lecture1_d3

MA 106 - Linear Algebra

Neela Nataraj

Department of Mathematics,Indian Institute of Technology Bombay,

Powai, Mumbai [email protected]

January 7, 2013

Neela Nataraj Lecture 1

Outline of the lecture

Matrices (A revision) & Gauss Elimination

MatrixAddition of MatricesMultiplication of MatricesProperties of Addition & MultiplicationTranspose & PropertiesSpecial MatricesInverse of a MatrixLinear SystemsGauss Elimination Method


Linear Algebra

One of the most important basic areas in Mathematics, havingat least as great an impact as Calculus.Provides a vital arena where the interaction of Mathematicsand machine computation is seen.Many of the problems studied in Linear Algebra are amenableto systematic and even algorithmic solutions, and this makesthem implementable on computers.Many geometric topics are studied making use of conceptsfrom Linear Algebra.Applications to Physics, Engineering, Probability & Statistics,Economics and Biology.


Matrix

A matrix is a rectangular array of numbers (or functions) enclosedin brackets with the numbers (or functions) called as entries orelements of the matrix.

Matrices are denoted by capital letters A, B, C , · · · .For integers m,n≥ 1, an m×n matrix is an array given by

a11 a12 . . . a1na21 a22 . . . a2n...

... . . ....

am1 am2 . . . amn

Denoting the matrix above as A= (aij)1≤i≤m, 1≤j≤n, (aij), i = 1, . . . ,mcorrespond to the m rows of the matrix and and (aij), j = 1, . . . ,ncorrespond to the n columns of the matrix.aij is called as the ij-th entry (or component) of the matrix.


Matrix

A matrix is a rectangular array of numbers (or functions) enclosedin brackets with the numbers (or functions) called as entries orelements of the matrix.Matrices are denoted by capital letters A, B, C , · · · .

For integers m,n≥ 1, an m×n matrix is an array given by

a11 a12 . . . a1na21 a22 . . . a2n...

... . . ....

am1 am2 . . . amn



Matrix

A matrix is a rectangular array of numbers (or functions) enclosedin brackets with the numbers (or functions) called as entries orelements of the matrix.Matrices are denoted by capital letters A, B, C , · · · .For integers m,n≥ 1, an m×n matrix is an array given by

a11 a12 . . . a1na21 a22 . . . a2n...

... . . ....

am1 am2 . . . amn



Matrix


a11 a12 . . . a1na21 a22 . . . a2n...

... . . ....

am1 am2 . . . amn

Denoting the matrix above as A= (aij)1≤i≤m, 1≤j≤n, (aij), i = 1, . . . ,mcorrespond to the m rows of the matrix and and (aij), j = 1, . . . ,ncorrespond to the n columns of the matrix.

aij is called as the ij-th entry (or component) of the matrix.


Matrix


a11 a12 . . . a1na21 a22 . . . a2n...

... . . ....

am1 am2 . . . amn



Row and Column vectors

A matrix consisting of a single row is called a row vector.

A row vector (x1, . . . ,xn) is a 1×n matrix.A matrix consisting of a single column is called a column vector.

A column vector

x1...xn

is an n×1 matrix.

Row and column vectors are denoted using lower case letters suchas a, b, and so on.A square matrix is a matrix with the same number of rows andcolumns.For n≥ 1, an n×n matrix is called a square matrix of order n.A matrix (aij) is called the zero matrix if all its entries are zeroesand is, denoted by O.The size of the zero matrix should be clear from the context.



A matrix consisting of a single row is called a row vector.A row vector (x1, . . . ,xn) is a 1×n matrix.

A matrix consisting of a single column is called a column vector.

A column vector

x1...xn

is an n×1 matrix.




A matrix consisting of a single row is called a row vector.A row vector (x1, . . . ,xn) is a 1×n matrix.A matrix consisting of a single column is called a column vector.

A column vector

x1...xn

is an n×1 matrix.





A column vector

x1...xn

is an n×1 matrix.

Row and column vectors are denoted using lower case letters suchas a, b, and so on.

A square matrix is a matrix with the same number of rows andcolumns.For n≥ 1, an n×n matrix is called a square matrix of order n.A matrix (aij) is called the zero matrix if all its entries are zeroesand is, denoted by O.The size of the zero matrix should be clear from the context.




A column vector

x1...xn

is an n×1 matrix.

Row and column vectors are denoted using lower case letters suchas a, b, and so on.A square matrix is a matrix with the same number of rows andcolumns.

For n≥ 1, an n×n matrix is called a square matrix of order n.A matrix (aij) is called the zero matrix if all its entries are zeroesand is, denoted by O.The size of the zero matrix should be clear from the context.




A column vector

x1...xn

is an n×1 matrix.

Row and column vectors are denoted using lower case letters suchas a, b, and so on.A square matrix is a matrix with the same number of rows andcolumns.For n≥ 1, an n×n matrix is called a square matrix of order n.

