First Exam Notes for ES 21

7/23/2019 First Exam Notes for ES 21

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First Exam CoverageES 21 Notes

ES 21 NotesPage 1 of 12

MATRICES

Definition

A matrix is a rectangular array of numbers or functionsarranged in rows and columns usually designated by acapital letter and enclosed by brackets, parentheses ordouble bars. A matrix may be denoted by:

mnmm

n

n

aaa

aaa

aaa

A

...

:::

...

...

21

22221

11211

Unless stated, we assume that all our matrices arecomposed of real numbers.

The horizontal groups of elements are called the rowsof thematrix. The ith row of A is

miaaa inii 1...21

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

nj

a

a

a

j

j

j

1:

3

2

1

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

We refer to aijas the entryor the elementin the ith row andjth column of the matrix.

We may often write a given matrix as

A = [aij].

SOME SPECIAL TYPES OF MATRICES

A. Row Matrix or Row Vector is a matrix consistsof only one row.

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

B. Column Matrix or Column Vector is a matrixconsists of only one column.

Example:

m

i

c

c

c

c

C

.

.

.

.

..

2

1

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

Order of a Square Matrix is the number of rowsor columns of the matrix. Thus, we can just referto a 3x3 matrix as a square matrix of order 3.

Principal Diagonal or Main Diagonal of a SquareMatrixconsists of the elements a11, a22, a33,

ann.

The Trace of a Square Matrixis the sum of theelements on the main diagonal of the matrix.

D. Upper Triangular Matrix a square matrix allelements of which below the principal diagonalare zero (aij= 0 for i>j).

Example:

33

2322

131211

00

0

u

uu

uuu

U

E. Lower Triangular Matrix a square matrix allelements of which above the principal diagonalare zero (aij= 0 for i


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EQUALITY OF MATRICES

Two matrices A = (aij) and B = (b ij) are equal if and only ifthe following conditions are satisfied:

a) They have equal number of rows.b) They have equal number of columns.c) All elements in A agree with the elements in

B. (aij=b

ij, for all i and j.)

Example: The matrices

143

221

112

3

21

21

c

bxBand

zb

ya

cx

A

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

ELEMENTARY OPERATIONS ON MATRICES

A. MATRIX ADDITION AND SUBTRACTION

If A = (aij) and B = (bij) are matrices of the same size m x n,then the sum A + B is another m x n matrix C = [c ij] wherecij= aij+ bijfor i = 1 to m and j = 1 to n. Matrix addition isaccomplished by adding algebraically correspondingelements in A and B.

Example:

114

223

123

421BA

211

204

11)1(243

)2(42231

114

223

123

421BA

B. SCALAR MULTIPLICATION

If A = (aij) is an m x n matrix and k is a real number (or ascalar), then the scalar multiple of A by k is the m x n matrixC = [cij] where cij= kaij for all i and j. In other words, thematrix C is obtained by multiplying each element of thematrix by the scalar k.Examples: We obtain

156912

9306

5234

31023

28168

4820

1248

742

125

312

4

C. MATRIX SUBTRACTION

If A and B are m x n matrices, the difference between A andB denoted as AB is obtained from the addition of A and (-1)B.

AB = A + (-1)BMatrix subtraction is accomplished by subtracting from theelements of the first matrix the elements of the secondmatrix correspondingly.

Example:

383

412

475

243BA

718

651

3487)3(5

42)1(423

383

412

475

243BA

Note: We can only add or subtract matrices with the same

number of rows and columns.

D. MATRIX MULTIPLICATION

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

n

k

kjikij bac1

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

The formula tells us that in order to get theelement cijof the matrix C, get the elements of the

ith row of A (the pre-multiplier) and the elementson the jth column of B (the post multiplier).Afterwards, obtain the sum of the products ofcorresponding elements on the two vectors.

Note: The product is defined only if thenumber of columns of the first factorA (pre-multiplier) is equal to thenumber of rows of the second factorB (post-multiplier). If this issatisfied, we say that the matricesare comformable in the order AB.

MATRIX EXPONENTIATION

The formula Anwill be defined as AAAA

Example:

13

42

43

21BA

20

1210

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

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

1618

68

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

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

BA

BA

AB

AB

Note: Although AB and BA are defined it is not necessarythat AB = BA.

