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Mathematics for Engineers Part I (ISE) Version 1.1/2007-01-29

9-1

9 Linear Algebra: Vectors, Matrices, Determinants, Linear

Systems of Equations.

9.1 Vector Algebra

Certain quantities in mathematics and physics and its engineering applications can-not be characterized alone by a real number; such quantities are represented by ar-rows. For example, a force is represented by an arrow; the direction of the arrow de-

scribes the direction in which the force is applied, andthe length or magnitude of the arrow indicates thestrength. The velocity is also represented by an arrowwhich points in the direction of the motion (Fig. 9.1.1),and whose length indicates the speed. We will call sucharrows vectors.

Not e 9.1.1:In mathematics and physics we use two kinds of quantities, scalarsand vectors. Anarrow (vecto r)is a quantity that is determined by both its magnitude and its direc-tion. Velocity, force and so on, which are represented by arrows, are vector q uant i-ties. A scalaris a quantity that is determined by its magnitude, its number of unitsmeasured on a suitable scale. Speed, weight, time, temperature, distance and so onare scalar quanti t ies.

An arrow (A,A) has a tail A, called its in i t ialpo in t, and a head A, called its terminal point.By a parallel shifting of the arrow (A,A) we get anew arrow (B,B) (Fig. 9.1.3) which has the samelength and the same direction. Therefore by par-allel shifting we get an infinite set of arrows

M = {(A,A'),(B,B'),(C,C'),...}.

Fig. 9.1.1

Fig 9.1.2 Arrow (A,A)

Velocity

vector

A

A

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9-2

Definiti on 9.1.1:

Arrows which have the same length and the same direction

are called parallel-equal.

Definiti on 9.1.2:The set M = {(A,A'),(B,B'),(C,C'),...} of all parallel-equalarrows is called an arrowclass a . An individual arrow of the class a is called a representativeof the arrow

class.

Since an arrow class consists of infinitely many arrows, it is impossible to draw sucha class. Therefore we designate a representative with the symbol of its arrow class

(Fig. 9.1.4).

Definiti on 9.1.3 (Vecto r sp ace):A non-empty set V of vectors is called a real vector space(or real linear space) if inV there are defined two algebraic operations (called vector addition and scalar multi-plication) as follows:

I. Vector addition associates with every pair of vectors u and v ofV an unique vec-torw ofV, called the sum ofu and v and denoted by w = u + v, such that the follow-ing axioms are satisfied:

V1 Commutativity:For any two vectors u and v ofV applies: u + v = v + u.

V2 Associativity:

For any three vectors u, v, wV applies: (u + v) + w = u + (v + w).

V3 Zero vector:There is a unique vector in V, called the zero vector and denoted by 0, such that forevery v in V applies:

v + 0 = 0 + v = v.

Fig. 9.1.3

Fig. 9.1.4: Representative of the arrow class a

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9-3

V4Inverse element:

For every v V there is an unique vector in V denoted by -v such that

v + (-v) = (-v) + v = 0.

II. Scalar multiplication: The real numbers are called scalars. Scalar multiplication

associates with every v in V and every scalar r Ran unique vector ofV, called theproduct of r and v and denoted by rv such that the following axioms are satisfied:

Distributivity:

For every scalar r, s R and vectors u and v in V applies:

V5 r(u + v) = ru + rv

V6 (r+ s)v = rv + sv

Associativity:

For all scalars r,s R and every v in V applies:

V7 r(sv) = (rs)v = s(rv)

Neutral element:For every v in V applies:

V8 1v = v

Remark 9.1.1:

A complex vector space is obtained if, instead of real numbers, we take complexnumbers as scalars.

Definiti on 9.1.4:

With Rn we denote the set of all ordered n-tuple (x1, x2,...,xn-1, xn),xiR , i = 1,...,n;

i.e.:

Rn = {(x1,x2,...,xn-1,xn):xiR, i = 1,...,n}.

Example 9.1.1:

Rn is a real vector space. Suppose a = (a1, a2,...,an) and b = (b1, b2,...,bn) are ele-

ments ofRn. Then ( )nn11 ba,...,ba ++=+ ba Rn and ( )n1 ra,...,rar =a Rn foreach real number r. It is easy to show that this so defined vector addition and scalarmultiplication satisfy V1 V8.

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9-4

Fig. 9.1.5 R2 Fig. 9.1.6 R3

Example 9.1.2:

If an experiment involves reading seven strategically placed thermometers in a giventimer interval (e.g. each hour) then a result can be recorded as a 7-dimensional vec-

tor ( )721 T,...,T,T R7.

Definiti on 9.1.5:

A non-empty subset of vector space V that itself forms a vector space with respect tothe two algebraic operations defined for the vector space V is called a subspaceofthe vector space V.

Definiti on 9.1.6:

Given any set ofm vectors1

v ,2

v , ,m

v in a vector space V. A l inear comb ina-

t ionof these vectors is an expression of the form

c1 1v +c2 2v ++cm mv ,

where c1, c2,, cm are any scalars.

Definiti on 9.1.7:

The vectors 1v , 2v , , mv are called l inearlyindependentif

c1 1v +c2 2v ++cm mv = 0

implies that c1= 0, c2= 0, , cm= 0.

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9-5

Example 9.1.3:

The vectors (1,0,3) and (0,1,8) of the R3 are linearly independent. Since for

(1,0,3) + (0,1,8) = (0,0,0)

we get: = 0 and = 0 and 3 + 8 = 0. Therefore these vectors are linearly in-

dependent.

