UNIVERSITY OF DUHOKFACLTY OF EDUATIONAL SCIENCE SCHOOL OF BASE EDUCATION DEPARTMWNT OF MATHMATICSMatrices and its Applications to Solve Some Methods of Systems of Linear Equations

A project submitted to the council of Department of Mathematics, School of Basic Education, University of Duhok, in partial fulfillment of the requirement for B.Sc. degree of mathematics

Prepared by: Supervisedby:Abdalla Haji Muwafaq Mahdi Salih

Karwan Hatm 1436 .A.H 2015.A.D 2715.K

Acknowledgement First of all, thanks to Allah throughout all his almighty kindness, and loveliness for letting us to finish our project. We would like to express our thanks to our supervisor Muwafaq Mahdi Salih for giving us opportunity to write this research under his friendly support. He made our research smoothly by his discerning ideas and suggestions. Also, we would like to thank all our friends and those people who helped us during our work.

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ContentsChapter OneBasic Concepts in Matrix.(1.1) Matrix…………………………………………………………….….....1(1.2) Some Special Types of Matrices…………………..………....2(1.3) Operations of Matrices…………………….……………..... ....11(1.4) The Invers of a Square Matrix……………………………….17(1.5) Some Properties of Determinants………………….…..….25Chapter TwoSystem of Linear Equations(2.1) Linear Equation………………………………………………………29(2.2) Linear System……………………………………………………….....30(2.2.1) Homogeneous System……………………………………….…..33(2.2.2) Gaussian Elimination………………………………………….…34(2.2.3) Gaussian-Jordan Elimination…………………………………35

(2.2.4) Cramer’s Rule………………………………………………….…...37References…………………………………………………………….…………39

AbstractIn this research, we have tried to introduce matrix, its types and finding inverse and determinant of matrix has involved. Then the applications of matrices to some methods of solving systems of linear equations, such as Homogeneous, Gaussian Elimination, Gaussian –jordan Elimination and Cramer’s Rule, have been illustrated.

Introduction Information in science and mathematics is often organized into rows and columns to form rectangular arrays called "Matrices" (plural of matrix). Matrices are often tables of numerical data that arise from physical observation, but they also occur in various mathematical contexts. Linear algebra is a subject of crucial important to mathematicians and users of mathematics. When mathematics is used to solve a problem it often becomes necessary to find a solution to a so-called system of linear equations. Applications of linear algebra are found in subjects as divers as economic, physics, sociology, and management consultants use linear algebra to express ideas solve problems, and model real activities. The aim of writing this subject is to applying matrices to solve some types of system of linear equations. In the first chapter of this work matrices have introduced. Then operations on matrices (such as addition and multiplication) where defined and the concept of the matrix inverse was discussed. In the second chapter theorems were given which provided additional insight into the relationship between matrices and solution of linear systems. Then we apply matrices to solve some methods of linear system such as Homogeneous, Gaussian Elimination, Gaussian –jordan Elimination and Cramer’s Rule.

CHAPTER ONEBasic Concepts in Matrix In this chapter we begin our study on matrix, some special types of matrix, operation of matrix, and finding invers and the determinant of matrices.

(1.1) MatrixAn m ×nmatrix A is rectangular array of m ×nnumbers arranged in m rows and n columns: A=[

a11 a12⋯ a1 j⋯ a1n

a21 a22⋯ a2 j⋯ a2n

⋮ ⋮ ⋮ ⋮ai1 ai2⋯ aij⋯ a¿

⋮ ⋮ ⋮ ⋮am1 am2⋯ amj⋯ amn

]The ijth component of A denoted a ij, is the number appearing in the ithrow and jthcolumn of A we will some time write matrix A as A=(a ij).An m×n matrix is said to have the size m×n.Examples:

(1) [1 23 45 6]3× 2

m=3 , n=2 , (2) [1 4 3 2 23 5 6 4 35 1 2 0 71 2 1 9 8

]4×5

m=4 , n=5

(1.2) Some Special Types of Matrices1. Square MatrixThe matrix which its number of rows equals to numbers of columns is called square matrix. That is A=[ a11 a12⋯ a1n

a21 a22⋯ a2n

⋮ ⋮ ⋮am1 am2⋯ amn

]m×nWhen m=n then:

A= [ a11 a12 ⋯ a1n

a21 a22 ⋯ a2n

⋮ ⋮ ⋱ ⋮an1 an2 ⋯ ann

]n×nA is a square matrixExample: K=[1 6 9

2 5 83 4 7]3× 3

2. Unit (Identity) MatrixThen × n matrix I n =a ij , defined by a ij=1 if ¿ j , a ij=0 ifi≠ j, is called the n × n identity matrix . I=[1 0 ⋯ 0

0 1 ⋯ 0⋮ ⋮ ⋱ ⋮0 0 ⋯ 1

] Example: I 2=[1 0

0 1] , I 3=[1 0 00 1 00 0 1 ]

Note:IA=AI=A

Example: A=[2 34 5] , I=[1 0

0 1]AI=[2 3

4 5]×[1 00 1]=[2∗1+3∗0 2∗0+3∗1

4∗1+5∗0 4∗0+5∗1]=[ 2 34 5 ]

IA=[1 00 1]×[2 3

4 5]=[1∗2+0∗4 1∗3+0∗50∗2+1∗4 0∗3+1∗5]=[ 2 3

4 5]We note that IA=AI=A3. Null (Zero) MatrixA zero matrix is a matrix which its elements are zeros and is denoted by the symbol 0 .0

mn=¿[0 0… 00 ⋯ 0⋮ ⋱ ⋮0 ⋯ 0

]¿Example:Let A=[2 1 −34 5 8 ] and B=[ 2 1

−3 4]We see that A +023=[2 1 −3

4 5 8 ]+[0 0 00 0 0 ]=[2 1 −3

4 5 8 ]=A

B022=[ 2 1−3 3 ][0 0

0 0]=[0 00 0]=022

Property of Zero Matrix A+0=0+ A=A

A−A=0 0−A=−A

0 A=0 , A 0=0

4. Diagonal Matrix

Diagonal matrix is a square matrix in which all the elements not on the main diagonal are zeros. A=[a11 0 ⋯ 0