Example:

12

31

541

326

64

35

12

CBA

1. A x B is a 3 x 3 matrix while B x A isa 2 x 2 matrix.

2. A x C is a 3 x 2 matrix but C x A is notdefined.

3. B x C is not defined but C x B isdefined (2 x 3).

E. MATRIX TRANSPOSITION

If A = [aij] is an m x n matrix, then the transpose of A,denoted by A

T=[aij] is an n x m matrix defined by a ij=aji.


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The transpose of A is obtained by interchanging the rowsand the columns of A.

Example:

2342

3132

5211

235

312

431

221

TAthenAIf

Note: The transpose of a symmetric matrix is equal to itself.PROPERTIES AND THEOREMS ON MATRIXOPERATIONS:

MATRIX ADDITION

A + O = A Existence of Additive IdentityA + (-A) = O Existence of Additive InverseA + B = B + A Commutative Property(A + B) + C = A + (B + C) Associative Property

SCALAR MULTIPLICATION

0 x A = O1 x A = Ok l (A) = k (l A) = l (k A)(k + l) A = k A + l Ak(A + B) = k A + k B

MATRIX MULTIPLICATION

A(BC) = (AB)C Associative PropertyA(B + C) = AB + AC Left Distributive Property(A + B)C = AC + BC Right Distributive PropertyAI = IA = A Existence of Multiplicative

Identitykl (AB) = (k A)(l B) = (l A)(k B)

Note: In general, Matrix Multiplication is not

commutative. That is, AB BA.

MATRIX TRANSPOSITION(A

T)T= A

(A + B)T= A

T+ B

T

(k A)T= k A

T

(AB)T= B

TA

T

In general (A1A2A3An-1An)T= An

TAn-1

TA3

TA2

TA1

T

DETERMINANTS

Another very important number associated with a squarematrix A is the determinant of Awhich we will now define.This unique number associated to a matrix A is useful in thesolutions of linear equation.

Permutation:

Let S={1, 2, 3, , n} be the set of integers from 1 to n,arranged in increasing order. A rearrangement a1a2a3anofthe elements in S is called a permutation of S.

By the Fundamental Principle of Counting we can put anyone of the n elements of S in the first position, any one ofthe remaining (n-1) elements in the second position, anyone of the remaining (n-2) elements in the third position,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! = 24 elements of S4.

Odd and Even Permutations

A permutation a1a2a3an is said to have an inversion if alarger number precedes a smaller one. If the total number ofinversion in the permutation is even, then we say that thepermutation is even, otherwise it is odd.Examples: ODD and EVEN Permutation

S1has only one permutation; that is 1, which is even sincethere are no inversions.

In the permutation 35241, 3 precedes 2 and 1, 5 precedes2, 4 and 1, 2 precedes 1 and 4 precedes 1. There is a totalof 7 inversions, thus the permutation is odd.

S3has 3! = 6 permutations: 123, 231and 312 are even while132, 213, and 321 are odd.

S4has 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 even permutationsand n!/2 odd permutations.

DEFINITION: DETERMINANT

Let A = [aij] be a square matrix of order n. The determinant

of A denoted by det(A) or Ais defined by

nnjjj

aaaAA ...det21 21

where the summation is over all permutations j1j2jnof theset S = {1,2,,n}. The sign is taken as (+) if the permutationis even and () if the permutation is odd.

Examples:

If A = [a11] is a 1 x 1 matrix then det(A) or A= a11.

If A =

2221

1211

aa

aa , then to get Awe write down the terms a1-

a2-b and replace the dashes with the all-possiblepermutations of S = {1, 2}, namely 12 (even) and 21 (odd).

Thus A= a11a22 - a12a21.

If A =

333231

232221

131211

aaa

aaa

aaa , then to compute the A we write

down the six terms a1-a2-a3-, a1-a2-a3-, a1-a2-a3-, a1-a2-a3-, a1-a2-a3-, a1-a2-a3-. Replace the dashes with all the elements of S3,affix a (+) or (-) sign and get the sum of the six terms.

If A is a square matrix of order n, there will be n! terms inthe determinant of A with n!/2 positive terms and n!/2negative terms.

9

620098

)2)(1)(3()1)(1)(2()3)(2)(0()1)(1)(0()3)(1)(3()2)(2)(2(

210

123

312

METHODS IN GETTING THE DETERMINANT

A. DIAGONAL METHOD

This method is applicable to matrices with sizeless than or equal to 3.

