Discrete Mathematics..

Gauss–Jordan Elimination 345Solution⎛⎝ 1 2 −3 0 ⎞3 −1 1 3 ⎠ (augmented matrix)2 −3 4 1⎛∼ ⎝ 1 2 −3 0 ⎞0 −7 10 3 ⎠ [R 2 → (R 2 − 3R 1 )]0 −7 10 1 [R 3 → (R 3 − 2R 1 )]⎛⎞1 2 −3 0⎜∼ ⎝ 0 1 − 107− 3 ⎟7 ⎠ [R 2 → (R 2 ÷ (−7))]0 −7 10 1⎛1 0 − 1 ⎞67 7[R 1 → (R 1 − 2R 2 )]⎜∼ ⎝ 0 1 − 107− 3 ⎟7⎠0 0 0 −2 [R 3 → (R 3 + 7R 2 )]⎛1 0 − 1 ⎞67 7⎜∼ ⎝ 0 1 − 107− 3 ⎟7⎠0 0 0 1 [R 3 → (R 3 ÷ (−2))]⎛1 0 − 1 ⎞70 [R 1 → (R 1 − 6 7⎜∼ ⎝ 0 1 − 10 ⎟R 3)]70 ⎠ [R 2 → (R 2 + 3 7 R 3)].0 0 0 1The bottom row of the reduced row echelon matrix represents the equation0x + 0y + 0z = 1, which clearly has no solution. Hence the system of equationsis inconsistent. (Note that A is again a singular square matrix.)In each of the last three examples, A, the matrix of coefficients, was a squaresingular matrix and so any process of applying elementary row operations to theaugmented matrix will not reduce that part of the matrix to an identity matrix.Hence a system of equations where A is singular will not have a unique solution.Depending on the form of the reduced row echelon matrix, such a system haseither an infinite number of solutions or no solution at all.The method of Gauss–Jordan elimination can be used just as easily in the casewhere A is not a square matrix, i.e. where there are more equations than variablesor more variables than equations. We consider each of these cases in the twoexamples below.