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Minor dari matriks  adalah ...

Pertanyaan

Minor dari matriks space B equals open vertical bar table row 1 3 5 row 2 4 6 row 5 3 1 end table close vertical bar adalah ...

Pembahasan Soal:

Minor suatu matriks B dilambangkan dengan M subscript i j end subscript adalah determinan matriks bagian dari matriks B yang diperoleh dengan cara menghilangkan elemen - elemennya pada baris ke-i dan elemen elemen pada kolom ke-j. Pada Matriks B terdapat 9 minor, diantaranya:

M subscript 11 equals open vertical bar table row 4 6 row 3 1 end table close vertical bar equals open parentheses 4 minus 18 close parentheses equals negative 14 M subscript 12 equals open vertical bar table row 2 6 row 5 1 end table close vertical bar equals open parentheses 2 minus 30 close parentheses equals negative 28 M subscript 13 equals open vertical bar table row 2 4 row 5 3 end table close vertical bar equals open parentheses 6 minus 20 close parentheses equals negative 14 M subscript 21 equals open vertical bar table row 3 5 row 3 1 end table close vertical bar equals open parentheses 3 minus 15 close parentheses equals negative 12 M subscript 22 equals open vertical bar table row 1 5 row 5 1 end table close vertical bar equals open parentheses 1 minus 25 close parentheses equals negative 24 M subscript 23 equals open vertical bar table row 1 3 row 5 3 end table close vertical bar equals open parentheses 3 minus 15 close parentheses equals negative 12 M subscript 31 equals open vertical bar table row 3 5 row 4 6 end table close vertical bar equals open parentheses 18 minus 20 close parentheses equals negative 2 M subscript 32 equals open vertical bar table row 1 5 row 2 6 end table close vertical bar equals open parentheses 6 minus 10 close parentheses equals negative 4 M subscript 33 equals open vertical bar table row 1 3 row 2 4 end table close vertical bar equals open parentheses 4 minus 6 close parentheses equals negative 2 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Sutiawan

Mahasiswa/Alumni Universitas Pasundan

Terakhir diupdate 05 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui , elemen baris ke-3 kolom 1 adalah

Pembahasan Soal:

Diketahui matriks M equals open parentheses table row 4 3 5 row 2 cell negative 1 end cell 6 row 1 1 cell negative 2 end cell end table close parentheses. pada matriks M. Baris ke 3 elemennya diantaranya 1 space space 1 space space minus 2, kemudian dari ketiga elemen tadi baris 3 yang merupakan kolom ke 1 adalah 1, sebagai berikut:

M equals open parentheses table row 4 3 5 row 2 cell negative 1 end cell 6 row cell open square brackets 1 close square brackets end cell 1 cell negative 2 end cell end table close parentheses 

Jadi elemen baris ke-3 kolom ke-1 pada matriks tersebut adalah 1

Roboguru

Diberikan matriks  Jika  maka nilai  adalah ....

Pembahasan Soal:

Kita cari masing-masing nilai begin mathsize 14px style C subscript 11 comma C subscript 22 comma C subscript 33 end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript 11 end cell equals cell open vertical bar table row cell negative 2 end cell b row 0 c end table close vertical bar equals left parenthesis negative 2 right parenthesis times c minus b times 0 equals negative 2 c minus 0 equals negative 2 c end cell row cell C subscript 11 end cell equals cell left parenthesis negative 1 right parenthesis to the power of 1 plus 1 end exponent times M subscript 11 end subscript equals left parenthesis negative 1 right parenthesis squared times left parenthesis negative 2 c right parenthesis equals 1 times left parenthesis negative 2 c right parenthesis equals negative 2 c end cell row cell M subscript 22 end cell equals cell open vertical bar table row 1 a row 1 c end table close vertical bar equals 1 times c minus a times 1 equals c minus a end cell row cell C subscript 22 end cell equals cell left parenthesis negative 1 right parenthesis to the power of 2 plus 2 end exponent times M subscript 22 end subscript equals left parenthesis negative 1 right parenthesis to the power of 4 times left parenthesis c minus a right parenthesis equals 1 times left parenthesis c minus a right parenthesis equals c minus a end cell row cell M subscript 33 end cell equals cell open vertical bar table row 1 0 row 0 cell negative 2 end cell end table close vertical bar equals 1 times left parenthesis negative 2 right parenthesis minus 0 times 0 equals negative 2 minus 0 equals negative 2 end cell row cell C subscript 33 end cell equals cell left parenthesis negative 1 right parenthesis to the power of 3 plus 3 end exponent times M subscript 33 end subscript equals left parenthesis negative 1 right parenthesis to the power of 6 times open parentheses negative 2 close parentheses equals 1 times open parentheses negative 2 close parentheses equals negative 2 end cell end table end style

