Roboguru

Diketahui persamaan matriks 3(−410​23​)+2(1−3​−4−1​)=(12​x5​)(24​y1​). Nilai 2y - 3x = ....

Pertanyaan

Diketahui persamaan matriks

begin mathsize 14px style 3 open parentheses table row cell negative 4 end cell 2 row 10 3 end table close parentheses plus 2 open parentheses table row 1 cell negative 4 end cell row cell negative 3 end cell cell negative 1 end cell end table close parentheses equals open parentheses table row 1 x row 2 5 end table close parentheses open parentheses table row 2 y row 4 1 end table close parentheses. end style

Nilai 2y - 3x = ....

  1. -9

  2. -7

  3. -4

  4. 8

  5. 11

Pembahasan Soal:

begin mathsize 14px style 3 open parentheses table row cell negative 4 end cell 2 row 10 3 end table close parentheses plus 2 open parentheses table row 1 cell negative 4 end cell row cell negative 3 end cell cell negative 1 end cell end table close parentheses equals open parentheses table row 1 x row 2 5 end table close parentheses open parentheses table row 2 y row 4 1 end table close parentheses  open parentheses table row cell negative 12 end cell 6 row 30 9 end table close parentheses plus open parentheses table row 2 cell negative 8 end cell row cell negative 6 end cell cell negative 2 end cell end table close parentheses equals open parentheses table row cell 2 plus 4 x end cell cell y plus x end cell row 24 cell 2 y plus 5 end cell end table close parentheses  open parentheses table row cell negative 10 end cell cell negative 2 end cell row 24 7 end table close parentheses equals open parentheses table row cell 2 plus 4 x end cell cell y plus x end cell row 24 cell 2 y plus 5 end cell end table close parentheses  minus 10 equals 2 plus 4 x  minus 12 equals 4 x  x equals negative 3    7 equals 2 y plus 5  2 equals 2 y  y equals 1    2 y minus 3 x equals 2 left parenthesis 1 right parenthesis minus 3 left parenthesis negative 3 right parenthesis equals 2 plus 9 equals 11 end style

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Intan

Mahasiswa/Alumni Institut Pertanian Bogor

Terakhir diupdate 04 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui persamaan matriks 2(a−3​21​)+(40​−1b​)=(3c​24​)(21​d3​)  Nilai dari a+b+c+d adalah ....

Pembahasan Soal:

Penjumlahan matriks pada ruas kiri persamaan matriks di atas dapat dihitung seperti berikut,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 2 open parentheses table row a 2 row cell negative 3 end cell 1 end table close parentheses plus open parentheses table row 4 cell negative 1 end cell row 0 b end table close parentheses end cell row blank equals cell open parentheses table row cell 2 times a end cell cell 2 times 2 end cell row cell 2 times left parenthesis negative 3 right parenthesis end cell cell 2 times 1 end cell end table close parentheses plus open parentheses table row 4 cell negative 1 end cell row 0 b end table close parentheses end cell row blank equals cell open parentheses table row cell 2 a end cell 4 row cell negative 6 end cell 2 end table close parentheses plus open parentheses table row 4 cell negative 1 end cell row 0 b end table close parentheses end cell row blank equals cell open parentheses table row cell 2 a plus 4 end cell cell 4 minus 1 end cell row cell negative 6 plus 0 end cell cell 2 plus b end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 2 a plus 4 end cell 3 row cell negative 6 end cell cell 2 plus b end cell end table close parentheses end cell end table end style  

Penyelesaian perkalian matriks pada ruas kanan yaitu,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row 3 2 row c 4 end table close parentheses open parentheses table row 2 d row 1 3 end table close parentheses end cell equals cell open parentheses table row cell 3 times 2 plus 2 times 1 end cell cell 3 times d plus 2 times 3 end cell row cell c times 2 plus 4 times 1 end cell cell c times d plus 4 times 3 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 6 plus 2 end cell cell 3 d plus 6 end cell row cell 2 c plus 4 end cell cell c d plus 12 end cell end table close parentheses end cell row blank equals cell open parentheses table row 8 cell 3 d plus 6 end cell row cell 2 c plus 4 end cell cell c d plus 12 end cell end table close parentheses end cell end table end style 

Diketahui hasil dari penjumlahan matriks di ruas kiri sama dengan perkalian matriks di ruas kanan, maka

begin mathsize 14px style open parentheses table row cell 2 a plus 4 end cell 3 row cell negative 6 end cell cell 2 plus b end cell end table close parentheses equals open parentheses table row 8 cell 3 d plus 6 end cell row cell 2 c plus 4 end cell cell c d plus 12 end cell end table close parentheses end style.

