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UPSC 2023 Maths Optional Paper 2 Q6b — Step-by-Step Solution

15 marks · Section B

Gauss-Seidel iteration · Numerical Analysis · asked 7× in 13 yrs · Read the full method →

Question

Solve the system of linear equations

correct up to 4 significant figures by the Gauss-Seidel iterative method. Take initially guessed solution as $x_1=x_2=x_3=0$.

Technique

Gauss-Seidel iteration with diagonal-dominance convergence guarantee.

Solution

Step 1 — Check diagonal dominance (convergence guarantee)

| Row | $|a_{ii}|$ | $\sum_{j\ne i}|a_{ij}|$ | |---|---|---| | 1 | $7$ | $|-1|+|2|=3$ | | 2 | $8$ | $|2|+|-1|=3$ | | 3 | $9$ | $|1|+|-2|=3$ |

Each row is strictly diagonally dominant ($|a_{ii}|>\sum_{j\ne i}|a_{ij}|$), so Gauss-Seidel converges from any starting guess. ✓

Step 2 — Iteration formula

Solve each equation for its diagonal variable and substitute the most recently updated values (the Gauss-Seidel rule):

Step 3 — Iterations (starting $(0,0,0)$)

Sample arithmetic (iteration 1):

• $x_1^{(1)}=(11+0-0)/7=1.5714$.
• $x_2^{(1)}=(9-2(1.5714)+0)/8=5.8572/8=0.7321$.
• $x_3^{(1)}=(7-1.5714+2(0.7321))/9=6.8928/9=0.7659$.

Sample arithmetic (iteration 2):

• $x_1^{(2)}=(11+0.7321-2(0.7659))/7=10.2003/7=1.4572$.
• $x_2^{(2)}=(9-2(1.4572)+0.7659)/8=6.8515/8=0.8564$.
• $x_3^{(2)}=(7-1.4572+2(0.8564))/9=7.2556/9=0.8062$.

Step 4 — Stopping criterion and answer

Between iterations 4 and 5 all three components agree to four significant figures, so terminate.

Answer