Roboguru

Diberikan matriks A=(12​23​). Jika matriks (A+kA−1) merupakan matriks singular, tentukan nilai k yang mungkin.

Pertanyaan

Diberikan matriks begin mathsize 14px style A equals open parentheses table row 1 2 row 2 3 end table close parentheses end style.

Jika matriks begin mathsize 14px style left parenthesis A plus k A to the power of negative 1 end exponent right parenthesis end style merupakan matriks singular, tentukan nilai begin mathsize 14px style k end style yang mungkin.

Pembahasan Soal:

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

Matriks singular jika determinan=0 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets open parentheses 1 minus 3 k close parentheses open parentheses 3 minus 1 k close parentheses close square brackets minus open square brackets open parentheses 2 plus 2 k close parentheses open parentheses 2 plus 2 k close parentheses close square brackets end cell equals 0 row cell open square brackets 3 minus 10 k plus 3 k squared close square brackets minus open square brackets 4 plus 8 k plus 4 k squared close square brackets end cell equals 0 row cell negative k squared minus 2 k minus 1 end cell equals 0 row cell k squared plus 2 k plus 1 end cell equals 0 row cell open parentheses k plus 1 close parentheses open parentheses k plus 1 close parentheses end cell equals 0 row k equals cell negative 1 end cell row blank blank blank end table 

 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Septianingsih

Mahasiswa/Alumni Universitas Gadjah Mada

Terakhir diupdate 07 Oktober 2021

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Pertanyaan yang serupa

Diberikan matriks A=(12​23​). Tentukan A+kA−1.

Pembahasan Soal:

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

0

Roboguru

Misal I=(10​01​); A=(42​−11​). Jika matriks (A−αI) adalah matriks singular, maka nilai α= ...

Pembahasan Soal:

Jika nilai determinan suatu matriks persegi equals 0, maka matriks tersebut tidak mempunyai matriks balikan/invers matriks. Dan matriks yang tidak mempunyai invers matriks disebut matriks singular.

Determinan matriks 2 cross times 2 berlaku:

open vertical bar table row a b row c d end table close vertical bar equals a d minus b c

Tentukan terlebih dahulu matriks A minus alpha I seperti berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell A minus alpha I end cell equals cell open parentheses table row 4 cell negative 1 end cell row 2 1 end table close parentheses minus alpha times open parentheses table row 1 0 row 0 1 end table close parentheses end cell row blank equals cell open parentheses table row 4 cell negative 1 end cell row 2 1 end table close parentheses minus open parentheses table row alpha 0 row 0 alpha end table close parentheses end cell row blank equals cell open parentheses table row cell 4 minus alpha end cell cell negative 1 minus 0 end cell row cell 2 minus 0 end cell cell 1 minus alpha end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 4 minus alpha end cell cell negative 1 end cell row 2 cell 1 minus alpha end cell end table close parentheses end cell end table

Kemudian diperoleh nilai alpha dengan menggunakan determinan matriks A minus alpha I sama dengan nol, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar A minus alpha I close vertical bar end cell equals 0 row cell open vertical bar table row cell 4 minus alpha end cell cell negative 1 end cell row 2 cell 1 minus alpha end cell end table close vertical bar end cell equals 0 row cell open parentheses 4 minus alpha close parentheses open parentheses 1 minus alpha close parentheses minus open parentheses negative 1 close parentheses open parentheses 2 close parentheses end cell equals 0 row cell 4 minus 4 alpha minus alpha plus alpha squared plus 2 end cell equals 0 row cell 6 minus 5 alpha plus alpha squared end cell equals 0 row cell 6 minus 3 alpha minus 2 alpha plus alpha squared end cell equals 0 row cell negative 3 open parentheses negative 2 plus alpha close parentheses plus alpha open parentheses negative 2 plus alpha close parentheses end cell equals 0 row cell open parentheses negative 3 plus alpha close parentheses open parentheses negative 2 plus alpha close parentheses end cell equals 0 end table

table row cell negative 3 plus alpha equals 0 end cell atau cell negative 2 plus alpha equals 0 end cell row cell alpha equals 3 end cell blank cell alpha equals 2 end cell end table

Maka nilai alpha equals2 atau 3.