Definiti on 9.1.8:

The vectors 1v , 2v , , mv are called l inearly dependentif

c1 1v +c2 2v ++cm mv = 0

holds with scalars not all zero.

Example 9.1.4:

The vectors (1,1,3) and (2,2,6) of the R3 are linearly dependent, since

2(1,1,3) + (-1)(2,2,6) = (0,0,0).

Remark 9.1.2:

The empty set is defined to be linearly independent.

Definiti on 9.1.9:

Let Sbe a non-empty subset of a vector space V. SspansV if every vector in V canbe written as a linear combination of (finitely many) elements from S. Sis then calleda spanning setorgenerating setofV. We can define the spanofS to be the set ofall linear combinations of elements ofS. Then Sspans Vif and only ifV is the spanofS; in general, however, the span ofSwill only be a subspace ofV.

Example 9.1.5:

The set {(1,0,0), (0,1,0), (1,1,0)} spans the space of all vectors in R3 whose last

component is zero.

Definiti on 9.1.10:

A linearly independent set in V consisting of a maximum possible number of vectorsin V is called a basisforV.

Example 9.1.6:

(i) The set {(1,0,0), (0,1,0), (0,0,1)} forms a basis forR3 (called the standard basis forthe vector space R

3).

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9-6

(ii) The real vector space R3 has {(1,0,0), (0,1,0), (0,0,1)} as spanning set. This

spanning set is actually a basis. Another spanning set for the same space is givenby {(1,2,3), (0,1,2), (1,1/2,3), (1,1,1)}, but this set is not a basis, because it is line-arly dependent.

Definiti on 9.1.11:The dimensionof a vector space V is the number of elements in a basis forV and isdenoted by dim(V).

Example 9.1.6:

(i) dim(Rn) = n.

(ii) The set Pk(R) of all polynomial functionspk: R R, k> 0, of the shape

( ) 011k

1k

k

kk axa...xaxaxp ++++=

, aiR, 1 i k,

is a vector space (proof this).

Vector addition:

Suppose ( ) 011k

1k

k


and ( ) 011k

1k

k

kk bxb...xbxbxq ++++=

are elements ofPk(R), then

( )( ) ( ) ( ) ( ) ( )00111k

1k1kk

kkkk baxba...xbaxbaxqp ++++++++=+

is an element ofPk(R).

Scalar multiplication:

Suppose ( ) 011k

1k

k


and rR, then

( )( ) 011k

1k

k

kk axa...xaxaxp rrrrr ++++=

is again in Pk(R). With these algebraic operations Pk(R) is a vector space.

The set {1, x,, xk} is a basis ofPk(R). Therefore dim(Pk(R)) = k+ 1.

Definiti on 9.1.12:

Let U and V be two vector spaces and let T: UV be a mapping. Then T is called al inear transformation(linear operator) if the following conditions are satisfied :

(i) T (u + w) = T (u) + T (w) , for all u , wU(ii) T (ru) = rT(u) , ris a scalar and uU

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9-7

Example 9.1.7:

(i) Suppose C1(a;b) is the set of all real valued functions on the open Interval (a;b)which have at least first order continuous derivatives.

This set is a vector space and the differential operatordx

dis a linear transformation.

(ii) Suppose Iis the set of all functions having an antiderivative. This set again is avector space and the integral operator dx is a linear transformation.

In particular our arrow classes defined above are vectors and the set of all arrowclasses embedded in a certain R

n with the vector addition and scalar multiplicationgeometrically defined below is a linear vector space.

Geometr ical vector addi t ion:Place the initial point of bat the terminal point of a ;

then the sum a+b is the vector drawn from the initial point of a to the terminal point

ofb .

Basic properties of vector addition:

(i) a+b= b+a (commutativity)

(ii) ( a+b ) + c = a+ (b+c ) (associativity)

(iii) a+ 0 = 0 + a = a

(iv) a+ (-a ) = (- a ) + a = 0

Where adenotes the vector having the length | a | and the direction opposite to that

ofa .

Fig 9.1.7

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9-8

Fig 9.1.8

Geometrical scalar mu ltipl ication: Suppose c is a real number. If a 0, then ca

with c> 0 has the same direction ofa and with c< 0 the direction opposite to a . The

length ofca is ctimes the length ofa .

Basic prop erties of s calar m ult ip l icat ion:

Suppose c, dR, then applies:(i) c(a+b ) = ca+ cb

(ii) (c+ d) a= ca+ da

(iii) c(da ) = (cd)a

(iv) 1 a= a

Since it is not convenient and also mathematically not exact to work with these geo-metrical operations we refer to arrow classes as elements of the vector space Rn

and introduce the components of an arrow (vector).

If a given arrow (vector) a has initial point (tail) P:(x1,,xn) and terminal point (head)Q:(y1,,yn), the n numbers

iii xya = , i= 1,,n,

are called the com ponents of the vector a with respect to that coordinate system,

and we write simply [ ]n1 a,...,a=a .