Apabila terdapat dua matriks yang sama, maka anggota-anggota dengan letak yang sama nilainya akan sama pula, sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 a plus 4 end cell equals 8 row cell 2 a end cell equals cell 8 minus 4 end cell row cell 2 a end cell equals 4 row a equals 2 end table end style     begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 3 d plus 6 end cell equals 3 row cell 3 d end cell equals cell 3 minus 6 end cell row cell 3 d end cell equals cell negative 3 end cell row d equals cell negative 1 end cell end table end style      begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 c plus 4 end cell equals cell negative 6 end cell row cell 2 c end cell equals cell negative 6 minus 4 end cell row cell 2 c end cell equals cell negative 10 end cell row c equals cell negative 5 end cell end table end style 

Didapatkan nilai begin mathsize 14px style a equals 2 comma space c equals negative 5 comma space dan space d equals negative 1 end style, maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 plus b end cell equals cell c d plus 12 end cell row cell 2 plus b end cell equals cell left parenthesis negative 5 right parenthesis times left parenthesis negative 1 right parenthesis plus 12 end cell row cell 2 plus b end cell equals cell 5 plus 12 end cell row cell 2 plus b end cell equals 17 row b equals cell 17 minus 2 end cell row b equals 15 end table end style 

Penjumlahan dari keempat variabel tersebut adalah

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a plus b plus c plus d end cell row blank equals cell 2 plus 15 plus left parenthesis negative 5 right parenthesis plus left parenthesis negative 1 right parenthesis end cell row blank equals 23 end table end style 

Jadi, nilai dari begin mathsize 14px style a plus b plus c plus d end style adalah begin mathsize 14px style 23 end style.

1

Roboguru

Jika diketahui A=(−1−1​a3​),B=(2−1​4−2​),C=(−61​−3b​),D=(2−6​−5−4​) dan AB=C+D. nilai a - b=···

Pembahasan Soal:

begin mathsize 14px style straight A equals open parentheses table row cell negative 1 end cell straight a row cell negative 1 end cell 3 end table close parentheses comma space straight B equals open parentheses table row 2 4 row cell negative 1 end cell cell negative 2 end cell end table close parentheses comma space straight C equals open parentheses table row cell negative 6 end cell cell negative 3 end cell row 1 straight b end table close parentheses comma space straight D equals open parentheses table row 2 cell negative 5 end cell row cell negative 6 end cell cell negative 4 end cell end table close parentheses  asterisk times AB equals open parentheses table row cell negative 1 end cell straight a row cell negative 1 end cell 3 end table close parentheses open parentheses table row 2 4 row cell negative 1 end cell cell negative 2 end cell end table close parentheses  space space space space space space equals open parentheses table row cell negative 2 minus straight a end cell cell negative 4 minus 2 straight a end cell row cell negative 2 minus 3 end cell cell negative 4 minus 6 end cell end table close parentheses  space space space space space space equals open parentheses table row cell negative 2 minus straight a end cell cell negative 4 minus 2 straight a end cell row cell negative 5 end cell cell negative 10 end cell end table close parentheses  asterisk times straight C plus straight D equals open parentheses table row cell negative 6 end cell cell negative 3 end cell row 1 straight b end table close parentheses plus open parentheses table row 2 cell negative 5 end cell row cell negative 6 end cell cell negative 4 end cell end table close parentheses  space space space space space space space space space space equals open parentheses table row cell negative 4 end cell cell negative 8 end cell row cell negative 5 end cell cell negative 4 plus straight b end cell end table close parentheses  asterisk times AB equals straight C plus straight D  space open parentheses table row cell negative 2 minus straight a end cell cell negative 4 minus 2 straight a end cell row cell negative 5 end cell cell negative 10 end cell end table close parentheses equals open parentheses table row cell negative 4 end cell cell negative 8 end cell row cell negative 5 end cell cell negative 4 plus straight b end cell end table close parentheses  asterisk times minus space 2 minus space straight a space equals negative 4 space  space space space space space space space space space minus space straight a equals negative 2 space  space space space space space space space space space space space space space straight a space equals 2  asterisk times minus 4 plus straight b equals negative 10  space space space space space space space space space space space straight b equals negative 6  asterisk times straight a minus straight b space equals 2 minus left parenthesis negative 6 right parenthesis  space space space space space space space space space space equals 2 plus 6 equals 8 end style

