Roboguru

Jika diketahui matriksZ=[2−2​−34​], invers matriks Z adalah ....

Pertanyaan

Jika diketahui matriksbegin mathsize 14px style Z equals open square brackets table row 2 cell negative 3 end cell row cell negative 2 end cell 4 end table close square brackets end style, invers matriks Z adalah ....

  1. begin mathsize 14px style open square brackets table row 4 3 row 2 2 end table close square brackets end style 

  2. begin mathsize 14px style open square brackets table row 2 1 row cell 3 over 2 end cell 1 end table close square brackets end style 

  3. begin mathsize 14px style 1 half open square brackets table row 4 3 row 2 2 end table close square brackets end style 

  4. begin mathsize 14px style 1 third open square brackets table row 4 3 row 2 2 end table close square brackets end style 

  5. begin mathsize 14px style 1 fourth open square brackets table row 4 3 row 2 2 end table close square brackets end style 

Pembahasan Soal:

Invers dari matriks Z adalah
begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell Z to the power of negative 1 end exponent end cell equals cell fraction numerator 1 over denominator open vertical bar Z close vertical bar end fraction a d j open parentheses Z close parentheses end cell row blank equals cell fraction numerator 1 over denominator 2 cross times 4 minus open parentheses negative 3 close parentheses cross times open parentheses negative 2 close parentheses end fraction open square brackets table row 4 3 row 2 2 end table close square brackets end cell row blank equals cell fraction numerator 1 over denominator 8 minus 6 end fraction open square brackets table row 4 3 row 2 2 end table close square brackets end cell row blank equals cell 1 half open square brackets table row 4 3 row 2 2 end table close square brackets end cell end table end style

Jadi jawaban yang tepat adalah C

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Sibuea

Mahasiswa/Alumni Universitas Riau

Terakhir diupdate 05 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Selesaikanlah setiap sistem persamaan linear dua variabel berikut. ⎩⎨⎧​x+y=1x−2y=−83x+y=−3​

Pembahasan Soal:

Menentukan penyelesaian SPLDV dengan Metode Invers Matriks adalah sebagai berikut:

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 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 b 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 equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank e 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 blank 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 d 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 equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank f end table  

Diubah ke dalam bentuk matriks didapatkan:

open parentheses table row a b row c d end table close parentheses open parentheses table row x row y end table close parentheses equals open parentheses table row e row f end table close parentheses space horizontal ellipsis space open parentheses 1 close parentheses  

Persamaan open parentheses 1 close parentheses dapat ditulis sebagai A X equals B, dengan:

A equals open parentheses table row a b row c d end table close parentheses comma space X equals open parentheses table row x row y end table close parentheses comma space dan space B equals open parentheses table row e row f end table close parentheses  

Penentuan X dapat dilakukan dengan sifat matriks, yaitu:

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

Diketahui A equals open parentheses table row a b row c d end table close parentheses, maka inversnya adalah:

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     

Diketahui sistem persamaan sebagai berikut:

begin mathsize 14px style open curly brackets table attributes columnalign left end attributes row cell x plus y equals 1 end cell row cell x minus 2 y equals negative 8 end cell row cell 3 x plus y equals negative 3 end cell end table close end style  

Ambil 2 persamaan yaitu:

 x plus y equals 1 x minus 2 y equals negative 8 

Dari 2 persamaan di atas diubah ke dalam matriks dan didapatkan:

open parentheses table row 1 1 row 1 cell negative 2 end cell end table close parentheses open parentheses table row x row y end table close parentheses equals open parentheses table row 1 row cell negative 8 end cell end table close parentheses   