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9-9

No te 9.1.2:

vector components = head coordinates - tail coordinates

Fig. 9.1.9

Definiti on 9.1.13:The lengthormagnitudeof the vectora, denoted by a , is the distance between its

initial point and terminal point. Thus if a given vectora has initial point P:(x1,,xn)and terminal point Q:(y1,,yn), then

(9.1.1) ( ) ( ) 2n2

1

2

nn

2

11 a...axy...xy ++=++=ar

Example 9.1.8:

The vectora with initial point P:(4, 0, 2) and terminal point P:(6, -1, 2) has the com-

ponents 461 =a = 2, 012 =a = -1, 223 =a = 0. Hence a = [2, -1, 0] and the

length of a is given by a = 50)1(2222 =++ .

Definiti on 9.1.14:

The posi t ion vectorrof a point A:(x1,,xn) is the vector which has initial point atthe origin O:(0,,0) and terminal point at A. Thus r = [x1,,xn].

point B:(2,4)

point A:(6,-1)

vectora = [-4,5]

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9-10

Fig. 9.1.10

Remark 9.1.3:

(i) We use the position vector as a standard representative of an arrow class.

(ii) Ifn > 3 then the n-dimensional vector [ ]n1 a,...,a=a cannot be pictured geometri-cally as an arrowor a point, but with the exception of the cross product in R

3 (seeSection 9.1.20) vector algebra will be the same whether the vector has 2, 3 or 100components.

Definiti on 9.1.15:

A uni t vectorornormalizedvectoris a vector of unit magnitude. The unit vector in

the direction ofa is denoted by a. Thus a=a

ar and 1=a .

The vectorsn1 e,...,e are the unit vectors in the positive directions of the axes of a

Cartesian coordinate system. Hence {

= ,...,010,...,

positionith

ier

.

Fig. 9.1.11

xy

z

a1 1e

a2 2e

a3 3e

point (x1,x2)

position vector [x1,x2]

yx

z

2e 1e

3e

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9-11

Propo sit ion 9.1.1:

Suppose a Rn and a = [a1

,...,an

], then applies:

a = a1 e1 + ... + an e n = =

n

i

ia1

ier

This representation is unique.

Example 9.1.9:

(i) The vector [7,8] can be written as

7 e1 + 8 e2 = 7[1,0] + 8[0,1] = [7,0] + [0,8] = [7 + 0,0 + 8] = [7,8]

(ii) The vector [4,3,9] can be written as

4 e1 + 3 e2 + 9 e3

= 4[1,0,0] + 3[0,1,0] + 9[0,0,1]

= [4,0,0] + [0,3,0] + [0,0,9]

= [4 + 0 + 0,0 + 3 + 0,0 + 0 + 9]

= [4,3,9].

Definiti on 9.1.16:

The sum a+b of two vectors a = [a1,,an] and b = [b1,,bn] is obtained by adding

the corresponding components. That is a+b = [ a1+b1, ,an+bn ].

Definiti on 9.1.17:

Scalar Multiplication: ra = [ra1,...,ran].

Example 9.1.10:

Let a= [ 4, 0, 1 ] and b= [ 2, 5, 3 ]. Then

(i) a + b = [ 6, 5, 4]

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9-12

(ii) a - b = a + (-b ) = [4, 0, 1] + [-2, -5, -3] = [2, -5, -2]

(iii) 2( a - b ) = [4, -10,-4] = 2 a 2b

Definiti on 9.1.18:

The scalar or dot product ba of two vectors aand b is the product of theirlengths times the cosine of their angle; i.e.:

(i) ba = a b cos() if a 0 and b 0

(ii) ba = 0 if a = 0 orb= 0

The angle , 0 , between aand b is measured when the vectors have thesame initial point.

In Fig. 9.1.12 there is shown a triangle with the sides

| a |, |b | and | ba |. By the law of cosines we get:

(9.1.2) ( )cosba2baba22 rrrrrr+=

2

If [ ]n1 a,...,a=a and [ ]n1 b,...,b=b then (9.1.2) becomes

(9.1.3) ( ) ( ) ( )cosbaban

i

ii

n

i

ii barr

21

22

1

2 += ==

.

Formula (9.1.3) simplifies to

(9.1.4) ( )cosba

n

iii ba

rr

==1

or, equivalently,

(9.1.5) ( )ba

barr

=cos .

Therefore we get

a

b

ba

Fig. 9.1.12

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9-13


If [ ]n1 a,...,a=a and [ ]n1 b,...,b=b are vectors of the Rn then the scalar product can be

computed by

(9.1.6) nn11

n

1i

ii ba...baba ++== =

barr

.

Fig 9.1.13

The sign of ( )cos determines whether is acute or obtuse. This sign is determined

by the sign of a b since the denominator in (9.1.5) is always positive. In particular(see also Fig. 9.1.13):

(9.1.7) Sign o f the scalar pro du ct:

(i) If a b > 0, then 0 < 90

(ii) If a b = 0, then = 90

(iii) If a b < 0, then 90 < 180

As a corollary of (9.1.7) we get for non-zero vectors

Cor ollary 9.1.1:

Suppose a , b Rn, a0 and b 0. Then applies:

a b = 0 if and only if a and b are perpendicular.

a

a

a

b b b

a b > 0 a b = 0 a b < 0

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9-14

Example 9.1.11:

If e i and ej are the standard unit vectors ofRn then applies:

(i) 0= ji ee ifij

(ii) 1= ji ee ifi=j

(9.1.7) General Properties o f the s calar prod uc t:

(i) a b =b a

(ii) a (b +c ) = a b+ a c

(iii) a a = |a |2

(iv) The dot product is a scalar quantity.

