ce26 lecture on matrices

8/9/2019 CE26 Lecture on Matrices

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CE 26

Lecture on Matrices

A. Matrix: Definition

B. Special Types of Matrices

C. Equality of Matrices

D. Elementary Operations on Matrices1. Matrix Addition

2. Scalar Multiplication

3. Matrix Subtraction

4. Matrix Multiplication

5. Transposition of a Matrix

E. Determinant of a Square Matrix

F. Adjoint of a Matrix

G. Inverse of a Matrix

H. Solution to System of Linear Equations(Direct Methods)

1. Inverse Method

2. Cramers Rule

3. LU Factorization

4. Gaussian Elimination

5. Gauss-Jordan Reduction Method

Matrix

A matrixis a rectangular array of numbers orfunctions arranged in rows and columnsusually designated by a capital letter andenclosed by brackets, parentheses or double

bars. A matrix may be denoted by:

=

mn2m1m

n22221

n11211

a...aa

:::

a...aa

a...aa

A

Unless stated, we assume that all ourmatrices are composed of real numbers.

The horizontal groups of elements are calledthe rowsof the matrix. The ith row of A is

[ ] ( )mi1a...aain2i1i


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The vertical groups of elements are called thecolumnsof the matrix. The jth column of A is

( )nj1

a

:

a

a

j3

j2

j1

The size of a matrixis denoted by m x n (mby n) where m is the number of rows and n is

the number of columns.

We refer to aij as the entryor the elementinthe ith row and jth column of the matrix.

We may often write a given matrix as

A = [aij].

Example:

4x38197

4110

5235

A

=

Columns

Rows

9 is element a32 located at3rd row and 2nd column ofmatrix A.

Size of Matrix A

Special Types of Matrices

A.A.A.A. Row Matrix or Row VectorRow Matrix or Row VectorRow Matrix or Row VectorRow Matrix or Row Vector is a matrixconsisting of only one row.

=

m

.

.

.i

.

.

.2

1

c

c

c

c

C

B = [b1 b2 . . . bj . . . bn]

B.B.B.B. Column Matrix or ColumnColumn Matrix or ColumnColumn Matrix or ColumnColumn Matrix or ColumnVectorVectorVectorVector is a matrix consistingof only one column.

C.C.C.C. Square MatrixSquare MatrixSquare MatrixSquare Matrix is a matrix in which the no.of rows equals the no. of columns.

Order of a Square Matrix is the number ofrows or columns of the matrix. Thus, we canjust refer to a 3x3 matrix as a square matrixof order 3.

3x3750

411

302

A

=


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Principal Diagonal or Main Diagonal of aSquare Matrix consists of the elements a11,a22, a33, ann.

3x3750

411302

A

=


The Trace of a Square Matrix is the sumof the elements on the main diagonal of thematrix. In the above matrix, trace = 2 + 1+ 7 = 10.

3x3750

411302

A

=

D.D.D.D. Upper Triangular MatrixUpper Triangular MatrixUpper Triangular MatrixUpper Triangular Matrix a square matrixall elements of which below the principaldiagonal are zero (aij = 0 for i>j).

=

33

2322

131211

u00

uu0

uuu

U

E.E.E.E. Lower Triangular MatrixLower Triangular MatrixLower Triangular MatrixLower Triangular Matrix a square matrixall elements of which above the principaldiagonal are zero (aij = 0 for i


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H.H.H.H. Identity MatrixIdentity MatrixIdentity MatrixIdentity Matrix represented by In, it is adiagonal matrix where all the elements

along the main diagonal are equal to 1 orunity.

=

100

010

001

I3

I.I.I.I. Null MatrixNull MatrixNull MatrixNull Matrix represented by is a matrix inwhich all the elements are zero.

=

0...00

:::

0...00

0...00

J.J.J.J. Symmetric MatrixSymmetric MatrixSymmetric MatrixSymmetric Matrix a square matrix whoseelement aij = aji.