Diketahui Error converting from MathML to accessible text.. Maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell C blank subscript 11 minus C blank subscript 22 plus C blank subscript 33 end cell equals 0 row cell negative 2 c minus left parenthesis c minus a right parenthesis plus left parenthesis negative 2 right parenthesis end cell equals 0 row cell negative 2 c plus a minus c minus 2 end cell equals 0 row cell a minus 3 c end cell equals 2 row a equals cell 2 plus 3 c end cell end table end style 

Jadi, jawabannya adalah C.

Roboguru

Invers dari matriks  adalah ....

Pembahasan Soal:

Diketahui undefined. Kita cari dulu masing-masing nilai minornya lalu kita tentukan kofaktornya.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight M subscript 11 end cell equals cell open vertical bar table row 2 cell negative 1 end cell row 3 5 end table close vertical bar equals 2 times 5 minus left parenthesis negative 1 right parenthesis times 3 equals 10 plus 3 equals 13 end cell row cell straight C subscript 11 end cell equals cell left parenthesis negative 1 right parenthesis to the power of 1 plus 1 end exponent times straight M subscript 11 end subscript equals left parenthesis negative 1 right parenthesis squared times 13 equals 1 times 13 equals 13 end cell row blank blank blank row cell straight M subscript 12 end cell equals cell open vertical bar table row 2 cell negative 1 end cell row cell negative 1 end cell 5 end table close vertical bar equals 2 times 5 minus left parenthesis negative 1 right parenthesis times left parenthesis negative 1 right parenthesis equals 10 minus 1 equals 9 end cell row cell straight C subscript 12 end cell equals cell left parenthesis negative 1 right parenthesis to the power of 1 plus 2 end exponent times straight M subscript 12 end subscript equals left parenthesis negative 1 right parenthesis cubed times 9 equals left parenthesis negative 1 right parenthesis times 9 equals negative 9 end cell row blank blank blank row cell straight M subscript 13 end cell equals cell open vertical bar table row 2 2 row cell negative 1 end cell 3 end table close vertical bar equals 2 times 3 minus 2 times left parenthesis negative 1 right parenthesis equals 6 plus 2 equals 8 end cell row cell straight C subscript 13 end cell equals cell left parenthesis negative 1 right parenthesis to the power of 1 plus 3 end exponent times straight M subscript 13 end subscript equals left parenthesis negative 1 right parenthesis to the power of 4 times 8 equals 1 times 8 equals 8 end cell row blank blank blank row cell straight M subscript 21 end cell equals cell open vertical bar table row 0 3 row 3 5 end table close vertical bar equals 0 times 5 minus 3 times 3 equals 0 minus 9 equals negative 9 end cell row cell straight C subscript 21 end cell equals cell left parenthesis negative 1 right parenthesis to the power of 2 plus 1 end exponent times straight M subscript 21 end subscript equals left parenthesis negative 1 right parenthesis cubed times left parenthesis negative 9 right parenthesis equals left parenthesis negative 1 right parenthesis times left parenthesis negative 9 right parenthesis equals 9 end cell row blank blank blank row cell straight M subscript 22 end cell equals cell open vertical bar table row 1 3 row cell negative 1 end cell 5 end table close vertical bar equals 1 times 5 minus 3 times left parenthesis negative 1 right parenthesis equals 5 plus 3 equals 8 end cell row cell straight C subscript 22 end cell equals cell left parenthesis negative 1 right parenthesis to the power of 2 plus 2 end exponent times straight M subscript 22 end subscript equals left parenthesis negative 1 right parenthesis to the power of 4 times 8 equals 1 times 8 equals 8 end cell row blank blank blank row cell straight M subscript 23 end cell equals cell open vertical bar table row 1 0 row cell negative 1 end cell 3 end table close vertical bar equals 1 times 3 minus 0 times left parenthesis negative 1 right parenthesis equals 3 plus 0 equals 3 end cell row cell straight C subscript 23 end cell equals cell left parenthesis negative 1 right parenthesis to the power of 2 plus 3 end exponent times straight M subscript 23 end subscript equals left parenthesis negative 1 right parenthesis to the power of 5 times 3 equals left parenthesis negative 1 right parenthesis times 3 equals negative 3 end cell row blank blank blank row cell straight M subscript 31 end cell equals cell open vertical bar table row 0 3 row 2 cell negative 1 end cell end table close vertical bar equals 0 times left parenthesis negative 1 right parenthesis minus 3 times 2 equals 0 minus 6 equals negative 6 end cell row cell straight C subscript 31 end cell equals cell left parenthesis negative 1 right parenthesis to the power of 3 plus 1 end exponent times straight M subscript 31 end subscript equals left parenthesis negative 1 right parenthesis to the power of 4 times left parenthesis negative 6 right parenthesis equals 1 times left parenthesis negative 6 right parenthesis equals negative 6 end cell row blank blank blank row cell straight M subscript 32 end cell equals cell open vertical bar table row 1 3 row 2 cell negative 1 end cell end table close vertical bar equals 1 times left parenthesis negative 1 right parenthesis minus 3 times 2 equals negative 1 minus 6 equals negative 7 end cell row cell straight C subscript 32 end cell equals cell left parenthesis negative 1 right parenthesis to the power of 3 plus 2 end exponent times straight M subscript 32 end subscript equals left parenthesis negative 1 right parenthesis to the power of 5 times left parenthesis negative 7 right parenthesis equals left parenthesis negative 1 right parenthesis times left parenthesis negative 7 right parenthesis equals 7 end cell row blank blank blank row cell straight M subscript 33 end cell equals cell open vertical bar table row 1 0 row 2 2 end table close vertical bar equals 1 times 2 minus 0 times 2 equals 2 minus 0 equals 2 end cell row cell straight C subscript 33 end cell equals cell left parenthesis negative 1 right parenthesis to the power of 3 plus 3 end exponent times straight M subscript 33 end subscript equals left parenthesis negative 1 right parenthesis to the power of 6 times 2 equals 1 times 2 equals 2 end cell end table end style  