Definiti on 9.1.19:

If aand b are non-zero vectors, then a and b are orthogonaliffa b = 0.

Example 9.1.12:

Find the angle between the vectors a = [1,2,3] and b= [-5,1,1].

Solution:

By Definition 9.1.18we get: ( )ba

barr

=cos

Now a b = [1,2,3] [-5,1,1] = 1(-5) + 21 + 31 = 0 which shows that a and b areorthogonal and so the angle between them is 90.

Definiti on 9.1.20:

The vector product (cross product) ab of two vectors a= [a1,a2 a3] and

b= [b1,b2,b3] is a vectorc= ab as follows:

(i) If a and b have the same or opposite direction or if one of these vectors is the

zero vector, then c = ab= 0.

(ii) In any other case, c= ab has the length c = a b sin().

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9-15

This is the area of the parallelogram with a and b as adjacent sides. The angle ,

0 180, between a and b is measured when the vectors have their initial point

coinciding.

(iii) The direction of c = ab is perpendicular to both aand b and such that a , b ,

c , in this order, form a right-handed system; i.e. c shows in the direction of the

thumb if the fingers of the right hand curl from a to b .

The componentsof the cross product vector

c= [c1,c2,c3] = ab are given by

c1 = a2b3 a3b2,

c2= a3b1 a1b3 and

c3 = a1b2 a2b1.

Therefore:

(9.1.8) ab= [a2b3 a3b2,a3b1 a1b3, a1b2 a2b1].

The formula in (9.1.8) looks formidable to memorize, but there are simple routines

for finding the cross product. It is convenient to use the determinant notation (seeSection 1.4). We define

(9.1.9) 211222112221

1211aaaa

aa

aa= ,

whereby ija , 21 i , 21 j , are real numbers.

Furthermore we set

(9.1.20)3231

2221

13

3331

2321

12

3332

2322

11

333231

232221

131211

aa

aaa

aa

aaa

aa

aaa

aaa

aaa

aaa

+= .

wherebyija , 31 i , 31 j , again real numbers.

Now the cross product can be written as a determinant and expanding this determi-nant across the first row:

Fig 9.1.14 Vector product

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9-16

(9.1.21) ab= 321 eeerrr

21

21

31

31

32

32

321

321

321

bb

aa

bb

aa

bb

aa

bbb

aaa

eee

+= ,

where 31 i,ie , is the i-th unit basis vector ofR3.

Note:There is a misusage in (9.1.21) of the determinant notation, since the ele-ments of a determinant are real numbers and not vectors. But this notation is verynice to compute the cross product.

The following example shows a short form of (9.1.21).

Example 9.1.13:

Ifa = [2,1,-4] and b = [3,-2,5] then

ab= 321523

412

523

412

523

412eeerrr

+

.

Therefore:

ab= ( ) ( ) ( ) [ ]722334121085 321 =++ ,,eee .

(9.1.22) General prop erties of th e vector prod uct:

(i) b a = (a b )

(ii) a(b + c) = (a b ) + (a c)

(iii) (a+ b ) c= (a c) + (b c)

(iv) a(b c) (a b )c)

(v) a (b ) = (a b ),where is any scalar

(vi) a b = 0 ifa and b are parallel.

Definiti on 9.1.21:

The scalar quantity a (b c ) is called the scalar tr ip le produ ctand has the follow-ing properties:

(i) If a = [a1, a2, a3], b= [b1, b2, b3] and c = [c1, c2, c3] then

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9-17

(9.1.23) a (bc ) =321

321

321

ccc

bbb

aaa

(ii) a (bc ) = b (c a ) = c ( ab )

(iii) If any two vectors are equal or parallel, then a (bc ) = 0

Geometrically, the scalar triple product a (bc ) is the volume of the parallelepiped

with a , b , c as edge vectors.

Fig 9.1.15 Parallelepiped

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9-18

9.2 Matrices

A matr ixis a rectangular array of real or complex numbers (or functions) enclosedby square brackets. These numbers (or functions) are called entries or elements ofthe matrix. We denote matrices by capital letters A, B, C, or by writing general en-try in square brackets; i. e.

(9.2.1) A =[ija ], A =[ ija ]mxn or A =[ ija ] njm,1i1 .

By a (mn)-matrix (read m by n matrix), we mean a matrix with m rows and n col-umns. Thus a (mn)-matrix is of the form

(9.2.2) A = [ ija ] =

mn1mnm2m1

2n12n2221

1n11n1211

aa...aa.

.

.

.

.

.

.

.....

aa...aa

aa...aa

, 1 i m, 1 j n.

If a matrix A has m rows and n columns then A is said to be a rectangu lar matrixofordermn. Ifm = n, A is said to be a squ are matrixof ordern. If A is a square ma-trix of ordern, then the entries a11, a22,, ann is called the maindiagonalof A.

Example 9.2.1:

(i) The matrix

=

402

531A is of order 23.

(ii) B =

4392515

462370

is a square matrix of order 3.

A matrix

(9.2.3) A = [a1, a2,.,an]

with a single row is a (1n)-matrix and is called a ro wvector.

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9-19

Similarly, asingle column

(9.2.4) B =

m

2

1

b.

.

.b

b

is a (m1)-matrix and is called a co lumnvector.

Definiti on 9.2.1:(i) A square matrix in which each element of the main diagonal is an element andall other elements are zero is called a scalarmatr ix. Thus

0000

....

....

0..00

0..00

is a scalar matrix of ordern.