A matrix (aij) is called the zero matrix if all its entries are zeroesand is, denoted by O.The size of the zero matrix should be clear from the context.




A column vector

x1...xn

is an n×1 matrix.

Row and column vectors are denoted using lower case letters suchas a, b, and so on.A square matrix is a matrix with the same number of rows andcolumns.For n≥ 1, an n×n matrix is called a square matrix of order n.A matrix (aij) is called the zero matrix if all its entries are zeroesand is, denoted by O.

The size of the zero matrix should be clear from the context.




A column vector

x1...xn

is an n×1 matrix.



(Addition of Matrices)

If A= (aij)1≤i≤m,1≤j≤n and B = (bij)1≤i≤m,1≤j≤n are two m×nmatrices, their sum, written as A+B, is obtained by adding thecorresponding entries.

That is, A+B = (aij +bij)1≤i≤m,1≤j≤n.

For any matrix A, O+A=A+O=A.

(Scalar multiplication)

The product of an m×n matrix A= (aij)1≤i≤m,1≤j≤n and any scalarc, written as cA, is the m×n matrix obtained by multiplying eachentry of A by c.That is, c(aij)1≤i≤m,1≤j≤n := (caij)1≤i≤m,1≤j≤n.

Write (−1)A=−A. Then A+ (−A)=O. The matrix −A is calledthe additive inverse of A.(−k)A is written as −kA, A+ (−B) is written as A−B and is calledthe difference of A and B .



If A= (aij)1≤i≤m,1≤j≤n and B = (bij)1≤i≤m,1≤j≤n are two m×nmatrices, their sum, written as A+B, is obtained by adding thecorresponding entries. That is, A+B = (aij +bij)1≤i≤m,1≤j≤n.










The product of an m×n matrix A= (aij)1≤i≤m,1≤j≤n and any scalarc, written as cA, is the m×n matrix obtained by multiplying eachentry of A by c.

That is, c(aij)1≤i≤m,1≤j≤n := (caij)1≤i≤m,1≤j≤n.








Write (−1)A=−A.

Then A+ (−A)=O. The matrix −A is calledthe additive inverse of A.(−k)A is written as −kA, A+ (−B) is written as A−B and is calledthe difference of A and B .







Write (−1)A=−A. Then A+ (−A)=O.

The matrix −A is calledthe additive inverse of A.(−k)A is written as −kA, A+ (−B) is written as A−B and is calledthe difference of A and B .







Write (−1)A=−A. Then A+ (−A)=O. The matrix −A is calledthe additive inverse of A.

(−k)A is written as −kA, A+ (−B) is written as A−B and is calledthe difference of A and B .







Write (−1)A=−A. Then A+ (−A)=O. The matrix −A is calledthe additive inverse of A.(−k)A is written as −kA,

A+ (−B) is written as A−B and is calledthe difference of A and B .


Matrix Addition Laws

If A, B and C are matrices of the same size m×n, 0 is the matrixof size m×n with all entries as zeroes, then

1 A+B =B +A (Commutative Law)

2 (A+B)+C =A+ (B +C ) (Associative Law)3 A+0=A (Existence of Identity)4 A+−A= 0 (Existence of additive inverse)




1 A+B =B +A (Commutative Law)2 (A+B)+C =A+ (B +C ) (Associative Law)

3 A+0=A (Existence of Identity)4 A+−A= 0 (Existence of additive inverse)




1 A+B =B +A (Commutative Law)2 (A+B)+C =A+ (B +C ) (Associative Law)3 A+0=A (Existence of Identity)

4 A+−A= 0 (Existence of additive inverse)




1 A+B =B +A (Commutative Law)2 (A+B)+C =A+ (B +C ) (Associative Law)3 A+0=A (Existence of Identity)4 A+−A= 0 (Existence of additive inverse)


Scalar Multiplication Laws

If A and B are matrices of the same size m×n, c , k are scalars,then

1 c(A+B)= cA+cB

2 (c +k)A= cA+kA3 c(kA)= (ck)A= ckA4 1 ·A=A




1 c(A+B)= cA+cB2 (c +k)A= cA+kA

3 c(kA)= (ck)A= ckA4 1 ·A=A




1 c(A+B)= cA+cB2 (c +k)A= cA+kA3 c(kA)= (ck)A= ckA

4 1 ·A=A




1 c(A+B)= cA+cB2 (c +k)A= cA+kA3 c(kA)= (ck)A= ckA4 1 ·A=A


Multiplication of matrices

(Multiplication of Matrices)

Let A= (aij) be an m×n matrix and B = (bjk) be an n× s matrix.Define the product AB to be the m× s matrix (cik), where

cik =n∑

j=1aijbjk .

