[PDF] Lecture Notes 1: Matrix Algebra Part C: Pivoting and Reduced

Lecture Notes 1: Matrix Algebra Part C: Pivoting and Reduced

The Intermediate Matrices and Pivot Steps After k 1 pivoting operations have been completed, and column ‘ k 1 (with ‘ k 1 k 1) was the last to be used: 1 The rst or \top" k 1 rows of the m n matrix form a (k 1) n submatrix in row echelon form 2 The last or \bottom" m k + 1 rows of the m n matrix form an (m k + 1) n submatrix whose rst ‘

Appendix A - University of Texas at Austin

of each row is called a pivot, and the columns in which pivots appear are called pivot columns If two matrices in row-echelon form are row-equivalent, then their pivots are in exactly the same places When we speak of the pivot columns of a general matrix A, we mean the pivot columns of any matrix in row-echelon form that is row-equivalent to A

Theorem 6) The pivot columns of a matrix A form a basis for

Theorem (6) The pivot columns of a matrix A form a basis for the column space Col(A) Proof The proof has two parts: show the pivot columns are linearly independent and show the pivot columns span the column space We use the reduced echelon matrix of A in the proof We designate it by B If A and B have r pivot

Pivoting in Maple/Matlab/Mathematica

Here we have just called the pivot command, but did not save the output of the command into a variable If I check the value of the matrix A (by typing matrix(A) and pressing enter), I will see that its unchanged So let’s just recall the command again, this time storing the resulting matrix in a matrix B: B := pivot(A,1,1);

Lecture Notes 1: Matrix Algebra Part C: Pivoting and Matrix

Aunitriangularmatrix is a triangular matrix (upper or lower) for which all elements on the principal diagonal equal 1 Theorem The determinant of any unitriangular matrix is 1 Proof The determinant of any triangular matrix is the product of its diagonal elements, which must be 1 in the unitriangular case when every diagonal elements is 1

Excessive Pivot Ratios Cheatsheet - MSC Software

grid point id degree of freedom matrix/factor diagonal ratio matrix diagonal 11077 R2 1 84205E+12 2 06674E+03 1198 T3 -1 73615E+14 6 82135E+05

Solutions to Section 1 2 Homework Problems S F

3 x6 matrix and the last column of this augmented matrix cannot be a pivot column Since there are only three rows , there can be at most three pivot columns and we are told that these three pivot columns are among the first five columns 2 7 If the coefficient matrix of a linear system has a pivot position in every row , then

A quick example calculating the column space and the

Determine the column space of A = A basis for col A consists of the 3 pivot columns from the original matrix A Thus basis for col A = Note the basis for col A consists of exactly 3 vectors

[PDF] pivot de gauss matrice 2x2

[PDF] livre des merveilles du monde de marco polo fiche lecture

[PDF] le livre des merveilles marco polo texte intégral pdf

[PDF] la fameuse invasion de la sicile par les ours questionnaire de lecture

[PDF] la fameuse invasion de la sicile par les ours film

[PDF] mobilisation de connaissances ses exemple

[PDF] la fameuse invasion de la sicile par les ours résumé

[PDF] la fameuse invasion de la sicile par les ours fiche de lecture

[PDF] la fameuse invasion de la sicile par les ours analyse

[PDF] l autonomie en crèche

[PDF] exemple ec2

[PDF] le pianiste personnages principaux

[PDF] le pianiste résumé complet du film

[PDF] le pianiste personnages principaux livre

[PDF] methodologie ec1

Lecture Notes 1: Matrix Algebra

Part C: Pivoting and Reduced Row Echelon Form

Peter J. Hammond

revised 2020 September 16th University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 1 of 92

Lecture Outline

Pivoting to Reach the Reduced Row Echelon Form

Example

The Row Echelon Form

The Reduced Row Echelon Form

Determinants and InversesProperties of Determinants

Eight Basic Rules for Determinants

Verifying the Product Rule

Cofactor Expansion

Expansion by Alien Cofactors and the Adjugate Matrix

Invertible MatricesDimensions, Rank, and Minors

Column and Row Rank

Solutions to Linear Equation Systems

Minor Determinants and Determinantal Rank

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 2 of 92

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

Dimensions, Rank, and Minors

Column and Row Rank

Solutions to Linear Equation Systems

Minor Determinants and Determinantal Rank

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 3 of 92

Three Simultaneous Equations

Consider the following system

of three simultaneous equations in three unknowns, which depends upon two \exogenous" constantsaandb: x+yz= 1 xy+ 2z= 2 x+ 2y+az=b It can be expressed, using an augmented 34 matrix, as : 1 111 11 22 1 2ab Perhaps even more useful is the doubly augmented 37 matrix:

1 1111 0 0

11 220 1 0

1 2ab0 0 1

whose last 3 columns are those of the 33 identity matrixI3. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 4 of 92

The First Pivot Step

Start with the doubly augmented 37 matrix:

1 1111 0 0

11 220 1 0

1 2ab0 0 1

First, we

pivot ab outthe element in ro w1 and column 1 to eliminate or \zeroize" the other elements of column 1. This elementa ryro wop eration requ iresus to subtr actro w1 from both rows 2 and 3. It is equivalent to multiplying by the lo wertriangula r matrix E1=0 @1 0 0 1 1 0

1 0 11

A Note: this is the result of applying the same row operations toI3.