=

354

522

421

S

K.K.K.K. Skew Symmetric MatrixSkew Symmetric MatrixSkew Symmetric MatrixSkew Symmetric Matrix a square matrixwhose element aij = -aji

=

054

502

420

T

Equality of Matrices

Two matrices A = (aij) and B = (bij) are equalif and only if the following conditions aresatisfied:

a) They have equal number of rows.

b) They have equal number of columns.

c) All elements in A agree with theelements in B. (aij=bij, for all i and j.)


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Example: The matrices

are equal if and only if x = 2, y = 1, z = 2, a =3, b = 4 and c = 1.

+

+

=

+

=

1c43

22b1x

112

Band

zb3

21ya

2c1x

A Elementary Operations onMatrices

A. MATRIX ADDITIONA. MATRIX ADDITIONA. MATRIX ADDITIONA. MATRIX ADDITION

If A = (aij) and B = (bij) are matrices of thesame size m x n, then the sum A + B isanother m x n matrix C = [cij] where cij = aij+ bij for i = 1 to m and j = 1 to n.

Matrix addition is accomplished by addingalgebraically corresponding elements in Aand B.

Example: Given two 2x3 matrices A and B

=

=

114

223B

123

421A

=

+++

+++=

+

=+

211

204

11)1(243

)2(42231

114

223

123

421BA

B. SCALAR MULTIPLICATIONB. SCALAR MULTIPLICATIONB. SCALAR MULTIPLICATIONB. SCALAR MULTIPLICATION

If A = (aij) is an m x n matrix and k is a realnumber (or a scalar), then the scalar multiple

of A by k is the m x n matrix C = [cij] where cij= k*aij for all i and j.

In other words, the matrix C is obtained bymultiplying each element of the matrix by thescalar k.

Examples:

=

156912

9306

5234

31023

=

28168

4820

1248

742

125

312

4


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C. MATRIX SUBTRACTIONC. MATRIX SUBTRACTIONC. MATRIX SUBTRACTIONC. MATRIX SUBTRACTION

If A and B are m x n matrices, the differencebetween A and B denoted as A B is obtainedfrom the addition of A and (-1)B.

A B = A + (-1)B

Matrix subtraction is accomplished bysubtracting from the elements of the firstmatrix the elements of the second matrixcorrespondingly.

Example:

Note: We can only add or subtract matriceswith the same number of rows and columns.

=

=

383

412B

475

243A

=

=

+

=

718

651

3487)3(5

42)1(423

383412

475243BA

D. MATRIX MULTIPLICATIOND. MATRIX MULTIPLICATIOND. MATRIX MULTIPLICATIOND. MATRIX MULTIPLICATION

If A = (aij) is an m x n matrix and B = (bij) isan n x p matrix, then the product of A and B,AB = C = [cij] is an m x p matrix where

for i = 1 to m and j = 1 to p

=

=n

1k

kjikijbac

The formula tells us that in order to get theelement cij of the matrix C, get the elementsof the ith row of A (the pre-multiplier) andthe elements on the jth column of B (the postmultiplier). Afterwards, obtain the sum of theproducts of corresponding elements on thetwo vectors.

Note:

The product is defined only if the number ofcolumns of the first factor A (pre-multiplier) is

equal to the number of rows of the secondfactor B (post-multiplier). If this is satisfied,we say that the matrices are conformable inthe order AB.

The formula An will be defined as A A A A

Examples:

=

=

13

42B

43

21A

=

++

++=

=

++

++=

20

1210BA

)4)(1()2(3)3)(1()1(3

)4)(4()2(2)3)(4()1(2BA

1618

68

AB

)1(4)4(3)3(4)2(3

)1(2)4(1)3(2)2(1AB

Note: Although AB andBA are defined it is notnecessary that AB = BA.


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Determinant

Another very important number associatedwith a square matrix A is the determinantdeterminantdeterminantdeterminant of

A which we will now define. This uniquenumber associated to a matrix A is useful inthe solutions of linear equation.

