Last Updated on October 26, 2026 by Maged kamel
Gauss Elimination with Scaled Pivoting- 2×2 Matrix
This Post demonstrates why scaled pivoting is necessary when solving a 2×2 linear system with Gaussian elimination using four-digit arithmetic and rounding. Direct elimination with the very small pivot 0.003 produces a huge multiplier, significant rounding errors, and an incorrect solution. The instructor then swaps the rows to use the larger pivot 5.291, performs elimination, and obtains the accurate solution x1 = 10 and x2 = 1. This Problem is taken from Numerical Analysis by Richard Burden.
Video for the solved Problem with Partial Pivoting: A Complete Guide- 2×2 Matrix
The following video illustrates this Post’s content.
Timestamps for Video:
- 00:00 Introduction to scaled pivoting for a 2×2 Matrix.
- 00:33 A solveProblemem# 2. 01:25 Find the absolute maximum Value S1 for a11 and a21.
- 02:26 Find the ratio of the absolute Value of a11 over S1.
- 03:14 Find the ratio of the absolute Value of a21 over S2 and swap rows.
- 05:10 Determine the Value of m21.
- 05:46 What are the values of the elements a21, a22, and b22?
- 07:02 The final values for x1 and x2.
Solved Problem for a 2×2 Matrix before scaled pivoting.
The following Problem includes two equations with two unknowns, and it requires using Gaussian elimination to find the Values of x1 and x2, with a given solution for the final Value.
Step 1: Find the maximum absolute Value for the first and Second rows.
First, the equations are: 0.003×1 + 59.14×2 = 59.17, and the Second equation is 5.291×1 – 6.130×2 = 46.78. We will check the maximum absolute Value for both the first and Second rows. For the first Row, the absolute maximum of 0.003 and 59.41 is 59.14, which is called S1.
For the Second Row, the absolute maximum of 5.29 and 6.13 is 6.13, which is called S2. The next step is to find the ratio of |a11| to S1, which is 0.507262*10^-4. Please review the following slide.
find the ratio of |a21 | to S2, which is 0.86313, tht is bigger than 0.507262* 10^-4. This indicates that the Second Row will be swapped with the first Row; we will rewrite the augmented Matrix as follows. The equations are: 5.291×1 – 6.130×2 = 46.78, and the Second equation is 0.003×1 + 59.14×2 = 59.17.
The First Row is left unchanged, and we multiply -1764*R1 and add it to R2. The Second Row will be (a21=0, while a22=-104300. In the next slide, the new Value of b22 equals -104353.22, which can be approximated as -104—the Value of x2 = +1.001. We back-substitute into the first equation to find x1.
Find the Value of 2m21 by dividing -0.003/5.291 = -0.000567. Please check the following slide image for more details.
Step 2: Find the values of a21, a22, and b21.
The following slide shows the elements we need to find: L22, L32, L33, and U23.
The Value of a21 is equal to zero. The Value of a22 is equal to 59.143. The Value of b21 is equal to 59.143.
Step 3: What are the values for x1 and x2?
To get this x2 Value, we divide 59.143 by 59.143, which gives x2 = +1. For x1, we back-substitute: x1 = (46.78 + 6.1381)/5.291 = 10.00. Both values of x1 and x2 match the exact solution. Please refer to the following slide.
You can view and download the PDF for the solved Problem from this Link.
Link to the Previous Post: 13-Gauss Elimination with Partial Pivoting: 2×2 Matrix
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