If A1, . . . ,Am are row vectors of A and B1, . . . ,Bs are column vectorsof B , then ik-th coordinate of the matrix AB is the product AiBk .




Let A= (aij) be an m×n matrix

and B = (bjk) be an n× s matrix.Define the product AB to be the m× s matrix (cik), where

cik =n∑

j=1aijbjk .





Let A= (aij) be an m×n matrix and B = (bjk) be an n× s matrix.

Define the product AB to be the m× s matrix (cik), where

cik =n∑

j=1aijbjk .






cik =

n∑

j=1aijbjk .






cik =n∑

j=1aijbjk .






cik =n∑

j=1aijbjk .

If A1, . . . ,Am

are row vectors of A and B1, . . . ,Bs are column vectorsof B , then ik-th coordinate of the matrix AB is the product AiBk .





cik =n∑

j=1aijbjk .

If A1, . . . ,Am are row vectors of A

and B1, . . . ,Bs are column vectorsof B , then ik-th coordinate of the matrix AB is the product AiBk .





cik =n∑

j=1aijbjk .

If A1, . . . ,Am are row vectors of A and B1, . . . ,Bs

are column vectorsof B , then ik-th coordinate of the matrix AB is the product AiBk .





cik =n∑

j=1aijbjk .

If A1, . . . ,Am are row vectors of A and B1, . . . ,Bs are column vectorsof B ,

then ik-th coordinate of the matrix AB is the product AiBk .





cik =n∑

j=1aijbjk .



a11 . . . a1k . . . a1p

.... . .

......

...

ai 1 . . . ai k . . . ai p

......

.... . .

...

an1 . . . ank . . . anp

A : n rows p columns

b11 . . . b1 j . . . b1q

.... . .

......

...

bk1 . . . bk j . . . bkq

......

.... . .

...

bp1 . . . bp j . . . bpq

B : p rows q columns

c11 . . . c1 j . . . c1q

.... . .

......

...

ci 1 . . . ci j . . . ci q

......

.... . .

...

cn1 . . . cnk . . . cnq

C = A×B : n rows q columns

a i1×b 1 j

a ik×b k j

a i p×b p j

+ . . .+

+ . . .+

Illustrations

Example

Let A=(2 1 −10 3 1

)

2×3

and B =1 6 0 22 −1 1 −22 0 −1 1

3×4

. Then

AB =(2 11 2 18 −3 2 −5

)

2×4.

ExampleConsider a system of two linear equations in three unknowns

3x −2y +3z = 1; −x +7y −4z =−5.

Let A=(3 −2 3−1 7 −4

); B =

(1−5

); and X =

xyz

. Then the system

of two equations can be written in the form AX =B .


Illustrations

Example

Let A=(2 1 −10 3 1

)

2×3and B =

1 6 0 22 −1 1 −22 0 −1 1

3×4

.

Then

AB =(2 11 2 18 −3 2 −5

)

2×4.


3x −2y +3z = 1; −x +7y −4z =−5.

Let A=(3 −2 3−1 7 −4

); B =

(1−5

); and X =

xyz

. Then the system



Illustrations

Example

Let A=(2 1 −10 3 1

)

2×3and B =

1 6 0 22 −1 1 −22 0 −1 1

3×4

. Then

AB =

(2 11 2 18 −3 2 −5

)

2×4.


3x −2y +3z = 1; −x +7y −4z =−5.

Let A=(3 −2 3−1 7 −4

); B =

(1−5

); and X =

xyz

. Then the system



Illustrations

Example

Let A=(2 1 −10 3 1

)

2×3and B =

1 6 0 22 −1 1 −22 0 −1 1

3×4

. Then

AB =(2

11 2 18 −3 2 −5

)

2×4.


3x −2y +3z = 1; −x +7y −4z =−5.

Let A=(3 −2 3−1 7 −4

); B =

(1−5

); and X =

xyz

. Then the system



Illustrations

Example

Let A=(2 1 −10 3 1

)

2×3and B =

1 6 0 22 −1 1 −22 0 −1 1

3×4

. Then

AB =(2 11

2 18 −3 2 −5

)

2×4.


3x −2y +3z = 1; −x +7y −4z =−5.

Let A=(3 −2 3−1 7 −4

); B =

(1−5

); and X =

xyz

. Then the system



Illustrations

Example

Let A=(2 1 −10 3 1

)

2×3and B =

1 6 0 22 −1 1 −22 0 −1 1

3×4

. Then

AB =(2 11 2

18 −3 2 −5

)

2×4.


3x −2y +3z = 1; −x +7y −4z =−5.

Let A=(3 −2 3−1 7 −4

); B =

(1−5

); and X =

xyz

. Then the system



Illustrations

Example

Let A=(2 1 −10 3 1

)

2×3and B =

1 6 0 22 −1 1 −22 0 −1 1

3×4

. Then

AB =(2 11 2 1

8 −3 2 −5)

2×4.