Permutation:

Let S={1, 2, 3, , n} be the set of integersfrom 1 to n, arranged in increasing order. Arearrangement a1a2a3an of the elements in Sis called a permutation of S.

By the Fundamental Principle of Counting wecan put any one of the n elements of S in thefirst position, any one of the remaining (n-1)elements in the second position, any one ofthe remaining (n-2) elements in the thirdposition, and so on until the nth position.Thus there are n(n-1)(n-2)3*2*1 = n!permutations of S. We refer to the set of allpermutations of S by Sn.

Examples:

If S = {1, 2, 3} then S3 = {123, 132, 213, 231,312, 321}

If S = {1, 2, 3, 4} then there are 4! = 24elements of S4.

Odd and Even Permutations

A permutation a1a2a3an is said to have aninversion if a larger number precedes asmaller one. If the total number of inversionin the permutation is even, then we say that

the permutation is even, otherwise it is odd.

Examples: ODD and EVEN Permutation

S1 has only one permutation; that is 1, which iseven since there are no inversions.

In the permutation 35241, 3 precedes 2 and 1,5 precedes 2, 4 and 1, 2 precedes 1 and 4precedes 1. There is a total of 7 inversions,thus the permutation is odd.

S3 has 3! = 6 permutations: 123, 231and 312are even while 132, 213, and 321 are odd.

S4 has 4! = 24 permutations: 1234, 1243,1324, 1342, 1423, 1432, 2134, 2143, 2314,

2341, 2413, 2431, 3124, 3142, 3214, 3241,3412, 3421, 4123, 4132, 4213, 4231, 4312,4321.

For any Sn, where n>1 it contains n!/2 evenpermutations and n!/2 odd permutations.


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METHODS IN GETTING THE DETERMINANTMETHODS IN GETTING THE DETERMINANTMETHODS IN GETTING THE DETERMINANTMETHODS IN GETTING THE DETERMINANT

B. COFACTOR METHODB. COFACTOR METHODB. COFACTOR METHODB. COFACTOR METHOD

Complementary Minor,Complementary Minor,Complementary Minor,Complementary Minor, detdetdetdet((((MijMijMijMij) or) or) or) or MijMijMijMij The complementary minor or simply minor of

an element aij of the matrix A is thatdeterminant of the sub-matrix Mij obtainedafter eliminating the ith row and jth columnof A



Algebraic Complement or Cofactor,Algebraic Complement or Cofactor,Algebraic Complement or Cofactor,Algebraic Complement or Cofactor, AAAAijijijij The algebraic complement or cofactor of an

element aij of the matrix A is that signedminor obtained from the formula (-1)i+j Mij



The determinant of a square matrix maybeobtained using expansion about a row orexpansion about a column. The followingformulas maybe used in getting thedeterminant:

(expansion about the ith row)

=

=+++=n

1kikikinin2i2i1i1i

AaAa...AaAa)Adet(



The determinant of a square matrix maybeobtained using expansion about a row orexpansion about a column. The followingformulas maybe used in getting thedeterminant:

(expansion about the jth column)

=

=+++=n

1k

kjkjnjnjj2j2j1j1AaAa...AaAa)Adet(



Note: We may choose any row or any column

in getting the determinant of a given matrix.

Example: Evaluate the given matrix

01300423

1412

0301

It is best to expand about the fourth row because it has the mostnumbers of zeros. The optimal course of action is to expandabout the row or column that has the largest number of zeros,because in that case the cofactors Aij of those aij which are zeroneed not be evaluated since the product of a ijAij = (0)Aij = 0.

4444343424241414AaAaAaAa +++=

24

42

242424M)1(aAa +==

( )( ) ( )( )012027021

130

423301

116

++=

=

13=


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Adjoint

The Adjoint of a square matrix A=[aij] of order n isthat square matrix with the same order n denoted

by adj(A)=[Aji] where Aij is the cofactor of theelement aij of matrix A. The adjoint of a matrix isthe transpose of the matrix of cofactors of theelements of A.