0 a22 ⋯ 0⋮ ⋮ ⋱ ⋮0 0 ⋯ a33

]The elements of a square matrix A where the subscripts are equal, namely a11, a22, … , ann, form the main diagonal.Example: A=[1 0 0

0 2 00 0 3 ] , main diagonal=1,2,3

5. Commutative MatrixWe say that the matrices A and B are commutative under the operation product if A and B are square matrices and A . B=B . Aand we say that A∧Bare invertible commutative if A and B are square matrices andA . B ¿ B . A .Example: A=[5 1

1 5 ] ,B=[2 44 2]

A . B=[5 11 5][ 2 4

4 2]=[10+4 20+22+20 4+10]=[14 22

22 14 ]B . A=[2 4

4 2 ][5 11 5]=[10+4 2+20

20+2 4+10]=[14 2222 14 ]Then A . B=B . ANote: A square matrix A is said to be invertible if there exists

B such that AB=BA=I . B is denotedA−1and is unique. If det ( A)=0 then a matrix is not invertible.6. Triangular Matrix

A square matrix A is n × n , (n≥3), is triangular matrix iff a ij =0 wheni≥ j+1 ,∨ j ≥i+1.The are two type of triangular matrices:i. Upper Triangular MatrixA square matrix is called an upper triangular matrix if all the elements below the main diagonal are zero.Example: A=[1 5 9

0 2 10 0 3 ]ii. Lower Triangular MatrixA square matrix is called lower triangular matrix if all the elements above the main diagonal are zero.Example:

A=[1 0 06 2 09 7 3 ]

Transpose of Matrix Transpose of m ×n matrix A ,denoted AT or A, is n × m matrix with ( Aij )

T=A ji A=[ a11 a12⋯ a1n

a21 a22⋯ a2n

⋮ ⋮ ⋮am1 am2⋯ amn

]m×n

, AT=[ a11 a21⋯ am1

a12 a22⋯ am2

⋮ ⋮ ⋮a1n a2n⋯ amn

]n× m

row and columns of A are transposed in AT

Example: A=[0 4

7 03 1 ] , AT=[0 7 3

4 0 1] Note: transpose converts row vectors to column vector , vice versa.

Properties of TransposeLet A and B be matrix andc be a scalar. Assume that the size of the matrix are such that the operations can be performed. ( A+B )T=AT+BT Transpose of the sum (cA )T=c AT Transpose of scalar multiple ( AB )T=BT AT Transpose of a product ( AT )T=A

7. Symmetric MatrixA real matrix A is called symmetric if AT=A. In other words A is square (n× n)anda ij=a ji for all 1≤i ≤ n ,1≤ j≤ n . Example: A=[1 0 5

0 2 65 6 3 ] AT=[1 0 5

0 2 65 6 3 ]Note: if A=AT then A is a symmetric matrix

8. Skew-Symmetric Matrix Areal matrix A is called Skew-Symmetric if AT=−A . In other words A is square (n× n) and a ji=−a ij for all 1≤i ≤ n ,1≤ j≤ nExample:A=[ 0 5 6

−5 0 8−6 −6 0] −A=[0 −5 −6

5 0 −86 8 0 ]

A=[ 0 5 6−5 0 8−6 −8 0] AT=[0 −5 −6

5 0 −86 8 0 ] ∴ AT=−A

Determinants of MatrixThe determinant of a square matrix A=[aij ] is a number denoted by |A| or det ( A) , through which important properties such as singularity can be briefly characterized .This number is defined as the following function of the matrix elements:¿ A∨¿det (A )=±∏ a1 j1

a2 j2…anjn

Where the column indices j1, j2, …, jn are taken from the set {1,2,…,n} with no repetitions allowed . The plus (minus) sign is taken if the permutation ( j1 j2 … jn) is even (odd).Some properties of determinants will be discussed later in this chapter9. Singular and Nonsingular MatrixA square matrix A is said to be singular if det ( A)=0. A is nonsingular if det ( A)≠0.Theorem:Let A be a square matrix. Then A is a singular if (a) all elements of a row (column) are zero.(b) two rows (column) are equal. (c) two rows(column) are proportional.Note: (b) is a special case of (c) , but we list it separately to give it special emphasis.Example: we show that the following matrices are singular.

(a) A=[ 2 0 −73 0 1

−4 0 9 ] (b) B=[2 −1 31 2 42 4 8 ]

(a) All the elements in column 2 of A are zero. Thus det=0.(b) Observe that every element in row 3 of B is twice the corresponding element in row 2. We write (row 3) = 2(row 2)Row 2 and 3 are proportional. Thus det (b)=0.10. Orthogonal Matrixwe say that a matrix A is orthogonal if A . AT=I=AT . AExample:A=[1 0 0

012

√32

0−√32

12

] , AT=[1 0 0

012

−√32

0√32

12

]A . AT=[1 0 0

012

√32

0−√32

12

] . [1 0 0

012

−√32

0 √32

12

]¿ [1 0 0

014+ 34

12

.−√32

+ √32

.12

0−√32

.12+12

. √32

34

.14

]=[1 0 00 1 00 0 1]=I

∴ A is orthogonal matrix11. Toeplitz Matrix A matrix A is said to beToeplislitz if it has common elements on their diagonals, that is a i , j=a i+1 , j+1

Example: A=[5 6 20 5 63 0 5 ]Where a11=a22=a33=5

a12=a23=6a21=a32=012. Nilpotent MatrixA square matrix A is said to nilpotent if there is a positive integer p such that AP=0. The integer p is called the degree of nilpotency of the matrix.Example:A=[ 1 −3 −4

−1 3 41 −3 −4] , p=2

A2=0 A . A=[ 1 −3 −4

−1 3 41 −3 −4 ] .[

1 −3 −4−1 3 41 −3 −4]=[0 0 0

0 0 00 0 0]

13. Periodic (Idempotent) MatrixA matrix A is said to be periodic , that period (order) K , if A is satisfy AK +1=A and if K=1 then A2=A so A is called idempotent matrix.Example:A=[ 1 −2 −6

−3 2 92 0 −3 ] ,K=2

AK +1=A2+1=A3 A . A=[−5 −6 −6

9 10 9−4 −4 −3] , A2 . A=[ 1 −2 −6

−3 2 92 0 3 ]

∴ A is idempotent matrix

14. Stochastic MatrixAn n × n matrix A is called stochastic if each element is a number between 0 and 1 and each column ofA adds up to 1.A=[

