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 ______.
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 ______.
\[ \begin{pmatrix} 1 & 2 & 1 \\ 0 & 2 & 2 \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} \]
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 ______.
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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).
Updated On: Jul 21, 2026
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Correct Answer: 4
Solution and Explanation
Step 1: Understanding the Question.
The Jacobi method solves \( Ax=b \) by updating each variable using only the OLD values of the other variables from the previous iterate, never the values just computed in the same round. We need to run this for two rounds starting from \( x^{(0)}=(0,0,0) \) and then add up the three components of \( x^{(2)} \).
Step 2: Key Formula or Approach.
Write out the three equations of the system:
\[ x_1 + 2x_2 + x_3 = 2, \qquad 2x_2 + 2x_3 = 2, \qquad x_1 + x_2 + x_3 = 2 \]
Solve each equation for the variable sitting on its own diagonal position, using old iterate values on the right:
\[ x_1^{(k+1)} = 2 - 2x_2^{(k)} - x_3^{(k)} \]
\[ x_2^{(k+1)} = \frac{2 - 2x_3^{(k)}}{2} = 1 - x_3^{(k)} \]
\[ x_3^{(k+1)} = 2 - x_1^{(k)} - x_2^{(k)} \]
Step 3: Compute the first iterate \( x^{(1)} \).
Using \( x_1^{(0)}=x_2^{(0)}=x_3^{(0)}=0 \):
\[ x_1^{(1)} = 2 - 0 - 0 = 2 \]
\[ x_2^{(1)} = 1 - 0 = 1 \]
\[ x_3^{(1)} = 2 - 0 - 0 = 2 \]
So \( x^{(1)} = (2, 1, 2) \).
Step 4: Compute the second iterate \( x^{(2)} \).
Now use \( x_1^{(1)}=2, x_2^{(1)}=1, x_3^{(1)}=2 \) on the right side of each formula:
\[ x_1^{(2)} = 2 - 2(1) - 2 = 2 - 2 - 2 = -2 \]
\[ x_2^{(2)} = 1 - 2 = -1 \]
\[ x_3^{(2)} = 2 - 2 - 1 = -1 \]
So \( x^{(2)} = (-2, -1, -1) \).
Final Answer:
\[ x_1^{(2)} + x_2^{(2)} + x_3^{(2)} = -2 - 1 - 1 = -4 \]
\[ \left| -4 \right| = 4 \]
\[ \boxed{4} \]
The Jacobi method solves \( Ax=b \) by updating each variable using only the OLD values of the other variables from the previous iterate, never the values just computed in the same round. We need to run this for two rounds starting from \( x^{(0)}=(0,0,0) \) and then add up the three components of \( x^{(2)} \).
Step 2: Key Formula or Approach.
Write out the three equations of the system:
\[ x_1 + 2x_2 + x_3 = 2, \qquad 2x_2 + 2x_3 = 2, \qquad x_1 + x_2 + x_3 = 2 \]
Solve each equation for the variable sitting on its own diagonal position, using old iterate values on the right:
\[ x_1^{(k+1)} = 2 - 2x_2^{(k)} - x_3^{(k)} \]
\[ x_2^{(k+1)} = \frac{2 - 2x_3^{(k)}}{2} = 1 - x_3^{(k)} \]
\[ x_3^{(k+1)} = 2 - x_1^{(k)} - x_2^{(k)} \]
Step 3: Compute the first iterate \( x^{(1)} \).
Using \( x_1^{(0)}=x_2^{(0)}=x_3^{(0)}=0 \):
\[ x_1^{(1)} = 2 - 0 - 0 = 2 \]
\[ x_2^{(1)} = 1 - 0 = 1 \]
\[ x_3^{(1)} = 2 - 0 - 0 = 2 \]
So \( x^{(1)} = (2, 1, 2) \).
Step 4: Compute the second iterate \( x^{(2)} \).
Now use \( x_1^{(1)}=2, x_2^{(1)}=1, x_3^{(1)}=2 \) on the right side of each formula:
\[ x_1^{(2)} = 2 - 2(1) - 2 = 2 - 2 - 2 = -2 \]
\[ x_2^{(2)} = 1 - 2 = -1 \]
\[ x_3^{(2)} = 2 - 2 - 1 = -1 \]
So \( x^{(2)} = (-2, -1, -1) \).
Final Answer:
\[ x_1^{(2)} + x_2^{(2)} + x_3^{(2)} = -2 - 1 - 1 = -4 \]
\[ \left| -4 \right| = 4 \]
\[ \boxed{4} \]
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