Misalkan A equals open parentheses table row 1 1 row 1 cell negative 2 end cell end table close parentheses comma space X equals open parentheses table row x row y end table close parentheses comma space dan space B equals open parentheses table row 1 row cell negative 8 end cell end table close parentheses, maka matriks X didapatkan:

table attributes columnalign right center left columnspacing 0px end attributes row cell A X end cell equals B row X equals cell A to the power of negative 1 end exponent B end cell row cell open parentheses table row x row y end table close parentheses end cell equals cell open parentheses table row 1 1 row 1 cell negative 2 end cell end table close parentheses to the power of negative 1 end exponent open parentheses table row 1 row cell negative 8 end cell end table close parentheses end cell row cell open parentheses table row x row y end table close parentheses end cell equals cell fraction numerator 1 over denominator open parentheses 1 close parentheses open parentheses negative 2 close parentheses minus open parentheses 1 close parentheses open parentheses 1 close parentheses end fraction open parentheses table row cell negative 2 end cell cell negative 1 end cell row cell negative 1 end cell 1 end table close parentheses open parentheses table row 1 row cell negative 8 end cell end table close parentheses end cell row cell open parentheses table row x row y end table close parentheses end cell equals cell fraction numerator 1 over denominator negative 2 minus 1 end fraction open parentheses table row cell negative 2 end cell cell negative 1 end cell row cell negative 1 end cell 1 end table close parentheses open parentheses table row 1 row cell negative 8 end cell end table close parentheses end cell row cell open parentheses table row x row y end table close parentheses end cell equals cell fraction numerator 1 over denominator negative 3 end fraction open parentheses table row 6 row cell negative 9 end cell end table close parentheses end cell row cell open parentheses table row x row y end table close parentheses end cell equals cell open parentheses table row cell negative 2 end cell row 3 end table close parentheses end cell end table      

Jadi, himpunan penyelesaiannya untuk setiap sistem persamaan linear dua variabel tersebut adalah open curly brackets open parentheses negative 2 comma 3 close parentheses close curly brackets.

0

Roboguru

Diketahui matriks A=[21​−13​]danB=[47​1−2​] tentukanlah: b. BT⋅A−1

Pembahasan Soal:

Ingat kembali aturan matriks dan transpose matriks sebagai berikut!

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

Dengan aturan di atas, didapat perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row A equals cell open square brackets table row 2 cell negative 1 end cell row 1 3 end table close square brackets end cell row cell A to the power of negative 1 end exponent end cell equals cell fraction numerator 1 over denominator 2 times 3 minus left parenthesis negative 1 right parenthesis times 1 end fraction open square brackets table row 3 1 row cell negative 1 end cell 2 end table close square brackets end cell row blank equals cell 1 over 7 open square brackets table row 3 1 row cell negative 1 end cell 2 end table close square brackets end cell row blank equals cell open square brackets table row cell 3 over 7 end cell cell 1 over 7 end cell row cell fraction numerator negative 1 over denominator 7 end fraction end cell cell 2 over 7 end cell end table close square brackets end cell row blank blank blank row B equals cell open square brackets table row 4 1 row 7 cell negative 2 end cell end table close square brackets end cell row cell B to the power of T end cell equals cell open square brackets table row 4 7 row 1 cell negative 2 end cell end table close square brackets end cell row blank blank blank row cell B to the power of T times A to the power of negative 1 end exponent end cell equals cell open square brackets table row 4 7 row 1 cell negative 2 end cell end table close square brackets times open square brackets table row cell 3 over 7 end cell cell 1 over 7 end cell row cell fraction numerator negative 1 over denominator 7 end fraction end cell cell 2 over 7 end cell end table close square brackets end cell row blank equals cell open square brackets table row cell fraction numerator 12 minus 7 over denominator 7 end fraction end cell cell fraction numerator 4 plus 14 over denominator 7 end fraction end cell row cell fraction numerator 3 plus 2 over denominator 7 end fraction end cell cell fraction numerator 1 minus 4 over denominator 7 end fraction end cell end table close square brackets end cell row blank equals cell open square brackets table row cell 5 over 7 end cell cell 18 over 7 end cell row cell 5 over 7 end cell cell fraction numerator negative 3 over denominator 7 end fraction end cell end table close square brackets end cell end table end style

Dengan demikian, begin mathsize 14px style B to the power of T times A to the power of negative 1 end exponent end style adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open square brackets table row cell 5 over 7 end cell cell 18 over 7 end cell row cell 5 over 7 end cell cell fraction numerator negative 3 over denominator 7 end fraction end cell end table close square brackets end cell end table.

