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Matrix Algebra Part C: Pivoting and Reduced Row Echelon Form

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Matrix Algebra Part C: Pivoting and Reduced Row Echelon Form Lecture Notes 1: Matrix Algebra Part C: Pivoting and Reduced Row Echelon Form Peter J. Hammond revised 2018 September 23rd University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 1 of 88 Lecture Outline Pivoting to Reach the Reduced Row Echelon Form Example The Row Echelon Form The Reduced Row Echelon Form Determinants and Inverses Properties of Determinants Eight Basic Rules for Determinants Verifying the Product Rule Cofactor Expansion Expansion by Alien Cofactors and the Adjugate Matrix Invertible Matrices Minors, Dimensions, and Rank Column Rank, Row Rank, and Determinantal Rank Minor Determinants Solutions to Linear Equations General Equation Systems University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 2 of 88 Outline Pivoting to Reach the Reduced Row Echelon Form Example The Row Echelon Form The Reduced Row Echelon Form Determinants and Inverses Properties of Determinants Eight Basic Rules for Determinants Verifying the Product Rule Cofactor Expansion Expansion by Alien Cofactors and the Adjugate Matrix Invertible Matrices Minors, Dimensions, and Rank Column Rank, Row Rank, and Determinantal Rank Minor Determinants Solutions to Linear Equations General Equation Systems University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 3 of 88 Three Simultaneous Equations Consider the following system of three simultaneous equations in three unknowns, which depends upon two \exogenous" constants a and b: x + y − z = 1 x − y + 2z = 2 x + 2y + az = b It can be expressed, using an augmented 3 × 4 matrix, as : 1 1 −1 1 1 −1 2 2 1 2 a b Perhaps even more useful is the doubly augmented 3 × 7 matrix: 1 1 −1 1 1 0 0 1 −1 2 2 0 1 0 1 2 a b 0 0 1 whose last 3 columns are those of the 3 × 3 identity matrix I3. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 4 of 88 The First Pivot Step Start with the doubly augmented 3 × 7 matrix: 1 1 −1 1 1 0 0 1 −1 2 2 0 1 0 1 2 a b 0 0 1 First, we pivot about the element in row 1 and column 1 to eliminate or \zeroize" the other elements of column 1. This elementary row operation requires us to subtract row 1 from both rows 2 and 3. It is equivalent to multiplying 0 1 0 01 by the lower triangular matrix E1 = @−1 1 0A. −1 0 1 Note: this is the result of applying the same row operations to I3. The resulting 3 × 7 matrix is: 1 1 −1 1 1 0 0 0 −2 3 1 −1 1 0 0 1 a + 1 b − 1 −1 0 1 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 5 of 88 The Second Pivot Step After augmenting again by the identity matrix, we have: 1 1 −1 1 1 0 0 1 0 0 0 −2 3 1 −1 1 0 0 1 0 0 1 a + 1 b − 1 −1 0 1 0 0 1 Next, we pivot about the element in row 2 and column 2. 1 Specifically, multiply the second row by − 2 , then subtract the new second row from the third to obtain: 1 1 −1 1 1 0 0 1 0 0 3 1 1 1 1 0 1 − 2 − 2 2 − 2 0 0 − 2 0 5 1 3 1 1 0 0 a + 2 b − 2 − 2 2 1 0 2 1 Again, the pivot operation is equivalent to multiplying 01 0 01 1 by the lower triangular matrix E2 = @0 − 2 0A, 1 0 2 1 which is the result of applying the same row operation to I3. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 6 of 88 Case 1: Dependent Equations 5 In case 1, when a + 2 = 0, the equation system reduces to: x + y − z = 1 3 1 y − 2 z = − 2 1 0 = b − 2 1 In case 1A, when b 6= 2 , neither the last equation, nor the system as a whole, has any solution. 1 In case 1B, when b = 2 , the third equation is redundant. In this case, the first two equations have a general solution 3 1 3 1 3 1 with y = 2 z − 2 and x = z + 1 − y = z + 1 − 2 z + 2 = 2 − 2 z, where z is an arbitrary scalar. In particular, there is a one-dimensional set of solutions 3 along the unique straight line in R that passes through both: 3 1 (i) ( 2 ; − 2 ; 0), when z = 0; (ii) (1; 1; 1), when z = 1. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 7 of 88 Case 2: Three Independent Equations 1 1 −1 1 1 0 0 1 0 0 3 1 1 1 1 0 1 − 2 − 2 2 − 2 0 0 − 2 0 5 1 3 1 1 0 0 a + 2 b − 2 − 2 2 1 0 − 2 1 5 Case 2 occurs when a + 2 6= 0, 5 and so the reciprocal c := 1=(a + 2 ) is well defined. 5 Now divide the last row by a + 2 , or multiply by c, to obtain: 1 1 −1 1 1 0 0 3 1 1 1 0 1 − 2 − 2 2 − 2 0 1 3 3 1 0 0 1 (b − 2 )c − 2 c 2 c 2 c The system has been reduced to row echelon form in which the leading zeroes of each successive row form the steps (in French, ´echelons, meaning rungs) of a ladder (or ´echelle in French) which descends steadily as one goes from left to right. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 8 of 88 Case 2: Three Independent Equations, Third Pivot 1 1 −1 1 1 0 0 3 1 1 1 0 1 − 2 − 2 2 − 2 0 1 3 3 1 0 0 1 (b − 2 )c − 2 c 2 c 2 c Next, we zeroize the elements in the third column above row 3. To do so, pivot about the element in row 3 and column 3. This requires adding 1 times the last row to the first, 3 and 2 times the last row to the second. In effect, one multiplies 01 1 11 3 by the upper triangular matrix E3 := @0 1 2 A 0 0 1 1 1 0 The first three columns of the result are 0 1 0 0 0 1 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 9 of 88 Case 2: Three Independent Equations, Final Pivot As already remarked, the first three columns of the matrix we are left with are 1 1 0 0 1 0 0 0 1 The final pivoting operation involves subtracting the second row from the first, so the first three columns become the identity matrix 1 0 0 0 1 0 0 0 1 This is a matrix in reduced row echelon form because, given the leading non-zero element of any row (if there is one), all elements above this element are zero. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 10 of 88 Final Exercise Exercise 1. Find the last 4 columns of each 3 × 7 matrix produced by these last two pivoting steps. 2. Check that the fourth column solves the original system of 3 simultaneous equations. 3. Check that the last 3 columns form the inverse of the original coefficient matrix. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 11 of 88 Outline Pivoting to Reach the Reduced Row Echelon Form Example The Row Echelon Form The Reduced Row Echelon Form Determinants and Inverses Properties of Determinants Eight Basic Rules for Determinants Verifying the Product Rule Cofactor Expansion Expansion by Alien Cofactors and the Adjugate Matrix Invertible Matrices Minors, Dimensions, and Rank Column Rank, Row Rank, and Determinantal Rank Minor Determinants Solutions to Linear Equations General Equation Systems University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 12 of 88 Definition of Row Echelon Form Definition An m × n matrix A is in row echelon form just in case: 1. The first r ≤ m rows i 2 Nr each have a non-zero leading entry ai;`i in column `i such that aij = 0 for all j < `i . 2. Each successive leading entry is in a column to the right of the leading entry in the previous row. That is, given the leading element ai;`i 6= 0 of row i, one has ahj = 0 for all h > i and all j ≤ `i . 3. If r < m, then any row i 2 fr + 1;:::; mg = Nm n Nr has no leading entry, because all its elements are zero. This row without a leading entry must be below any row with a leading entry. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 13 of 88 Examples Assuming that α; β; γ 2 R n f0g, here are three examples of matrices in row echelon form: 0α 2 0 01 0α 2 0 01 0α 01 0 0 β 0 A = 0 0 β 0 ; B = B C ; C = 0 β @ A B0 0 0 γC @ A 0 0 0 γ @ A 0 0 0 0 0 0 Here are three examples of matrices that are not in row echelon form 00 11 01 21 01 01 D = @1 0A ; E = @0 1A ; F = @0 0A 0 0 0 1 0 1 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 14 of 88 Pivoting to Reach a Generalized Row Echelon Form Any m × n matrix A can be transformed into row echelon form by applying a series of determinant preserving row operations involving non-zero pivot elements.
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