← 2015 Paper 2

UPSC 2015 Maths Optional Paper 2 Q8b — Step-by-Step Solution

15 marks · Section B

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

Question

Find the solution of the system

using Gauss-Seidel method (make four iterations).

Technique

Standard Gauss-Seidel iteration using diagonal-dominant rearrangement; each new $x_i^{(k+1)}$ uses already-updated $x_1^{(k+1)},\dots,x_{i-1}^{(k+1)}$ and old $x_{i+1}^{(k)},\dots,x_n^{(k)}$.

Solution

Setup. Rearrange each equation to solve for the diagonal variable:

Diagonal dominance check: Each diagonal has $|10|$ vs. sum of off-diagonals $|−2|+|−1|+|−1|=4$, so $10>4$ ✓. Gauss-Seidel converges.

Initial guess: $x_i^{(0)}=0$ for all $i$.

Iteration 1 (use most recent values)

$x_1^{(1)}=(3+0+0+0)/10=0.3$. $x_2^{(1)}=(15+2(0.3)+0+0)/10=(15+0.6)/10=1.56$. $x_3^{(1)}=(27+0.3+1.56+0)/10=(28.86)/10=2.886$. $x_4^{(1)}=(-9+0.3+1.56+2(2.886))/10=(-9+0.3+1.56+5.772)/10=(-1.368)/10=-0.1368$.

After iter 1: $(0.3, 1.56, 2.886, -0.1368)$.

Iteration 2

$x_1^{(2)}=(3+2(1.56)+2.886-0.1368)/10=(3+3.12+2.886-0.1368)/10=(8.8692)/10=0.88692$.

$x_2^{(2)}=(15+2(0.88692)+2.886-0.1368)/10=(15+1.77384+2.886-0.1368)/10=(19.52304)/10=1.952304$.

$x_3^{(2)}=(27+0.88692+1.952304+2(-0.1368))/10=(27+0.88692+1.952304-0.2736)/10=(29.565624)/10=2.9565624$.

$x_4^{(2)}=(-9+0.88692+1.952304+2(2.9565624))/10=(-9+0.88692+1.952304+5.9131248)/10=(-0.2476512)/10=-0.02476512$.

After iter 2: $(0.88692, 1.952304, 2.9565624, -0.02476512)$.

Iteration 3

$x_1^{(3)}=(3+2(1.952304)+2.9565624-0.02476512)/10=(3+3.904608+2.9565624-0.02476512)/10=(9.83640528)/10=0.983640528$.

$x_2^{(3)}=(15+2(0.983640528)+2.9565624-0.02476512)/10=(15+1.967281056+2.9565624-0.02476512)/10=(19.899078)/10=1.9899078$.

$x_3^{(3)}=(27+0.983640528+1.9899078+2(-0.02476512))/10=(27+0.983640528+1.9899078-0.04953024)/10=(29.92401809)/10=2.992401809$.

$x_4^{(3)}=(-9+0.983640528+1.9899078+2(2.992401809))/10=(-9+0.983640528+1.9899078+5.984803618)/10=(-0.041648054)/10=-0.0041648054$.

After iter 3: $(0.9836, 1.9899, 2.9924, -0.00416)$ (rounded).

Iteration 4

$x_1^{(4)}=(3+2(1.9899078)+2.992401809-0.0041648054)/10=(3+3.9798156+2.992401809-0.0041648054)/10=(9.96805260)/10=0.996805260$.

$x_2^{(4)}=(15+2(0.996805260)+2.992401809-0.0041648054)/10=(15+1.99361052+2.992401809-0.0041648054)/10=(19.98184752)/10=1.998184752$.

$x_3^{(4)}=(27+0.996805260+1.998184752+2(-0.0041648054))/10=(27+0.996805260+1.998184752-0.008329611)/10=(29.98666040)/10=2.998666040$.

$x_4^{(4)}=(-9+0.996805260+1.998184752+2(2.998666040))/10=(-9+0.996805260+1.998184752+5.99733208)/10=(-0.00767791)/10=-0.000767791$.

After iter 4: $(0.9968, 1.9982, 2.9987, -0.000768)$.

Convergence to exact solution

The iterates clearly converge to $(1,2,3,0)$. Verify:

• $10(1)-2(2)-3-0=10-4-3=3$ ✓
• $-2(1)+10(2)-3-0=-2+20-3=15$ ✓
• $-1-2+10(3)-2(0)=27$ ✓
• $-1-2-2(3)+10(0)=-9$ ✓

So exact solution is $(x_1,x_2,x_3,x_4)=(1,2,3,0)$.

Summary table

Answer