(ii) If = 1 the scalar matrix is called the identi ty matrixof ordern and is denoted byIn or simply by I.

Definiti on 9.2.2:An upp er (lower) tr iangular matrixis a matrix whose elements below (above) themain diagonal are all zero. Thus

mn

n222

n11211

a...00

...

...

a...a0

a...aa

is an upper triangular matrix.

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9-20

Definiti on 9.2.3A square matrix, all of whose elements are zero except those in the main diagonal,is called diagonal matr ix. Thus

0a,

a...000

....

....

0...0a0

0...00a

ii

nn

22

11

,

is a diagonal matrix.

Remark 9.2.1:Every scalar matrix is a diagonal matrix.

Definiti on 9.2.4:Two matrices A = [aij] and B = [bij] are equal, written A = B, if and only if they havesame order and the corresponding entries are equal, that is, aij = bij for each iandfor eachj.

Example 9.2.2:Let

A =

dc

baand B =

06

14

Then A = B if and only ifa = 4, b = 1, c= 6 and d= 0.

Definiti on 9.2.5:A matrix having all elements equals zero is called a zero matrix. If it has m rowsand n columns, we denote it by Omn or simply by O.

Definiti on 9.2.6:Two matrices A and B are said to be con formable for addi t ionif they have thesame number of rows and the same number of columns. Thus if A = [aij] and B = [bij]are (mn)-matrices then theirsumis the matrix

A + B = [aij+ bij]

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9-21

of order (mn). That is, we add the corresponding elements of A and B to obtain thesum A + B.

Definiti on 9.2.7:(i) Given a (mn)-matrix A = [aij], we define A = [-aij]. Then A is a (mn)-matrix andby definition

A + (-A) = (-A) + A = O

Thus -A is the inverse of A with respect to addition. That is, -A is the addit ive in-verseof A.

(ii) Also O + A = A + O = A that is, O is the neutral elementwith respect to addit ion.

Definiti on 9.2.8:We define subtract ionof two matrices A and B of same order as

A - B = A + (-B)

Example 9.2.3;

Let

A =

104

312and B =

010

211.

Then A B = A + (-B) =

104

312+

010

211

=

+++011004

231112

=

114

521

Definiti on 9.2.9:Scalar Multipl ication(Multiplication by a number): The product of any (mn)-matrix

A = [aij] and any scalarr, written rA, is the (mn)-matrix

rA = [raij]

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obtained by multiplying each entry in A by r.

Example 9.2.4:Let

A =

53

12.

Then 2A =

106

24.

Propo sit ion 9.2.1:Let A, B be (mn)-matrices and r, s are scalars then:

(i) (r+ s)A = rA + sA(ii) r(A +B) = rA + rB(iii) r(sA) = (rs)A(iv) 1A = A

Remark 9.2.2:The set of all (mn)-matrices together with the matrix addition and scalar multiplica-tion form a vector space Rmxn (orCmxn if the entries are complex numbers). The di-mension of this vector space is given by dim(R

mxn) = mn.

Definiti on 9.2.10:The productC = AB of a (mn)-matrix A = [aij] and a (pq)-matrix B = [bij] is definedif and only ifp = n, that is,

number of rows of 2nd factor B = Number of columns of 1st factor A,

and is then defined as the (mq)-matrix C = [cij] with entry

njin

n

1k

2ji21ji1kjikij ba........bababac =

+++== .

In other words: cij is the sum of the products of the corresponding elements of thei-th row of A andj-th column of B.

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Example 9.2.5:

Let A =

035

742

101

and B =

3240

16

.

Then AB=

035

742

101

3240

16

=

++++++++

++

)3(0)4(3)1(5)2(0)0(3)6(5)3(7)4(4)1(2)2(7)0(4)6(2

)3(1)4(0)1(1)2(1)0(0)6(1

=

1730

392

28

Hence A is (33)-matrix, B is a (32)-matrix and AB is (32)-matrix. Note that theproduct BA is not defined.

Not e 9.2.1:

Matrix multiplication is not commutative, AB BA in general.

Example 9.2.6:

Let A =

43

21and B =

20

11.

Then AB =

113

51and BA =

86

64.

Thus AB BA.

Not e 9.2.2:

We may have AB = O when neither A = O nor B = O.

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Example 9.2.7:

Let A =

041011

021

and B =

941000

000

.

Then AB =

000

000

000

.

But neither A = O nor B = O.

Not e 9.2.3:

The cancellation laws do not hold for matrices. That is, we may have AB = AC (orBA = CA) when B C.

Example 9.2.8:

Let A =

041011

021

, B =

222

111

321

and C =

111

111

321

.

Then AB =

723232

143

and AC =

723232

143

.

Thus AB = AC but B C .

Propo sit ion 9.2.2:If the matrices A, B and C are conformable for the indicated sums and productsthen:

(i) A(BC) = (AB)C (Associative Law)(ii) A(B + C) = AB + AC (Left Distributive Law)(iii) (A + B)C = AC + BC (Right Distributive Law)(iv) r(AB) = (rA)B = A(rB), where ris any scalar.

Definiti on 9.2.11:The t ransposeof a (mn)-matrix A = [aij], denoted by A

T, is a (nm)-matrix obtainedby interchanging rows and columns of A.

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Example 9.2.9:

Let A =

035742

101

and B =

3240

16

.

Then AT =

071340

521

and BT =

341

206.

