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Pertanyaan

Invers matriks A = ⎝ ⎛ ​ 4 − 2 3 ​ − 1 3 − 1 ​ 5 1 4 ​ ⎠ ⎞ ​ adalah:

Invers matriks  adalah:

  1. open parentheses table row 13 cell negative 1 end cell cell negative 16 end cell row 11 1 cell negative 14 end cell row cell negative 7 end cell 1 10 end table close parentheses 

  2. open parentheses table row cell negative 13 end cell 1 16 row cell negative 11 end cell cell negative 1 end cell 14 row 7 cell negative 1 end cell cell negative 10 end cell end table close parentheses

  3. open parentheses table row cell 13 over 6 end cell cell negative 1 over 6 end cell cell negative 16 over 6 end cell row cell 11 over 6 end cell cell 1 over 6 end cell cell negative 14 over 6 end cell row cell negative 7 over 6 end cell cell 1 over 6 end cell cell 10 over 6 end cell end table close parentheses

  4. open parentheses table row cell negative 13 over 6 end cell cell 1 over 6 end cell cell 16 over 6 end cell row cell negative 11 over 6 end cell cell negative 1 over 6 end cell cell 14 over 6 end cell row cell 7 over 6 end cell cell negative 1 over 6 end cell cell negative 10 over 6 end cell end table close parentheses 

  5. Bukan A, B, C, dan D

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A. Rahma

Master Teacher

Mahasiswa/Alumni Institut Teknologi Bandung

Jawaban terverifikasi

Pembahasan

Invers matriks berorodo Terlebih dahulu mencari determinan dari A dengan menggunakan metode ekspansi kofaktor pada baris pertama. Kemudian, mencari matriks adjoin dari A. Diketahui, maka, Kemudian ubah kof(A) menjadi adj(A) Sehingga, invers dari matriks A adalah Oleh karena itu, jawaban yang benar adalah B.

Invers matriks berorodo left parenthesis 3 cross times 3 right parenthesis

A to the power of negative 1 end exponent equals fraction numerator 1 over denominator d e t space A end fraction a d j o i n t left parenthesis A right parenthesis

Terlebih dahulu mencari determinan dari A dengan menggunakan metode ekspansi kofaktor pada baris pertama.

A equals open parentheses table row 4 cell negative 1 end cell 5 row cell negative 2 end cell 3 1 row 3 cell negative 1 end cell 4 end table close parentheses

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 4 cell negative 1 end cell 5 row cell negative 2 end cell 3 1 row 3 cell negative 1 end cell 4 end table close vertical bar end cell row blank equals cell 4 open vertical bar table row 3 1 row cell negative 1 end cell 4 end table close vertical bar minus left parenthesis negative 2 right parenthesis open vertical bar table row cell negative 2 end cell 1 row 3 4 end table close vertical bar plus 5 open vertical bar table row cell negative 2 end cell 3 row 3 cell negative 1 end cell end table close vertical bar end cell row blank equals cell 4 open vertical bar table row 3 1 row cell negative 1 end cell 4 end table close vertical bar plus 2 open vertical bar table row cell negative 2 end cell 1 row 3 4 end table close vertical bar plus 5 open vertical bar table row cell negative 2 end cell 3 row 3 cell negative 1 end cell end table close vertical bar end cell row blank equals cell 4 left parenthesis 12 plus 1 right parenthesis plus 2 left parenthesis negative 8 minus 3 right parenthesis plus 5 left parenthesis 2 minus 9 right parenthesis end cell row blank equals cell 4 left parenthesis 13 right parenthesis plus 2 left parenthesis negative 9 right parenthesis plus 5 left parenthesis negative 7 right parenthesis end cell row blank equals cell 52 minus 18 minus 35 end cell row blank equals cell negative 1 end cell end table


Kemudian, mencari matriks adjoin dari A.

Adjoin space A space equals space left parenthesis k o f left parenthesis A right parenthesis right parenthesis to the power of T

k o f left parenthesis A right parenthesis equals left parenthesis left parenthesis negative 1 right parenthesis to the power of i plus j end exponent M subscript i j end subscript right parenthesis k o f left parenthesis A right parenthesis equals open parentheses table row cell M subscript 11 end cell cell M subscript 12 end cell cell M subscript 13 end cell row cell M subscript 21 end cell cell M subscript 22 end cell cell M subscript 23 end cell row cell M subscript 31 end cell cell M subscript 32 end cell cell M subscript 33 end cell end table close parentheses 


Diketahui, A equals open parentheses table row 4 cell negative 1 end cell 5 row cell negative 2 end cell 3 1 row 3 cell negative 1 end cell 4 end table close parentheses maka,

table attributes columnalign right center left columnspacing 0px end attributes row cell k o f left parenthesis A right parenthesis end cell equals cell open parentheses table row cell open vertical bar table row 3 1 row cell negative 1 end cell 4 end table close vertical bar end cell cell negative open vertical bar table row cell negative 2 end cell 1 row 3 4 end table close vertical bar end cell cell open vertical bar table row cell negative 2 end cell 3 row 3 cell negative 1 end cell end table close vertical bar end cell row cell negative open vertical bar table row cell negative 1 end cell 5 row cell negative 1 end cell 4 end table close vertical bar end cell cell open vertical bar table row 4 5 row 3 4 end table close vertical bar end cell cell negative open vertical bar table row 4 cell negative 1 end cell row 3 cell negative 1 end cell end table close vertical bar end cell row cell open vertical bar table row cell negative 1 end cell 5 row 3 1 end table close vertical bar end cell cell negative open vertical bar table row 4 5 row cell negative 2 end cell 1 end table close vertical bar end cell cell open vertical bar table row 4 cell negative 1 end cell row cell negative 2 end cell 3 end table close vertical bar end cell end table close parentheses end cell row blank equals cell open parentheses table row 13 11 cell negative 7 end cell row cell negative 1 end cell 1 1 row cell negative 16 end cell cell negative 14 end cell 10 end table close parentheses end cell end table


Kemudian ubah kof(A) menjadi adj(A)

table attributes columnalign right center left columnspacing 0px end attributes row cell Adj space A end cell equals cell left parenthesis k o f left parenthesis A right parenthesis right parenthesis to the power of T end cell row blank equals cell open parentheses table row 13 11 cell negative 7 end cell row cell negative 1 end cell 1 1 row cell negative 16 end cell cell negative 14 end cell 10 end table close parentheses to the power of T end cell row blank equals cell open parentheses table row 13 cell negative 1 end cell cell negative 16 end cell row 11 1 cell negative 14 end cell row cell negative 7 end cell 1 10 end table close parentheses end cell row blank blank blank end table

Sehingga, invers dari matriks A adalah

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 d e t space A end fraction a d j o i n t left parenthesis A right parenthesis end cell row blank equals cell fraction numerator 1 over denominator negative 1 end fraction open parentheses table row 13 cell negative 1 end cell cell negative 16 end cell row 11 1 cell negative 14 end cell row cell negative 7 end cell 1 10 end table close parentheses end cell row blank equals cell negative 1 open parentheses table row 13 cell negative 1 end cell cell negative 16 end cell row 11 1 cell negative 14 end cell row cell negative 7 end cell 1 10 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 13 end cell 1 16 row cell negative 11 end cell cell negative 1 end cell 14 row 7 cell negative 1 end cell cell negative 10 end cell end table close parentheses end cell end table


Oleh karena itu, jawaban yang benar adalah B.

 

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