1. 2 x 2 Matrices


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12212211

2221

1211aaaa

aa

aa s

2. 3 x 3 Matrices

B. METHOD OF COFACTORS

Complementary Minor, det(Mij) or MijThe complementary minor or simply minor of an element a ijof the matrix A is that determinant of the sub-matrix Mijobtained after eliminating the ith row and jth column of A.

Algebraic Complement or Cofactor, Aij

The algebraic complement or cofactor of an element a ij ofthe matrix A is that signed minor obtained from the formula

(-1)i+j

Mij

DETERMINANT USING THE COFACTOR METHOD

The determinant of a square matrix maybe obtained usingexpansion about a row or expansion about a column. Thefollowing formulas maybe used in getting the determinant:

n

k

ikikininiiii AaAaAaAaA1

2211 ...)det(

(expansion about the ith row)

and

n

k

kjkjnjnjjjjj AaAaAaAaA1

2211 ...)det(

(expansion about the ith row)

Note: We may choose any row or any column in getting thedeterminant of a given matrix.

Example: To evaluate

0130

0423

1412

0301

It is best to expand about the fourth row because it has themost numbers of zeros. The optimal course of action is toexpand about the row or column that has the largestnumber of zeros, because in that case the cofactors Aijof

those aijwhich are zero need not be evaluated since theproduct of aijAij= (0)Aij= 0.

THEOREMS ON DETERMINANTS

1. If a square matrix A = [aij] contains a row (or a column)

that has elements all equal to zero, then A= 0.

Example:

0

042

032

021

2. The determinant of a square matrix A = [aij] is equal to

the determinant of its transpose AT = [aij].(i.e.A=AT)

Example:

233

422

141

241

324

321

3. If a row (or column) of a square matrix A = [aij] ismultiplied by a constant k, then the determinant of theresulting matrix B = [bij] is equal to k times the

determinant of A (i.e. B= kA).

Example:

10)5(2

224

224

1

5

224

2

1

ba

yx

ba

then

ba

yx

ba

If

4. As a corollary to the third theorem, if A has a row (orcolumn) that has a common factor l, then this k may befactored out of the determinant of A, where a simplified

matrix B is formed. (i.e. .A= kB.

Example:

xy

yyx

xxy

xy

xy

yyx

xxy

xy

xyxy

yyyx

xxxy

218343

23

)2(22183862

23

22186864

232

2

2

5. If two row (or columns) of a square matrix A = [aij] were

interchanged to form a new matrix B = [b ij], then B=-A.

Example:

1

02

002

1

002

02

yzyx

xxy

x

yzyx

x

xxy

6. If two rows (or columns) of a matrix A = [aij] are

identical then A= 0.Example:

0

xyzxy3z4

xyzz5yx3x2

xz2x3yx2

xzyz5yx3x2

x3

xyzxy3z4

yzx3xz15xy3x9x6

xz2x3yx2

xyzz5yx3x2

22

222

7. As a corollary to the sixth theorem, if the elements in a

row (or column) of a square matrix A = [a ij] aremultiples of the corresponding elements of another row

or column of the matrix A, then A= 0.Example:

0

216

464

232

xyy

8. If B = [bij] is a square matrix of order n that is derivedfrom another square matrix A = [a ij] of order n, byadding correspondingly the elements of a row (orcolumn) to a multiple of the elements of another row

(or column), then B=A.Example:

333231

232221

331332123111

333231

232221

131211

aaa

aaa

kaakaakaa

aaa

aaa

aaa

9. If the elements of one row (or column) of a squarematrix A = [aij] of order n may be expressed as

binomials such that two square matrices B = [bij] and C= [cij] both of order n, are formed after splitting the

binomial elements, then A=B+C.Example:

3231

2221

1211

333231

232221

131211

aa

aa

aa

aaa

aaa

aaa


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333231

232221

131211

333231

232221

131211

333231

232322222121

131211

aaa

ccc

aaa

aaa

bbb

aaa

aaa

cbcbcb

aaa

10. The determinant of the product of two square matrices A= [aij] and B = [bij] of the same order n is equal to the

product of the determinant of A and the determinant of B.Example:

If 2

216

464

232

xyy

z

x

and

222

56

543

bbaa

ba

a

= 3

then

xyy

z

x

216

464

232

222

56

543

bbaa

ba

a

= 32 = 6 .