Propo sit ion 9.2.3:If the matrices A and B are conformable for the sum A + B and the product AB, then:

(i) (A + B)T = AT + BT(ii) (AT)T = A(iii) (rA)T = rAT, where ris a scalar.(iv) (AB)T = BTAT

Definiti on 9.2.12:(i) A square matrix A for which Ak+1 = A, (kbeing a positive integer), is called peri-od ic . Ifk is the least positive integer for which Ak+1 = A, then A is said to be ofpe-r iod k.

(ii) A square matrix A for which A2

= A is called idempotent.

(iii) A square matrix A for which Ap = O, (p being a positive integer), is called ni lpo-tent. Ifp is the least positive integer for which Ap = O, then A is said to be ni lpotentof index p.

(iv) A square matrix A for which A2 = Iis called involutorymatrix (an involutory ma-trix is its own matrix inverse).

(v) A square matrix A for which AT = A is called symmetr icmatrix.

(vi) A square matrix A for which AT = -A is called skew-symmetr icmatrix.

(vii) If A is a matrix over C and its elements are replaced by their complex conju-

gates, then the resulting matrix is called con jugateof A denoted by A (to be read Aconjugate).

(viii) A square matrix A such that (A )T = A is called Hermitianmatrix (the namecomes from an old French mathematician by the name of Hermite).

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(ix) A square matrix A such that (A )T = -A is called skew Hermitianmatrix.

(x) The inverseof a (nn)-matrix A is denoted by A-1 and is a (nn)-matrix such that

AA-1 = A-1A = I, where Iis the (nn)-unit matrix. If A has an inverse, then A is callednon-singularmatrix. If A has no inverse, then A is called singularmatrix.

(xi) A square matrix A is called orthogonalif AT = A-1.

(xii) A square matrix A is called uni taryif(A )T = A-1 .

Not e 9.2.4:

(i) Only square matrices can have inverses.

(ii) A real matrix is Hermitian iff it is symmetric.

Remark 9.2.2:The inverse of a matrix, if it exists, is unique. Suppose A has two inverses, say Band C. Then AB = BA = Iand AC = CA = I, so that we obtain the uniqueness fromB = IB = (CA)B = C(AB) = CI= C.

Example 9.2.8:

(i) If A =

+

10

021

0211

j

jj

j

, then A =

+

10

021

0211

j

jj

j

and

( A )T =

+

10

021

0211

j

jj

j

. Therefore A is a Hermitian matrix.

(iii) Let A =

111

011

326

and B =

421

232321

232121

.

Now AB =

100

010

001

= I3 and BA =

100

010

001

= I3 .

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Hence B = A-1. Also A = B-1

(iv) A =

212

221

122

3

1is an orthogonal matrix and

B =

j

jj

00

02

1

2

1

02

1

2

1

is a unitary matrix.

Propo sit ion 9.2.4:Let A and B be non-singular matrices of the same ordern, then AB is non-singularand (AB)

-1= B

-1A

-1

Proof:If we show that (AB) (B-1A-1) = I= (B-1A-1) (AB) then we prove that AB is non-singular

and that its inverse is B

-1

A

-1

. Now with Proposition 9.2.2(AB) (B

-1

A

-1

) = A(B B

-1

)A

-1

=AIA-1 = A A-1 = Iand (B-1A-1)(AB) = B-1(A-1 A)B = B-1IB = B-1B = I. Thus the productAB of two non-singular matrices A and B is non-singular and (AB)-1 = B-1A-1.

(9.2.5) Elementary Row Oper ation s:The following operations on a matrix are called elementary row operations :

(i) Interchange of any two rows(ii) Multiplication of a row by any non-zero scalar(iii) Addition of any multiple of one row to another row.

Any (mn)-matrix B is called row equivalentto a (mn)-matrix A if B is obtainedfrom A by performing a finite sequence of elementary row operations on A. We write

BR

~ A

to denote B is row equivalent to A.

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Definiti on 9.2.13:A (mn)-matrix A is said to be an echelon m atr ixif it has the following properties:

(i) The first krows of A are non-zero; the remaining (m - k) rows are zero (k m).(ii) The number of zeros preceding the first non-zero entry in each non-zero row is

larger than the number of zeros that appear before the first non-zero element inany preceding row. In other words: Each non-zero row in the matrix starts withmore zeros then the previous row.

In an echelon matrix, the first non-zero entry of a row is called p ivo t. A column con-taining pivot is called a p ivo t co lumn. If a matrix in echelon form has the additionalproperty that each pivot is 1 and every other entry of the pivot column is zero, then itis said to be in reduced echelon form.

Not e 9.2.5:

There is at least one pivot in each row and in each column of an echelon matrix.

Example 9.2.9:

The matrices

A =

6000

1900

3250

and B =

1000

0100

0031

are in echelon form. The second matrix is in reduced echelon form.

Notat ion:

Thefollowing notations prove useful in numerical problems:

(i) Rij denotes interchange ofith andjth rows of a matrix(ii) rRi denotes the multiplication ofith row of a matrix by a non-zero scalarr(iii) Ri +rRj denotes addition ofrtimes the elements ofjth row in the corresponding

elements ofith row.

Theor em 9.2.1:Any matrix is row equivalent to a matrix in echelon form (reduced echelon form).

Propo sit ion 9.2.5:A square matrix A of ordern is non-singular if and only if it is row equivalent to theidentity matrix In.

