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Question 940697: Solve the following systems of equations using Cramer Rule. [10 marks]
X+Y+Z=2
Z+2Y+X=1
Y+4X+2Z=0
Answer by krutarthas(22)   (Show Source):
You can put this solution on YOUR website! x + y + z = 2 -----(1)
x + 2y + z = 1 -----(2)
4x +y+2z = 0 -----(3)
Δ =
| 1 1 1 |
| 1 2 1 |
| 4 1 2 |
= 1(4 - 1) – 1(2 - 4) + 1(1 - 8)
= 3+2-7
= -2
Here Determinant is not equal to zero, Hence Cramer’s rule can be applicable.
X = ∆x / ∆ , y = ∆y / ∆ , z = ∆z / ∆
Δx =
| 2 1 1 |
| 1 2 1 |
| 0 1 2 |
=2(4-1)-1(2-0)+1(1-0)
=6-2+1
∆x= 5
Δy =
| 1 2 1 |
| 1 1 1 |
| 4 0 2 |
= 1(2-0)-2(2-4)+1(0-4 )
=2+4-4
∆y= 2
Δz =
| 1 1 2 |
| 1 2 1 |
| 4 1 0 |
= 1(0-1)-1(0-4)+2(1-8)
= -1+4-14
∆z= -11
By using Cramer’s rule:
x = Δx / ∆
x = 5/-2
x = -2.5
y = Δy / ∆
y = 2 / -2
y = -1
z = Δz / ∆
z = -11/-2
z = 5.5
Therefore the required x = -2.5, y = -1 & z = 5.5
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