Input: Square MatrixOutput: Square Matrix (with the same size as theoriginal matrix)

Notation: adj A, adj B

Step 1:Step 1:Step 1:Step 1: Get the cofactors of all the elementsin the original matrix.

Recall: the cofactor of an element aij can bedenoted as Aij and is defined by:

ij

ji

ijM)1(A +=

Step 2:Step 2:Step 2:Step 2: Set up the adjoint matrix by takingthe transpose of the matrix of cofactors.

Example:

If A = then adj(A) =

[ ]Tij

AAadj =

dc

ba

ac

bd

Inverse

The inverse of a square matrix A = [aij] oforder n is that matrix B = [bij] of the sameorder n such that AB = BA = In. We denotethe inverse matrix of A by A-1. Thus, we

define the inverse of A as that matrix A-1

such that

A(A-1) = (A-1)A = In.


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Not all matrices has its inverse. However, ifthe inverse of a matrix exists, it is unique.

If the inverse of a matrix exists, we say thatthe matrix is invertible or non-singular.Otherwise, we say that the matrix is non-invertible or singular.

Matrix Inversion applies only to square

matrices and can be produced using theadjoint matrix and the determinant.

Notation: A-1, B-1

A

adjAA 1 =

From the above formula for inverse, it ishighly suggested that the determinant becomputed first. If it so happened that thematrix is singular (i.e., the determinant iszero), then the inverse of the matrix is said tobe non-existent.

Note that it is a waste of effort to stillproduce the adjoint if the matrix is singular.Therefore, it is advised that you first check

for singularity.

Example 1Example 1Example 1Example 1: Find A-1 if

Compute first the determinant using thediagonal method

Since matrix A is singular, as evidenced by itszero determinant, it can thus be concludedthat the Inverse of A (or A-1) does not exist.

=

771

252

111

A

01414514235A =+++=

Example 2:Example 2:Example 2:Example 2: Set up the inverse of the given matrix

Compute first the determinant using the diagonalmethod

Since the determinant is not zero, then matrix Ais said to be non-singular. In this case, theinverse exists and there is a need to set up theadjoint.

=

324

321

111

A

73682126A =++=

ij

ji

ijM)1(A +=

1232

32

)1(A

2

11=

=

934

31

)1(A

3

12== 10

24

21)1(A 413 =

=

132

11)1(A 3

21=

= 1

34

11)1(A 422 == 224

11)1(A 5

23=

=

532

11)1(A 4

31=

= 2

31

11)1(A 5

32== 3

21

11)1(A 6

33=

=

Getting the cofactors of all theelements in the original matrix.

=

324

321

111

A

[ ]

=

325

211

10912

A ijThus, the cofactormatrix Aij is


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Cofactor Matrix from the previous slide

[ ]

=

325

21110912

A ij

Therefore the adjoint of A is

=

3210

219

5112

AadjTij

AAadj =

Consequently, the inverse of A is

AAadjA 1 =

=

3210

219

5112

71A 1

To check if A-1 is correct, use n1 IAA =

3

1 I

100

010

001

324

321

111

3210

219

5112

7

1AA =

=

=

)IAAor(n

1 =

Solution to System ofLinear Equations(Direct Methods)

In general, we can think of a system of linearequations as a set of m equations thatcontains n unknowns. There are severalforms by which a system of equations can bewritten.