11. The determinant of a triangular matrix is equal to theproduct of the elements in its principal diagonal.

Example:

4

1

02

0022

x

yzyx

xxy

x

12. The determinant of an Identity Matrix is equal to 1.

ADJOINT OF A MATRIX

The Adjoint of a square matrix A=[aij] of order n is thatsquare matrix with the same order n denoted by adj(A)=[Aji]where Aijis the cofactor of the element aijof matrix A. Theadjoint of a matrix is the transpose of the matrix of cofactors

of the elements of A.

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

Notation: adj A, adj B

Step 1: Get the cofactors of all the elements in theoriginal matrix.

Recall: the cofactor of an element aij can be denoted as Aijand is defined by:

ijji

ij MA )1(

Step 2: Set up the adjoint matrix by taking the transposeof the matrix of cofactors.

TijAAadj

Example:

If A =

dc

bathen adj(A) =

ac

bd.

Inverse of a Matrix

The inverse of a square matrix A = [aij] of order n is that

matrix B = [bij] of the same order n such that AB = BA = In.We denote the 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.

Not all matrices has its inverse. However, if the inverse of amatrix exists, it is unique.

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

Matrix Inversion Using the Adjoint

and the Inverse

Matrix Inversion applies only to square matrices and can beproduced using the adjoint matrix and thedeterminant.

Notation: A-1, B

-1

The proof of this needs the knowledge of the followingtheorem:

The sum of the products of the elements in one row (or

column) and the cofactors of the elements of another row(or column) of a given square matrix is zero.

From the above formula for inverse, it is highly suggestedthat the determinant be computed first. If it sohappened that the matrix is singular (i.e., thedeterminant is zero), then the inverse of thematrix is said to be non-existent.

Note that it is a waste of effort to still produce the adjoint if

the matrix is singular. Therefore, it is advised

that you first check for singularity.Example: Set up the Inverse of the given matrix.

Using the diagonal method to compute for the determinantof the given matrix:

Since matrix A is singular, as evidenced by its zerodeterminant, it can thus be concluded that theInverse of A (or A-1) does not exist.

Example 2: Set up the Inverse of the given matrix

324

321

111

A 73682126 A

Since the determinant is not zero, then matrix A is said to be non-singular. In this case, the inverse exists and there is a need to set upthe adjoint.

Getting the cofactors of all the elements in the original matrix.

ijji

ij MA )1(

1232

32)1( 211

A 934

31)1( 312 A

10

24

21)1( 413

A 1

32

11)1( 321

A

134

11)1( 422 A

224

11)1( 523

A

771

252

111

A

A

adjAA 1

01414514235 A


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532

11)1( 431

A 2

31

11)1( 532 A

321

11)1( 633

A

325

211

10912

ijA ,thus

Consequently,A

AadjA 1

3210

219

5112

7

11A

SOLUTION TO SYSTEM OF LINEAR EQUATIONS

In general, we can think of a system of linearequations as a set of m equations that contains nunknowns. There are several forms by which a system ofequations can be written.

We can have the equation form:

mnmnmmm

nn

nn

nn

bxaxaxaxa

bxaxaxaxa

bxaxaxaxa

bxaxaxaxa

332211

33333232131

22323222121

11313212111

Or we can transform that to the matrix form:

mnmnmmm

n

n

n

b

b

b

b

x

x

x

x

aaaa

aaaa

aaaa

aaaa

3

2

1

3

2

1

321

3333231

2232221

1131211

Referring to the matrix form, we can actually rewrite thesystem of equations as a compact matrix operation:

AX = B.

Where:

A Coefficient Matrix

X

Column Matrix of Unknowns/VariablesB Column Matrix of Constants

SOLUTION TO SYSTEM OF n-LINEAR EQUATIONSWITH n UNKNOWNS

A. USING THE INVERSE METHOD

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

If the matrix of coefficients A is non-singular, thesolution to the system is unique. On the other hand, if A issingular, either the system has a unique solution or nosolution at all.

Derivation of the Solution forxis:

BAX

BAXI

BAXAA

BAAXA

BAX

1

1

11

11

*

**)(

*

Example: Determine the values of x1, x2 and x3 in thefollowing system of equations.

5324

632

1

321

321

321

xxx

xxx

xxx

Solution:

The above system of equations can be written in matrixform:

56

1

324321

111

3

2

1

xx

x

We can write this in matrix form AX = B and let X = A-1B,

where:

324

321

111

A

Getting A-1

3210

219

5112

7

11A

To getx1,x2andx3, multiplyA-1to B:

5

6

1

*

3210

219

5112

7

1

3

2

1

x

x

x

X

Performing the operationA-1B will yield the solution

matrix:

1

1

1

3

2

1

x

x

x

X

Make it a habit to check if all the computed values of the

unknowns satisfy all the given equations. Checking is doneby substituting the valuesx1 = 1,x2= 1 andx3= 1 to theoriginal equations.

