Question

If Jacobi method is used to solve the following system of linear equations
\[ \begin{pmatrix} 1 & 2 & 1 \\ 0 & 2 & 2 \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} \]
with the initial guess \( x^{(0)} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \) and \( x^{(i)} = \begin{pmatrix} x_1^{(i)} \\ x_2^{(i)} \\ x_3^{(i)} \end{pmatrix} \), \( i = 1,2,3,\ldots \), denotes the \( i^{th} \) iterate, then the value of \( \left| x_1^{(2)} + x_2^{(2)} + x_3^{(2)} \right| \) is equal to ______.

Hint:

Update each variable using only the previous iterate's values (never the newly computed ones in the same round) and repeat twice from (0,0,0).

Answer 1

Correct Answer: 4

Step 1: Split the matrix into diagonal and off diagonal parts. Write $A = D + R$ where $D$ holds only the diagonal entries of $A$ and $R$ holds everything else:
\[ D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \qquad R = \begin{pmatrix} 0 & 2 & 1 \\ 0 & 0 & 2 \\ 1 & 1 & 0 \end{pmatrix} \]
The Jacobi method is the fixed point recursion $x^{(k+1)} = D^{-1}\left(b - R x^{(k)}\right)$.

Step 2: Build the iteration matrix $T = -D^{-1}R$ and vector $c = D^{-1}b$. Since $D^{-1} = \text{diag}(1, \frac{1}{2}, 1)$, scaling each row of $-R$ by the matching diagonal entry of $D^{-1}$ gives
\[ T = \begin{pmatrix} 0 & -2 & -1 \\ 0 & 0 & -1 \\ -1 & -1 & 0 \end{pmatrix}, \qquad c = D^{-1}b = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} \]
so that $x^{(k+1)} = Tx^{(k)} + c$.

Step 3: Apply the recursion once to get $x^{(1)}$. Starting from $x^{(0)} = (0,0,0)$:
\[ x^{(1)} = T(0,0,0)^T + c = c = (2, 1, 2) \]

Step 4: Apply the recursion again to get $x^{(2)}$. Multiply $T$ by $x^{(1)} = (2,1,2)$:
\[ Tx^{(1)} = \begin{pmatrix} 0(2) - 2(1) - 1(2) \\ 0(2) + 0(1) - 1(2) \\ -1(2) - 1(1) + 0(2) \end{pmatrix} = \begin{pmatrix} -4 \\ -2 \\ -3 \end{pmatrix} \]
Add $c = (2,1,2)$:
\[ x^{(2)} = (-4+2,\ -2+1,\ -3+2) = (-2, -1, -1) \]
This matches the componentwise computation exactly.

Final Answer.
\[ x_1^{(2)} + x_2^{(2)} + x_3^{(2)} = -2-1-1 = -4, \qquad \left|-4\right| = 4 \]
\[ \boxed{4} \]