A. Equation Form

B. Matrix Form

We can have the equation form

where aij are constant coefficients of theunknowns xj and bj are the constants

mnmn33m22m11m

3nn3333232131

2nn2323222121

1nn1313212111

bxaxaxaxa

bxaxaxaxa

bxaxaxaxa

bxaxaxaxa

=++++

=++++

=++++

=++++

L

MMMMM

L

L

L

Or we can transform that to the matrix form:

=

m

3

2

1

n

3

2

1

mn3m2m1m

n3333231

n2232221

n1131211

b

b

b

b

x

x

x

x

aaaa

aaaa

aaaa

aaaa

MM

L

MMMMM

L

L

L


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Referring to the matrix form, we can actuallyrewrite the system of equations as a compact

matrix operation:

A X = BWhere:

A Coefficient MatrixX Column Matrix of

Unknowns/Variables

B Column Matrix of Constants

Solution to System ofLinear Equations(Direct Methods)

A. Inverse MethodB. Cramers RuleC. LU FactorizationD. Gaussian EliminationE. Gauss-Jordan Reduction

Inverse Method

The Inverse Method maybe applied only to asystem of linear equations in which the numberof independent equations is equal to thenumber of unknowns. If the number ofequations is equal to the number of unknowns,the equation AX = B will have a matrix ofcoefficients that is square.

If the matrix of coefficients A is non-singular,the solution to the system is unique. On theother hand, if A is singular, either the systemhas a non-unique solution or no solution at all.

Derivation of the Solution for xis :

BAX

BAX*I

B*AX*)AA(

B*AAXA

BAX

1

1

11

11

=

=

==

= Take note thatthe derivationassumes that A-1

exists. If A-1 doesnot exist, we cannot find thesolution to thesystemAX = B.

Example:Example:Example:Example: Determine thevalues ofx1, x2and x3 inthe following system ofequations. 5x3x2x4

6x3x2x

1xxx

321

321

321

=+

=++

=+

Solution:Solution:Solution:Solution:

The above system ofequations can be writtenin matrix form AX = B:

=

5

6

1

x

x

x

324

321

111

3

2

1

A X B


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Consider matrix A

=

324

321

111

A

=

3210

219

5112

7

1A 1

Getting A-1

==

=

5

6

1

*

3210

219

5112

7

1BA

x

x

x

X 1

3

2

1

To get x1, x2 and x3 , multiply A-1 to B:

Note: Refer to previousexample for A-1

Performing the operationA-1B will yield the solutionsolutionsolutionsolution

matrixmatrixmatrixmatrix:

=

=

1

1

1

x

x

x

X

3

2

1

Make it a habit to check if all the computed valuesof the unknowns satisfy all the given equations.Checking is done by substituting the values x1 = 1,x2 = 1 and x3 = 1 to the original equations.

Equation 1 1(1) 1(1) + 1(1) =? 1 Satisfied

Equation 2 1(1) + 2(1) + 3(1) =? 6 Satisfied

Equation 3 4(1) 2(1) + 3(1) =? 5 Satisfied

Since all the equations were satisfied, then (x1, x2,

x3) = (1, 1, 1) is indeed the solution to the system.

Cramers Rule

Recall that A system of equation nequations in n unknowns can be modeledas a matrix operation AX = B.

=

n

3

2

1

n

3

2

1

nn3n2n1n

n3333231

n2232221

n1131211

b

b

b

b

x

x

x

x

aaaa

aaaa

aaaa

aaaa

MM

L

MMMMM

L

L

L

Let:

AAAA coefficient matrixxxxxiiii i

th variableBBBB right hand side constantsAAAAiiii matrix resulting from

replacing the ith column ofAby the column vector ofconstants B

Solution is given by:

A

Ax

i

i=

Notice that regardless of the variable i that iscomputed, the denominator of the aboveformula is fixed at |A|. Therefore, it issuggested that the determinant of the

coefficient matrix be the first to be computed. Example:Example:Example:Example: Using Cramer's

Rule, determine the valuesofx1, x2and x3thatsimultaneously satisfy thefollowing system ofequations.

=

3

2

1

x

x

x

121

133

121

3

2

1


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Solution:Solution:Solution:Solution: Compute the determinant of A first.

