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Math 3191 Applied Linear Algebra Lecture 2: Row Reduction/Vector Equations

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Math 3191 Applied Linear Algebra Lecture 2: Row Reduction/Vector Equations Math 3191 Applied Linear Algebra Lecture 2: Row Reduction/Vector Equations Stephen Billups University of Colorado at Denver Math 3191Applied Linear Algebra – p.1/25 Announcements Please discard pages 15–25 from Lecture 1. (These slides have been redone to be consistent with the textbook). Math 3191Applied Linear Algebra – p.2/25 Outline Review Row Reduction Vector Equations Math 3191Applied Linear Algebra – p.3/25 Review Last time, we introduced elementary row operations. Key Fact: Elementary row operations do not alter the solution set. Tonight: we look out how to use elementary row operations to solve a linear system. Key Concepts: Reduced echelon form. (RREF) Row reduction. Math 3191Applied Linear Algebra – p.4/25 Gauss-Jordan Elimination Apply a sequence of elementary row operations to get the matrix into a form that is trivial to solve. Example: A series of elementary row operations yields the following transformation 2 1 1 1 3 3 2 1 0 0 1 3 2 −3 −1 −8 =) 0 1 0 4 6 7 6 7 6 7 6 7 4 −1 2 2 3 5 4 0 0 1 −2 5 The righthand matrix corresponds to the system of equations: x1 = 1 x2 = 4 x3 = −2 Trivial to solve!! The above matrix is in a special form called the reduced row echelon form. Math 3191Applied Linear Algebra – p.5/25 Reduced Row Echelon Form: Zero rows are at the bottom. The leftmost nonzero entry of each nonzero row, (called its pivot) equals 1. Each pivot is further to the right than the pivot in the row above it. Each pivot is the only nonzero entry in its column. Examples: 2 1 2 0 0 3 2 1 0 2 2 3 0 0 1 0 0 1 −1 3 6 7 6 7 6 7 6 7 4 0 0 0 1 5 4 0 0 0 0 5 Math 3191Applied Linear Algebra – p.6/25 Examples–NOT RREF 1 2 1 0 0 0 0 0 2 0 0 1 0 3 2 0 1 0 3 3 0 0 0 1 0 0 1 −2 6 7 6 7 4 5 4 5 Nonzero in pivot col. Zero row not at bottom. 1 0 2 2 2 0 2 −1 3 3 0 0 0 0 6 7 4 5 Pivot =6 1 Math 3191Applied Linear Algebra – p.7/25 Exercise Which of the following are in RREF? 1 0 7 1 0 3 0 1 −8 A. 0 1 −2 . C. 2 3 : 2 3 0 0 0 0 0 1 6 7 6 7 6 0 0 0 7 4 5 6 7 1 0 −2 1 2 4 5 B. : 1 0 0 1 " 0 1 −1 3 1 # D. 2 0 0 1 −1 3 : 0 0 0 0 6 7 4 5 Math 3191Applied Linear Algebra – p.8/25 Interpretting RREF (for Augmented Matrix) Inconsistent system: If there is a pivot in the right-hand side, the system is inconsistent (i.e., it has no solution). 1 2 0 0 x = −2x =) 1 2 " 0 0 0 1 # 0 = 1 Impossible! NOTE: You can usually detect inconsistency before reaching RREF: 2 4 6 2 4 3 =) (Inconsistent) " 3 6 8 # " 0 0 −1 # Math 3191Applied Linear Algebra – p.9/25 Free Variables Non-pivot columns correspond to free variables. Free variables can take on any values, and each choice of values yields a solution. The (general) solution to the linear system is written in terms of the free variables. Example: 2 1 2 0 3 0 7 3 0 0 1 0 0 1 6 7 6 7 6 0 0 0 0 1 2 7 6 7 6 7 4 0 0 0 0 0 0 5 Free variables = x2 and x4. Two ways to write the solution: x1 = −2s − 3t + 7 x1 = −2x2 − 3x4 + 7 x2 = s x3 = 1 or x3 = 1 x5 = 2 x4 = t x2; x4 free x5 = 2 Math 3191Applied Linear Algebra – p.10/25 Exercise What is the general solution of the system whose augmented matrix is as follows? 