Equation11(1)1(1) + 1(1) =? 1Satisfied

Equation 21(1) + 2(1) + 3(1) =? 6Satisfied

Equation 34(1)2(1) + 3(1)=? 5Satisfied

Since all the equations were satisfied, then (x1,x2,x3) = (1,1, 1) is indeed the solution to the system.

SOLUTION TO SYSTEM OF EQUATIONS USINGCRAMER'S RULE

Recall that A system of equation n equations in nunknowns can be modeled as a matrix operation AX = B.

3210

219

5112

Aadj

where aijareconstantcoefficients of theunknowns xjand bI

Take note that thederivation assumesthat A

-1exists. If A

-1

does not exist, wecan not find thesolution to the

system AX = B.


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nnnnnnn

n

n

n

b

b

b

b

x

x

x

x

aaaa

aaaa

aaaa

aaaa

3

2

1

3

2

1

321

3333231

2232221

1131211

Let:A

coefficient matrixxi ithvariable

B right hand side constantsA I matrix resultingfrom replacing the

ithcolumn ofAby the column

vector of constants B

The solution of the system of equations can be determinedby using the formula:

A

Ax

i

i

Notice that regardless of the variable ithat is computed, thedenominator of the above formula is fixed at |A|. Therefore,it is suggested that the determinant of the coefficient matrixbe the first to be computed.

Example:Using Cramer's Rule, determine the values ofx1,x2andx3that simultaneously satisfy the following system ofequations.

3

2

1

121

133

121

3

2

1

x

x

x

Solution:

Compute for the determinant ofA first:

6623623

121

133

121

A

A

Now, let us compute for the value ofx1by using the formula

A

Ax

11

The right hand side matrix B is

3

2

1

To set up the matrix A1, all you just have to do is to replacethe first column ofAby b. Doing what has just beendescribed will result in:

0429463

123132

121

1

1

A

A 06

011

A

A

x

Applying the same process to solvex2 and x3:

1

321

233

121

*6

11

131

123

111

*6

1 33

22

A

Ax

A

Ax

SOLUTION TO SYSTEM OF LINEAR EQUATIONS USINGL-U FACTORIZATION

Direct L-U Factorization:

In theory any square matrix A may be factored into aproduct of lower and upper triangular matrices.

Let us take the case of a 4thorder matrix:

1000

100

10

1

*0

00

000

34

2423

141312

44434241

333231

2221

11

44434241

34333231

24232221

14131211

u

uu

uuu

llll

lll

ll

l

aaaa

aaaa

aaaa

aaaa

Notice that the diagonal elements of the upper triangularmatrix have been set to values of 1 for reason of simplicity.(L-U Factorization is not unique.)

From matrix multiplication, we know that:

)0(0)0(0)0(0)1(1111 la or 1111 al

)0(0)0(0)0()1( 222121 lla or 2121 al

)0(0)0()0()1( 33323131 llla or 3131 al

)0()0()0()1( 4443424141 lllla or

4141 al

)0(0)0(0)1(0)( 121112 ula or 111212 / lau

)0(0)1(0)(0)( 23131113 uula or 111313 / lau

)1(0)(0)(0)( 3424141114 uuula or

111414 / lau

)0(0)0(0)1()( 22122122 lula or

)( 12212222 ulal

)0(0)0()1()( 3332123132 llula or

)( 12313232 ulal

)0()0()1()( 444342124142 lllula or

)( 12414242 ulal

)0(0)1(0)()( 2322132123 ulula or

2213212323 /)]([ lulau

)1(0)(0)()( 232422142124 uulula or

2214212424 /)]([ lulau

)0(0)1()()( 332332133133 lulula or

)()( 233213313333 ululal

)0()1()()( 44432342134143 llulula or

)()( 234213414343 ululal

)1(0)()()( 34332432143134 ululula or

33243214313434 /)]()([ lululau

)1()()()( 4434432442144144 lululula or

)()()( 3443244214414444 ulululal

How to get the solution to a system of equations usingL-U Decomposition Method?

Recall:A system of equations can be written as a compactmatrix operationAX = B

If we factor out the coefficient matrix A as L*U andsubstitute to AX = B, we can generate the equationL(UX)=B.