6623623A

121133

121

A

=+++=

=

Now, let us compute for the value of x1 byusing the formula

A

Ax

1

1 =

The right hand side matrix B is

To set up the matrix A1, all you just have to do isto replace the first column of A by B. Doing whathas just been described will result in:

3

2

1

0429463A

123

132

121

A

1

1

=+++=

= 06

0

A

Ax 1

1===

Applying the same process to solve x2 and x3:

1

321

233

121

*6

1

A

Ax1

131

123

111

*6

1

A

Ax

3

3

2

2 ====

==

LU Factorization

Direct LDirect LDirect LDirect L----U Factorization:U Factorization:U Factorization:U Factorization:In theory any square matrix A may be factored intoa product of lower and upper triangular matrices.

Let us take the case of a 4 th order matrix:

Notice that the diagonal elements of the uppertriangular matrix have been set to values of 1 forreason of simplicity. (L-U Factorization is notunique.)

=

1000

u100

uu10

uuu1

*

llll

0lll

00ll

000l

aaaa

aaaa

aaaa

aaaa

34

2423

141312

44434241

333231

2221

11

44434241

34333231

24232221

14131211

From matrix multiplication, we know that:

=

1000

u100

uu10

uuu1

*

llll

0lll

00ll

000l

aaaa

aaaa

aaaa

aaaa

34

2423

141312

44434241

333231

2221

11

44434241

34333231

24232221

14131211

)0(0)0(0)0(0)1(la1111

+++= 1111 al =

)0(0)0(0)0(l)1(la 222121 +++= 2121 al =

)0(0)0(l)0(l)1(la33323131

+++=3131

al =

)0(l)0(l)0(l)1(la 4443424141 +++= 4141 al =

or

1. Get First Column Elements of Matrix L (1 unknown per equation)


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=

1000

u100uu10

uuu1

*

llll

0lll00ll

000l

aaaa

aaaaaaaa

aaaa

34

2423

141312

44434241

333231

2221

11

44434241

34333231

24232221

14131211

or

2. Get First Row Elements of Matrix U (1 unknown per equation)

)0(0)0(0)1(0)u(la 121112 +++=

)0(0)1(0)u(0)u(la 23131113 +++=

)1(0)u(0)u(0)u(la3424141114+++=

111212l/au =

111313l/au =

111414l/au =


=

1000

u100uu10

uuu1

*

llll

0lll00ll

000l

aaaa

aaaaaaaa

aaaa

34

2423

141312

44434241

333231

2221

11

44434241

34333231

24232221

14131211

or

3. Get Second Column Elements of Matrix L (1 unknown per equation)

)0(0)0(0)1(l)u(la 22122122 +++=

)0(0)0(l)1(l)u(la3332123132

+++=

)0(l)0(l)1(l)u(la444342124142

+++=

)u(lal12212222

=

)u(lal12313232

=

)u(lal12414242

=


=

1000

u100

uu10

uuu1

*

llll

0lll

00ll

000l

aaaa

aaaa

aaaa

aaaa

34

2423

141312

44434241

333231

2221

11

44434241

34333231

24232221

14131211

4. Get Second Row Elements of Matrix U (1 unknown per equation)

)0(0)1(0)u(l)u(la2322132123

+++=

2213212323l/)]u(la[u =

)1(0)u(0)u(l)u(la232422142124+++=

2214212424l/)]u(la[u =


=

1000

u100

uu10

uuu1

*

llll

0lll

00ll

000l

aaaa

aaaa

aaaa

aaaa

34

2423

141312

44434241

333231

2221

11

44434241

34333231

24232221

14131211

5. Get Third Column Elements of Matrix L (1 unknown per equation)

)0(0)1(l)u(l)u(la332332133133

+++=)u(l)u(lal

233213313333=

)0(l)1(l)u(l)u(la 44432342134143 +++=

)u(l)u(lal234213414343

=


=

1000

u100

uu10

uuu1

*

llll

0lll

00ll

000l

aaaa

aaaa

aaaa

aaaa

34

2423

141312

44434241

333231

2221

11

44434241

34333231

24232221

14131211

6. Get Third Row Element of Matrix U (1 unknown per equation)

)1(0)u(l)u(l)u(la 34332432143134 +++=

33243214313434l/)]u(l)u(la[u =


=

1000

u100

uu10

uuu1

*

llll

0lll

00ll

000l

aaaa

aaaa

aaaa

aaaa

34

2423

141312

44434241

333231

2221

11

