Matakuliah
Tahun
: Kalkulus II
: 2008 / 2009
Soal Pertemuan 6-7
Tentukan penyelesaian SPL dibawah ini dengan cara :
a. Eliminasi
b. OBE
c. Kofaktor/Adjoint
d. Aturan Cramer
1. 3x1 – x2 2x3 = 20
2x1 3x2 – x1 = 15
x1 – 2x2 – 3x3 = 11
x1 = 5
jawab : x2 = 3
x3 = 4
2. 4x1 – 3x2
x3 = 11
3x1 2x2 – x3 = 9
x1 – 4x2 – 3x3 = 5
x1 = 3
jawab : x2 = 1
x3 = 2
2x1 – 5x2 – 2x3 = 7
x1 2x2 – 4x3 = 3
3x1 – 4x2 – 6x3 = 5
x1 = 5
jawab : x2 = 1
x3 = 1
3.
Bina Nusantara University
3
3x1 – 2x2 – x3 = -15
5x1 3x2 3x3 = 0
11x1 7x2
= -30
jawab:
5.
2x1 – 3x2 – 4x3 = 17
-5x1 – 2x2 – x3 = -17
4x1
x2 2x3 = 8
x1 = 2
jawab : x2 = -3
x3 = 1
6.
2x1 – x2 3x3 = 16
x1 – 2x2 – x3 = 3
3x1 – x2 2x3 = 17
x1 = 4
jawab : x2 = 1
x3 = 3
4.
Bina Nusantara University
x1 = 45
x2 = -75
x3 = 0
4
7.
3x1 – 2x2 + x3 = 15
2x1 + x2 – 3x3 = 5
x1 – x2 + 2x3 = 8
x1 = 5
jawab : x2 = 1
x3 = 2
8.
3x1 – x2 – 2x3 = 13
x1 – 2x2
x3 = 7
2x1 – x2 - 3x3 = 4
x1 = 6
jawab : x2 = 1
x3 = 3
9.
2x1 – x2 – 3x3 – x4 = 17
x1 2x2 – x3 2x4 =11
3x1 – x2 -4x2 + x4 = 4
4x1 – 3x2 – 2x3 3x4 = 9
Bina Nusantara University
jawab
x1 = 5
x2 = 3
x3 = 4
x4 = 2
5