14

13

0

12

23

34

14

014] ,

∑ column¿¿1) =1 ,∑ column(¿2)=1¿ , ∑ column(¿3)=1¿

15. Trace MatrixLet A be a square matrix, the trace of A denoted tr (A ) is the sum of the diagonal elements of A .Thus if A is an n × n matrix.tr (A )=a11+a22+…+ann Example:The trace of the matrix A =[4 1 −2

2 −5 67 3 0 ].is, tr (A )=4+(−5)+0=−1Properties of TraceLet A and B be matrix and c be a scalar, assume that the sizes of the matrices are such that the operations can be performed.

tr (A+B)=tr (A )+ tr(B)

tr (AB)=tr (BA) tr (cA)=ctr ( A) tr ( A )T=tr ( A)

Note: if A is not square that the trace is not defined. 16. Hermitian Matrix A square matrix A is said to be hermitian if AT=A.Note: The conjugate of a complex number z=a+ib is defined and written z=a-ib .Example:A=[ 3 7+ j2

7− j2 −2 ] ,is hermitian Taking the complex conjugates of each of the elements in A gives A=[ 3 7− j2

7+ j 2 −2 ] Now taking the transposes of A , we getAT=[ 3 7+ j2

7− j2 −2 ] So we can see that AT=A (1.3) Operations of Matrices1. Addition If A and B are m ×n matrices such that

A=[ a11 a12⋯ a1n

a21 a22⋯ a2n

⋮ ⋮ ⋮am1 am2⋯ amn

]m×n

and B= [ b11 b12⋯ b1n

b21 b22⋯ b2n

⋮ ⋮ ⋮bm1 bm2⋯ bmn

]m×n

Then A+B=[ a11+b11 a12+b12 ⋯ a1n+b1n

a21+b21 a22+b22 ⋯ a2n+b2n

⋮ ⋮ ⋱ ⋮am1+bm1 am2+bm2 ⋯ amn+bmn

]Note: Addition of matrices of different sizes is not defined.Example: [0 4

7 03 1]+[1 2

2 30 4]=[1 6

9 33 5] Properties of Matrix Addition

A+B=B+ A (commutative) ( A+B)+C=A+(B+C ) (associative), so we can write as

A+B+C A+0=0+ A=A

( A+B )T=AT+BT

2. Subtraction Matrix subtraction is defined for two matrix A=[aij ] and B=[bij ] of the same size in the usual way; that is A−B=[a ij]−[bij ]=[aij−b ij] . If A and B m ×n matrix such thatA=[ a11 a12⋯ a1n

a21 a22⋯ a2n

⋮ ⋮ ⋮am1 am2⋯ amn

]m×n

and B= [ b11 b12⋯ b1n

b21 b22⋯ b2n

⋮ ⋮ ⋮bm1 bm2⋯ bmn

]m×n

Then A−B=[ a11−b11 a12−b12 ⋯ a1n−b1n

a21−b21 a22−b22 ⋯ a2n−b2n

⋮ ⋮ ⋱ ⋮am1−bm1 am2−bm2 ⋯ amn−bmn

]

Note: Subtraction of matrices of different sizes is not defined.Example: [0 4

7 03 1]-[1 2

2 30 4]=[−1 2

5 −33 −3]3. NegativeConsider C to be a matrix, the negative of C denoted by −C, which defined as (−1)C, Where each element in C is multiplied by (−1 ) .Example:

C=[3 −2 47 −3 0 ]. Then – C=[−3 2 −4

−7 3 0 ].4. MultiplicationWe can product two matrices A and B if the number of column in a matrix A be equal to the number of rows in a matrix B.The element in row i and column jof AB is obtained by multiplying the corresponding element of rowiof A and column j of B and adding the products. [ . ¿

. ¿

. ¿ ] B is 3×n

A is m ×3 ¿ ¿ AB is m ×nNote: The product of A and B con not be obtained if the number of columns in A does not equal the number of rows in B . Let A have n columns and B have n rows .The ith row of A is [a i1ai2 …a¿] and the jth column of B is[b1 j

b2 j

⋮bnj

].Thus if ¿ AB , then c ij=a i1b1 j+a i2b2 j+…+a¿bnj.

Properties of Matrix Multiplication 0 A=0 , A 0=0 (here 0 can be scalar, or a compatible matrix) IA=AI=A

( AB)C=A(BC), so we can write asABC

α(AB)=( αA)B , where α is a scalar A(B+C )=AB+ AC , (A+B)C=AC+BC

( AB )T=BT AT

Scalar Multiplicationlet A be a matrix and c be a scalar, the scalar multiple of A by c, denoted cA , is the matrix obtained by multiplying every element of A by c, the matrix cA will be the same size of A.Thus if B=CA, then b ij =ca ij.Example: let A=[1 −2 4

7 −3 0],determine 3 A.Then Now multiple every element of A by 3 we get 3 A=[ 3 −6 12

21 −9 0 ] . Observe that A and 3 A are both 2×3matrixRemark: If A is a square matrix, thenAmultiplied by itself k times is written Ak. Ak=AA … A , K times

Familiar rules of exponents of real numbers hold for matrices.Theorem:If A is an n × nsquare matrix and r and s are nonnegative integers, then 1. Ar A s=A r+ s2. ( A r )s=A rs3. A0=In (by definition)We verify the first rule, the proof of the second rule is similar Ar A s= A … A⏟

r׿ A … A⏟s׿= A… A⏟

s+r ׿= Ar+ s¿

¿

¿

Example: If ¿ [ 1 −2−1 0 ] , compute A4.This example illustrates how the above rules can be used to reduce the amount of computation involved in multiplying matrices. We know that A4=AAAA. We could perform three matrix multiplication to arrive atA4.However we con apply rule 2 above to write A4=( A2)2 and thus arrive at the result using two products. We get

A2=[ 1 −2−1 0 ] [ 1 −2

−1 0 ]=[ 3 −2−1 2 ]

