Find the first two iteration of the Jacobi method for the following linear system, using $x^{(0)} = 0$ $$3x_{1} - x_{2} + x_{3} = 1,$$
$$3x_{1} + 6x_{2} + 2x_{3} = 0,$$
$$3x_{1}+3x_{2}+7x_{3} = 4$$
So, from this I got 
T =  \begin{bmatrix}
       0  & \frac{-1}{3} & \frac{1}{3}           \\[0.3em]
       \frac{1}{2} & 0           & \frac{1}{3} \\[0.3em]
       \frac{3}{7}          & \frac{3}{7} & 0
     \end{bmatrix}
$C= [-1/3, 0, -4/7]$
Then, I got $x_{1}^1 = 0.33$, $x_{2}^1 = 0, x_{3}^1 = -0.57$ and 
$x_{1}^2 = 0.33, x_{2}^2 = 0.833, x_{3}^2 = 0.142$
But, the answer in the book is $(0.1428571, -0.3571429, 0.4285714)$. Did I do something wrong? I am confused here. Can someone please help me understand this? Thanks. 

First iteration:
$$0.333333, \text{      }0.,\text{       }0.571429$$
Second iteration:
$$0.142857,\text{   }-0.357143\text{   },0.428571$$
So, it looks like somehow the sign of your $x_3$ on the first iteration got corrupted.
We rewrite the system as:
$$x_1 = \dfrac{1}{3} x_2 -\dfrac{1}{3}x_3 + \dfrac{1}{3} \\ x_2 = -\dfrac{1}{2}x_1 + 0 x_2 - \dfrac{1}{3}x_3 \\ x_3 = - \dfrac{3}{7}x_1 - \dfrac{3}{7}x_2 + \dfrac{4}{7}$$
Initial start point: $x^{(0)} = (x_1, x_2, x_3) = (0,0,0)$.
First iteration (substitute $x^{(0)}$ into the system to find the next set of values):
$$x_1 = 0 + 0 + \dfrac{1}{3}, x_2 = 0 + 0 + 0, x_ 3 = 0 + 0 +\dfrac{4}{7}$$
Now, you try the second iteration, just substitute the first iterations values into the three equations.