3x −2y +3z = 1; −x +7y −4z =−5.

Let A=(3 −2 3−1 7 −4

); B =

(1−5

); and X =

xyz

. Then the system



Illustrations

Example

Let A=(2 1 −10 3 1

)

2×3and B =

1 6 0 22 −1 1 −22 0 −1 1

3×4

. Then

AB =(2 11 2 18 −3 2 −5

)

2×4.


3x −2y +3z = 1; −x +7y −4z =−5.

Let A=(3 −2 3−1 7 −4

); B =

(1−5

); and X =

xyz

. Then the system



Illustrations

Example

Let A=(2 1 −10 3 1

)

2×3and B =

1 6 0 22 −1 1 −22 0 −1 1

3×4

. Then

AB =(2 11 2 18 −3 2 −5

)

2×4.


3x −2y +3z = 1; −x +7y −4z =−5.

Let A=(3 −2 3−1 7 −4

);

B =(1−5

); and X =

xyz

. Then the system



Illustrations

Example

Let A=(2 1 −10 3 1

)

2×3and B =

1 6 0 22 −1 1 −22 0 −1 1

3×4

. Then

AB =(2 11 2 18 −3 2 −5

)

2×4.


3x −2y +3z = 1; −x +7y −4z =−5.

Let A=(3 −2 3−1 7 −4

); B =

(1−5

)

; and X =xyz

. Then the system



Illustrations

Example

Let A=(2 1 −10 3 1

)

2×3and B =

1 6 0 22 −1 1 −22 0 −1 1

3×4

. Then

AB =(2 11 2 18 −3 2 −5

)

2×4.


3x −2y +3z = 1; −x +7y −4z =−5.

Let A=(3 −2 3−1 7 −4

); B =

(1−5

); and X =

xyz

.

Then the system



Illustrations

Example

Let A=(2 1 −10 3 1

)

2×3and B =

1 6 0 22 −1 1 −22 0 −1 1

3×4

. Then

AB =(2 11 2 18 −3 2 −5

)

2×4.


3x −2y +3z = 1; −x +7y −4z =−5.

Let A=(3 −2 3−1 7 −4

); B =

(1−5

); and X =

xyz

. Then the system

of two equations can be written in the form

AX =B .


Illustrations

Example

Let A=(2 1 −10 3 1

)

2×3and B =

1 6 0 22 −1 1 −22 0 −1 1

3×4

. Then

AB =(2 11 2 18 −3 2 −5

)

2×4.


3x −2y +3z = 1; −x +7y −4z =−5.

Let A=(3 −2 3−1 7 −4

); B =

(1−5

); and X =

xyz

. Then the system



Matrix Multiplication - 3 ways

Example

Let A=(2 −2 41 3 5

)

and B =2 −14 −26 3

.

Then, AB =(20 1444 8

).

(i) Each entry of AB is the product of a row of A with a column ofB .ij th entry of AB = the product of i th row of A & j th column of B .

(ii) Each column of AB is the product of a matrix and a column.Column j of AB is A times column j of B .(iii) Each row of AB is the product of a row and a matrix.Row i of AB is row i of A times B .



Example

Let A=(2 −2 41 3 5

)and B =

2 −14 −26 3

.

Then, AB =(20 1444 8

).





Example

Let A=(2 −2 41 3 5

)and B =

2 −14 −26 3

.

Then, AB =(20 1444 8

).

(i) Each entry of AB is the product of a row of A with a column ofB .

ij th entry of AB = the product of i th row of A & j th column of B .(ii) Each column of AB is the product of a matrix and a column.Column j of AB is A times column j of B .(iii) Each row of AB is the product of a row and a matrix.Row i of AB is row i of A times B .



Example

Let A=(2 −2 41 3 5

)and B =

2 −14 −26 3

.

Then, AB =(20 1444 8

).





Example

Let A=(2 −2 41 3 5

)and B =

2 −14 −26 3

.

Then, AB =(20 1444 8

).


(ii) Each column of AB is the product of a matrix and a column.

Column j of AB is A times column j of B .(iii) Each row of AB is the product of a row and a matrix.Row i of AB is row i of A times B .



Example

Let A=(2 −2 41 3 5

)and B =

2 −14 −26 3

.

Then, AB =(20 1444 8

).




Properties of Matrix Multiplication

Let A,B ,C be matrices;

B ,C of same size, such that AB isdefined. ThenA(B +C )=AB +AC . (Distributive Law)(kA)B = k(AB)=A(kB), where k is a scalar.Let A,B ,C be matrices such that AB and BC is defined.Then, A(BC )= (AB)C . (Associative Law)

AB 6=BA (Commutative law doesn’t hold true in general)AB = 0 doesn’t necessarily imply that A= 0 or B = 0 or BA= 0.AC =AD doesn’t necessarily imply that C =D.