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Propo sit ion 9.2.6:If a square matrix A is reduced to the identity matrix by a sequence of elementaryrow operations, the same sequence of operations performed on the identity matrixproduces the inverse matrix A-1of A.

Remark 9.2.3:Proposition 9.2.6explains a methodto evaluate the inv erseof a given matrix. Thisis illustrated in the following example.

Example 9.2.10:

Find the inverse of the matrix

A =

031

142

301

Solution:Performing single row operation at a time and exhibit the successive matrices rowequivalent to A in left hand column and successive matrices row equivalent to Iinthe right hand column.

A I3

031

142

301

100

010

001

031

540

301

100

012

001

R2 2R1

330

540

301

101

012

001

R3 R1

110

540

301

31031

012

001

(-1/3)R3

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110

010

301

31031

35131

001

R2 + 5 R3

100

010

301

34132

35131

001

R3 R2

100

010

001

34132

35131

431

R1 3R3

100

010

001

34132

35131

431

(1)R2

Hence A-1 =

34132

35131

431

.

Definiti on 9.2.14:

The maximum number of linearly independent row vectors of a matrix A=[aij] is calledthe rankof A and is denoted by rank(A).

Propo sit ion 9.2.7:The non-zero rows of a matrix in echelon form are linearly independent.

Remark 9.2.4:(i) The rank of a matrix A equals the maximum number of linearly independent col-

umn vectors of A. Hence A and its transpose AT have the same rank.

(ii) Row equivalent matrices have the same rank. The rank of a matrix is equal to thenumber of non-zero rows in its echelon form.

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Example 9.2.11:

Find the rank of

A =

351

653

395

.

Solution:

AR

~

395

653

351

R13

R

~

395

653

351

(1)R1

R

~

1216015200

351

R2 + 3R1 and R3 5R1

R

~

121604310

351

(1/20)R2

R

~

000

4310

351

R3 + 16R2

This is an echelon form of matrix A and the number of its non-zero rows is 2.Hence rank(A) = 2.

Note: A general algorithm to reduce a given Matrix to an echelon form is the Gaussel iminat ion m ethodwhich will be introduced in the next section.

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In Example 9.2.11 the rank of a matrix has been found by reducing it to an echelonform. To reduce a matrix A to an echelon form many row operations are needed andin the process fractions creep in which make the computations awkward and cum-bersome (see the above example) excluded you use a computer program. The fol-lowing new method is very elegant involving no fractions and it also yields an eche-lon matrix that is row equivalent to A and whose non-zero rows constitute a basis forthe row space of a given matrix A.

Propo sit ion 9.2.8:LetA = [aij] be a (mn)-matrix with a11 0. We define determinants dij of order 2 asfollows:

For 2 im and 2 jn, we set

dij= i11jij11iji1

1j11aaaa

aa

aa= .

Then applies:

( )

+=

mn3m2m

n33332

n22322

d......dd

...

...

d......dd

d......dd

rankrank 1A .

This method is illustrated by the following example.

Example 9.2.12:

Find the rank of the matrix

A =

6138

4027

2153

.

Also write an echelon matrix row equivalent to A.

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9-33

Solution:

rank(A) = 1 + rank

68

23

18

13

38

53

4723

0713

2753

= 1 + rank

2531

2741

= 2 + rank

231

241

531

741

= 2 + rank([12 144])

= 3

An echelon matrix row equivalent to A is

1441200

27410

2153

.

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9.3 Systems of linear equations

The theory of matrices has been usefully employed in various branches of pure aswell as applied mathematics. In what follows, we shall apply this theory to solution ofm linear equations in n unknowns. Consider the m equations:

(9.3.1)

=+++

=+++=+++

mnmn22m11m

2nn2222121

1nn1212111

bxa...xaxa.

.

.bxa...xaxa

bxa...xaxa

in n unknowns x1 , x2 , ., xn, where aij and bi are scalars, i = 1, 2,, m;j= 1, 2, , n. Using the matrix notation, the system (9.3.1) can be written as

(9.3.2) Ax = b,

where A, x, b are the following matrices.:

(9.3.3) A =

mn2m1m

n22221

n11211

a......aa

...

...

a......aa

a......aa

, x =

n

2

1

x

.

.

x

x

, b =

m

2

1

b

.

.

b

b

.

The matrix A is called the matr ix of the coeff ic ientsof the system of equations, thecolumns of constants bi forms a column vectorb of orderm and the unknowns xj

form the column vectorx of ordern. Thus equation (9.3.2) can be written as:

(9.3.4)

=

m

2

1

n

2

1

mn2m1m

n22221

n11211

b.

.

.b

b

x.

.

.x

x

a...aa.

.

.

.

.

.

.

.

.a...aa

a...aa

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Definiti on 9.3.1:An equation of the type given in (9.3.1) above with b 0, is called a system ofnon-homogeneous l inear equat ions. Ifb = 0, then the system of equations (9.3.1) isknown as a system ofhomogeneous l inear equat ions.

Definiti on 9.3.2:(i) A solution of (9.3.1) is a set of real numbers {x1, x2, , xn} that satisfies all the mequations.

(ii) A solution depending on parameters i R, i= 1, ,k, is called a general solu-tion of (9.3.1). For instance

( ) ( ) ( ) ( )3,0,1,42,1,0,0,4,,32x,x,x,x21221214321

+== , 1, 2 R

is a general solution of a linear system in 4 unknowns. Each special choice of the

parameters leads to a particularsolution. Setting 1 = 1 and 2 = 0 in our examplewe get theparticular solution 0xx1,x2,x 4321 ==== .