Momentarily define UX = Y which suggests LY = B. Fromthis transformation, we have actually decomposedAX = b totwo systems of equations.

Two-stage solution:


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I. Solve for Y in the equation LY = B using forwardsubstitution.

II. Solve for X in the equation UX = Y using backsubstitution.

Example: Determine the values ofxi'sin :

8

4

3

121

153

124

3

2

1

x

x

x

Knowing that

121

153

124

A

, therefore

25251

0273

004

L

and

100211041

211

U

Stage 1: Forward substitution using LY = B

8

43

25

251

02

73004

3

2

1

y

yy

Note that the computed values of yi'shere are not yet thesolution since the original system of equations is in terms ofxi's.

Stage 2: Back substitution using UX = Y

32

14

3

1002

110

41

211

3

2

1

x

x

x

This time (x1,x2,x3) = (1, 2, 3) is the solution to the originalsystem of equations.

AUGMENTED MATRIX OF A AND B

If A is an m x n matrix and B is a p x n matrix, then theaugmented matrix of A and B denoted by [A : B] is thematrix formed by the elements of A and B separated bypipes.

Example:

If

833

617

521

A

and

47

20

12

Bthen A : B is

47

20

12

|

|

|

833

617

521

The augmented matrix associated to a system of linearequation AX=B is the matrix [A : B]. For example, we cannow rewrite the system of equation:

1

1

1

142

621

312

z

y

xas simply

1

1

1

|

|

|

142

621

312.

ECHELON FORM OF A MATRIX

An m x n matrix A is said to be in row echelon form if itsatisfies the following properties:

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

2. If at least one element on a row is not equalto zero, the first non-zero element is 1, andthis is 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 theright of the leading entry of the other row.

An m x n matrix A is said to be in reduced row echelon formif added to the first three properties it satisfies a fourthproperty:

4. If a column contains a leading entry ofsome row, then all the other entries mustbe zero.

Example:

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

0010

8100

7610

4221

0000

1000

0100

0210

0001

100000

000000

291000

531010

231021

CBA

The following matrices are in row echelon form but not inreduce row echelon form.

0000

0000

0100

3010

0801

00100

80010

26501

1000

2100

3210

4321

FED

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

000

000

000

010

301

01000

30100

20010

1000

0100

0010

0001

JHG

ELEMENTARY ROW (COLUMN) OPERATIONS ONMATRICES

An elementary row (column) operation on a matrix A is anyone of the following operations:

Type I. Interchange any two rows (columns).Type II. Multiply a row (column) by a non-zero

constant k.Type III. Add to elements of a row k times of the elementsof another row the correspondingly.

Example: Let

2282

4613

0201

A

Interchanging rows 1 and 3 of A (R1R3) obtain

0201

4613

2282

B

Multiplying row 3 by (R3R3), we obtain

34 1 y 43

1 y

42

7321

yy2

12y

82

52

5321

yyy 33y

33 x

21

21

32 xx 22 x

43

41

21

321 xxx 11x

33 x


9/12



1141

4613

0201

C

Adding 3 times the elements in row 1 to the elements in row

2 (R2R2 + 3R1), we obtain

22824010

0201

A

ELEMENTARY ROW OPERATIONS AS APPLIED TOTHE A SYSTEM OF EQUATION A:B

As a applied to the augmented matrix [A : B] as a system ofequation, the three elementary row operation willcorrespond to the following:

TYPE I rearranging the order of the equationsTYPE II multiplying both side of the equation by

a constant

TYPE III working with two equations

From this observation, we could see that as applied to aoperations does not alter the solution of the system.

ROW (COLUMN) EQUIVALENT MATRICES

An m x n matrix A is row (column) equivalent to an m x nmatrix B if B can be obtained from A by applying a finitesequence of elementary row operations.

THEOREMS ON MATRIX EQUIVALENCE

1. Every nonzero m x n matrix A = [aij] is row (column)equivalent to a matrix in row (column) echelon form.

2. Every nonzero m x n matrix A = [aij] is row (column)equivalent to a matrix in reduced row (column) echelonform.

3. Let AX = B and CX = D be two systems of m linearequations in n unknowns. If the augmented matrices[A : B] and [C : D] are row equivalent, then the linearsystems are equivalent (i.e. they have exactly thesame solutions).

4. As a corollary to the third theorem, if A and B are rowequivalent matrices, then the homogeneous systemsAX = 0 and BX = 0 are equivalent.

