Agar sistem persamaan linear
ax + by - 3z = -3
-2x - by + cz = -1
ax + 3y - cz = -3
mempunyai penyelesaian x = 1, y=-1, dan z=2, maka nilai a+b+c adalah ....
a.-1
b.1
c.2
d.3
e.4

2
Eleminasi pers i dan iii
a - b - 6c = -3
a - 3 -2c = -3 _
-b+3-6c+2c = 0
b + 4c = 3 ... (iv)
" -b+3-6c+2c = 0" menjadi "b" + 4c = 3 kenapa "b" semula negatif baru berubah menjadi positif ?????
pindah ruas
semua kali -1 jadikan b berubah jadi positif

Jawabanmu

2014-07-01T11:53:40+07:00
Masukin aja dulu x, y, z
a - b - 6c = -3 ...(i)
-2 + b + 2c = -1 ...(ii)
a - 3 -2c = -3 ..(iii)

Eleminasi pers i dan iii
a - b - 6c = -3
a - 3 -2c = -3  _
-b+3-6c+2c = 0
b + 4c = 3 ... (iv)

Eleminasi pers ii dan iv
-2 + b + 2c = -1
b + 4c = 3        _
-2-2c=-4
-2c = -2
c = 1

Subtitusi ke iv
b + 4 . 1 = 3
b = 3-4
= -1

Subtitusi ke iii
a - 3 - 2.1 = -3
a -5 = -3
a = 2

a+b+c = 2 + (-1) + 1
= 2

:)



2014-07-01T12:18:04+07:00
A - b - 6=-3 -> a-b= 3
-2 + b + 2c = -1 -> b + 2c = 1 ->b = 1 - 2c
a - 3 - 2c = -3 -> a - 2c = 0 -> a = 2c

a - b = 3
2c - (1 - 2c) = 3 -> 2c-1+2c = 3 -> c = 1
                                                 a = 2c -> a = 2
                                                 b = 1 - 2c -> b = -1
a + b + c = 2 +  (-1) + 1 = 2