Roboguru

Diberikan matriks A=(11​−24​) dan P=(1−1​21​). Tentukan: a. X=P−1⋅A⋅P  b. P−1⋅Xn⋅P

Pertanyaan

Diberikan matriks begin mathsize 14px style A equals open parentheses table row 1 cell negative 2 end cell row 1 4 end table close parentheses end style dan begin mathsize 14px style P equals open parentheses table row 1 2 row cell negative 1 end cell 1 end table close parentheses end style.

Tentukan:

a. undefined 

b. begin mathsize 14px style P to the power of negative 1 end exponent times X to the power of n times P end style 

Pembahasan Soal:

a. 

table attributes columnalign right center left columnspacing 0px end attributes row X equals cell fraction numerator 1 over denominator 1 plus 2 end fraction open parentheses table row 1 cell negative 2 end cell row 1 1 end table close parentheses open parentheses table row 1 cell negative 2 end cell row 1 4 end table close parentheses open parentheses table row 1 2 row cell negative 1 end cell 1 end table close parentheses end cell row X equals cell 1 third open parentheses table row cell negative 1 end cell cell negative 10 end cell row 2 2 end table close parentheses open parentheses table row 1 2 row cell negative 1 end cell 1 end table close parentheses end cell row X equals cell 1 third open parentheses table row 9 cell negative 12 end cell row 0 6 end table close parentheses end cell row X equals cell open parentheses table row 3 cell negative 4 end cell row 0 2 end table close parentheses end cell end table 

b. 

P to the power of negative 1 end exponent. X to the power of n. P equals 1 third open parentheses table row cell negative 1 end cell cell negative 10 end cell row 2 2 end table close parentheses. open parentheses table row 3 cell negative 4 end cell row 0 2 end table close parentheses to the power of n. open parentheses table row 1 2 row cell negative 1 end cell 1 end table close parentheses 

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

Dengan menggunakan konsep OBE, tentukan nilai (x+y+z) dari persamaan matriks berikut: ⎝⎛​213​−10−1​114​⎠⎞​⎝⎛​xyz​⎠⎞​=⎝⎛​300​⎠⎞​

Pembahasan Soal:

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Roboguru

Diketahui matriks A=[24​34​] dan B=[4−4​36​]. Tentukan AB−1!

Pembahasan Soal:

  • Menentukan B to the power of negative 1 end exponent:

B to the power of negative 1 end exponent equals fraction numerator 1 over denominator d e t left parenthesis B right parenthesis end fraction open parentheses table row 6 cell negative 3 end cell row 4 4 end table close parentheses B to the power of negative 1 end exponent equals fraction numerator 1 over denominator left parenthesis 4 right parenthesis left parenthesis 6 right parenthesis minus left parenthesis 3 right parenthesis left parenthesis negative 4 right parenthesis end fraction open parentheses table row 6 cell negative 3 end cell row 4 4 end table close parentheses B to the power of negative 1 end exponent equals 1 over 36 open parentheses table row 6 cell negative 3 end cell row 4 4 end table close parentheses B to the power of negative 1 end exponent equals open parentheses table row cell 6 over 36 end cell cell fraction numerator negative 3 over denominator 36 end fraction end cell row cell 4 over 36 end cell cell 4 over 36 end cell end table close parentheses B to the power of negative 1 end exponent equals open parentheses table row cell 1 over 6 end cell cell fraction numerator negative 1 over denominator 12 end fraction end cell row cell 1 over 9 end cell cell 1 over 9 end cell end table close parentheses

  • Menentukan A B to the power of negative 1 end exponent:

AB1=[2434][619112191]AB1=[62+9364+94122+93124+94]AB1=[329106191]

Dengan demikian, AB1=[329106191].