0

Roboguru

Diketahui matriks A=(43​−9−4p​), B=(5p1​−53​), dan C=(−10−4​86p​). Jika matriks A−B=C−1, maka nilai 4p= ...

Pembahasan Soal:

Rumus selisih dua buah matriks 2 cross times 2 yaitu:

open parentheses table row a b row c d end table close parentheses minus open parentheses table row p q row r s end table close parentheses equals open parentheses table row cell a minus p end cell cell b minus q end cell row cell c minus r end cell cell d minus s end cell end table close parentheses

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 cell A minus B end cell equals cell C to the power of negative 1 end exponent end cell row cell open parentheses table row 4 cell negative 9 end cell row 3 cell negative 4 p end cell end table close parentheses minus open parentheses table row cell 5 p end cell cell negative 5 end cell row 1 3 end table close parentheses end cell equals cell open parentheses table row cell negative 10 end cell 8 row cell negative 4 end cell cell 6 p end cell end table close parentheses to the power of negative 1 end exponent end cell row cell open parentheses table row cell 4 minus 5 p end cell cell negative 9 minus open parentheses negative 5 close parentheses end cell row cell 3 minus 1 end cell cell negative 4 p minus 3 end cell end table close parentheses end cell equals cell fraction numerator 1 over denominator negative 10 open parentheses 6 p close parentheses minus 8 open parentheses negative 4 close parentheses end fraction open parentheses table row cell 6 p end cell cell negative 8 end cell row 4 cell negative 10 end cell end table close parentheses end cell row cell open parentheses table row cell 4 minus 5 p end cell cell negative 9 plus 5 end cell row 2 cell negative 4 p minus 3 end cell end table close parentheses end cell equals cell fraction numerator 1 over denominator negative 60 p plus 32 end fraction open parentheses table row cell 6 p end cell cell negative 8 end cell row 4 cell negative 10 end cell end table close parentheses end cell row cell open parentheses table row cell 4 minus 5 p end cell cell negative 4 end cell row 2 cell negative 4 p minus 3 end cell end table close parentheses end cell equals cell open parentheses table row cell fraction numerator 6 p over denominator negative 60 p plus 32 end fraction end cell cell fraction numerator negative 8 over denominator negative 60 p plus 32 end fraction end cell row cell fraction numerator 4 over denominator negative 60 p plus 32 end fraction end cell cell fraction numerator negative 10 over denominator negative 60 p plus 32 end fraction end cell end table close parentheses end cell row cell open parentheses table row cell 4 minus 5 p end cell cell negative 4 end cell row 2 cell negative 4 p minus 3 end cell end table close parentheses end cell equals cell open parentheses table row cell fraction numerator 3 p over denominator negative 30 p plus 16 end fraction end cell cell fraction numerator negative 2 over denominator negative 15 p plus 8 end fraction end cell row cell fraction numerator 1 over denominator negative 15 p plus 8 end fraction end cell cell fraction numerator negative 5 over denominator negative 30 p plus 16 end fraction end cell end table close parentheses end cell end table

Perhatikan pada perhitungan matriks terdapat persamaan-persamaan.

Misal diambil persamaan 2 equals fraction numerator 1 over denominator negative 15 p plus 8 end fraction, maka nilai p yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row 2 equals cell fraction numerator 1 over denominator negative 15 p plus 8 end fraction end cell row cell 2 open parentheses negative 15 p plus 8 close parentheses end cell equals 1 row cell negative 30 p plus 16 end cell equals 1 row cell negative 30 p end cell equals cell 1 minus 16 end cell row cell negative 30 p end cell equals cell negative 15 end cell row p equals cell fraction numerator negative 15 over denominator negative 30 end fraction end cell row p equals cell 1 half end cell end table

Sehingga diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 p end cell equals cell 4 times 1 half end cell row blank equals 2 end table

Maka nilai 4 p equals2.