Sehingga kita peroleh matriks kofaktornya adalah

begin mathsize 14px style open parentheses table row 13 cell negative 9 end cell 8 row 9 8 cell negative 3 end cell row cell negative 6 end cell 7 2 end table close parentheses end style  

Jadi, adjoinnya adalah

begin mathsize 14px style open parentheses table row 13 cell negative 9 end cell 8 row 9 8 cell negative 3 end cell row cell negative 6 end cell 7 2 end table close parentheses to the power of T equals open parentheses table row 13 9 cell negative 6 end cell row cell negative 9 end cell 8 7 row 8 cell negative 3 end cell 2 end table close parentheses end style  

Selanjutnya, kita cari determinan matriks D  dengan metode Sarrus.

Kita peroleh

diagonal kanan = 1 ∙ 2 ∙ 5 + 0 ∙ (-1) ∙ (-1) + 3 ∙ 2 ∙ 3
                         = 10 + 0 + 18
                         = 28

diagonal kiri = 3 ∙ 2 ∙ (-1) + 1 ∙ (-1) ∙ 3 + 0 ∙ 2 ∙ 5
                    = - 6 - 3 + 0
                    = - 9

Sehingga determinannya adalah

determinan = diagonal kanan - diagonal kiri
                   = 28 - (-9)
                   = 37

Jadi, invers dari matriks D  adalah

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight D to the power of negative 1 end exponent end cell equals cell fraction numerator 1 over denominator vertical line straight D vertical line end fraction times adj blank straight D end cell row blank equals cell 1 over 37 open parentheses table row 13 9 cell negative 6 end cell row cell negative 9 end cell 8 7 row 8 cell negative 3 end cell 2 end table close parentheses end cell end table end style  

Dengan demikian, jawaban yang tepat adalah B.

Roboguru

Invers dari matriks  adalah ....

Pembahasan Soal:

Diketahui undefined.

Terlebih dahulu, cari masing-masing nilai minornya, lalu tentukan kofaktornya.

Hilangkan elemen pada baris 1 dan kolom 1 sehingga didapat hasil sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript 11 end cell equals cell open vertical bar table row 1 cell negative 4 end cell row 2 cell negative 1 end cell end table close vertical bar end cell row blank equals cell 1 times open parentheses negative 1 close parentheses minus open parentheses negative 4 close parentheses times 2 end cell row blank equals cell negative 1 plus 8 end cell row blank equals 7 row cell C subscript 11 end cell equals cell open parentheses negative 1 close parentheses to the power of 1 plus 1 end exponent times M subscript 11 end cell row blank equals cell open parentheses negative 1 close parentheses squared times 7 end cell row blank equals cell 1 times 7 end cell row blank equals 7 end table

