end cell 4 end table close parentheses end cell equals A row cell open parentheses table row 6 7 row 19 18 end table close parentheses times 1 fifth open parentheses table row 2 cell negative 3 end cell row cell negative 1 end cell 4 end table close parentheses end cell equals A row cell open parentheses table row 6 7 row 19 18 end table close parentheses times open parentheses table row cell 2 over 5 end cell cell negative 3 over 5 end cell row cell negative 1 fifth end cell cell 4 over 5 end cell end table close parentheses end cell equals A row cell open parentheses table row cell 6 open parentheses 2 over 5 close parentheses plus 7 open parentheses negative 1 fifth close parentheses end cell cell 6 open parentheses negative 3 over 5 close parentheses plus 7 open parentheses 4 over 5 close parentheses end cell row cell 19 open parentheses 2 over 5 close parentheses plus 18 open parentheses negative 1 fifth close parentheses end cell cell 19 open parentheses negative 3 over 5 close parentheses plus 18 open parentheses 4 over 5 close parentheses end cell end table close parentheses end cell equals A row cell open parentheses table row cell 12 over 5 minus 7 over 5 end cell cell negative 18 over 5 plus 28 over 5 end cell row cell 38 over 5 minus 18 over 5 end cell cell negative 57 over 5 plus 72 over 5 end cell end table close parentheses end cell equals A row cell open parentheses table row cell 5 over 5 end cell cell 10 over 5 end cell row cell 20 over 5 end cell cell 15 over 5 end cell end table close parentheses end cell equals A row cell open parentheses table row 1 2 row 4 3 end table close parentheses end cell equals A end table

Kemudian tentukan invers dari matriks A, yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row A equals cell open parentheses table row 1 2 row 4 3 end table close parentheses end cell row cell A to the power of negative 1 end exponent end cell equals cell open parentheses table row 1 2 row 4 3 end table close parentheses to the power of negative 1 end exponent end cell row blank equals cell fraction numerator 1 over denominator 1 open parentheses 3 close parentheses minus 2 open parentheses 4 close parentheses end fraction open parentheses table row 3 cell negative 2 end cell row cell negative 4 end cell 1 end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator 3 minus 8 end fraction open parentheses table row 3 cell negative 2 end cell row cell negative 4 end cell 1 end table close parentheses end cell row blank equals cell negative 1 fifth open parentheses table row 3 cell negative 2 end cell row cell negative 4 end cell 1 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 3 over 5 end cell cell 2 over 5 end cell row cell 4 over 5 end cell cell negative 1 fifth end cell end table close parentheses end cell end table

Matriks A to the power of negative 1 end exponent equalsopen parentheses table row cell negative 3 over 5 end cell cell 2 over 5 end cell row cell 4 over 5 end cell cell negative 1 fifth end cell end table close parentheses.

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

Jika  , maka x + 2y = ….

Pembahasan Soal:

open square brackets table row 3 cell negative 2 end cell row 4 4 end table close square brackets open square brackets table row x row y end table close square brackets equals open square brackets table row 2 row 0 end table close square brackets  space space space space space space space space space space space space space space space space open square brackets table row x row y end table close square brackets equals open square brackets table row 3 cell negative 2 end cell row 4 4 end table close square brackets to the power of negative 1 end exponent open square brackets table row 2 row 0 end table close square brackets  space space space space space space space space space space space space space space space space open square brackets table row x row y end table close square brackets equals fraction numerator 1 over denominator 12 plus 8 end fraction open square brackets table row 4 2 row cell negative 4 end cell 3 end table close square brackets to the power of space open square brackets table row 2 row 0 end table close square brackets    space space space space space space space space space space space space space space space open square brackets table row x row y end table close square brackets equals 1 over 20 open square brackets table row 8 row cell negative 8 end cell end table close square brackets  open square brackets table row x row y end table close square brackets equals open square brackets table row cell 8 over 20 end cell row cell negative 8 over 20 end cell end table close square brackets      x plus 2 y equals 8 over 20 plus 2 open parentheses negative 8 over 20 close parentheses equals negative 8 over 20 equals negative 2 over 5

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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