Gauss-Jordan Method Explained for Solving Systems of Equations
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The Gauss-Jordan method is an algorithm for solving systems of linear equations. It transforms the initial matrix of coefficients into an identity matrix with the solutions along the main diagonal through a series of row operations. The method works by choosing a pivot element from each row and performing elimination to clear all other elements in the column, with the goal of leaving only non-zero elements along the main diagonal at the end. An example applying the Gauss-Jordan method to a system of 4 equations with 4 unknowns is shown step-by-step.
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Gauss-Jordan Method Explained for Solving Systems of Equations
- 1. The Gauss-Jordan method Algorithm The Gauss-Jordan method transforms the initial expanded matrix into a identity matrix and a right side equal to the solution that is sought: ↔ nn cx ..... cx cx = = = 22 11 ( ) ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = nnnnn n n b ... b b a...aa ..... a...aa a...aa b|A 2 1 21 22221 11211 ↔ ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ nc ... c c ... ..... ... ... 2 1 100 010 001 Example 1. Solve the system using the Gauss-Jordan method with a chosen pivot element from a row: 7 03253 10524 932 432 4321 321 421 −=−+− =−++ −=+− =+− xxx xxxx xxx xxx . Solution: We work the same way as with the Gauss method by choosing a pivot element from a row but the unknowns are excluded under the main diagonal as well as above it. The aim is to be left with non-zero elements only along the main diagonal. )4(: ↔↔ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ − − ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ −− − − − 7 0 9 10 1110 3253 3012 052 ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ − − 4 ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ 0 3 −− − − − 7 0 10 9 111 325 052 3012 4 . (-2) .(-3) ↔ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ − − ⎜⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ −− −− − − 7 14 1110 30 300 01 2 15 2 5 4 7 2 5 4 5 2 1 ↔ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ − − ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ − − 7 0 9 11 32 30 0 2 5 4 51 − − − 10 53 12 1 2 : )( 2 13 ↔ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ − − ⎜⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ −− −− −− − 7 14 1110 300 10 01 13 15 2 5 2 5 13 6 26 7 4 5 2 1 . )( .(1) ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ − − ⎜⎜ ⎜ ⎜ −0 00 ⎜ ⎜ ⎝ ⎛ − −− −− − 7 14 111 3 30 01 2 15 2 5 2 5 4 7 4 5 2 1 4 4 2 13 2 1 ↔13/2
- 2. ⎟⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ − − ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ − −− − 13 76 13 15 13 25 13 19 26 19 13 6 26 7 13 3 26 29 14 00 300 10 01 : )( 2 5 − ↔ -5/2 ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ − − − ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ − − −− − 13 76 5 28 13 15 13 25 13 19 26 19 5 6 13 6 26 7 13 3 26 29 00 100 10 01 . )( 26 7 . )( 26 29− . )( 26 19 − ↔ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ − − − ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ − − − 65 114 5 28 65 23 65 281 65 38 5 6 65 51 65 72 000 100 010 001 : )( 65 38 − ↔ ⎟⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ − − ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ − − 31000 100 010 001 5 28 65 23 65 281 5 6 65 51 65 72 . )( 5 6 . )(65 51 . )( 65 72 − ↔ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ − ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ 3 2 2 1 1000 0100 0010 0001 ↔ 3 2 2 1 4 3 2 1 = −= = = x x x x By Iliya Makrelov, ilmak@uni-plovdiv.bg