Oleh karena itu, jawaban yang benar adalah C.

0

Roboguru

Diberikan matriks A=(23​41​). Jumlah nilai-nilai m yang memenuhi, agar matriks (A+mI) menjadi matriks singular adalah ...

Pembahasan Soal:

Penjumlahan matriks 2 cross times 2 yaitu:

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

Perkalian skalar dengan matriks 2 cross times 2 yaitu:

k times open parentheses table row a b row c d end table close parentheses equals open parentheses table row cell k times a end cell cell k times b end cell row cell k times c end cell cell k times d end cell end table close parentheses

Penjumlahan matriks A dengan perkalian skalar m terhadap matriks identitas 2 cross times 2, yaitu I equals open parentheses table row 1 0 row 0 1 end table close parentheses.

table attributes columnalign right center left columnspacing 0px end attributes row cell A plus m I end cell equals cell open parentheses table row 2 4 row 3 1 end table close parentheses plus m open parentheses table row 1 0 row 0 1 end table close parentheses end cell row blank equals cell open parentheses table row 2 4 row 3 1 end table close parentheses plus open parentheses table row cell m times 1 end cell cell m times 0 end cell row cell m times 0 end cell cell m times 1 end cell end table close parentheses end cell row blank equals cell open parentheses table row 2 4 row 3 1 end table close parentheses plus open parentheses table row m 0 row 0 m end table close parentheses end cell row blank equals cell open parentheses table row cell 2 plus m end cell cell 4 plus 0 end cell row cell 3 plus 0 end cell cell 1 plus m end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 2 plus m end cell 4 row 3 cell 1 plus m end cell end table close parentheses end cell end table

Matriks open parentheses A plus m I close parentheses menjadi matriks singular, maka nilai determinannya adalah nol open parentheses 0 close parentheses.

Rumus determinan 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 open vertical bar A close vertical bar end cell equals cell a times d minus b times c end cell end table

Sehingga diperoleh penyelesaiannya yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar A plus m I close vertical bar end cell equals 0 row cell open vertical bar table row cell 2 plus m end cell 4 row 3 cell 1 plus m end cell end table close vertical bar end cell equals 0 row cell open parentheses 2 plus m close parentheses open parentheses 1 plus m close parentheses minus 4 open parentheses 3 close parentheses end cell equals 0 row cell 2 plus 2 m plus m plus m squared minus 12 end cell equals 0 row cell m squared plus 3 m minus 10 end cell equals 0 row cell m squared plus 5 m minus 2 m minus 10 end cell equals 0 row cell m open parentheses m plus 5 close parentheses minus 2 open parentheses m plus 5 close parentheses end cell equals 0 row cell open parentheses m minus 2 close parentheses open parentheses m plus 5 close parentheses end cell equals 0 end table

dengan

table attributes columnalign right center left columnspacing 0px end attributes row cell m minus 2 end cell equals 0 row cell m subscript 1 end cell equals 2 end table

atau

table attributes columnalign right center left columnspacing 0px end attributes row cell m plus 5 end cell equals 0 row cell m subscript 2 end cell equals cell negative 5 end cell end table

Jumlah nilai-nilai m yang memenuhi, yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell m subscript 1 plus m subscript 2 end cell equals cell 2 plus open parentheses negative 5 close parentheses end cell row blank equals cell 2 minus 5 end cell row blank equals cell negative 3 end cell end table

Jumlah nilai-nilai m yang memenuhi, agar matriks open parentheses A plus m I close parentheses menjadi matriks singular adalah negative 3.

