Menyelesaikan Persamaan Matriks – 2

Selasa, 11 Oktober 2022

Sistem Persamaan Linier Tiga Variable

 

Untuk lebih jelanya, ikutolah contoh soal berikut ini:

 

1.     Tentukanlah himpunan penyelesaian sistem persamaan linier

 

  

Alternatif Pembahasan :

 

  

D = (1)(1)(1) + (–2)(1)(2) + (0)(0)(0) – (0)(1)(2) – (1)(1)(0) – (–2)(0)(1)

D = (1) + (–4) + (0) – (0) – (0) – (0)

D = 1 – 4 + 0 + 0 + 0 + 0

D = –3

 

  

D_x = (–3)(1)(1) + (–2)(1)(1) + (0)(1)(0) – (0)(1)(1) – (–3)(1)(0) – (–2)(1)(1)

D_x = (–3) + (–2) + (0) – (0) – (0) – (–2)

D_x = –3 – 2 + 0 – 0 – 0 + 2

D_x = –3

 

  

D_y = (1)(1)(1) + (–3)(1)(2) + (0)(0)(1) – (0)(1)(2) – (1)(1)(1) – (–3)(0)(1)

D_y = ( 1) + (–6) + (0) – (0) – (1) – (0)

D_y = 1 – 6 + 0 + 0 – 1 – 0

D_y = –6

 

  

D_z = (1)(1)(1) + (–2)(1)(2) + (–3)(0)(0) – (–3)(1)(2) – (1)(1)(0) – (–2)(0)(1)

D_z = (1) + (–4) + (0) – (–6) – (0) – (0)

D_z = 1 – 4 + 0 + 6 – 0 – 0

D_z = 3

 

  

Sumber

Labels: Matematika

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