0

Roboguru

Diketahui matriks A=[32​21​], B=[10​21​] dan A=[12​31​]. Hasil dari 2A−3B+C adalah...

Pembahasan Soal:

Penjumlahan dan pengurangan pada matriks,

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 A minus 3 B plus C end cell equals cell 2 open square brackets table row 3 2 row 2 1 end table close square brackets minus 3 open square brackets table row 1 2 row 0 1 end table close square brackets plus open square brackets table row 1 3 row 2 1 end table close square brackets end cell row blank equals cell open square brackets table row 6 4 row 4 2 end table close square brackets minus open square brackets table row 3 6 row 0 3 end table close square brackets plus open square brackets table row 1 3 row 2 1 end table close square brackets end cell row blank equals cell open square brackets table row cell 6 minus 3 plus 1 end cell cell 4 minus 6 plus 3 end cell row cell 4 minus 0 plus 2 end cell cell 2 minus 3 plus 1 end cell end table close square brackets end cell row blank equals cell open square brackets table row 4 1 row 6 0 end table close square brackets end cell end table

Jadi, hasil dari 2 A minus 3 B plus C adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open square brackets table row 4 1 row 6 0 end table close square brackets end cell end table.

0

Roboguru

Nilai a+b+c yang memenuhi persamaan matriks (1−2​23​)(c3c​a2a​)=(8a16b​49c​)−(a2b​−65c​) adalah ....

Pembahasan Soal:

Pengurangan pada matriks merupakan pengurangan elemen-elemen matriks yang seletak, dan dengan menggunakan aturan perkalian pada matriks, didapatkan

(1223)(c3ca2a)(c+6c2c+9ca+4a2a+6a)(7c7c5a4a)5aa4ac14b2bb===========(8a16b49c)(a2b65c)(7a14b104c)(7a14b104c)1024ca27cc1

Dengan demikian, nilai dari a+b+c=5.

0

Roboguru

Diketahui persamaan matriks sebagai berikut. (a−1​4c​)+(2d​b−3​)=(13​−34​)(01​10​) maka nilai a+b+c+d=...

Pembahasan Soal:

Ingatlah operasi perkalian dan invers matriks berikut:

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

Berdasarkan operasi di atas, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row a 4 row cell negative 1 end cell c end table close parentheses plus open parentheses table row 2 b row d cell negative 3 end cell end table close parentheses end cell equals cell open parentheses table row 1 cell negative 3 end cell row 3 4 end table close parentheses open parentheses table row 0 1 row 1 0 end table close parentheses end cell row cell open parentheses table row cell a plus 2 end cell cell 4 plus b end cell row cell negative 1 plus d end cell cell c minus 3 end cell end table close parentheses end cell equals cell open parentheses table row cell negative 3 end cell 1 row 4 3 end table close parentheses end cell end table

Sehingga diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row blank rightwards double arrow cell a plus 2 equals negative 3 end cell row a equals cell negative 5 end cell row blank rightwards double arrow cell 4 plus b equals 1 end cell row b equals cell negative 3 end cell row blank rightwards double arrow cell c minus 3 equals 3 end cell row c equals 6 row blank rightwards double arrow cell negative 1 plus d equals 4 end cell row d equals 5 end table

dengan demikian, nilai 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 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 plus 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 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 equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 5 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 3 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 6 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 5 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 3 end table.

Jadi, jawaban yang tepat adalah D.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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