Roboguru

Diketahui  dan  Jika  maka matriks B adalah...

Pertanyaan

Diketahui straight A equals open square brackets table row 2 cell negative 1 end cell row cell negative 3 end cell 3 end table close square brackets comma space straight C equals open square brackets table row 1 2 row 2 5 end table close square brackets comma dan straight D equals open square brackets table row 3 0 row 6 9 end table close square brackets. Jika straight A times straight B times straight C equals straight D maka matriks B adalah...space 

Pembahasan Soal:

Perhatikan penghitungan berikut!

 straight A times straight B times straight C equals straight D straight A to the power of negative 1 end exponent times straight A times straight B times straight C equals straight A to the power of negative 1 end exponent times straight D straight I times straight B times straight C equals straight A to the power of negative 1 end exponent times straight D straight B times straight C equals straight A to the power of negative 1 end exponent times straight D straight B times straight C times straight C to the power of negative 1 end exponent equals straight A to the power of negative 1 end exponent times straight D times straight C to the power of negative 1 end exponent straight B times straight I equals straight A to the power of negative 1 end exponent times straight D times straight C to the power of negative 1 end exponent straight B equals straight A to the power of negative 1 end exponent times straight D times straight C to the power of negative 1 end exponent straight A equals open square brackets table row 2 cell negative 1 end cell row cell negative 3 end cell 3 end table close square brackets space rightwards arrow straight A to the power of negative 1 end exponent equals fraction numerator 1 over denominator open parentheses 2 cross times 3 close parentheses minus open parentheses negative 1 cross times negative 3 close parentheses end fraction open square brackets table row 3 1 row 3 2 end table close square brackets straight A to the power of negative 1 end exponent equals 1 third open square brackets table row 3 1 row 3 2 end table close square brackets  straight C equals open square brackets table row 1 2 row 2 5 end table close square brackets space rightwards arrow straight C to the power of negative 1 end exponent equals fraction numerator 1 over denominator open parentheses 1 cross times 5 close parentheses minus open parentheses 2 cross times 2 close parentheses end fraction open square brackets table row 5 cell negative 2 end cell row cell negative 2 end cell 1 end table close square brackets straight C to the power of negative 1 end exponent equals 1 over 1 open square brackets table row 5 cell negative 2 end cell row cell negative 2 end cell 1 end table close square brackets equals open square brackets table row 5 cell negative 2 end cell row cell negative 2 end cell 1 end table close square brackets  straight B equals straight A to the power of negative 1 end exponent times straight D times straight C to the power of negative 1 end exponent straight B equals 1 third open square brackets table row 3 1 row 3 2 end table close square brackets times open square brackets table row 3 0 row 6 9 end table close square brackets times open square brackets table row 5 cell negative 2 end cell row cell negative 2 end cell 1 end table close square brackets straight B equals 1 third open square brackets table row 15 9 row 21 18 end table close square brackets times open square brackets table row 5 cell negative 2 end cell row cell negative 2 end cell 1 end table close square brackets straight B equals 1 third open square brackets table row 57 cell negative 21 end cell row 69 cell negative 24 end cell end table close square brackets straight B equals open square brackets table row 19 cell negative 7 end cell row 23 cell negative 8 end cell end table close square brackets                

Jadi, nilai matriks B adalah open square brackets table row 19 cell negative 7 end cell row 23 cell negative 8 end cell end table close square brackets .space      space 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Fatimatuzzahroh

Mahasiswa/Alumni Institut Pertanian Bogor

Terakhir diupdate 02 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Ditentukan matriks  dan . Jika matriks  kali invers matriks , maka  ...

Pembahasan Soal:

Diketahui: Matriks A equals k kali invers matriks B.

Sehingga diperoleh penyelesainnya yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row A equals cell k times B to the power of negative 1 end exponent end cell row cell open parentheses table row x 2 row 5 y end table close parentheses end cell equals cell k times open parentheses table row 4 cell negative 2 end cell row cell negative 5 end cell 3 end table close parentheses to the power of negative 1 end exponent end cell row cell open parentheses table row x 2 row 5 y end table close parentheses end cell equals cell k times fraction numerator 1 over denominator 4 open parentheses 3 close parentheses minus open parentheses negative 2 close parentheses open parentheses negative 5 close parentheses end fraction open parentheses table row 3 2 row 5 4 end table close parentheses end cell row cell open parentheses table row x 2 row 5 y end table close parentheses end cell equals cell k times fraction numerator 1 over denominator 12 minus 10 end fraction open parentheses table row 3 2 row 5 4 end table close parentheses end cell row cell open parentheses table row x 2 row 5 y end table close parentheses end cell equals cell k times 1 half open parentheses table row 3 2 row 5 4 end table close parentheses end cell row cell open parentheses table row x 2 row 5 y end table close parentheses end cell equals cell k times open parentheses table row cell 3 over 2 end cell cell 2 over 2 end cell row cell 5 over 2 end cell cell 4 over 2 end cell end table close parentheses end cell row cell open parentheses table row x 2 row 5 y end table close parentheses end cell equals cell k times open parentheses table row cell 3 over 2 end cell 1 row cell 5 over 2 end cell 2 end table close parentheses end cell end table

Pada perhitungan matriks, diperoleh persamaan:

table row cell x equals k times 3 over 2 end cell cell... open parentheses 1 close parentheses end cell row cell 2 equals k times 1 end cell cell... open parentheses 2 close parentheses end cell row cell y equals k times 2 end cell cell... open parentheses 3 close parentheses end cell end table

Pada persamaan open parentheses 2 close parentheses diperoleh nilai k yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row 2 equals cell k times 1 end cell row 2 equals k end table

Substitusi k equals 2 ke persamaan open parentheses 2 close parentheses untuk memperoleh nilai x.

table attributes columnalign right center left columnspacing 0px end attributes row x equals cell k times 3 over 2 end cell row x equals cell down diagonal strike 2 times fraction numerator 3 over denominator down diagonal strike 2 end fraction end cell row x equals 3 end table

Sehingga diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell k plus x end cell equals cell 2 plus 3 end cell row blank equals 5 end table

Maka k plus x equals5.

