--- title: "Gauss elimination: 4x4 system" subject: "Numerical Method" chapter: 3 key: "semester_4/numerical_method:gauss-elimination-4x4-eleven" source: https://www.bctengineer.com/q/semester_4/numerical_method/gauss-elimination-4x4-eleven site: BCT Engineering --- # Gauss elimination: 4x4 system Numerical Method · Chapter 3: Solution of system of linear algebraic equations - Asked in: 2068 Magh, 8 marks - Video: [Gaussian Elimination | Solving a 4x4 System of Linear Equations](https://www.youtube.com/watch?v=i_SzpPqKTf4) — John's Maths Book - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/numerical_method/gauss-elimination-4x4-eleven - Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-3 ## Question 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 ## 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. ## Answer **Gauss elimination:** forward elimination to upper triangular form, then back substitution. #### Step 1: System ```text 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]: ```text 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: ```text 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: ```text 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: ```text 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 ```text 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** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the first elimination, what multiplier is used for R4? A. 2 B. 5/2 = 2.5 C. 5 D. 0.4 Answer: B. Multiplier = coefficient to remove / pivot = 5/2 = 2.5. (Key point: First multiplier) 2. Which multiplier is used to remove 4.25 below the pivot 2.5 in the last stage? A. 4.25/2.5 = 1.7 B. 2.5/4.25 = 0.588 C. 3.125 D. 1.25 Answer: A. Multiplier = 4.25/2.5 = 1.7. (Key point: Third-stage multiplier) 3. From the last row 3.0 x4 = 13.2 and the third row 2.5 x3 + 2.5 x4 = 2.75, what are x4 and x3? A. x4 = 4.4, x3 = 3.3 B. x4 = 13.2, x3 = -3.3 C. x4 = 4.4, x3 = -3.3 D. x4 = 3, x3 = -2.2 Answer: C. x4 = 13.2/3 = 4.4; x3 = (2.75 - 2.5 × 4.4)/2.5 = -3.3. (Key point: Back substitution) ## Close variants set in other exams - [Gauss elimination: x+2y+3z=6](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Agauss-elimination-x-2y-3z) — Numerical Method, chapter 3, asked 3× - [Gauss elimination, partial pivoting, 4 unknowns](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Agauss-elimination-four-pivoting) — Numerical Method, chapter 3, asked 4× ## Related in Numerical Method - [LU factorization: 5x1+2x2+3x3](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Alu-factorization-5x1) — Numerical Method, chapter 3, asked 2× - [Gauss-Seidel: four unknowns](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Agauss-seidel-four-unknowns) — Numerical Method, chapter 3 - [Gauss elimination/Jordan with pivoting](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Agauss-jordan-partial-pivoting) — Numerical Method, chapter 3, asked 2× - [Gauss-Seidel: 10x+y-z=11.19](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Agauss-seidel-10x-y-z) — Numerical Method, chapter 3 ## Related in other subjects - [Two-mass system equations, X2/F, F-I and F-V analogies](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Ade-tf-fi-fv-m1-m2) — Control Systems, chapter 2 - [Transfer function from state equations, output x2](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-from-de-3x1-x2-y) — Control Systems, chapter 8, asked 2×