A4=[ 3 −2−1 2 ][ 3 −2

−1 2 ]=[ 11 −10−5 6 ] The usual index laws hold provided AB=BA

( AB )n ¿ An Bn

Am Bn=Bn Am

( A+B )2=A2+2 AB+B2

( A+B )n=∑i=0

n

(ni ) Ai Bn−i

( A+B)( A−B)=A2– B2

We now state basic of the natural numbers.EqualityThe matrix A and B are said to be equal if A and B have the same size and corresponding element are equal; that is A andB∈M (m× n) and A=[aij ] , B=[b ij]with a ij =b ij for 1≤i ≤ m,1≤ j ≤ n .Example:A=[1 2 3

4 5 67 8 9 ] ,B=[3 1 2

4 5 67 8 9] ,C=[1 1+1 3

4 5122

7 5+3 3∗3]

A=C

Minor Let A be an n × n square matrix obtained from A by deleting the ith row and jth column of A M ijis called the ijth minor of A .A=[a11 a12 a13

a21 a22 a23a31 a32 a33

] M 11=[a22 a23

a32 a33 ] , M 12=[a21 a23a31 a33 ] , M 13=[a21 a22

a31 a32 ] CofactorsThe cofactor c ij is defined as the coefficient of a ij in the determinant A If is given by the formulaC ij=(−1)i+ jmijWhere the minor is the determinant of order (n−1)×(n−1) formed by deleting the column and row containing a ij.

c11=(−1)1+1m11=+1. [a11 a12 a13a21 a22 a23a31 a32 a33

] =+1. |a22 a23a32 a33|=a22a33-a32a23

c12=(−1)1+2m12= -1. [a11 a12 a13a21 a22 a23a31 a32 a33

]=-1. |a21 a23a31 a33|=a21 a33-a31a23

c13=(−1)1+3m13=+1.[a11 a12 a13a21 a22 a23a31 a32 a33

]=+1. |a21 a22a31 a32|=a21a32-a31a22

Definition: Let A be n × nmatrix and c ij be the cofactor of a ij.The matrix whose (i , j)th element is c ij is called the matrix of the cofactors of A. The transpose of this matrix is called the adjoint of A and is denoted adj( A).[ c11 c12 ⋯ c1n

c21 c22 ⋯ c2n

⋮ ⋮ ⋱ ⋮cn1 cn2 ⋯ cnn

] [ c11 c12 ⋯ c1n

c21 c22 ⋯ c2n

⋮ ⋮ ⋱ ⋮cn1 cn2 ⋯ cnn

]T

¿cofactors Adjoint ¿Example: give the matrix of the cofactors and the adjoint matrix of the following matrix A. A=[ 2 0 3

−1 4 −21 −3 5 ] The cofactor of A are as follows.

c11=| 4 −2−3 5 |=14 , c12=−|−1 −2

1 5 |=3 , c13=|−1 41 −3|=-1

c21=−| 0 3−3 5|=-9 , c22=|2 3

1 5|=7 , c23=−|2 01 −3|=6

c31=|0 34 −2|=-12 , c32=−| 2 3

−1 −2|=1 , c33=| 2 0−1 4|=8

The matrix of cofactors of A is [ 14 3 −1−9 7 6−12 1 8 ]

The adjoint of A is the transpose of this matrix adj( A)=[ 14 −9 −12

3 7 1−1 6 8 ]

(1.4) The Inverse of a Square MatrixThe inverse a−1 of a scalar (¿anumber) a is defined bya a−1=1.For square matrices we use a similar definition the inverseA−1 of a n × n matrix A fulfils the relation

A A−1=Iwhere I is then × n unit matrix defined earlier.Example: we show that, as usual, the identity matrix of the appropriate size. A=[−1 1−2 0] , B=1

2 [0 −12 −1]

All we need do is to check that AB=BA=I

AB=[−1 1−2 0]× 12 [0 −1

2 −1]=12 [0 −12 −1]×[−1 1

−2 0]=12 [2 00 2]

¿ [1 00 1] The reader should check that AB=I alsoNote: if A−1exists then

det ( A)det (A−1)=det ( A A−1)=det(I )=1 Hence det ( A−1)= (detA )−1

Example: A=[3 72 6]

det ( A)=18−14=4 , ( det ( A ) )−1=14

A−1=14 [ 6 −7

−2 3 ]=[ 64 −74

−24

34

] det ( A−1)=18

16−1416

= 416

=14

∴det( A−1)=(detA )−1 , 14=14

Remark: Non-square matrices do not have an inverse. The inverse of A is usually written A−1. Not all square matrices have an inverse. A square matrix A is invertible if and only if det ( A)≠0. A−1 exists if and only if A is nonsingular.

Finding Inverse of Matrix1-The Inverse of a 2×2 MatrixIf ad−bc≠0 then the 2×2 matrix A=[a b

c d ] has a (unique)inverse given by A−1= 1ad−bc [ d −b

−c a ]Note: ad−bc=0 that Ahas no inverse.In words: to find the inverse of a 2×2matrix A we effectively interchange the diagonal elements a and d, change the sign of the other two elements and then divide by the determinant of A. Example: A=[3 7

2 6]

det ( A)=18−14=4 , A−1=14 [ 6 −7

−2 3 ]=[ 64 −74

−24

34

] 2-The Inverse of a 3×3 Matrix-The Determinant Method Given a square matrixA: Find det ( A), if det ( A)=0 then ,as we know A−1does not exists. If det ( A)≠0 we can proceed to find the inverse matrix. Replace each element ofA by its cofactor

C11=[a22 a23a32 a33] , C12=[a21 a23

a31 a33] , C13=[a21 a22a31 a32]

C21=[a12 a13a32 a33] , C22=[a11 a13

a31 a33] , C23=[a11 a12a31 a32]

C31=[a12 a13a22 a23] , C32=[a11 a13

a21 a23] , C33=[a11 a12a21 a22]

Cofactor =[C11 C12 C13

C21 C22 C23

C31 C32 C33] =A

Transpose the result to form theadjoint ¿ , adj( A) .

Adj( A)=[C11 C21 C31

C12 C22 C32

C13 C23 C33] , adj(A )=AT

Then A−1= 1det( A)

adj( A).Example: find the inverse of A=[ 1 −1 2

−3 1 23 −2 −1]

det ( A)=1| 1 2−2 −1|−(−1)|−3 2

3 −1|+2|−3 13 −2|

¿1×3+1× (−3 )+2×3=6 Since the determinant is non-zero an inverse exists.