Hilangkan elemen pada baris 1 dan kolom 2 sehingga didapat hasil sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript 12 end cell equals cell open vertical bar table row 0 cell negative 4 end cell row 0 cell negative 1 end cell end table close vertical bar end cell row blank equals cell 0 times open parentheses negative 1 close parentheses minus open parentheses negative 4 close parentheses times 0 end cell row blank equals cell 0 minus 0 end cell row blank equals 0 row cell C subscript 12 end cell equals cell open parentheses negative 1 close parentheses to the power of 1 plus 2 end exponent times M subscript 12 end cell row blank equals cell open parentheses negative 1 close parentheses cubed times 0 end cell row blank equals cell open parentheses negative 1 close parentheses times 0 end cell row blank equals 0 end table

Hilangkan elemen pada baris 1 dan kolom 3 sehingga didapat hasil sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript 13 end cell equals cell open vertical bar table row 0 1 row 0 2 end table close vertical bar end cell row blank equals cell 0 times 2 minus 1 times 0 end cell row blank equals cell 0 plus 0 end cell row blank equals 0 row cell C subscript 13 end cell equals cell open parentheses negative 1 close parentheses to the power of 1 plus 3 end exponent times M subscript 13 end cell row blank equals cell open parentheses negative 1 close parentheses to the power of 4 times 0 end cell row blank equals cell 1 times 0 end cell row blank equals 0 end table

Hilangkan elemen pada baris 2 dan kolom 1 sehingga didapat hasil sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript 21 end cell equals cell open vertical bar table row 0 cell negative 2 end cell row 2 cell negative 1 end cell end table close vertical bar end cell row blank equals cell 0 times open parentheses negative 1 close parentheses minus open parentheses negative 2 close parentheses times 2 end cell row blank equals cell 0 plus 4 end cell row blank equals 4 row cell C subscript 21 end cell equals cell open parentheses negative 1 close parentheses to the power of 2 plus 1 end exponent times M subscript 21 end cell row blank equals cell open parentheses negative 1 close parentheses cubed times 4 end cell row blank equals cell open parentheses negative 1 close parentheses times 4 end cell row blank equals cell negative 4 end cell end table

Hilangkan elemen pada baris 2 dan kolom 2 sehingga didapat hasil sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript 22 end cell equals cell open vertical bar table row 1 cell negative 2 end cell row 0 cell negative 1 end cell end table close vertical bar end cell row blank equals cell 1 times open parentheses negative 1 close parentheses minus open parentheses negative 2 close parentheses times 0 end cell row blank equals cell negative 1 minus 0 end cell row blank equals cell negative 1 end cell row cell C subscript 22 end cell equals cell open parentheses negative 1 close parentheses to the power of 2 plus 2 end exponent times M subscript 22 end cell row blank equals cell open parentheses negative 1 close parentheses to the power of 4 times open parentheses negative 1 close parentheses end cell row blank equals cell 1 times open parentheses negative 1 close parentheses end cell row blank equals cell negative 1 end cell end table

Hilangkan elemen pada baris 2 dan kolom 3 sehingga didapat hasil sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript 23 end cell equals cell open vertical bar table row 1 0 row 0 2 end table close vertical bar end cell row blank equals cell 1 times 2 minus 0 times 0 end cell row blank equals cell 2 minus 0 end cell row blank equals 2 row cell C subscript 23 end cell equals cell open parentheses negative 1 close parentheses to the power of 2 plus 3 end exponent times M subscript 23 end cell row blank equals cell open parentheses negative 1 close parentheses to the power of 5 times 2 end cell row blank equals cell open parentheses negative 1 close parentheses times 2 end cell row blank equals cell negative 2 end cell end table

Hilangkan elemen pada baris 3 dan kolom 1 sehingga didapat hasil sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript 31 end cell equals cell open vertical bar table row 0 cell negative 2 end cell row 1 cell negative 4 end cell end table close vertical bar end cell row blank equals cell 0 times open parentheses negative 4 close parentheses minus open parentheses negative 2 close parentheses times 1 end cell row blank equals cell 0 plus 2 end cell row blank equals 2 row cell C subscript 31 end cell equals cell open parentheses negative 1 close parentheses to the power of 3 plus 1 end exponent times M subscript 31 end cell row blank equals cell open parentheses negative 1 close parentheses to the power of 4 times 2 end cell row blank equals cell 1 times 2 end cell row blank equals 2 end table