44434241

34333231

24232221

14131211

7. Get Fourth Column Element of Matrix L (1 unknown per equation)

)1(l)u(l)u(l)u(la4434432442144144

+++=

)u(l)u(l)u(lal3443244214414444

=


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Recall:Recall:Recall:Recall: A system of equations can be writtenas a compact matrix operation AX = B

If we factor out the coefficient matrix A asL*Uand substitute to AX = B, we cangenerate the equation L(UX)=B.

Momentarily define UX = Ywhich suggestsLY = B. From this transformation, we haveactually decomposed AX = Bto two systemsof equations.

TwoTwoTwoTwo----Stage SolutionStage SolutionStage SolutionStage Solution::::

Stage 1: Solve for Yin the equation LY = Busing forward substitution.

Stage 2: Solve for Xin the equation UX = Yusing back substitution.

Example:Example:Example:Example: Determine the solution X in

=

8

4

3

x

x

x

121

153

124

3

2

1

Knowing that A = LU, determining L and U asbelow

=

121

153

124

A

=

25

251

02

73

004

L

=

1002

1104

12

11

U

Stage 1:Stage 1:Stage 1:Stage 1: Forward substitution using LY = B

=

8

4

3

y

y

y

25

251

02

73

004

3

2

1

4

3y1 =

2

1y2 =

3y3 =

Note that the computed values of yi's hereare not yet the solution since the originalsystem of equations is in terms of xi's.

Stage 2:Stage 2:Stage 2:Stage 2: Back substitution using UX = Y

=

321

43

xx

x

1002110

41

211

3

2

1

This time (x1, x2, x3) = (1, 2, 3) is the solutionto the original system of equations.

1x1 =

2x2 =

3x3=

If A is an m x n matrix and B is a p x nmatrix, then the augmented matrix of A andB denoted by [A : B] is the matrix formed bythe elements of A and B separated by pipes.

Example:Example:Example:Example:

=

833

617

521

A

=

47

20

12

B

=

47

20

12

|

|

|

833

617

521

B:A


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The augmented matrix associated to a systemof linear equation AX=B is the matrix [A : B].

For example, we can now rewrite the systemof equation:

=

1

1

1

z

y

x

142

621

312

1

1

1

|

|

|

142

621

312

An m x n matrix A is said to be in row echelonform if it satisfies the following properties:

1. All rows whose elements are all zeros, if exist,are at the bottom of the matrix.

2. If at least one element on a row is not equal tozero, the first non-zero element is 1, and thisis called the leading entry of the row.

3. If two successive rows of the matrix haveleading entries, the leading entry of the rowbelow the other row must appear to the right ofthe leading entry of the other row.

An m x n matrix A is said to be in reduced rowechelon form if added to the first threeproperties it satisfies a fourth property:

4. If a column contains a leading entry of somerow, then all the other entries must be zero.

ExamplesExamplesExamplesExamples::::

=

=

=

0010

8100

7610

4221

C

0000

1000

0100

0210

0001

B

100000

000000

291000

531010

231021

A

The following matrices are not in row echelon form.(Why not?)

Did not satisfy

property 1

Did not satisfy

property 2

Did not satisfy

property 3

=

=

=

0000

0000

0100

3010

0801

F

00100

80010

26501

E

1000

2100

3210

4321

D


The following matrices are in row echelon form but notin reduce row echelon form.

All properties are satisfied except property 4.

Must all be zeroes to satisfy the 4th property.


The following matrices are in reduced row echelonform. (Hence, in row echelon form.)