SOLUTIONS TO A SYSTEM OF m EQUATIONS in nUNKNOWNS

In general a system of m equations inn unknowns maybe written in matrix form:

mnmnmmm

n

n

n

b

b

b

b

x

x

x

x

aaaa

aaaa

aaaa

aaaa

3

2

1

3

2

1

321

3333231

2232221

1131211

.

This system may now be represented by the augmentednotation:

mmnmmm

n

n

n

b

b

b

b

aaaa

aaaa

aaaa

aaaa

3

2

1

321

3333231

2232221

1131211

|

|

|

|

Applying the theorems on equivalent matrices we now havethe following methods of solution:

GAUSSIAN ELIMINATION METHOD

The objective of the Gaussian Elimination Method is totransform the augmented matrix [A : B] to the matrix [A* :B*] in row echelon form by applying a series of elementaryrow transformations. Getting the solution of the system [A* :B*] using back substitution will also give the solution to theoriginal system [A : B].

To reduce any matrix to row echelon form, apply thefollowing steps:

1. Find the leftmost non-zero column.2. If the 1

st row has a zero in the column of step 1,

interchange it with one that has a non-zero entry in

the same column.3. Obtain zeros below the leading entry by adding

suitable 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 leftover submatrix.Repeat this process with the rest of the rows.

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

Example: The linear system

33

82

932

zx

zyx

zyx

has the augmented matrix associated to the system

3|103

8|112

9|321

:BA

which can be transformed as a matrix in row echelon form

3|100

2|110

9|321

: ** BA

using back substitution we have

2)3(3)1(29932

1322

3

xzyx

yzy

z

thus we have the solution (x, y, z) = (2, -1, 3).

GAUSS-JORDAN REDUCTION METHOD

On the other hand a second method called the Gauss-Jordan Reduction Method gets rid of the back substitutionphase. The objective of the Gauss-Jordan ReductionMethod is to transform the augmented matrix [A : B] to thematrix [A* : B*] in reduced row echelon form by applying aseries of elementary row transformations. Doing this willautomatically give the solution of the system [A* : B*] whichalso provides the solution to the original system [A : B].

To reduce any matrix to reduced row echelon form, apply

the following steps (SINE):

1. Search search the ith column of the augmentedmatrix from the ith row to the nth row for the


10/12



maximum pivot, i.e. element with the largest absolutevalue.

2. Interchange assuming the maximum pivot occursin thejthrow, interchange the ithrow and thejthrowso that the maximum pivot will now occur in thediagonal position.

3. Normalizenormalize the new ithrow by dividing itby the maximum pivot on the diagonal position.

4. Eliminateeliminate the ithcolumn from the first upto the nthequation, except in the ithequation itselfusing the transformations.

Example: The linear system

33

82

932

zx

zyx

zyx

has the augmented matrix associated to the system

3|103

8|112

9|321

: BA

which can be transformed as a matrix in row echelon form

3|100

1|010

2|001

: ** BA

thus we have the solution (x, y, z) = (2, -1, 3).

SUBMATRIX AND RANK

SUBMATRIXA submatrix of A=[aij] is any matrix obtained by eliminatingsome rows and/or columns of the matrix A.

Example: Let

275114

25961

28303

81032

A

The following are some submatrices of A:

)remaining

columnond(sec28303

)remaining

columnfourth(

7

5

8

1

)columnfifthand

third,firstremove(

2511

296

230

803

)columnfifthandthird

androwondsecremove(

7114

561

132

)rowthirdremove(

275114

28303

81032

RANK OF A MATRIX

The rank of a matrix A = [a ij] is the order of the largestsquare submatrix of A with a non-zero determinant. Wedenote the rank of A by rank(A) or simply r(A).

Example: What is the rank of A?

10503

3412

4321

A

Solution:

Checking out first the determinants of 3x3 submatrices:

060

1050

341

432

014

1053

342

431

0

1003

312

421

0

503

412

321

Since at least one 3x3 submatrix of A has a non-zerodeterminant, then r(A) = 3.

Example: What is the rank of B?

161284

12963

8642

4321

B

Solution:

The determinant of B is equal to zero (THEOREM:Proportional rows). And it can also be shown that 3x3submatrices of B will have determinants equal to zero.

0

1684

1263

842

.g.e

(Rows are proportional)

But at least one 2x2 submatrix has non-zero determinant.

02463

42.g.e

Therefore r(B) = 2.