Calculate the matrix of minorsM=||

1 2−2 −1| |−3 2

3 −1| |−3 13 −2|

|−1 2−2 −1| |1 2

3 −1| |1 −13 −2|

|−1 21 2| | 1 2

−3 2| | 1 −1−3 1 || =[ 3 −3 3

5 −7 1−4 8 −2]

Modify the signs according to whether i+ j is even or odd to calculate the matrix of cofactors C=| 3 3 3

−5 −7 −1−4 −8 −2|

It follows that A−1=16

CT=16 |3 −5 −43 −7 −83 −1 −2|To check that we have made no mistake we can compute

A−1 . A=16|3 −5 −43 −7 −83 −1 −2|. [ 1 −1 2

−3 1 23 −2 −1] =[1 0 0

0 1 00 0 1 ] .This way of computing the invers is only useful for hand calculations in the cases of 2×2 or 3×3 matrices.

Definition:A matrix is said to be in row-echelon form if 1. If there are any rows of all zeros then they are at the bottom of the matrix.2. If a row does not consist of all zeros then its first non-zero entry (i.e. the left most nonzero entry) is a 1. This 1 is called a leading 1.3. In any two successive rows, neither of which consists of all zeroes, the leading 1 of the lower row is to the right of the leading 1 of the higher row.

Example:The following matrices are all in row echelon form.[1 −6 9 1 00 0 1 −4 −50 0 0 1 2 ] [1 0 5

0 1 30 0 1] [1 −8 10 5 −3

0 1 13 9 120 0 0 1 10 0 0 0 0

]Definition:A matrix is in reduced row−echelon form if1. Any rows consisting entirely of zeros are grouped at the bottom of the matrix.2. The first nonzero element of each other row is 1. This element is called a leading 1.3. The leading 1 of each row after the first is positioned to the right of the leading 1 of the previous row.4. All other elements in a column that contains a leading 1 are zero.Example:The following matrices are all in reduced echelon form.[1 0 80 1 20 0 0 ] [1 0 0 7

0 1 0 30 0 1 9] [1 4 0 0

0 0 1 00 0 0 1] [1 2 3 0

0 0 0 10 0 0 0]

The following matrices are not in reduced echelon form.

[1 2 0 40 0 0 00 0 1 3 ]

row of zerosnot at bottom

of ¿¿

[1 2 0 3 00 0 3 4 00 0 0 0 1]

first nonzeroelement∈row2is not 1

[1 0 0 20 0 1 40 1 0 3]

leading1∈¿ row3not ofthe¿

of ¿

leading1∈¿ row 2¿¿ [1 7 0 80 1 0 30 0 1 20 0 0 0

]nonzero

element aboveleading1∈¿ row2

3-The Inverse of a 3×3 Matrix – Gauss− jordan Elimination Method Let A be an n × n matrix1. Adjoin the identityn × n matrix I n to A to form the matrix [ A : I n] .2. Compute the reduced echelon form of[ A : I n] .If the reduced echelon form is of the type [I n :B ], then B is the inverse of A. If the reduced echelon form is not of the type

[I n :B ], in that the first n × nsub matrix is notI n, then A has no inverse.Some Notes for Operation of The Method Inter changing two rows. Adding a multiple of on row to on other row. Multiplying one row by a non-zero constant.The following example illustrate the method Example: determine the inverse of the matrix

A=[ 1 −1 −22 −3 −5

−1 3 5 ]

Applying the method of Gauss− jordan elimination ,we get [ A : I n] = [ 1 −1 −2 ⋮ 1 0 02 −3 −5 ⋮ 0 1 0

−1 3 5 ⋮ 0 0 1] r 2 :R2+ (−2 ) R1r 3:R3+R1

→[1 −1 −2 ⋮ 1 0 00 −1 −1 ⋮ −2 1 00 2 3 ⋮ 1 0 1]

r 2 : (−1 ) R2→ [1 −1 −2 ⋮ 1 0 0

0 1 1 ⋮ 2 −1 00 2 3 ⋮ 1 0 1] r 1 :R1+R2

r 3 :R3+(−2 ) R2→

[1 0 −1 ⋮ 3 −1 00 1 1 ⋮ 2 −1 00 0 1 ⋮ −3 2 1]

r 1:R1+R3r 2 :R2+ (−1 ) R3

→[1 0 0 ⋮ 0 1 10 1 0 ⋮ 5 −3 −10 0 1 ⋮ −3 2 1 ] = [I n :B ]

B =[ 0 1 15 −3 −1

−3 2 1 ]=A−1

The following example illustrates the application of the method for a matrix that does not have an inverse.Example: determine the inverse of the following matrix ,if it exists.A=[1 1 5

1 2 72 −1 4 ]

Applying the method of Gauss− jordan elimination. we get[ A : I 3]= [1 1 5 ⋮ 1 0 0

1 2 7 ⋮ 0 1 02 −1 4 ⋮ 0 0 1] r 2 :R2+ (−1 ) R1

r 3 :R3+(−2 ) R1→

[1 1 5 ⋮ 1 0 00 1 2 ⋮ −1 1 00 −3 −6 ⋮ −2 0 1]

r 1:R1+(−1 ) R2r 3:R3+3 R2

→[1 0 3 ⋮ 2 −1 00 1 2 ⋮ −1 1 00 0 0 ⋮ −5 3 1]

Properties of Inverse If AB=Iand CA=I, then B=C. Consequently A has at most one inverse.

( AB )−1=B−1 A−1 (assuming A , B are invertible ). ( AT )−1=( A−1 )T (assuming A is invertible ). ( A−1 )−1=A, i .e. inverse of inverse is original matrix (assuming A is invertible ). I−1=I . (∝ A )−1=( 1∝ ) A−1 (assuming A invertible , ∝≠0 ). If y=AX,where X∈Rnand A invertible , then X=A−1 y

A−1 y=A−1 AX=IX=X . If A1 , A2 ,…, A k ,are all invertible , so is their product A1 , A2 ,…, A k , and ( A1 A2… Ak )−1=Ak

−1… A2−1 A1

−1. If A is invertible , so is Ak for k ≥1, and ( A k )−1=( A−1 )k.