Let A,B ,C be matrices; B ,C of same size, such that AB isdefined.

ThenA(B +C )=AB +AC . (Distributive Law)(kA)B = k(AB)=A(kB), where k is a scalar.Let A,B ,C be matrices such that AB and BC is defined.Then, A(BC )= (AB)C . (Associative Law)




Let A,B ,C be matrices; B ,C of same size, such that AB isdefined. ThenA(B +C )=AB +AC . (Distributive Law)

(kA)B = k(AB)=A(kB), where k is a scalar.Let A,B ,C be matrices such that AB and BC is defined.Then, A(BC )= (AB)C . (Associative Law)




Let A,B ,C be matrices; B ,C of same size, such that AB isdefined. ThenA(B +C )=AB +AC . (Distributive Law)(kA)B = k(AB)=A(kB), where k is a scalar.

Let A,B ,C be matrices such that AB and BC is defined.Then, A(BC )= (AB)C . (Associative Law)




Let A,B ,C be matrices; B ,C of same size, such that AB isdefined. ThenA(B +C )=AB +AC . (Distributive Law)(kA)B = k(AB)=A(kB), where k is a scalar.Let A,B ,C be matrices such that AB and BC is defined.

Then, A(BC )= (AB)C . (Associative Law)




Let A,B ,C be matrices; B ,C of same size, such that AB isdefined. ThenA(B +C )=AB +AC . (Distributive Law)(kA)B = k(AB)=A(kB), where k is a scalar.Let A,B ,C be matrices such that AB and BC is defined.Then, A(BC )= (AB)C . (Associative Law)

AB 6=BA (Commutative law doesn’t hold true in general)

AB = 0 doesn’t necessarily imply that A= 0 or B = 0 or BA= 0.AC =AD doesn’t necessarily imply that C =D.




AB 6=BA (Commutative law doesn’t hold true in general)AB = 0 doesn’t necessarily imply that A= 0 or B = 0 or BA= 0.

AC =AD doesn’t necessarily imply that C =D.


Transpose of a matrix

(Transpose of a matrix)

Let A= (aij)1≤i≤m,1≤j≤n be a m×n matrix. The n×m matrix(bji )1≤j≤n,1≤i≤m, where bji = aij is called the transpose of A and isdenoted by AT .

Taking transpose of a matrix amounts to changing rows intocolumns and vice versa.

Example

Let A=(1 2 −2−3 0 1

). Then AT =

1 −32 0−2 1

.




Let A= (aij)1≤i≤m,1≤j≤n be a m×n matrix. The n×m matrix(bji )1≤j≤n,1≤i≤m,

where bji = aij is called the transpose of A and isdenoted by AT .


Example

Let A=(1 2 −2−3 0 1

). Then AT =

1 −32 0−2 1

.






Example

Let A=(1 2 −2−3 0 1

). Then AT =

1 −32 0−2 1

.






Example

Let A=(1 2 −2−3 0 1

).

Then AT =1 −32 0−2 1

.






Example

Let A=(1 2 −2−3 0 1

). Then AT =

1 −32 0−2 1

.


Properties of transpose

1 (A+B)T =AT +BT (A and B are matrices of the same size).

2 (AT )T =A3 (kA)T = kAT (k is a scalar).4 (AB)T =BTAT (A and B are compatible for multiplication).

If A= AT , then A is called a symmetric matrix.Note: A symmetric matrix is always a square matrix.



1 (A+B)T =AT +BT (A and B are matrices of the same size).2 (AT )T =A

3 (kA)T = kAT (k is a scalar).4 (AB)T =BTAT (A and B are compatible for multiplication).




1 (A+B)T =AT +BT (A and B are matrices of the same size).2 (AT )T =A3 (kA)T = kAT (k is a scalar).

4 (AB)T =BTAT (A and B are compatible for multiplication).




1 (A+B)T =AT +BT (A and B are matrices of the same size).2 (AT )T =A3 (kA)T = kAT (k is a scalar).4 (AB)T =BTAT (A and B are compatible for multiplication).





If A= AT , then A is called a

symmetric matrix.Note: A symmetric matrix is always a square matrix.




If A= AT , then A is called a symmetric matrix.

Note: A symmetric matrix is always a square matrix.


Special Matrices

(Upper triangular Matrices)

Upper triangular matrices are square matrices that can havenon-zero entries only on and above the main diagonal,

whereas anyentry below the main diagonal must be zero.

Example

A=(1 30 2

), B =

1 4 20 3 20 0 6

, C =

4 2 2 00 3 −5 10 0 0 −60 0 0 5

are examples of

upper triangular matrices.


Special Matrices


Upper triangular matrices are square matrices that can havenon-zero entries only on and above the main diagonal, whereas anyentry below the main diagonal must be zero.