THEOREMS ON RANKS

1. The rank of a matrix is not altered by anysequence of elementary row (column)transformations.

2. Let A = [aij] and B = [bij] be two mxn matrices, ifrank(A) = rank(B) then A and B are equivalent.

3. If A = [aij] and B = [bij] are mxn matrices, andrank(A) = rank(B) = n, then rank(AB) = rank(BA) =n.

Example: What is the rank of C?

98765

87654

76543

65432

54321

C

Solution:Operating on rows of matrix C, we obtain the equivalentmatrix C

455

344

233

122

RR'R

RR'R

RR'R

RR'R

11111

11111

11111

11111

54321

'C

We could easily see that all 5x5, 4x4 and 3x3 submatricesof C have determinants equal to zero (THEOREM: Identicalrows). But for at least one 2x2 submatrix of C has a non-zero determinant.

(e.g. 0111

21 )


11/12



Consequently r(C) = 2. But C and C are equivalentmatrices and hence they have equal ranks. Therefore r(C) isalso equal to 2.

RANKS AND THE TYPES OF SOLUTION TO A SYSTEMOF EQUATION

Recall that for the system of m linear equations in nunknowns AX = B. We can associate the system ofequation to the augmented matrix of the system [A:B].

The type of solution may be classified as unique, non-unique or inconsistent. Applying the concept of rank to theaugmented matrix [A:B], we have the following propositions:

1. If r(A) = r([A:B]) = n then the solution to thesystem is unique.

Example:

3:100

1:010

3:001

1:111

27:139

1:111

2. If r(A) = r([A:B]) < n, then the solution to the

system is non-unique.

Example:

0:000

1:010

1:001

6:224

3:112

0:111

3. If r(A) < r([A:B]), then the system has no solutionor inconsistent.

Example:

3:000

23:01045:001

6:116

2:224

3:112

Example 1: Rank and the Type of Solution to a System

For what values of k will the system of equations havea) a unique solutionb) a non-unique solutionc) no solution

2

k8z5y

k5z4y2x

k2z3y3x2

Solution: In augmented matrix form, we have:

28|510

5|421

2|332

k

k

k

Performing Gaussian Elimination Method:

kk

k

kRRR

k

k

kRRR

k

k

kRR

k

k

k

88|000

8|510

5|421

8|510

8|510

5|4212

8|510

2|332

5|421

8|510

5|421

2|332

2

23

'

3

2

12

'

2

2

21

2

Therefore we have the following conclusions:

a) For a unique solution, r(A) = r[A:B] = n

There will be no value of k that will satisfy thissince r(A) = 2 < n =3.

b) For non-unique solutions, r(A) = r[A:B] < n.

This will be satisfied if r[A:B] is also equal to 2.This will happen when the last element in the third row ofthe augmented matrix is also equal to zero.

8k2

- 8k = 0 k = 0, 1.

c) For the system to be inconsistent, r(A) < r[A:B]. This willbe satisfied if r[A:B] = 3 > 2. This will happen when the lastelement in the third row of the augmented matrix is notequal to zero.

8k2- 8k 0 k 0, 1.

Example 2:

For what values of m will the system of equations haved) a unique solutione) a non-unique solutionf) no solution

4mc2mba

0ca)2m(2cba

Solution: In augmented matrix form, we have:

4|11

0|102

2|111

2mm

m

Performing Gaussian Elimination Method:

2|100

42|120

2|111)2(

4|11

0|102

2|111

213

'

3

12

'

2

2

mm

mmm

RRR

RmRR

mm

m



a) For a unique solution, r(A) = r[A:B] = n

This will be satisfied if m2-10 andm-20. Thuswe have a unique solution if m 1,2.

b) For non-unique solutions, r(A) = r[A:B] < n.


12/12



This will happen when the last element in the thirdrow of matrix A and the augmented matrix are both equal tozero.

m2- 1 = 0 and m+2 = 0.

There is no value of m that will satisfy bothequation.

The other value of m to be checked is when m = -2.Substituting this to the system gives the system:

0|300

0|100

2|111

Clearly, we could see that the resulting systemgives a non-unique solution because, r(A) = r[A:B]=2 2. This willhappen when the last element in the third row of theaugmented matrix is not equal to zero but the last elementof the third row of A is equal to zero.

m21 = 0 and m+2 0.

Thus, the system is inconsistent when m = 1 .

Documents

First Exam Notes for ES 21