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UPSC 2021 Maths Optional Paper 2 Q7b — Step-by-Step Solution

15 marks · Section B

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

Question

Solve $3x_1+9x_2-2x_3=11$, $4x_1+2x_2+13x_3=24$, $4x_1-2x_2+x_3=-8$ correct to 4 sig figs by Gauss-Seidel, after verifying applicability in transformed form.

Technique

Rearrange rows to achieve diagonal dominance; standard Gauss-Seidel iteration with new ordering; iterate until convergence.

Solution

Step 1 — Check diagonal dominance

For Gauss-Seidel convergence (sufficient condition): strict diagonal dominance, $|a_{ii}|>\sum_{j\ne i}|a_{ij}|$ for each row.

Row 1: $|3|$ vs $9+2=11$. $3<11$, fails. Row 2: $|2|$ vs $4+13=17$. Fails. Row 3: $|1|$ vs $4+2=6$. Fails.

Not diagonally dominant in given order. Need to rearrange.

Step 2 — Rearrange rows

Look for permutations making diagonal dominant:

• Choose row with large coeff of $x_1$ for first equation. Coefficients of $x_1$: 3, 4, 4. Take 4 (rows 2 or 3) — both candidates.
• Choose row with large coeff of $x_2$ for second equation: coefficients 9, 2, -2. Take 9 (row 1).
• Choose row with large coeff of $x_3$ for third equation: 13.

Re-order: row 2 → first equation, row 1 → second, row 2 again? No, each row used once.

Try: equation 1 (uses row 2): $4x_1+2x_2+13x_3=24$ — coefficient 4 for $x_1$, 13 for $x_3$. Place this as the third equation (so $x_3$ on diagonal).

Equation 1 (target $x_1$): need large coefficient for $x_1$. Row 3: $4x_1-2x_2+x_3=-8$. Coefficient 4 vs $2+1=3$ — diagonally dominant for row 3.

Equation 2 (target $x_2$): row 1: $3x_1+9x_2-2x_3=11$. Coefficient 9 vs $3+2=5$ — dominant ✓.

Equation 3 (target $x_3$): row 2: $4x_1+2x_2+13x_3=24$. Coefficient 13 vs $4+2=6$ — dominant ✓.

Rearranged system:

All three rows diagonally dominant ✓.

Step 3 — Iterative formulas

$x_1=(-8+2x_2-x_3)/4$. $x_2=(11-3x_1+2x_3)/9$. $x_3=(24-4x_1-2x_2)/13$.

Step 4 — Iterate (initial guess $x_1=x_2=x_3=0$)

Iter 1: $x_1=(-8+0-0)/4=-2$. $x_2=(11-3(-2)+0)/9=(11+6)/9=17/9\approx 1.889$. $x_3=(24-4(-2)-2(1.889))/13=(24+8-3.778)/13=28.222/13\approx 2.171$.

Iter 2: $x_1=(-8+2(1.889)-2.171)/4=(-8+3.778-2.171)/4=-6.393/4\approx -1.598$. $x_2=(11-3(-1.598)+2(2.171))/9=(11+4.795+4.343)/9=20.138/9\approx 2.238$. $x_3=(24-4(-1.598)-2(2.238))/13=(24+6.391-4.476)/13=25.915/13\approx 1.993$.

Iter 3: $x_1=(-8+2(2.238)-1.993)/4=(-8+4.476-1.993)/4=-5.517/4\approx -1.379$. $x_2=(11-3(-1.379)+2(1.993))/9=(11+4.137+3.986)/9=19.124/9\approx 2.125$. $x_3=(24-4(-1.379)-2(2.125))/13=(24+5.517-4.25)/13=25.267/13\approx 1.944$.

Iter 4: $x_1=(-8+2(2.125)-1.944)/4=(-8+4.250-1.944)/4=-5.694/4\approx -1.424$. $x_2=(11-3(-1.424)+2(1.944))/9=(11+4.271+3.887)/9=19.158/9\approx 2.129$. $x_3=(24-4(-1.424)-2(2.129))/13=(24+5.694-4.257)/13=25.437/13\approx 1.957$.

Iter 5: $x_1=(-8+2(2.129)-1.957)/4=(-8+4.257-1.957)/4=-5.700/4\approx -1.425$. $x_2=(11-3(-1.425)+2(1.957))/9=(11+4.275+3.914)/9=19.189/9\approx 2.132$. $x_3=(24-4(-1.425)-2(2.132))/13=(24+5.700-4.263)/13=25.437/13\approx 1.957$.

Iterates have stabilised.

Answer