Corollary 1: If AB=Iand CA=I, then B=C, consequently A has at most one inverse.Proof: If AB=I and CA=I, then B=IB=CAB=CI=C, if B and C are both inverses ofA, then , by definition, AB=BA=I and AC=CA=I, in particular AB=I and CA=I, so that B=C.Corollary 2: If A and B are both invertible, then so is AB and ( AB )−1=B−1 A−1.Proof: We have a guess for ( AB )−1, to check that the guess is correct, we merely need to check the requirements of the definition

( AB)(B−1 A−1)=AB B−1 A−1=AI A−1=A A−1=I(B−1 A−1)(AB)=B−1 A−1 AB=B−1 IB=B−1B=ICorollary 3: If A is invertible, then so is AT and ( AT )−1=( A−1 )T.Proof: Let’s use B to denote the inverse of A (so there won’t be so many superscripts around) by definition AB=BA=IThese three matrices are the same, so their transposes are the same. Since ( AB )T=AT BT, ( BA )T=AT BT and I T=I, we have

BT AT=AT BT=IT ¿ IWhich is exactly the definition of “the inverse of AT is BT”. (1.5) Some Properties of Determinants

1. Rows and columns can be interchanged without affecting the value of a determinant. Consequently det ( A)=det ( AT ). Example: A=|3 4

1 2|, AT=|3 14 2|

det ( A)=2 , det ( A T )=2 ∴det ( A)=det ( AT )2. If two rows, or two columns, are interchanged the sign of the determinant is reversed. Example: if A=|3 41 −2| , then det (|3 41 −2|)=−10 , det (|1 −2

3 4 |)=103. If a row(or column) is changed by adding to or subtracting from its elements the corresponding elements of any other row (or column) the determinant remains unaltered.Example: det (|3 41 −2|)=det (|3+1 4−2

1 −2 |)=det (|4 21 −2|)=−104. If the elements in any row (or column) have a common factor α then the determinant equals the determinant of the corresponding matrix in which α = 1, multiplied by α. Example: A=|6 8

1 −2|, α=2det (|6 8

1 −2|)=−20 , det (|6 81 −2|)=2det (|3 4

1 −2|)¿2×(−10)=−20 5. The determinant of an upper triangular or lower triangular matrix is the product of the main diagonal entries. Example: A upper triangular, B lower triangular

A=|2 2 10 2 −10 0 4 |=det ( A )=2×2×4=16

B=|2 0 03 −3 04 1 4|=det (B )=2×(−3)×4=−24

This rule is easily verified from the definition det ( A)=±∏ a1 j1a2 j2

…anjn because all terms vanish except j1= 1, j2= 2, . . . jn = n, which is the product of the main diagonal entries. Diagonal matrices are a particular case of this rule.6. The determinant of the product of two square matrices is the product of the individual determinants: det ( AB)=det ( A)det(B) .Example: A=|6 8

1 −2| , B=|6 01 −2|,

det ( A )=6× (−2 )−8×1=−20det (B)=6×(−2)−0×1=−12

det ( AB)=det (|6 81 −2|.|6 0

1 −2|)=det (|44 −164 4 |)=240

det ( A)det (B)=−20×−12=240 This rule can be generalized to any number of factors. One immediate application is to matrix powers: |A2|=|A||A|=|A|2, and more generally |An|=|A|n for integer n.7. Let A be an n × n matrix and c be a scalar then, det (cA )=cn det (A )Example : For the given matrix below we compute both det ( A) and det (2 A ).

A=[ 4 −2 5−1 −7 100 1 −3 ] First the scalar multiple.

2A=[ 8 −4 10−2 −14 200 2 −6 ] The determinants.

det ( A )=45 , det (2 A )=360=(8 ) (45 )=23det (A )8. Suppose that A is an invertible matrix then, det ( A−1)= 1det (A )

Example: For the given matrix we compute det ( A)∧det ( A−1)

A=[8 −92 5 ]

A−1=[ 558 958

−129

429

] Here are the determinants for both of these matrices.det ( A)=58det (A−1)= 1

589. Suppose that A is an n× n triangular matrix then,det ( A)=a11a22…ann

Finding Determinant1-The Determinant of a 2×2 MatrixA=[a11 a12

a21 a22 ], is writtendet ( A)=a11a22-a12a21

Example: A=[1 24 −7 ],det ( A)=−7−8=−152-The Determinant of a3×3 Matrix

det ( A)= [a11 a12 a13a21 a22 a23a31 a32 a33

] =a11[a22 a23a32 a33] −a12[a21 a23

a31 a33 ] +a13[a21 a22a31 a32 ]

=a11(a22a33-a32a23)−a12(a21a33-a31a23)+a13(a21a32-a31a22)That is the 3×3determinants is defined in terms of determinants of 2×2sub-matrices of .

Or[a11 a12 a13a21 a22 a23a31 a32 a33

]a11 a12a21 a22a31 a32

det ( A )=a11a22a33+a12a23 a31+a13a21a32−a13 a22a31−a12a21a33−a11a23a32.Example: find the determinant of A=[1 2 3

1 0 23 2 1]

det ( A)=1× det|0 22 1|−2× det|1 2

3 1|+3× det|1 03 2| ¿1×(0−4)−2×(1−6)+3×(2−0)=12Or

det ( A)=(1×0×1)+(2×2×3)+(1×2×3)-(3×0×3)-(2×1×1)-(2×2×1) =0+12+6-0-2-4=18-6=123-Cofactor ExpansionThe determinant of an n × n matrix may be found by choosing a row (or column) and summing the products of the entries: det ( A)=a1 j c1 j+a2 jc2 j+…+anj cnj (cofactor expansion along the jthcolumn)

det ( A)=ai1c i1+ai2 c i2+…+a¿ c¿(cofactor expansion along the ith row)Where c ij is the determinant of A with row i and column j deleted, multiplied by (−1)i+ j. The matrix of elements c ij is called the cofactors matrix.Example: cofactor expansion along the first column A=[ 3 1 0

−2 −4 35 4 −2] . evaluate det ( A) by cofactor expansion along the column of A .

det ( A)=[ 3 1 0−2 −4 35 4 −2]

=3|−4 34 −2|+(-2)|1 0

4 −2|+5| 1 0−4 3| =3(-4)+(-2)(-2)+5(3)=6

CHAPTER TWOSystem of Linear EquationsIn this chapter we study the system of linear equation. Then we illustrate using of matrices to solve system linear equation in terms of the methods of Homogeneous Systems, Gaussian Elimination, Gauss-Jordan Elimination and Cramer’s Rule