Example

A=(1 30 2

), B =

1 4 20 3 20 0 6

, C =

4 2 2 00 3 −5 10 0 0 −60 0 0 5

are examples of



Special Matrices



Example

A=(1 30 2

),

B =1 4 20 3 20 0 6

, C =

4 2 2 00 3 −5 10 0 0 −60 0 0 5

are examples of



Special Matrices



Example

A=(1 30 2

), B =

1 4 20 3 20 0 6

,

C =

4 2 2 00 3 −5 10 0 0 −60 0 0 5

are examples of



Special Matrices



Example

A=(1 30 2

), B =

1 4 20 3 20 0 6

, C =

4 2 2 00 3 −5 10 0 0 −60 0 0 5

are examples of



Special Matrices (Contd..)

(Lower Triangular Matrices)

These are square matrices that can have non-zero entries only onand below the main diagonal.

Example

A=(5 02 3

), B =

2 0 08 −1 07 6 8

, C =

3 0 0 09 −3 0 01 0 2 01 9 3 6

are examples of

lower triangular matrices.





Example

A=(5 02 3

),

B =2 0 08 −1 07 6 8

, C =

3 0 0 09 −3 0 01 0 2 01 9 3 6

are examples of






Example

A=(5 02 3

), B =

2 0 08 −1 07 6 8

,

C =

3 0 0 09 −3 0 01 0 2 01 9 3 6

are examples of






Example

A=(5 02 3

), B =

2 0 08 −1 07 6 8

, C =

3 0 0 09 −3 0 01 0 2 01 9 3 6

are examples of




(Diagonal Matrices)

These are square matrices than can have non-zero entries only onthe main diagonal.

Any entry below or above the main diagonalmust be zero.

Example

A=2 0 00 −3 00 0 2

is an example of a diagonal matrix.



(Diagonal Matrices)

These are square matrices than can have non-zero entries only onthe main diagonal. Any entry below or above the main diagonalmust be zero.

Example

A=2 0 00 −3 00 0 2

is an example of a diagonal matrix.



(Scalar Matrices)

Scalar matrices are diagonal matrices with diagonal entries, allequal, say c.

Examplec 0 00 c 00 0 c

is an example of a scalar matrix.

It is called as scalar matrix because multiplication of any squarematrix A by a scalar matrix S of the same size has the same effectas multiplication by a scalar.That is, AS = SA= cA .



(Scalar Matrices)




It is called as scalar matrix because multiplication of any squarematrix A by a scalar matrix S of the same size has the same effectas multiplication by a scalar.

That is, AS = SA= cA .



(Scalar Matrices)




It is called as scalar matrix because multiplication of any squarematrix A by a scalar matrix S of the same size has the same effectas multiplication by a scalar.That is, AS = SA= cA .



(Identity Matrix (Unit Matrix))

Identity matrices are scalar matrices whose entries in the maindiagonal are all equal to 1.

Example

I2 =(1 00 1

)is the identity matrix of size 2×2.

I3 =1 0 00 1 00 0 1

is the identity matrix of size 3×3.


Inverse of a Matrix

(Inverse of a matrix)

Let A be a square matrix of size n×n. The inverse of A (if itexists), is a matrix B (of the same size) such that AB =BA= In.

Remark : Inverse of a square matrix need not exist always.

TheoremIf the inverse of a matrix exists, it is unique.

Proof : If possible, let B and C be inverses of A.Then, AB =BA= I and AC =CA= I . Hence,B =BI =B(AC )= (BA)C = IC =C . ä• The inverse of A (if exists) is denoted by A−1.

Exercise : The transpose of an inverse is the inverse of thetranspose, i.e. (A−1)T = (AT )−1.EXERCISE : TUTORIAL SHEET 1.


Inverse of a Matrix


Let A be a square matrix of size n×n.

The inverse of A (if itexists), is a matrix B (of the same size) such that AB =BA= In.






Inverse of a Matrix








Inverse of a Matrix





Proof : If possible, let B and C be inverses of A.

Then, AB =BA= I and AC =CA= I . Hence,B =BI =B(AC )= (BA)C = IC =C . ä• The inverse of A (if exists) is denoted by A−1.



Inverse of a Matrix





Proof : If possible, let B and C be inverses of A.Then, AB =BA= I

and AC =CA= I . Hence,B =BI =B(AC )= (BA)C = IC =C . ä• The inverse of A (if exists) is denoted by A−1.



Inverse of a Matrix





Proof : If possible, let B and C be inverses of A.Then, AB =BA= I and AC =CA= I .

Hence,B =BI =B(AC )= (BA)C = IC =C . ä• The inverse of A (if exists) is denoted by A−1.



Inverse of a Matrix





Proof : If possible, let B and C be inverses of A.Then, AB =BA= I and AC =CA= I . Hence,B =BI

=B(AC )= (BA)C = IC =C . ä• The inverse of A (if exists) is denoted by A−1.