1 2 0 1 2 0 0 1 −1 3 : 0 0 0 0 6 7 4 5 Math 3191Applied Linear Algebra – p.11/25 Uniqueness of RREF Theorem 1: Each matrix is row equivalent to one and only one reduced row echelon matrix. The order of row operations doesn’t matter–you always get the same RREF. The book also talks about row echelon form (REF). Look over this definition on your own. Row echelon form is not unique. Math 3191Applied Linear Algebra – p.12/25 0 0 −2 0 7 12 originally row 1 0 0 5 0 −17 −29 row 3 plus (-1) times the pivot row 0 0 −4 0 15 26 row 4 plus (-4) times the pivot row Row Reduction Algorithm − 2 0 0 2 0 7 12 3 2 4 −10 6 12 28 6 7 Example: 6 7 6 2 4 −5 6 −5 −1 7 6 7 6 7 4 8 16 −44 24 63 138 5 Step 1: Locate the pivot column (the leftmost column that does not consist entirely of zeros). (Ex: Column 1). Step 2: Choose a row with a nonzero entry in the pivot column. Interchange this row with the top row. (For consistency, choose the topmost row with a nonzero entry. (Ex: row 2 ). Step 3: Create zeros in the pivot column below the pivot by adding multiples of the pivot row to the remaining rows: − 2 2 4 10 6 12 28 3 Originally row 2 6 7 6 7 6 7 6 7 6 7 4 5 Math 3191Applied Linear Algebra – p.13/25 0 0 5 0 −17 −29 row 3 plus (-1) times the pivot row 0 0 −4 0 15 26 row 4 plus (-4) times the pivot row Row Reduction Algorithm − 2 0 0 2 0 7 12 3 2 4 −10 6 12 28 6 7 Example: 6 7 6 2 4 −5 6 −5 −1 7 6 7 6 7 4 8 16 −44 24 63 138 5 Step 1: Locate the pivot column (the leftmost column that does not consist entirely of zeros). (Ex: Column 1). Step 2: Choose a row with a nonzero entry in the pivot column. Interchange this row with the top row. (For consistency, choose the topmost row with a nonzero entry. (Ex: row 2 ). Step 3: Create zeros in the pivot column below the pivot by adding multiples of the pivot row to the remaining rows: − 2 2 4 10 6 12 28 3 Originally row 2 0 0 −2 0 7 12 originally row 1 6 7 6 7 6 7 6 7 6 7 4 5 Math 3191Applied Linear Algebra – p.13/25 0 0 −4 0 15 26 row 4 plus (-4) times the pivot row Row Reduction Algorithm − 2 0 0 2 0 7 12 3 2 4 −10 6 12 28 6 7 Example: 6 7 6 2 4 −5 6 −5 −1 7 6 7 6 7 4 8 16 −44 24 63 138 5 Step 1: Locate the pivot column (the leftmost column that does not consist entirely of zeros). (Ex: Column 1). Step 2: Choose a row with a nonzero entry in the pivot column. Interchange this row with the top row. (For consistency, choose the topmost row with a nonzero entry. (Ex: row 2 ). Step 3: Create zeros in the pivot column below the pivot by adding multiples of the pivot row to the remaining rows: − 2 2 4 10 6 12 28 3 Originally row 2 0 0 −2 0 7 12 originally row 1 6 7 6 7 6 0 0 5 0 −17 −29 7 row 3 plus (-1) times the pivot row 6 7 6 7 4 5 Math 3191Applied Linear Algebra – p.13/25 Row Reduction Algorithm − 2 0 0 2 0 7 12 3 2 4 −10 6 12 28 6 7 Example: 6 7 6 2 4 −5 6 −5 −1 7 6 7 6 7 4 8 16 −44 24 63 138 5 Step 1: Locate the pivot column (the leftmost column that does not consist entirely of zeros). (Ex: Column 1). Step 2: Choose a row with a nonzero entry in the pivot column. Interchange this row with the top row. (For consistency, choose the topmost row with a nonzero entry. (Ex: row 2 ). Step 3: Create zeros in the pivot column below the pivot by adding multiples of the pivot row to the remaining rows: − 2 2 4 10 6 12 28 3 Originally row 2 0 0 −2 0 7 12 originally row 1 6 7 6 7 6 0 0 5 0 −17 −29 7 row 3 plus (-1) times the pivot row 6 7 6 7 4 0 0 −4 0 15 26 5 row 4 plus (-4) times the pivot row Math 3191Applied Linear Algebra – p.13/25 0 0 0 0 1=2 1 row 3 plus 5/2 times the pivot row 0 0 0 0 1 2 row 4 plus (-2) times the pivot row Row Reduction Alg. (cont). Step 4: Cover up top row and repeat Steps 1-3 on the remaining part of the matrix. − 2 2 4 10 6 12 28 3 0 0 −2 0 7 12 6 7 6 7 6 0 0 5 0 −17 −29 7 6 7 6 7 4 0 0 −4 0 15 26 5 − 2 2 4 10 6 12 28 3 cover (ignore) this row 0 0 −2 0 7 12 pivot column = third column 6 7 6 7 6 7 6 7 6 7 4 5 Math 3191Applied Linear Algebra – p.14/25 0 0 0 0 1 2 row 4 plus (-2) times the pivot row Row Reduction Alg.
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