Title Solving system of linear equations_02.pdf ✅

Page | 1 Contents:  Reduced Row Echelon Form (RREF)  Gauss–Jordan Elimination Method (An extension of Gaussian Elimination)  Solving System of Linear Equations by Gauss–Jordan Elimination  Inverse of a matrix  Properties of inverse matrices  Finding inverse of a matrix by Gauss–Jordan Elimination  Solving system of linear equations by using the inverse matrix _________________________________________________________________________________________________________
Reduced Row Echelon Form Page | 2 We recall that A matrix is in row echelon form if it satisfies the following properties: 1. Any rows consisting entirely of zeros are at the bottom. 2. In each nonzero row, the first nonzero entry (called the leading entry) is in a column to the left of any leading entries below it. Examples of matrices in row echelon form: ൥ 2 0 0 4 െ1 0 1 2 0 ൩, ൥ 1 0 0 0 1 0 1 5 4 ൩, ൥ 1 0 0 1 0 0 2 1 0 1 3 0 ൩, ൦ 0 0 0 0 2 0 0 0 0 െ1 0 0 1 1 0 0 െ1 2 4 0 3 2 0 5 ൪ Definition A matrix is in reduced row echelon form if it satisfies the following properties: 1. It is in row echelon form. 2. The leading entry in each nonzero row is a 1 (called a leading 1). 3. Each column that has a leading 1 has zeros in every position above and below its leading 1. For example, the following matrix is in reduced row echelon form: ⎣ ⎢ ⎢ ⎢ ⎡ 1 0 0 0 0 3 0 0 0 0 0 1 0 0 0 0 0 1 0 0 െ2 5 2 0 0 1 െ3 െ2 0 0 0 0 0 1 0⎦ ⎥ ⎥ ⎥ ⎤ For 2 ൈ 2 matrices, the possible reduced row echelon forms are ቂ 1 0 0 1 ቃ, ቂ 1 0 ∗ 0ቃ, ቂ 0 0 1 0 ቃ, and ቂ 0 0 0 0 ቃ where ∗ can be any number. Note. Unlike the row echelon form, the reduced row echelon form of a matrix is unique.
Gauss–Jordan Elimination Page | 3 In Gauss–Jordan elimination, we proceed as in Gaussian elimination (with leading 1 in each nonzero row) but reduce the augmented matrix to reduced row echelon form. Steps for Gauss‐Jordan Elimination: 1. Write the augmented matrix of the system of linear equations. 2. Use elementary row operations to reduce the augmented matrix to reduced row echelon form. 3. If the resulting system is consistent, solve for the leading variables in terms of any remaining free variables. Example 1. Solve by using Gauss–Jordan elimination 2xଵ ൅ 5xଶ ൅ 3xଷ ൌ 11 െxଵ ൅ 3xଶ ൅ xଷ ൌ 5 xଵ ൅ xଶ െ 2xଷ ൌ െ3 Solution. The system has its matrix form Ax ൌ b where, A ൌ ൥ 2 െ1 1 5 3 1 3 1 െ2 ൩, x ൌ ൥ xଵ xଶ xଷ ൩ and b ൌ ൥ 11 5 െ3 ൩ The augmented matrix of the above system is ሾA|bሿ ൌ ൥ 2 െ1 1 5 3 1 3 1 െ2 อ 11 5 െ3 ൩ ோభ↔ோయ ሱ⎯⎯⎯ሮ ൥ 1 െ1 2 1 3 5 െ2 1 3 อ െ3 5 11 ൩ ோమାோభ→ோమ ோయିଶோభ→ோయ ሱ⎯⎯⎯⎯⎯⎯⎯ሮ ൥ 1 0 0 1 4 3 െ2 െ1 7 อ െ3 2 17 ൩ ோమିோయ→ோమ ሱ⎯⎯⎯⎯⎯⎯ሮ ൥ 1 0 0 1 1 3 െ2 െ8 7 อ െ3 െ15 17 ൩
Gauss–Jordan Elimination Page | 4 ோయିଷோమ→ோయ ሱ⎯⎯⎯⎯⎯⎯⎯ሮ ൥ 1 0 0 1 1 0 െ2 െ8 31 อ െ3 െ15 62 ൩ ଵ ଷଵ ோయ→ோయ ሱ⎯⎯⎯⎯⎯ሮ ൥ 1 0 0 1 1 0 െ2 െ8 1 อ െ3 െ15 2 ൩ ோమା଼ோయ→ோమ ோభାଶோయ→ோభ ሱ⎯⎯⎯⎯⎯⎯⎯ሮ ൥ 1 0 0 1 1 0 0 0 1 อ 1 1 2 ൩ ோభିோమ→ோభ ሱ⎯⎯⎯⎯⎯⎯ሮ ൥ 1 0 0 0 1 0 0 0 1 อ 0 1 2 ൩ Therefore, the required solution is ൥ xଵ xଶ xଷ ൩ ൌ ൥ 0 1 2 ൩. which is the unique solution to the given system. ∎ Example 2. Solve the system using Gauss–Jordan elimination xଵ െ xଶ െ xଷ ൅ 2xସ ൌ 1 2xଵ െ 2xଶ െ xଷ ൅ 3xସ ൌ 3 െxଵ ൅ xଶ െ xଷ ൌ െ3 Solution. The augmented matrix is ൥ 1 2 െ1 െ1 െ2 1 െ1 െ1 െ1 2 3 0 อ 1 3 െ3 ൩ R2െ2R1→R2 R3൅R1→R3 ሱ⎯⎯⎯⎯⎯⎯ሮ ൥ 1 0 0 െ1 0 0 െ1 1 െ2 2 െ1 2 อ 1 1 െ2 ൩ ଶோమାோయ→ோయ ሱ⎯⎯⎯⎯⎯⎯⎯ሮ ൥ 1 0 0 െ1 0 0 െ1 1 0 2 െ1 0 อ 1 1 0 ൩ ோభାோమ→ோభ ሱ⎯⎯⎯⎯⎯⎯ሮ ൥ 1 0 0 െ1 0 0 0 1 0 1 െ1 0 อ 2 1 0 ൩