1. Soal
P1 3X1 -2X2 -3X3 = 10
P2 -3X1 4X2 4X3 = -17
P3 -4X1 7X2 3X3 = -19
Penyelesaian
a) Substitusi
P1 3X1 -2X2 -3X3 = 10
3X1 = 10 +2X2 +3X3
X1 = + + X3 ---------------------------------------------------- P4
P4 disubstitusi ke- P2
P2 -3X1 +4X2 +4X3 = -17
-3( + + X3 ) +4X2 +4X3 = -17
-10 -2X2 -3X3 +4X2 +4X3 = -17
2X2 + X3 = - 7 --------------------------------------------------- P5
P4 disubstitusi ke- P3
P3 -4X1 7X2 3X3 = -19
-4( + + X3 ) +7X2 +3X3 = -19
- - 4X3 +7X2 +3X3 = -19
- X3 = -19 +
( - X3 = - )x3
13X2 - 3X3 = -17 --------------------------------------------------- P6
2X2 + X3 = - 7 disubstitusi
13X2 - 3X3 = -17
P5 2X2 + X3 = - 7
X3 = - 7 - 2X2 --------------------------------------------------- P7
P7 disubstitusi ke- P6
P6 13X2 - 3X3 = -17 X2 disubstitusi ke- P5
13X2 - 3 (- 7 - 2X2 ) = -17 P5 2X2 + X3 = - 7
13X2 +21 +6X2 = -17 2(-2) + X3 = - 7
19X2 = - 38 -4 + X3 = - 7
X2 =-2 X3 = - 3
X2 dan X3 disubstitusi ke- P1
P1 3X1 -2X2 -3X3 = 10
3X1 -2(-2) -3(-3) = 10
3X1 +4 +9 = 10
3X1 = -3
X1 = -1 HP
P2 -3X1 +4X2 +4X3 = -17 P3 -4X1 7X2 3X3 = -19
-3(-1) +4X2 +4(-3) = -17 -4(-1) +7(-2) 3X3 = -19
3 +4X2 - 12 = -17 4 -14 3X3 = -19
4X2 = -8 3X3 = -9
X2 = -2 X3 = -3

2. b) Eliminasi
P1 3X1 -2X2 -3X3 = 10
P2 -3X1 +4X2 +4X3 = -17 +
2X2 + X3 = - 7 --------------------------------------------------- P4
P2 -3X1 +4X2 +4X3= -17 |.4| -12X1 +16X2 +16X3= -68
P3 -4X1 +7X2 +3X3 = -19 |.3| -12X1 +21X2 + 9X3 = -57 +
-5X2 + 7X3 = -11 ------------------------------ P5
P4 & P5 dieliminasi
P4 2X2 + X3 = - 7 |.7| 14X2 + 7X3 = - 49
P5 -5X2 + 7X3 = -11 |.1| -5X2 + 7X3 = -11 -
19X2 = - 38
X2 = - 2
X2 disubstitusi ke- P4 X2 & X3 disubstitusi ke- P1
P4 2X2 + X3 = -7 P1 3X1 -2X2 -3X3 = 10
2(-2) + X3 = -7 3X1 -2(-2) -3(-3) = 10
-4 + X3 = -7 3X1 +4 +9 = 10
X3 = -3 3X1 = -3
X1 = -1
HP
P2 -3X1 +4X2 +4X3 = -17 P3 -4X1 7X2 3X3 = -19
-3(-1) +4X2 +4(-3) = -17 -4(-1) +7(-2) 3X3 = -19
3 +4X2 - 12 = -17 4 -14 3X3 = -19
4X2 = -8 3X3 = -9
X2 = -2 X3 = -3
c) Metode Gauss – Versi Segitiga Bawah
b1 3 -2 -3 10 b1. 3 - -
b2 -3 4 4 -17 -3 4 4 -17 b2 + 3.b1
b3 -4 7 3 -19 -4 7 3 -19 b3 + 4.b1
3 - - 3 - -
0 2 1 -7 b2. 0 1 -
0 -1 - 0 -1 - b3 - .b2
3 - - 3 - - b1 X1 - X2 - X3 =
0 1 - 0 1 - b2 X2 + X3 = -
0 0 - b2.- 0 0 1 -3 b3 X3 = -3
b3 disubstitusi ke b2 b3 & X2 disubstitusi ke b1
b2 X2 + X3 = - b1 X1 - X2 - X3 =
X2 + (-3) = - X1 - (-2) - (-3) =
X2 - = - X1 + + =
X2 = -2 X1 = -1 HP
d) Metode Gauss Jordan
3 - - b1+ b3 1 - 0 b1 + b2 1 0 0 -1
0 1 - b2 - b3 0 1 0 -2 0 1 0 -2
0 0 1 -3 0 0 1 -3 0 0 1 -3

3. e) Invers Matrik A-1
3 -2 -3 X1 = 10 B =
-3 4 4 X X2 = -17 B = C. A-1
-4 7 3 X3 = -19 A-1 = adj.A
A x B = C
- Determinan (Sarraus)
3 -2 -3 3 -2
-3 4 4 -3 4
-4 7 3 -4 7
Det.A = (3.4.3) + (-2.4.4) + (-3.-3.7)
- (-4.4.-3) – (7.4.3) – (3.-3.-2)
= 36 + 32 + 63 – 48 – 84 – 18
= 131 – 150
= -19
- Cotactor Cij = (-1)i-j . Mij
C11 C12 C13 + - + 3 -2 -3
C21 C22 C23 = - + - => -3 4 4
C31 C32 C33 + - + -4 7 3
C11= + 4 4 C12= - -3 4 C13= + -3 4
7 3 -4 3 -4 7
= (4.3) – (7.4) =- (-3.3) – (-4.4) = (-3.7) – (-4.4)
= 12 – 28 =- (-9 + 16) = -21 + 16
= -16 = -7 = -5
C21= - -2 -3 C22= + 3 -3 C23= - 3 -2
7 3 -4 3 -4 7
=- (-6 + 21) = (9 - 12) =- (21 - 8)
= -15 = -3 = -13
C31= + -2 -3 C32= - 3 -3 C33= + 3 -2
4 4 -3 4 -3 4
= (-18 + 12) =- (12 - 9) = (12 - 6)
= 4 = -3 = 6
Cij = -16 -7 -5 -16 -15 4
-15 -3 -13 T -7 -3 -3
4 -3 6 -5 -13 6
A-1= -16 -15 4
-7 -3 -3 =
-5 -13 6
Cari X1, X2, X3
B= 10 -1
X -17 = = = -2
-19 -3