--- title: "Gauss elimination/Jordan with pivoting" subject: "Numerical Method" chapter: 3 key: "semester_4/numerical_method:gauss-jordan-partial-pivoting" source: https://www.bctengineer.com/q/semester_4/numerical_method/gauss-jordan-partial-pivoting site: BCT Engineering --- # Gauss elimination/Jordan with pivoting Numerical Method · Chapter 3: Solution of system of linear algebraic equations - Asked in: 2071 Bhadra, 8 marks - Video: [Gauss Elimination With Partial Pivoting Example | Numerical Methods](https://www.youtube.com/watch?v=hc1VMzVVtMg) — StudySession - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/numerical_method/gauss-jordan-partial-pivoting - Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-3 ## Question Solve the following linear equations using Gauss Elimination or Gauss Jordan method using partial pivoting. 2x + 3y + 2z = 2 10x + 3y + 4z = 16 3x + 6y + z = 6 ## Short answer Applying Gauss-Jordan elimination with partial pivoting, the augmented matrix reduces to the identity, yielding the solution x = 2.0909, y = 0.1818, z = -1.3636. ## Answer **Gauss-Jordan method with partial pivoting:** at every stage the largest element (in absolute value) of the pivot column is moved to the pivot position by a row swap, the pivot row is divided by the pivot to make it 1, and the other entries of that column (above and below) are made 0. The final matrix is the identity, so the right column is the solution. #### Step 1: System ```text 2x + 3y + 2z = 2 10x + 3y + 4z = 16 3x + 6y + z = 6 ``` #### Step 2: Reduce to the identity form Augmented matrix [A | b]: ```text 2.0000 3.0000 2.0000 | 2.0000 10.0000 3.0000 4.0000 | 16.0000 3.0000 6.0000 1.0000 | 6.0000 ``` Column 1: largest pivot is in row 2, so swap R1 ↔ R2: ```text 10.0000 3.0000 4.0000 | 16.0000 2.0000 3.0000 2.0000 | 2.0000 3.0000 6.0000 1.0000 | 6.0000 ``` R1 ← R1 / 10.0000: ```text 1.0000 0.3000 0.4000 | 1.6000 2.0000 3.0000 2.0000 | 2.0000 3.0000 6.0000 1.0000 | 6.0000 ``` R2 ← R2 − (2.0000) R1; R3 ← R3 − (3.0000) R1: ```text 1.0000 0.3000 0.4000 | 1.6000 0.0000 2.4000 1.2000 | -1.2000 0.0000 5.1000 -0.2000 | 1.2000 ``` Column 2: largest pivot is in row 3, so swap R2 ↔ R3: ```text 1.0000 0.3000 0.4000 | 1.6000 0.0000 5.1000 -0.2000 | 1.2000 0.0000 2.4000 1.2000 | -1.2000 ``` R2 ← R2 / 5.1000: ```text 1.0000 0.3000 0.4000 | 1.6000 0.0000 1.0000 -0.0392 | 0.2353 0.0000 2.4000 1.2000 | -1.2000 ``` R1 ← R1 − (0.3000) R2; R3 ← R3 − (2.4000) R2: ```text 1.0000 0.0000 0.4118 | 1.5294 0.0000 1.0000 -0.0392 | 0.2353 0.0000 0.0000 1.2941 | -1.7647 ``` R3 ← R3 / 1.2941: ```text 1.0000 0.0000 0.4118 | 1.5294 0.0000 1.0000 -0.0392 | 0.2353 0.0000 0.0000 1.0000 | -1.3636 ``` R1 ← R1 − (0.4118) R3; R2 ← R2 − (-0.0392) R3: ```text 1.0000 0.0000 0.0000 | 2.0909 0.0000 1.0000 0.0000 | 0.1818 0.0000 0.0000 1.0000 | -1.3636 ``` #### Step 3: Solution From the final matrix (identity on the left), the unknowns are read directly: ```text x = 2.0909 y = 0.1818 z = -1.3636 ``` **Answer: x = 2.0909, y = 0.1818, z = -1.3636** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In column 1 of the augmented matrix the first step swaps R1 and R2. Why? A. Row 2 has the largest right-hand side B. 10 is the largest absolute value in that column, so it becomes the pivot C. Row 1 is always swapped first D. To make the matrix symmetric Answer: B. Partial pivoting moves the largest absolute value in the pivot column to the pivot position. (Key point: Partial pivoting rule) 2. After eliminating column 1, the column-2 entries below the pivot are 2.4 and 5.1. What is done? A. R2 and R3 are swapped so that 5.1 is the pivot B. Nothing; 2.4 stays as the pivot C. R1 and R2 are swapped D. Both rows are added Answer: A. 5.1 is larger in absolute value than 2.4, so row 3 is moved up as pivot row. (Key point: Second pivot) 3. How does Gauss-Jordan differ from plain Gauss elimination in this answer? A. It stops at upper triangular form and back-substitutes B. It never uses row swaps C. It solves by iteration D. Entries above the pivot are also made 0, ending in an identity matrix so the right column is the solution Answer: D. Gauss-Jordan clears the column above and below each pivot, so no back substitution is needed. (Key point: Gauss-Jordan finish) 4. What is the solution? A. x = 2.0909, y = 0.1818, z = -1.3636 B. x = 1.6, y = 0.2353, z = -1.7647 C. x = 2.0909, y = -1.3636, z = 0.1818 D. x = 0.3, y = 1.0, z = 1.2941 Answer: A. The final identity form gives x = 2.0909, y = 0.1818, z = -1.3636. 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