sistem persamaan liniar tiga variabel a. x,y,z adlah penyelesaian sistem persamaan : 3x + 4y - 5z = 12 2x + 5y + z = 17 6x - 2y + 3z = 17 b. x,y,z adlah penyele

3x + 4y - 5z = 12 ... (i)
2x + 5y + z = 17 ... (ii)
6x - 2y + 3z = 17 ... (iii)
Eliminasi (i) dan (ii)
3x + 4y - 5z = 12 (x1) --> 3x + 4y - 5z = 12
2x + 5y + z = 17 (x5) --> 10x + 25y + 5z = 85 +
                                       13x + 29y = 97 ... (iv)
Eliminasi (ii) dan (iii)
2x + 5y + z = 17 (x3) --> 6x + 15y + 3z = 51
6x - 2y + 3z = 17 (x1) --> 6x - 2y + 3z = 17 _
                                             17y = 34 --> y = 2
substitusi y = 2 ke (iv)
13x + 29y = 97
13x +29.2 = 97
13x = 97 - 58
13x = 39
x = 3

subst ke (i)
3x + 4y - 5z = 12
3.3 + 4.2 - 5z = 12
9 + 8 - 5z = 12
-5z = 12 - 8 - 9
-5z = -5
z = 1
Hp = {3,2,1}


x + 2y = -4 .. (i)
2x + z = 5 ... (ii)
y - 3z = -6 ... (iii)
Eliminasi (i) dan (ii)
x + 2y = -4 (x2) --> 2x + 4y = -8
2x + z = 5 (x1) --> 2x + z = 5  _
                             4y - z = -13 ... (iv)
Eliminasi (iii) dan 9iv)
y - 3z = -6 (x4) --> 4y - 12z = -24
4y - z = -13 (x1) --> 4y - z = -13  _
                                 -11z = -11 --> z = 1
Substitusi z = 1 ke (iv)
4y - z = -13
4y - 1 = -13
4y = -13 + 1
4y = -12
y = -3
Subst ke (ii)
2x + z = 5
2x + 1 = 5
2x = 4
x = 2
Hp = {2,-3,1}


misal 1/x = a, 1/y = b, dan 1/z = c maka pers mjdi:
4a + 3b + c = 9 ... (i)
3a - 4b + 2c = 3 ... (ii)
2a + 5b - c = 5 ... (iii)
Eliminasi (i) dan (ii)
4a + 3b + c = 9 (x2) --> 8a + 6b + 2c = 18
3a - 4b + 2c = 3 (x1) --> 3a - 4b + 2c = 3    _
                                     5a + 10b = 15 ... (iv)
Eliminasi (i) dan 9iii)
4a + 3b + c = 9
2a + 5b - c = 5  +
6a + 8b = 14 ... (v)
Eliminasi (iv) dan (v)
5a + 10b = 15 (x4) --> 20a + 40b = 60
6a + 8b = 14 (x5) --> 30a + 40b = 70 _
                                   -10a = -10 --> a = 1
Subst a = 1 ke (iv)
5a + 10b = 15
5.1  +10b = 15
10b = 15 - 5
b = 10/10
b = 1
Subst ke (i)
4a + 3b + c = 9
4.1 + 3.1 + c = 9
c = 9 - 4 - 3
c = 2
a = 1 --> 1/x = 1 --> x = 1
b = 1 --> 1/y = 1 --> y = 1
c = 2 --> 1/z = 2 --> z = 1/2
Hp = {1,1,1/2}