=

=

=

000

000

000

010301

J

01000

30100

20010

H

1000

0100

0010

0001

G


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2

An elementary row (column) operation on amatrix A is any one of the following

operations:

Type I. Interchange any two rows (columns).

Type II. Multiply a row (column) by a non-zeroconstant k.

Type III. Add to elements of a row k times of theelements of another row thecorrespondingly.

Example:Example:Example:Example: Let

=

2282

4613

0201

A

=

0201

4613

2282

B

=

1141

4613

0201

C

Type I:Type I:Type I:Type I: Interchanging rows 1 and 3 of A (R1R3)obtain

Type II:Type II:Type II:Type II: Multiplying row 3 by (R3 R3), we obtain

Example:Example:Example:Example: Let

=

2282

4613

0201

A

=

2282

4010

0201

D

Type III:Type III:Type III:Type III: Adding 3 times the elements in row 1 to the

elements in row 2 (R2R2 + 3R1), we obtain

As a applied to the augmented matrix [A:B] asa system of equation, the three elementary rowoperation will correspond to the following:

TYPE I rearranging the order of theequations

TYPE II multiplying both side of theequation by a constant

TYPE III working with two equations

From this observation, we could see that asapplied to a operations does not alter thesolution of the system.

In general a system of m equations in nunknowns may be written in matrix form:

=

m

3

2

1

n

3

2

1

mn3m2m1m

n3333231

n2232221

n1131211

b

b

b

b

x

x

x

x

aaaa

aaaa

aaaa

aaaa

MM

L

MMMMM

L

L

L

m

3

2

1

mn3m2m1m

n3333231

n2232221

n1131211

b

b

b

b

|

|

|

|

aaaa

aaaa

aaaa

aaaa

MM

L

MMMMM

L

L

L

This system may now be represented by theaugmented notation:

Gaussian Elimination Method


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2

The objective of the Gaussian EliminationGaussian EliminationGaussian EliminationGaussian EliminationMethodMethodMethodMethod is to transform the augmented matrix

[A : B] to the matrix [A* : B*] in row echelonform by applying a series of elementary rowtransformations. Getting the solution of thesystem [A* : B*] using back substitution willalso give the solution to the original system[A : B].

1. Find the leftmost non-zero column.2. If the 1st row has a zero in the column of step 1,

interchange it with one that has a non-zero entryin the same column.

3. Obtain zeros below the leading entry by addingsuitable multiples of the top row and to the rowsbelow that.

4. Cover the top row and repeat the same processstarting with step 1 applied to the leftoversubmatrix. Repeat this process with the rest ofthe rows.

5. For each row obtain leading entry 1 by dividingeach row by their corresponding leading entry

Example:Example:Example:Example: The linear system

3zx3

8zyx2

9z3y2x

=

=+

=++

[ ]

=

3|103

8|112

9|321

B:A

Has the augmented matrix associated to the system

TransformTransformTransformTransform in row echelon form

[ ]

=

3|103

8|112

9|321

B:A

Element in the 1st row 1st column is non-zero and alreadyis equal to 1 (leading entry).

Transform into zeroes all elements below the leadingentry.

[ ]

=

24|1060

10|550

9|321

*B:*A122 R2R'R

133R3R'R

Consider now the transformation of column 2elements

Element at 2nd row 2nd column is non-zero and will betransformed to 1 (leading entry). Afterwards, transformelements below to zero.

[ ]

=

24|1060

10|550

9|321

*B:*A

22 R5

1'R

[ ]

=

24|1060

2|110

9|321

*B:*A

[ ]

=

12|400

2|110

9|321

*B:*A233 R6R'R +

Consider now the transformation of column 3elements

Element at 3rd row 3rd column is non-zero and will betransformed to 1 (leading entry).

[ ]

=

3|100

2|110

9|321

*B:*A

[ ]

=

12|400

2|110

9|321

*B:*A

Augmented Matrix in Row Echelon Form!

33R

4

1'R


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THE END

ce26 lecture on matrices

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