(2.1) Linear Equations Definition:Let a1 , . . . , an, y be elements of R, and let x1, . . . , xn be unknowns(also called variables or indeterminates). Then an equation of the form a1 x1+ … + an xn= yis called a linear equation in n unknowns (over R), the scalars a i are called the coefficients of the unknowns , and y is called the constant term of the equation.Example:x1+2x1=5 a1=1 , a2=2 , y=5 (This equation are linear )But: x21+3√ x2 =5 (This equation is not linear) Note: not all a are zero .Definition:A linear equation in the two variables x1and x2 is an equation that can in the form a1 x1+a2 x2=b

Where a1, a2 , and b are numbers, in general a linear equation in the n variables x1,x2,…,xnis an equation that can be written in the form a1 x1+a2 x2+…+an xn=bWhere the coefficients a1, a2, …, an and the constant term b are numbers. Example: x1+7x2=3 , x1+x2+…+xn=4Some example of equations that are not linear are:x12+x1 x3=3, 1x1+x2+x3=5

2 , e(x1 )+x2=12 , x1+x2

x3+x4=x5+7

(2.2) Linear System In general, a system of linear equations (also called a linear system) in the variables x1 , x2 ,…, xn consists of a finite number of linear equations in these variables. The general form of a system of m equations in n unknowns is a11 x1 +¿a12 x2 +¿⋯ +¿a1n xn ¿ b1a21 x1 +¿a22 x2 +¿⋯ +¿a2n xn ¿ b2⋮ ¿ ⋮ ¿ ¿ ⋮ ¿ ¿

+¿ am2 x2 ¿+¿⋯ ¿+¿amn xn¿=¿bm¿

We will call such a system an m ×n (m by n) linear system. Example:6 x1 +¿2 x2 −¿ x3 ¿ 53 x1 +¿ x2 −¿4 x3 ¿ 9−x1 +¿3 x2 +¿2 x3 ¿ 0Definition:The coefficients of the variables form a matrix called the matrix of coefficient of the system.

[ a11 a12⋯ a1n

a21 a22⋯ a2n

⋮ ⋮ ⋮am1 am2⋯ amn

]m×n

Definition: The coefficients, together with the constant terms, form a matrix called the augmented matrix of the system. aug A=[ a11 a12⋯ a1n y1

a21 a22⋯ a2n y2⋮ ⋮ ⋮ ⋮

am1 am2⋯ amn yn]m× n

Example: The matrix of coefficients and the augmented matrix of the following system of linear equations are as shown:x1 +¿ x2 +¿x3 ¿ 22 x1 +¿3 x2 +¿x3 ¿ 3x1 −¿ x2 −¿2x3 ¿ −6

[1 1 12 3 11 −1 −2]⏟

¿coefficients

[1 1 1 22 3 1 31 −1 −2 −6]⏟

augmented ¿¿

In solving systems of equations we are allowed to perform operations of the following types:1. Multiply an equation by a non-zero constant.2. Add one equation (or a non-zero constant multiple of one equation) to another equation.These correspond to the following operations on the augmented matrix :1. Multiply a row by a non-zero constant.2. Add a multiple of one row to another row.

3. We also allow operations of the following type : Interchange two rows in the matrix (this only amounts to writing down the equations of the system in a different order).

Definition:Operations of these three types are called Elementary Row Operations (ERO's) on a matrix.Example: [ 1 2 −1 53 1 −2 9

−1 4 2 0] x +¿2 y −¿ z ¿ 53 x +¿ y −¿2 z ¿ 9−x +¿ 4 y +¿2 z ¿ 0

R3→ R3+R1[1 2 −1 53 1 −2 90 6 1 5]

x +¿2 y −¿ z ¿ 53 x +¿ y −¿2 z ¿ 96 y +¿ z ¿ ¿ ¿

R2→ R2−3R1[1 2 −1 50 −5 1 −60 6 1 5 ] x +¿2 y −¿ z ¿ 5

−¿5 y +¿ z ¿ ¿ ¿+¿ z ¿=¿5¿

R2→ R2+R3 [1 2 −1 50 1 2 −10 6 1 5 ] x +¿2 y −¿ z ¿ 5

y +¿2 z ¿ ¿ ¿+¿ z¿=¿5¿

R3→ R3−6 R2[1 2 −1 50 1 2 −10 0 −11 11 ] x +¿2 y −¿ z ¿ 5

y +¿2 z ¿ ¿ ¿=¿11¿

R3×(−111 )[1 2 −1 50 1 2 −10 0 1 −1] x +¿2 y −¿ z ¿ 5… ( A )

y +¿2 z ¿ ¿ ¿=¿−1… (C )¿

We have produced a new system of equations, this is easily solved:Backsubstitution

{ (C ) z=−1(B) y=−1−2 z → y=−1−2(−1)=1

( A ) x=5−2 y+z→ x=5−2 (1 )+(−1 )=2∴ x=2, y=1 , z=−1 (2.2.1) Homogeneous Systems A system of linear equations is said to be homogeneous if all the constant terms are zeros, a system of homogeneous linear equations is a system of the form

a11 x1 +¿a12 x2 +¿⋯ +¿a1m xm ¿ 0a21 x1 +¿a22 x2 +¿⋯ +¿a2m xm ¿ 0⋮ ¿ ⋮ ¿ ¿ ⋮ ¿ ¿

+¿an2 x2 ¿+¿⋯ ¿+¿anm xm ¿=¿0¿

Such a system is always consistent as x1=0,x2=0 , …. , xm=0is a solution, this solution is called the trivial (or zero) solution, any other solution is called a non-trivial solution.Example:x− y=0x+ y=0Has only the trivial solution, whereas the homogeneous system

x− y+z=0x+ y+z=0Has the complete solution x=−z ; y=0 ; z arbitrary. In particular, takingz=1 gives the non-trivial solution x=−1 ; y=0 ; z=1.There is simple but fundamental theorem concerning homogeneous systems.