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

Matriks x yang memenuhi (21​34​)x=(1113​1711​) adalah ...

Pembahasan Soal:

Gunakan sifat invers pada persamaan matriks, yaitu:

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

Invers matriks 2 cross times 2 berlaku:

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 open vertical bar A close vertical bar 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 row blank equals cell 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 end cell end table

Sehingga diperoleh penyelesaiannya yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row 2 3 row 1 4 end table close parentheses x end cell equals cell open parentheses table row 11 17 row 13 11 end table close parentheses end cell row x equals cell open parentheses table row 2 3 row 1 4 end table close parentheses to the power of negative 1 end exponent times open parentheses table row 11 17 row 13 11 end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator 2 open parentheses 4 close parentheses minus 3 open parentheses 1 close parentheses end fraction open parentheses table row 4 cell negative 3 end cell row cell negative 1 end cell 2 end table close parentheses open parentheses table row 11 17 row 13 11 end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator 8 minus 3 end fraction open parentheses table row cell 4 open parentheses 11 close parentheses plus open parentheses negative 3 close parentheses open parentheses 13 close parentheses end cell cell 4 open parentheses 17 close parentheses plus open parentheses negative 3 close parentheses open parentheses 11 close parentheses end cell row cell negative 1 open parentheses 11 close parentheses plus 2 open parentheses 13 close parentheses end cell cell negative 1 open parentheses 17 close parentheses plus 2 open parentheses 11 close parentheses end cell end table close parentheses end cell row blank equals cell 1 fifth open parentheses table row cell 44 minus 39 end cell cell 68 minus 33 end cell row cell negative 11 plus 26 end cell cell negative 17 plus 22 end cell end table close parentheses end cell row blank equals cell 1 fifth open parentheses table row 5 35 row 15 5 end table close parentheses end cell row blank equals cell open parentheses table row 1 7 row 3 1 end table close parentheses end cell end table

Matriks x yang memenuhi open parentheses table row 2 3 row 1 4 end table close parentheses x equals open parentheses table row 11 17 row 13 11 end table close parentheses adalah open parentheses table row 1 7 row 3 1 end table close parentheses.

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

Diketahui matriks A=(23​34​), B=(−53​−22​) dan matriks AB=C. Matriks C−1 adalah invers matriks C. Matriks = . . . .

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell A B end cell equals C row cell open parentheses table row 2 3 row 3 4 end table close parentheses open parentheses table row cell negative 5 end cell cell negative 2 end cell row 3 2 end table close parentheses end cell equals cell open parentheses table row cell open parentheses 2 close parentheses open parentheses negative 5 close parentheses plus open parentheses 3 close parentheses open parentheses 3 close parentheses end cell cell open parentheses 2 close parentheses open parentheses negative 2 close parentheses plus open parentheses 3 close parentheses open parentheses 2 close parentheses end cell row cell open parentheses 3 close parentheses open parentheses negative 5 close parentheses plus open parentheses 4 close parentheses open parentheses 3 close parentheses end cell cell open parentheses 3 close parentheses open parentheses negative 2 close parentheses plus open parentheses 4 close parentheses open parentheses 2 close parentheses end cell end table close parentheses end cell row cell open parentheses table row 2 3 row 3 4 end table close parentheses open parentheses table row cell negative 5 end cell cell negative 2 end cell row 3 2 end table close parentheses end cell equals cell open parentheses table row cell negative 1 end cell 2 row cell negative 3 end cell 2 end table close parentheses end cell row blank blank blank end table end style

sehingga, invers matriks begin mathsize 14px style C end style,

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

Jadi, jawaban yang tepat adalah A.

 

7

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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