The resulting 37 matrix is:

1 1111 0 0

02 311 1 0

0 1a+ 1b11 0 1

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 5 of 92

The Second Pivot Step

After augmenting again by the identity matrix, we have:

1 1111 0 01 0 0

02 311 1 00 1 0

0 1a+ 1b11 0 10 0 1

Next, we pivot about the element in row 2 and column 2.

Specically, multiply the second row by12

then subtract the new second row from the third to obtain:

1 1111 0 01 0 0

0 132 121
2 12 0012 0

0 0a+52b12

32
12 10 12 1 Again, the pivot operation is equivalent to multiplying by the lo wertriangula r matrix E2=0 @1 0 0 012 0 0 12 11 A which is the result of applying the same row operation toI3. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 6 of 92

Case 1: Dependent Equations

In case 1 , whena+52 = 0, the equation system reduces to: x+yz= 1 y32 z=12

0 =b12

In case 1A , whenb6=12 , neither the last equation, nor the system as a whole, has any solution. In case 1B , whenb=12 , the third equation is redundant. In this case, the rst two equations have a general solution withy=32 z12 andx=z+ 1y=z+ 132 z+12 =32 12 z, wherezis an arbitrary scalar. In particular, there is a one-dimensional set of solutions along the unique straight line inR3that passes through both: (i) ( 32
;12 ;0), whenz= 0; (ii) (1;1;1), whenz= 1. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 7 of 92

Case 2: Three Independent Equations

1 1111 0 01 0 0

0 132 121
2 12 0012 0

0 0a+52b12

32
12 1012
1

Case 2

o ccurswhen a+52 6= 0, and so the reciprocalc:= 1=(a+52 ) is well dened.

Now divide the last row bya+52

, or multiply byc, to obtain:

1 1111 0 0

0 132 121
2 12

00 0 1(b12

)c 32
c32 c12 cThe system has been reduced toro wechelon fo rm in which the leading zeroes of each successive row form the steps (in French,echelons, meaning rungs) of a ladder (orechellein French) which descends steadily as one goes from left to right. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 8 of 92

Case 2: Three Independent Equations, Third Pivot

1 1111 0 0

0 132 121
2 12

00 0 1(b12

)c 32
c32 c12 cNext, 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 rst, and 32
times the last row to the second.

In eect, one multiplies

by the upp ertriangula r matrix E3:=0 @1 1 1 0 1 32

0 0 11

A

The rst three columns of the result are

1 1 0 0 1 0 0 0 1 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 9 of 92

Case 2: Three Independent Equations, Final Pivot

As already remarked, the rst three columns of the matrix we are left with are1 1 0 0 1 0 0 0 1 The nal pivoting operation involves subtracting the second row from the rst, so the rst three columns become the identity matrix 1 0 0 0 1 0 0 0 1

This is a matrix in

reduced ro we chelonfo rm b ecause, 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 92

Final Exercise

Exercise

1.Find the last 4 columns of each37matrix

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 coecient matrix. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 11 of 92

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

Dimensions, Rank, and Minors

Column and Row Rank

Solutions to Linear Equation Systems

Minor Determinants and Determinantal Rank

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 12 of 92

Denition of Row Echelon Form

Denition

AnmnmatrixAis inro wechelon fo rmjust in case:

1.

The rst rmrowsi2Nr

each have a non-zero leading entry ai;`iin column`i such thataij= 0 for allj< `i. 2. Each success iveleading entry is in a column to the right of the leading entry in the previous row. That is, given the leading elementai;`i6= 0 of rowi, one hasahj= 0 for allh>iand allj`i. 3.

If r 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 92

Examples

Assuming that;;

2Rn f0g,

here are three examples of matrices in row echelon form: A=0 @2 0 0 0 00 0 0 0 1 A ;B=0 B

B@2 0 0

0 00 0 0 0

0 0 0 01

C

CA;C=0

@0 0 0 01 A

Here are three examples of matrices

that are not in ro wechelon fo rm D=0 @0 1 1 0 0 01 A ;E=0 @1 2 0 1 0 11 A ;F=0 @1 0 0 0 0 11 A University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 14 of 92

Pivoting to Reach a Generalized Row Echelon Form

AnymnmatrixAcan be transformed into row echelon form by applying a series of determinant preserving row operations involving non-zero pivot elements 1.

Lo okfo rthe rst o r

leading non-zero column `1in the matrix. 2.

Find within column `1an elementai1`16= 0

quotesdbs_dbs13.pdfusesText_19