- Asked 3/13
- 2068 Magh · 8 marks
Gauss elimination: 4x4 system
Short answer
Forward elimination reduces the system to upper triangular form; back substitution yields the solution x1=1.1000, x2=-2.2000, x3=-3.3000, x4=4.4000.
Solve the following system of linear algebraic equations using the Gauss Elimination Method. 2x1 + 3x2 + 2x3 + 5x4 = 11 4x1 + 2x2 + 2x3 + 4x4 = 11 4x1 + x2 + 4x3 + 5x4 = 11 5x1 - 5x2 + 3x3 + x4 = 11
Gauss elimination: forward elimination to upper triangular form, then back substitution.
Step 1: System
2x1 + 3x2 + 2x3 + 5x4 = 11
4x1 + 2x2 + 2x3 + 4x4 = 11
4x1 + x2 + 4x3 + 5x4 = 11
5x1 - 5x2 + 3x3 + x4 = 11
Step 2: Forward elimination (reduce to upper triangular form)
Augmented matrix [A | b]:
2.0000 3.0000 2.0000 5.0000 | 11.0000
4.0000 2.0000 2.0000 4.0000 | 11.0000
4.0000 1.0000 4.0000 5.0000 | 11.0000
5.0000 -5.0000 3.0000 1.0000 | 11.0000
R2 ← R2 − (2.0000) R1; R3 ← R3 − (2.0000) R1; R4 ← R4 − (2.5000) R1:
2.0000 3.0000 2.0000 5.0000 | 11.0000
0.0000 -4.0000 -2.0000 -6.0000 | -11.0000
0.0000 -5.0000 0.0000 -5.0000 | -11.0000
0.0000 -12.5000 -2.0000 -11.5000 | -16.5000
R3 ← R3 − (1.2500) R2; R4 ← R4 − (3.1250) R2:
2.0000 3.0000 2.0000 5.0000 | 11.0000
0.0000 -4.0000 -2.0000 -6.0000 | -11.0000
0.0000 0.0000 2.5000 2.5000 | 2.7500
0.0000 0.0000 4.2500 7.2500 | 17.8750
R4 ← R4 − (1.7000) R3:
2.0000 3.0000 2.0000 5.0000 | 11.0000
0.0000 -4.0000 -2.0000 -6.0000 | -11.0000
0.0000 0.0000 2.5000 2.5000 | 2.7500
0.0000 0.0000 0.0000 3.0000 | 13.2000
Step 3: Back substitution
x4 = (13.2000) / 3.0000 = 4.4000
x3 = (2.7500 − (2.5000)(4.4000)) / 2.5000 = -3.3000
x2 = (-11.0000 − (-2.0000)(-3.3000) − (-6.0000)(4.4000)) / -4.0000 = -2.2000
x1 = (11.0000 − (3.0000)(-2.2000) − (2.0000)(-3.3000) − (5.0000)(4.4000)) / 2.0000 = 1.1000
Answer: x1 = 1.1000, x2 = -2.2000, x3 = -3.3000, x4 = 4.4000
✓ Answer checked by Claude Opus 5.5 (high effort)
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