3x1 x2 x3 1 3x1 6x2 2x3 0 3x1 3x2 7x3 4 Use the Jac

3x₁ - x₂ + x₃ = 1,

3x₁ + 6x₂ + 2x₃ = 0,

3x₁ + 3x₂ + 7x₃ = 4.

Use the Jacobi method to solve the linear systems with T O L = 10^-3 in the L infinite norm. 3x1 - x2 + x3 = 1, 3x1 + 6x2 + 2x3 = 0, 3x1 + 3x2 + 7x3 = 4.

Solution

We can solve by taking x1 = 1(say)

-x2+x3 = -2

6x2+2x3 = -3

8x2 = 1 or x2 =1/8 , x3 = 17/8, x1 = 1

So as this satisfies our assumption let us check

for second equation

3+3/4+1/4 not equals 0

hence x1 =1 is not right

Like this we try in all ways

7x2+x3 = -1 and

3x2-5x3 =-4

21x2+3x3 = -3

21x2-35x3 = -28

38x3 = 25

x3 =25/38

x2 = -9/38 and

x1 - 1-x2+x3 = 72/38