Inverse of a Matrix





Proof : If possible, let B and C be inverses of A.Then, AB =BA= I and AC =CA= I . Hence,B =BI =B(AC )

= (BA)C = IC =C . ä• The inverse of A (if exists) is denoted by A−1.



Inverse of a Matrix





Proof : If possible, let B and C be inverses of A.Then, AB =BA= I and AC =CA= I . Hence,B =BI =B(AC )= (BA)C

= IC =C . ä• The inverse of A (if exists) is denoted by A−1.



Inverse of a Matrix





Proof : If possible, let B and C be inverses of A.Then, AB =BA= I and AC =CA= I . Hence,B =BI =B(AC )= (BA)C = IC

=C . ä• The inverse of A (if exists) is denoted by A−1.



Inverse of a Matrix





Proof : If possible, let B and C be inverses of A.Then, AB =BA= I and AC =CA= I . Hence,B =BI =B(AC )= (BA)C = IC =C . ä

• The inverse of A (if exists) is denoted by A−1.



Inverse of a Matrix





Proof : If possible, let B and C be inverses of A.Then, AB =BA= I and AC =CA= I . Hence,B =BI =B(AC )= (BA)C = IC =C . ä• The inverse of A (if exists)

is denoted by A−1.



Inverse of a Matrix








Inverse of a Matrix






Exercise : The transpose of an inverse is the inverse of thetranspose,

i.e. (A−1)T = (AT )−1.EXERCISE : TUTORIAL SHEET 1.


Inverse of a Matrix






Exercise : The transpose of an inverse is the inverse of thetranspose, i.e. (A−1)T = (AT )−1.

EXERCISE : TUTORIAL SHEET 1.


Inverse of a Matrix








System of linear equations

A system of m linear equations in n variables has the form

a11x1+a12x2+·· ·+a1nxn = b1

a21x1+a22x2+·· ·+a2nxn = b2...

......

......

... = ... (1)am1x1+am2x2+·· ·+amnxn = bm

Here, aij ’s, 1≤ i ≤m, 1≤ j ≤ n are the coefficients;x1,x2, · · · ,xn are the unknown variables,and [b1, b2, , · · ·bn]

t is the right hand side vector.


Matrix form

The linear system (1) can be written as AX =B , where

A=

a11 a12 . . . a1na21 a22 . . . a2n...

... . . ....

am1 am2 . . . amn

is the coefficient matrix;

B =

b1...bm

is the right hand side vector,

and X =

x1...xn

is a column vector of unknown variables.


Some definitions

(Homogeneous & Non-homogeneous Systems)

The linear system (1) is said to be homogeneous ifbi = 0 ∀i = 1,2, · · · ,m.

If at least one bi 6= 0 for 1≤ i ≤m, then the system is called anon-homogeneous system.

(Solution)

A solution is a set of numbers x1,x2, · · · ,xn that satisfies all the mequations of the system.

(Solution Vector)

A solution vector is a vector x whose components constitute asolution of the system.


Some definitions

(Homogeneous & Non-homogeneous Systems)

The linear system (1) is said to be homogeneous ifbi = 0 ∀i = 1,2, · · · ,m.If at least one bi 6= 0 for 1≤ i ≤m, then the system is called anon-homogeneous system.

(Solution)

A solution is a set of numbers x1,x2, · · · ,xn that satisfies all the mequations of the system.

(Solution Vector)

A solution vector is a vector x whose components constitute asolution of the system.



A homogeneous system of m linear equations in n variables :

a11x1+a12x2+·· ·+a1nxn = 0a21x1+a22x2+·· ·+a2nxn = 0

......

......

...... = 0

am1x1+am2x2+·· ·+amnxn = 0

has at least the trivial solution x1 = 0,x2 = 0, · · ·xn = 0.Any other solution, if it exists, is called a non-zero or non-trivialsolution.





......

......

...... = 0


has at least the trivial solution x1 = 0,x2 = 0, · · ·xn = 0.

Any other solution, if it exists, is called a non-zero or non-trivialsolution.





......

......

...... = 0


has at least the trivial solution x1 = 0,x2 = 0, · · ·xn = 0.Any other solution, if it exists, is called a non-zero or non-trivialsolution.


Consistent & Inconsistent Systems

The linear system (1) may haveNO SOLUTION,

PRECISELY ONE SOLUTIONor INFINITELY MANY SOLUTIONS.

A linear system isCONSISTENT, if it has at least one solution;INCONSISTENT if it has no solution.

How to find the solution(s) of a CONSISTENT system?



The linear system (1) may haveNO SOLUTION,PRECISELY ONE SOLUTION

or INFINITELY MANY SOLUTIONS.





The linear system (1) may haveNO SOLUTION,PRECISELY ONE SOLUTIONor INFINITELY MANY SOLUTIONS.






