The document contains solutions to several systems of linear equations. It can be summarized as: 1) It solves the systems of equations for variables X, Y, and Z. The solutions are: X=3/11, Y=2/11, Z=-1/11. 2) It solves another system and finds the solutions: X=-4, Y=2, Z=7. 3) It solves a third system and obtains the solutions: D=2, A=-1, B=0, C=1.

1. Nama : RIZKIBUDIMAN
Kelas: TIF 2 A
Nim: 11551100287
A. 4X + 5Y = 2
11X + Y + 2Z = 3
X + 5Y +2Z = 1
Jawaban:
B1 ><B3
[
1 5 2 1
11 1 2 3
4 5 0 2
]
[
1 5 2 1
0 −54 −20 −8
0 −15 −8 −2
]
[
1 5 2 1
0 1
20
54
1
54
0 −15 −8 −2
]
[
1 5 2 1
0 1
20
54
8
54
0 0 −
22
9
−
1
11
]
[
1 5 2 1
0 1
20
54
8
54
0 0 1 −
1
11
]
Y +
20
54
Z =
8
54
X + 5Y + 2Z = 1
Y +
20
54
(-
1
11
) =
8
54
X + 5(
2
11
)+ 2(-
1
11
) = 1
Y –
20
594
=
8
54
X +
10
11
-
2
11
= 1
Y –
10
297
=
4
27
X = 1 -
8
11
Y =
4
27
+
10
297
Y =
1188+270
8019
Y =
1458
8019
:
729
729
B1 (-11) + B2
B1 (-4) + B3
B2 (-
1
54
)
B2 (15) + B3
B3 (-
9
22
Z = -
1
11
X =
3
11
Y =
2
11

2. B. 3X + 2Y – Z = -15
5X + 3Y + 2Z = 0
3X + Y + 3Z = 11
-6X – 4Y + 2Z = 30
Jawaban:
[
3 2 −1 −15
5 3 2 0
3 1 3 11
−6 −4 2 30
]
[
1
2
3
−
1
3
−5
5 3 2 0
3 1 3 11
−6 −4 2 30]
[
1
2
3
−
1
3
−5
0 −
1
3
11
3
−75
0 −1 4 26
0 0 0 0 ]
[
1
2
3
−
1
3
−5
0 1 −11 −75
0 −1 4 26
0 0 0 0 ]
[
1
2
3
−
1
3
−5
0 1 −11 −75
0 0 −7 −49
0 0 0 0 ]
[
1
2
3
−
1
3
−5
0 1 −11 −75
0 0 1 7
0 0 0 0 ]
Y – 11Z = -75 X +
2
3
Y -
1
3
Z = -5
Y – 11(7) = -75 X +
2
3
(2) -
1
3
(7)= -5
Y – 77 = -75 X +
4
3
-
7
3
= -5
Y = -75 + 77 X – 1 = -5
X = -5 + 1
B1 (
1
3
)
B1 (-5) + B2
B1 (-3) + B3
B1 (6) + B4
B2 (-3)
B2 (1) + B3
B3 (-
1
7
)
Z=7
Y =2
X = -4

3. C. 2A – B + 3C + 4D = 9
A -2C + 7D = 11
3A – 3B + 3C + 5D = 8
2A + B + 4C + 4D = 10
Jawaban:
B1 >< B2
[
1 0 −2 7 11
2 −1 3 4 9
3 −3 1 5 8
2 1 4 4 10
]
[
1 0 −2 7 11
0 −1 7 −10 −13
0 −3 7 −16 −25
0 1 8 −10 −12
]
[
1 0 −2 7 11
0 −1 −7 10 13
0 −3 7 −16 −25
0 1 8 −10 −12
]
[
1 0 −2 7 11
0 1 −7 10 13
0 0 −14 14 14
0 0 15 15 −25
]
[
1 0 −2 7 11
0 1 −7 10 13
0 0 1 −1 −1
0 0 15 −20 −25
]
[
1 0 −2 7 11
0 1 −7 10 13
0 0 1 −1 −1
0 0 0 −5 −10
]
[
1 0 −2 7 11
0 1 −7 10 13
0 0 1 −1 −1
0 0 0 1 2
]
C – D = -1 B – 7C + 10D = 13 A – 2C + 7D = 11
C – 2 = -1 B – 7(1) + 10(2) = 13 A – 2(1) + 7(2) = 11
C = -1 + 2 B – 7 + 20 = 13 A – 2 + 14 = 11
B = 13 – 13 A = 11 - 12
B1 (-2) + B2
B1 (-3) + B3
B1 (-2) + B4
B2 (-1)
B2 (3) + B3
B2 (-1) +B4
B3 (-
1
14
)
B3 (-15) + B4
B4 (-
1
5
)
D = 2
A = -1B = 0
C = 1