Hilangkan elemen pada baris 3 dan kolom 2 sehingga didapat hasil sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript 32 end cell equals cell open vertical bar table row 1 cell negative 2 end cell row 0 cell negative 4 end cell end table close vertical bar end cell row blank equals cell 1 times open parentheses negative 4 close parentheses minus open parentheses negative 2 close parentheses times 0 end cell row blank equals cell negative 4 minus 0 end cell row blank equals cell negative 4 end cell row cell C subscript 32 end cell equals cell open parentheses negative 1 close parentheses to the power of 3 plus 2 end exponent times M subscript 32 end cell row blank equals cell open parentheses negative 1 close parentheses to the power of 5 times open parentheses negative 4 close parentheses end cell row blank equals cell open parentheses negative 1 close parentheses times open parentheses negative 4 close parentheses end cell row blank equals 4 end table

Hilangkan elemen pada baris 3 dan kolom 3 sehingga didapat hasil sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript 33 end cell equals cell open vertical bar table row 1 0 row 0 1 end table close vertical bar end cell row blank equals cell 1 times 1 minus 0 times 0 end cell row blank equals cell 1 minus 0 end cell row blank equals 1 row cell C subscript 33 end cell equals cell open parentheses negative 1 close parentheses to the power of 3 plus 3 end exponent times M subscript 33 end cell row blank equals cell open parentheses negative 1 close parentheses to the power of 6 times 1 end cell row blank equals cell 1 times 1 end cell row blank equals 1 end table

Oleh karena itu, diperoleh matriks kofaktornya sebagai berikut.

begin mathsize 14px style open parentheses table row 7 0 0 row cell negative 4 end cell cell negative 1 end cell cell negative 2 end cell row 2 4 1 end table close parentheses end style

Dengan demikian, adjoinnya adalah sebagai berikut.

begin mathsize 14px style open parentheses table row 7 0 0 row cell negative 4 end cell cell negative 1 end cell cell negative 2 end cell row 2 4 1 end table close parentheses to the power of T equals open parentheses table row 7 cell negative 4 end cell 2 row 0 cell negative 1 end cell 4 row 0 cell negative 2 end cell 1 end table close parentheses end style

Selanjutnya, cari determinan matriks A dengan metode Sarrus.

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar A close vertical bar end cell equals cell open vertical bar table row 1 0 cell negative 2 end cell 1 0 row 0 1 cell negative 4 end cell 0 1 row 0 2 cell negative 1 end cell 0 2 end table close vertical bar end cell row blank equals cell open parentheses open parentheses 1 close parentheses open parentheses 1 close parentheses open parentheses negative 1 close parentheses plus open parentheses 0 close parentheses open parentheses negative 4 close parentheses open parentheses 0 close parentheses plus open parentheses negative 2 close parentheses open parentheses 0 close parentheses open parentheses 2 close parentheses close parentheses minus end cell row blank blank cell open parentheses open parentheses negative 2 close parentheses open parentheses 1 close parentheses open parentheses 0 close parentheses plus open parentheses 1 close parentheses open parentheses negative 4 close parentheses open parentheses 2 close parentheses plus open parentheses 0 close parentheses open parentheses 0 close parentheses open parentheses negative 1 close parentheses close parentheses end cell row blank equals cell open parentheses negative 1 plus 0 plus 0 close parentheses minus open parentheses 0 minus 8 plus 0 close parentheses end cell row blank equals cell negative 1 plus 8 end cell row blank equals 7 end table

Dengan demikian, invers dari matriks A adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell A to the power of negative 1 end exponent end cell equals cell fraction numerator 1 over denominator vertical line A vertical line end fraction times space Adjoin blank A end cell row blank equals cell 1 over 7 open parentheses table row 7 cell negative 4 end cell 2 row 0 cell negative 1 end cell 4 row 0 cell negative 2 end cell 1 end table close parentheses end cell end table

Jadi, jawaban yang tepat adalah B.

Roboguru

Diketahui matriks  nilai dari  adalah ....

Pembahasan Soal:

open vertical bar M subscript m n end subscript close vertical bar adalah determinan dari matriks seterlah baris ke-m dan kolom ke-n dihapus.

table attributes columnalign right center left columnspacing 0px end attributes row A equals cell open vertical bar table row 2 3 cell negative 1 end cell row 2 4 2 row cell negative 1 end cell cell negative 1 end cell 3 end table close vertical bar end cell row cell M subscript 13 end cell equals cell open vertical bar table row 2 4 row cell negative 1 end cell cell negative 1 end cell end table close vertical bar equals negative 2 minus open parentheses negative 4 close parentheses equals 2 end cell end table   

Jadi, nilai dari open vertical bar M subscript 13 close vertical bar adalah 2.

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Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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