A linear system isCONSISTENT, if it has at least one solution;

INCONSISTENT if it has no solution.How to find the solution(s) of a CONSISTENT system?


Gauss Elimination

Gauss Elimination is a standard method for solving linear systemsof the form AX =B , where

A=

a11 a12 . . . a1na21 a22 . . . a2n...

... . . ....

am1 am2 . . . amn

is the coefficient matrix;

B =

b1...bm

is the right hand side vector,

and X =

x1...xn

is a column vector of unknown variables.


Basic Strategy

Replace the system of linear equations with an equivalent system

(one with the same solution set) which is easier to solve.


Basic Strategy

Replace the system of linear equations with an equivalent system(one with the same solution set) which is easier to solve.


How to obtain equivalent systems?

We state the elementary row operations for matrices andequations which yield row-equivalent linear systems.

Elementary Row Operations for Matrices Elementary Row Operations for Equations1 Interchange two rows Interchange two equations

(Ri ↔Rj ) interchange ith and jth rows2 Addition of a constant multiple of Addition of a constant multiple of

one row to another row one equation to another equationRi →Ri +cRj (j 6= i)

3 Multiplication of a row by Multiplication of an equation bya non-zero constant c ( 6= 0) a non-zero constant c (6= 0)Ri → cRi

Definition (Row-equivalent systems)

A linear system S1 is row-equivalent to a linear system S2 (S1 ∼ S2),if S2 can be obtained from S1 by finitely many row operations.





(Ri ↔Rj ) interchange ith and jth rows

2 Addition of a constant multiple of Addition of a constant multiple ofone row to another row one equation to another equationRi →Ri +cRj (j 6= i)









one row to another row one equation to another equation

Ri →Ri +cRj (j 6= i)3 Multiplication of a row by Multiplication of an equation by

a non-zero constant c ( 6= 0) a non-zero constant c (6= 0)Ri → cRi









3 Multiplication of a row by Multiplication of an equation bya non-zero constant c ( 6= 0) a non-zero constant c (6= 0)

Ri → cRi




Augmented matrix

We illustrate the Gauss-elimination method for some examplesbefore formalizing the method.

Since the linear system is completely determined by its augmentedmatrix, defined below, the Gauss elimination process can be downby merely considering the matrices.The matrix A obtained by augmenting A by column b, written as

A=

a11 a12 . . . a1n b1a21 a22 . . . a2n b2...

... . . ....

...am1 am2 . . . amn bm

, or

A=

a11 a12 . . . a1n... b1

a21 a22 . . . a2n... b2

...... . . .

......

...

am1 am2 . . . amn... bm

is called the augmented

matrix. We will work out the examples in the next lecture.


Augmented matrix

We illustrate the Gauss-elimination method for some examplesbefore formalizing the method.Since the linear system is completely determined by its augmentedmatrix, defined below, the Gauss elimination process can be downby merely considering the matrices.

The matrix A obtained by augmenting A by column b, written as

A=

a11 a12 . . . a1n b1a21 a22 . . . a2n b2...

... . . ....

...am1 am2 . . . amn bm

, or

A=

a11 a12 . . . a1n... b1

a21 a22 . . . a2n... b2

...... . . .

......

...

am1 am2 . . . amn... bm




Augmented matrix

We illustrate the Gauss-elimination method for some examplesbefore formalizing the method.Since the linear system is completely determined by its augmentedmatrix, defined below, the Gauss elimination process can be downby merely considering the matrices.The matrix A obtained by augmenting A by column b, written as

A=

a11 a12 . . . a1n b1a21 a22 . . . a2n b2...

... . . ....

...am1 am2 . . . amn bm

, or

A=

a11 a12 . . . a1n... b1

a21 a22 . . . a2n... b2

...... . . .

......

...

am1 am2 . . . amn... bm




Augmented matrix


A=

a11 a12 . . . a1n b1a21 a22 . . . a2n b2...

... . . ....

...am1 am2 . . . amn bm

,

or

A=

a11 a12 . . . a1n... b1

a21 a22 . . . a2n... b2

...... . . .

......

...

am1 am2 . . . amn... bm




Augmented matrix


A=

a11 a12 . . . a1n b1a21 a22 . . . a2n b2...

... . . ....

...am1 am2 . . . amn bm

, or

A=

a11 a12 . . . a1n... b1

a21 a22 . . . a2n... b2

...... . . .

......

...

am1 am2 . . . amn... bm




Augmented matrix


A=

a11 a12 . . . a1n b1a21 a22 . . . a2n b2...

... . . ....

...am1 am2 . . . amn bm

, or

A=

a11 a12 . . . a1n... b1

a21 a22 . . . a2n... b2

...... . . .

......

...

am1 am2 . . . amn... bm


matrix. We will work out the examples in the next lecture.Neela Nataraj Lecture 1

lecture1_d3

Documents