Oleh karena itu, jawaban yang benar adalah C.

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Roboguru

Diketahui matriks , , jika  maka nilai ...

Pembahasan Soal:

Gunakan konsep invers matriks, begin mathsize 14px style straight A equals open parentheses table row 3 2 row 6 x end table close parentheses end style maka invers matriks A adalah

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight A to the power of negative 1 end exponent end cell equals cell fraction numerator 1 over denominator det space straight A end fraction times adj space straight A end cell row blank equals cell fraction numerator 1 over denominator 3 times x minus 2 times 6 end fraction times open parentheses table row x cell negative 2 end cell row cell negative 6 end cell 3 end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator 3 x minus 12 end fraction times open parentheses table row x cell negative 2 end cell row cell negative 6 end cell 3 end table close parentheses end cell row blank equals cell open parentheses table row cell fraction numerator x over denominator 3 x minus 12 end fraction end cell cell fraction numerator negative 2 over denominator 3 x minus 12 end fraction end cell row cell fraction numerator negative 6 over denominator 3 x minus 12 end fraction end cell cell fraction numerator 3 over denominator 3 x minus 12 end fraction end cell end table close parentheses end cell end table end style 

Jika begin mathsize 14px style straight A to the power of negative 1 end exponent equals straight B to the power of straight t end style, gunakan konsep transpose matriks sehingga:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight A to the power of negative 1 end exponent end cell equals cell straight B to the power of straight t end cell row cell open parentheses table row cell fraction numerator x over denominator 3 x minus 12 end fraction end cell cell fraction numerator negative 2 over denominator 3 x minus 12 end fraction end cell row cell fraction numerator negative 6 over denominator 3 x minus 12 end fraction end cell cell fraction numerator 3 over denominator 3 x minus 12 end fraction end cell end table close parentheses end cell equals cell open parentheses table row cell x over 3 end cell cell negative 2 over 3 end cell row cell negative 2 end cell 1 end table close parentheses end cell end table end style 

Berdasarkan konsep kesamaan matriks maka diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator negative 6 over denominator 3 x minus 12 end fraction end cell equals cell negative 2 end cell row cell negative 6 end cell equals cell negative 2 open parentheses 3 x minus 12 close parentheses end cell row cell negative 6 end cell equals cell negative 6 x plus 24 end cell row cell 6 x end cell equals cell 24 plus 6 end cell row cell 6 x end cell equals 30 row x equals cell 30 over 6 end cell row x equals 5 end table end style  

Dengan demikian, nilai dari begin mathsize 14px style x end style adalah 5.

0

Roboguru

Diketahui matriks , matriks , dan matriks . Nilai  yang memenuhi  ( invers matriks ) adalah ....

Pembahasan Soal:

0

Roboguru

Harga  dan  berturut-turut yang memenuhi persamaan:   adalah ....

Pembahasan Soal:

0

Roboguru

Jika  dengan  menyatakan invers matriks P, maka

Pembahasan Soal:

Menentukan invers P.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight P to the power of negative 1 end exponent end cell equals cell fraction numerator 1 over denominator det space straight P end fraction Adj space straight P end cell row blank equals cell fraction numerator 1 over denominator open parentheses 1 cross times 3 close parentheses minus open parentheses 1 cross times 2 close parentheses end fraction open parentheses table row 3 cell negative 2 end cell row cell negative 1 end cell 1 end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator 3 minus 2 end fraction open parentheses table row 3 cell negative 2 end cell row cell negative 1 end cell 1 end table close parentheses end cell row blank equals cell open parentheses table row 3 cell negative 2 end cell row cell negative 1 end cell 1 end table close parentheses end cell end table end style 

sehingga dengan kesamaan dua matriks diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row x y row cell negative z end cell z end table close parentheses end cell equals cell 2 straight P to the power of negative 1 end exponent end cell row cell open parentheses table row straight x straight y row cell negative straight z end cell straight z end table close parentheses end cell equals cell 2 open parentheses table row 3 cell negative 2 end cell row cell negative 1 end cell 1 end table close parentheses end cell row cell open parentheses table row straight x straight y row cell negative straight z end cell straight z end table close parentheses end cell equals cell open parentheses table row 6 cell negative 4 end cell row cell negative 2 end cell 2 end table close parentheses end cell end table end style 

sehingga diperoleh:

Error converting from MathML to accessible text.   

Dengan demikian, nilai dari table attributes columnalign right center left columnspacing 0px end attributes row blank blank size 14px x 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 y end table table attributes columnalign right center left columnspacing 0px end attributes row blank size 14px equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank size 14px 2 end table.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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