(2.2.2) Gaussian EliminationThis method is used to solve the linear system AX=B by transforms the augmented matrix [ A :B] to row echelon form and then uses substitution to obtain the solution this procedure some what more efficient than Gauss-Jordan reduction. Consider a linear system1. Construct the augmented matrix for the system.2. Use elementary row operation to transform the augmented matrix into a triangular one.3. Write down the new linear system for which the triangular matrix is the associated augmented matrix.4. Solve the new system, you may need to assign some parametric values to some unknowns, and then apply the method of back substitution to solve the new system.Example: We use Gaussian elimination to solve the system of linear equations.2 x2+x3=−8

x1−2 x2−3x3=0

−x1+ x2+2 x3=3The augmented matrix is[ 0 2 1 ⋮ −81 −2 −3 ⋮ 0

−1 1 2 ⋮ 3 ]Swap Row1 and Row2

[ 1 −2 −3 ⋮ 00 2 1 ⋮ −8

−1 1 2 ⋮ 3 ]Add Row1 to Row3[1 −2 −3 ⋮ 00 2 1 ⋮ −80 −1 −1 ⋮ 3 ]

Swap Row2 and to Row3[1 −2 −3 ⋮ 00 −1 −1 ⋮ 30 2 1 ⋮ −8]Add twice Row2 to Row3[1 −2 −3 ⋮ 00 −1 −1 ⋮ 30 0 −1 ⋮ −2]Multiply Row3 by -1[1 −2 −3 ⋮ 00 −1 −1 ⋮ 30 0 1 ⋮ 2]

x1−2 x2−3x3=0…….(a)−x2−x3=3……..(b)x3=2……..(c)Then x3=2 and we put (c) in (b)Then x2=−5 and we put (b)and (c) in (a)Then x1=¿-4.(2.2.3) Gauss-Jordan EliminationThe Gauss-Jordan elimination method to solve a system of linear equations is described in the following steps.1. Write the augmented matrix of the system.

2. Use row operations to transform the augmented matrix in the form described below, which is called the reduced row echelon form (RREF).3. Stop process in step 2 if you obtain a row whose elements are all zeros except the last one on the right. In that case, the system is inconsistent and has no solutions. Otherwise, finish step 2 and read the solutions of the system from the final matrix.Note: When doing step 2, row operations can be performed in any order. Try to choose row operations so that as few fractions as possible are carried through the computation. This makes calculationeasier when working by hand.Example: we Solve the following system by using the Gauss-Jordan elimination method.{ x+ y+z=52x+3 y+5 z=84 x+5 z=2The augmented matrix of the system is the following.

[1 1 1 ⋮ 52 3 5 ⋮ 84 0 5 ⋮ 2]We will now perform row operations until we obtain a matrix in reduced row echelon form.

[1 1 1 ⋮ 52 3 5 ⋮ 84 0 5 ⋮ 2]R2−2R1

→ [1 1 1 ⋮ 50 1 3 ⋮ −24 0 5 ⋮ 2 ]

R3−4 R1→ [1 1 1 ⋮ 5

0 1 3 ⋮ −20 −4 1 ⋮ −18] R3+4 R2

→ [1 1 1 ⋮ 50 1 3 ⋮ −20 0 13 ⋮ −26]

113 R3→

[1 1 1 ⋮ 50 1 3 ⋮ −20 0 1 ⋮ −2] R2−3 R3

→ [1 1 1 ⋮ 50 1 0 ⋮ 40 0 1 ⋮ −2]

R1−R3→ [1 1 0 ⋮ 7

0 1 0 ⋮ 40 0 1 ⋮ −2] R1−R2

→ [1 0 0 ⋮ 30 1 0 ⋮ 40 0 1 ⋮ −2]

From this final matrix, we can read the solution of the system. It isx=3 , y=4 , z=−2 .

(2.2.4) Cramer’s RuleIf AX=B is a system of linear equation in n-unknown, such that det ( A)≠0 then the system has a unique solution, this solution is

X1=det ( A1)det ( A)

, X2=det (A2)det ( A)

, X3=det ( A3)det (A )Where Ai is the matrix obtained by replacing the entries in the ith column of A by the entries of b th column. The system

a11 x1 +¿a12 x2 +¿⋯ +¿a1m xm ¿ b1a21 x1 +¿a22 x2 +¿⋯ +¿a2m xm ¿ b2⋮ ¿ ⋮ ¿ ¿ ⋮ ¿ ¿

+¿an2 x2 ¿+¿⋯ ¿+¿anm xm ¿=¿bn¿

[ a11 a12⋯ a1m b1a21 a22⋯ a2m b2⋮ ⋮ ⋮ ⋮

an1 an2⋯ anm bn]n × m

A=[ a11 a12⋯ a1m

a21 a22⋯ a2m

⋮ ⋮ ⋮an1 an2⋯ anm

]n× m

A1=[b1 a12⋯ a1m

b2 a22⋯ a2m

⋮ ⋮ ⋮bn an2⋯ anm

]n ×m

, A2=[ a11 b1⋯ a1m

a21 b2⋯ a2m

⋮ ⋮ ⋮an1 bn⋯ anm

]n ×m

, … , Am=[ a11 a12⋯ bm

a21 a22⋯ bm

⋮ ⋮ ⋮an1 an2⋯ bm

]n ×m

Example: we use crammer’s Rule to solve:3 x1+4 x2−3 x3=53 x1−2x2+4 x3=73x1+2 x2− x3=2In matrix form AX=b or [a1a2a3] X=b this is

[3 4 −33 −2 43 2 −1] [x1x2

x3]=[573].

Cramer’s Rule says that x1=det ( A1)det ( A) , x2=det ( A2)

det ( A) , x3=det ( A3)det ( A)

A=[3 4 −33 −2 43 2 −1] ,det ( A)=6

A1=[5 4 −37 −2 43 2 −1] , det ( A1)=−14

A2=[3 5 −33 7 43 3 −1] ,det ( A2)=54

A3=[3 4 53 −2 73 2 3] , det ( A3)=48

x1=−146

=−73

, x2=546

=9 , x3=486

=8.

References

1.Dennis M. Schneider, Manfred Steeg ,Frank H. Young. Linear Algebra a Concrete Introduction. Macmillan Co. Unites of America, 1982.2.S. Barry and S. Davis, Essential Math. Skills, National Library of Australia, 2002.3.Garth Williams . Linear Algebra With Applications . Jones And Bartlett , Canada , 2001.4.P. Dawkins . Linear Algebra.

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