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Untuk A, B, dan X matriks berordo 2×2. Buktikan bahwa: (A−1)−1=A

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell Misal space straight A end cell equals cell open parentheses table row straight a straight b row straight c straight d end table close parentheses end cell row cell straight A to the power of negative 1 end exponent end cell equals cell fraction numerator 1 over denominator ad minus bc end fraction open parentheses table row straight d cell negative straight b end cell row cell negative straight c end cell straight a end table close parentheses end cell row cell straight A to the power of negative 1 end exponent end cell equals cell open parentheses table row cell fraction numerator straight d over denominator ad minus bc end fraction end cell cell fraction numerator negative straight b over denominator ad minus bc end fraction end cell row cell fraction numerator negative straight c over denominator ad minus bc end fraction end cell cell fraction numerator straight a over denominator ad minus bc end fraction end cell end table close parentheses end cell row cell open parentheses straight A to the power of negative 1 end exponent close parentheses to the power of negative 1 end exponent end cell equals cell fraction numerator 1 over denominator open parentheses fraction numerator straight d over denominator ad minus bc end fraction close parentheses open parentheses fraction numerator straight a over denominator ad minus bc end fraction close parentheses minus open parentheses fraction numerator negative straight c over denominator ad minus bc end fraction close parentheses open parentheses fraction numerator negative straight b over denominator ad minus bc end fraction close parentheses end fraction open parentheses table row cell fraction numerator straight a over denominator ad minus bc end fraction end cell cell fraction numerator straight b over denominator ad minus bc end fraction end cell row cell fraction numerator straight c over denominator ad minus bc end fraction end cell cell fraction numerator straight d over denominator ad minus bc end fraction end cell end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator begin display style ad over open parentheses ad minus bc close parentheses squared end style minus begin display style bc over open parentheses ad minus bc close parentheses squared end style end fraction open parentheses table row cell fraction numerator straight a over denominator ad minus bc end fraction end cell cell fraction numerator straight b over denominator ad minus bc end fraction end cell row cell fraction numerator straight c over denominator ad minus bc end fraction end cell cell fraction numerator straight d over denominator ad minus bc end fraction end cell end table close parentheses end cell row blank equals cell fraction numerator open parentheses ad minus bc close parentheses squared over denominator open parentheses ad minus bc close parentheses end fraction open parentheses table row cell fraction numerator straight a over denominator ad minus bc end fraction end cell cell fraction numerator straight b over denominator ad minus bc end fraction end cell row cell fraction numerator straight c over denominator ad minus bc end fraction end cell cell fraction numerator straight d over denominator ad minus bc end fraction end cell end table close parentheses end cell row blank equals cell fraction numerator open parentheses ad minus bc close parentheses over denominator 1 end fraction open parentheses table row cell fraction numerator straight a over denominator ad minus bc end fraction end cell cell fraction numerator straight b over denominator ad minus bc end fraction end cell row cell fraction numerator straight c over denominator ad minus bc end fraction end cell cell fraction numerator straight d over denominator ad minus bc end fraction end cell end table close parentheses end cell row blank equals cell open parentheses table row straight a straight b row straight c straight d end table close parentheses end cell row cell open parentheses straight A to the power of negative 1 end exponent close parentheses to the power of negative 1 end exponent end cell equals cell straight A space left parenthesis terbukti right parenthesis end cell row blank blank blank end table 

0

Roboguru

Untuk A, B, dan X matriks berordo 2×2. Buktikan bahwa: A(BA)−1⋅B=I

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell Ingat space sifat space open parentheses AB close parentheses to the power of negative 1 end exponent end cell equals cell straight B to the power of negative 1 end exponent. straight A to the power of negative 1 end exponent end cell row cell dan space straight A. straight A to the power of negative 1 end exponent end cell equals straight I end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell A open parentheses B A close parentheses to the power of negative 1 end exponent B end cell equals I row cell A. A to the power of negative 1 end exponent. B to the power of negative 1 end exponent. B end cell equals I row cell I. I end cell equals I row I equals I row blank blank cell left parenthesis Terbukti right parenthesis end cell end table 

0

Roboguru

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