TUGAS KLASIBER KALKULUS 2, Sabtu 29 Maret 2014Pecahkanlah sistem persamaan linear (CARI HARGA KOEFISIEN) dari persamaan linear berikut ini :

2 x1 x2 + 3 x3 = 2

x1 + 3 x2 x3 = 11

2 x1 2 x2 + 5x3 = 3

Jawab:

2-13

213-1 =

11

2-25

3

= ((2.3.5)+(-1.-1.2)+(3.1.-2))-((2.3.3)+(-2.-1.2)+(5.1.-1))

= ((30+2-6)-(18+4-5))

= 9

2-13

113-1

3-25

1= ((2.3.5)+(-1.-1.3)+(3.11.-2))-((3.3.3)+(-2.-1.2)+(5.11.-1))

=((30+3-66)-(27+4-55))

= -9

223111-1

235

2= ((2.11.5)+(2.-1.2)+(3.1.3))-((2.11.3)+(3.-1.2)+(5.1.2))

= ((110-4+9)-(66-6+10))

= 452-12

1311

2-23

3= ((2.3.3)+(-1.11.2)+(2.1.-2))-((2.3.2)+(-2.11.2)+(3.1.-1))

= ((18-22-4)-(12-44-3))

= 27x1 = 1

x2 = 2

x3 = 3 -9

45 27 = 9

= 9

= 9 =-1

= 5

= 3

CROSS CEK 2 x1 x2 + 3 x3= 2

((2.-1) (5) + (3.3)= 2-2-5+9

= 2 benar