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Untuk A, B, dan X matriks berordo 2×2. Buktikan bahwa: (kA)−1=k1​⋅A−1, k bilangan bulat positif

Pembahasan Soal:

Misal space straight A equals open parentheses table row straight a straight b row straight c straight d end table close parentheses 

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses kA close parentheses to the power of negative 1 end exponent end cell equals cell 1 over straight k straight A to the power of negative 1 end exponent end cell row cell open square brackets straight k. open parentheses table row straight a straight b row straight c straight d end table close parentheses close square brackets to the power of negative 1 end exponent end cell equals cell 1 over straight k. 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 open square brackets open parentheses table row ak bk row ck dk end table close parentheses close square brackets to the power of negative 1 end exponent end cell equals cell fraction numerator 1 over denominator straight k left parenthesis ad minus bc right parenthesis 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 fraction numerator 1 over denominator adk squared minus bck squared end fraction open parentheses table row dk cell negative bk end cell row cell negative ck end cell ak end table close parentheses end cell equals cell fraction numerator 1 over denominator straight k left parenthesis ad minus bc right parenthesis 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 fraction numerator 1 over denominator straight k squared left parenthesis ad minus bc right parenthesis end fraction open parentheses table row dk cell negative bk end cell row cell negative ck end cell ak end table close parentheses end cell equals cell fraction numerator 1 over denominator straight k left parenthesis ad minus bc right parenthesis 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 fraction numerator straight k over denominator straight k squared left parenthesis ad minus bc right parenthesis 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 equals cell fraction numerator 1 over denominator straight k left parenthesis ad minus bc right parenthesis 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 fraction numerator 1 over denominator straight k left parenthesis ad minus bc right parenthesis 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 equals cell fraction numerator 1 over denominator straight k left parenthesis ad minus bc right parenthesis 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 blank blank cell left parenthesis terbukti right parenthesis end cell row blank blank blank row blank blank blank end table 

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Matriks-matriks A, B, dan X berordo 2×2 dengan  dan  masing-masing mempunyai invers matriks A−1 dan B−1. Tentukan bentuk sederhana untuk matriks , jika: (A+BXA)⋅B=BAB

Pembahasan Soal:

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

0

Roboguru

Hasil kali matriks A⋅(50​−36​)=(−1035​3037​). Matriks A sama dengan ....

Pembahasan Soal:

Jika diketahui matriks A equals open parentheses table row a b row c d end table close parentheses, maka dapat ditentukan determinan dan invers matriks sebagai berikut.

text det  end text A equals a d minus b c

A to the power of negative 1 end exponent equals fraction numerator 1 over denominator text det end text space A end fraction open parentheses table row d cell negative b end cell row cell negative c end cell a end table close parentheses

Pada persamaan matriks berlaku rumus berikut.

A X equals B space rightwards double arrow space X equals A to the power of negative 1 end exponent B X A equals B space rightwards double arrow space X equals B A to the power of negative 1 end exponent

Matriks A pada soal di atas dapat ditentukan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell A times open parentheses table row 5 cell negative 3 end cell row 0 6 end table close parentheses end cell equals cell open parentheses table row cell negative 10 end cell 30 row 35 37 end table close parentheses end cell row A equals cell open parentheses table row cell negative 10 end cell 30 row 35 37 end table close parentheses open parentheses table row 5 cell negative 3 end cell row 0 6 end table close parentheses to the power of negative 1 end exponent end cell row A equals cell open parentheses table row cell negative 10 end cell 30 row 35 37 end table close parentheses times fraction numerator 1 over denominator 5 times 6 minus open parentheses negative 3 close parentheses times 0 end fraction open parentheses table row 6 3 row 0 5 end table close parentheses end cell row A equals cell 1 over 30 open parentheses table row cell negative 10 end cell 30 row 35 37 end table close parentheses open parentheses table row 6 3 row 0 5 end table close parentheses end cell row A equals cell 1 over 30 open parentheses table row cell negative 60 end cell 120 row 210 290 end table close parentheses end cell row A equals cell open parentheses table row cell negative 2 end cell 4 row 7 cell 29 over 3 end cell end table close parentheses end cell end table

Oleh karena itu, tidak ada pilihan jawaban yang tepat.

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Roboguru

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