Example 9.3.1:

(i) The linear system

==+=

=+

0

0573

0532

024

432

4321

4321

321

xxxxxxx

xxxx

xxx

has the general solution

( ) ( ) ( )4,1,0,12,1,1,0x,x,x,x 214321 += , 1, 2 R.

For1 = 1 and 2 = 0 we get the particular solution 0x1,x1,x2,x 4321 ==== . Us-ing other values for the both parameters you can find infinitely many particular solu-

tions.

(ii) The matrix notation of the system of equations

=++=+=+

3x7x5x3

2xx

1xx3x2

321

21

321

is given by

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9-36

=

3

2

1

753

011

132

3

2

1

x

x

x

.

Definiti on 9.3.3:

The matrix

(1.3.5) Ab =

mmn2m1m

2n22221

1n11211

ba...aa

ba...aa

ba...aa

MMMM

is called the augmented m atr ixof the system (9.3.1).

The augmented matrix is quite important in the sense that its rank and rank of A (thematrix of coefficients) determine whether the system of equations Ax = b does ordoes not have a solution.

The Gauss elim ination(named after its inventor, the German mathematician C. F.Gauss (1777-1855)) is a standard method forsolving linear systems. It is a system-atic elimination process, a method of great importance that works in practice.

For instance, to solve the system

=+=+

183x4x

25x2x

21

21

we multiply the first equation by (-2) and add it to the second, obtaining

==+

147x-

25x2x

2

21

This is Gauss elimination for 2 two equations. The solution now follows by back sub-stitution:

2-x2 = and ( )( ) 6/2252x1 ==

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Remark 9.3.1:Since a linear system is completely determined by its augmented matrix, the elimina-tion process can be done by merely considering this matrix. We apply row opera-tions on the augmented matr ixto reduce it to an echelon formfrom which we shallthen readily obtain the values of the unknowns by back substitution.

The general Gauss -el im ination-method

First elimination step: If11a = 0 interchange rows until 0a11 . Multiply the first row

by11

21

a

a and add it to the second row. Multiply the first row by

11

31

a

a and add it to

the third row etc. By these elementary row operations we get a new matrix that is

row equivalent to the starting matrix (9.3.5) and has the shape

mmn2m

2n222

1n11211

cd...d0

cd...d0

ba...aa

MMMM.

Second elimination step: For the new submatrix

mmn3m2m

3n33332

2n22322

cd...dd

cd...dd

cd...dd

MMMM

we perform again the elementary row operations described above. That is: If22d = 0

interchange rows until 0d22 . If such a row does not exist the second step is fin-

ished. If yet, multiply the first row by22

32

d

d and add it to the second row etc. After the

second step the starting matrix (9.3.5) has the shape

mmn3m

2n22322

1n1131211

c~e...e00

...0

c~e...ee0

ba...aaa

MMMM

M

After at most (m -1) steps we get matrix of the shape

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(9.3.7) ( )

rm

r

d

d

d

d

d

......

...

......

*...*...

****

******

*...******

AG

m

r

r

b

=

+

MMMMM

MMM

1

2

1

000000

000000

0000

00

0

,

which is row equivalent to the augmented matrix Ab. The places marked by * are realnumbers calculated by the Gauss elimination method and r= rank(A). Ifm > rand a

scalar m1r d,...,d+ in (9.3.7) is different from zero then the system of equations Ax = bhas nosolution. If e.g. 0d 1r + , then we get the contradiction

0dx0...x0x00 1rn21 =+++= + .

If 0d....d m1r ===+ , we can compute the solution of (9.3.1) from (9.3.7) by means ofback substitution. Ifkcoefficients in the r-th equation are different from zero, thenthere are (k-1) degrees of freedom; i.e. we can choose (k- 1) unknowns freely. Afterchoosing these free variables the k-th variable can be computed. Again by means ofback substitution we get the solution. If (k 1) > 0, then the system (9.3.1) has infi-

nitely many solutions.


Let Ax = b be a system ofm linear equations in n unknowns.

(i) Ifm = n, and the matrix A is non-singular, then the system has the unique solution

(9.3.8) x = A-1b.

(ii) The non-homogeneous system has a solution if and only ifrank(A) = rank(Ab);i.e. the rank of the coefficient matrix A is equal to the rank of augmented matrix Ab.

(iii) The homogeneous system Ax = 0 has a non-trivial solution ifm < n.

(iv) The homogeneous linear system Ax = 0 has the unique solution

0x...xx n21 ==== if and only ifrank(A) = n.

(v) The homogeneous system Ax = 0 has a non-trivial solution if and only ifrank(A) < n.

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(vi) The solution of the homogeneous linear system Ax = 0 has (n rank(A)) un-knowns that can be chosen freely. In other words: The general solution of Ax = 0has (n rank(A)) parameters.

(vii) If the system Ax = b has a solution then the general solution is given by

( ) ( ) ( )n21n21 u,...,u,ux,...,x,xx,...,x,x n21 +=000

where ( )000n21

x,...,x,x is a particular solution of the inhomogeneous system Ax = b

and ( )n21 u,...,u,u is the general solution of the assigned homogeneous systemAx = 0.

Proof:We give only the proof for (i). Since A is non-singular, its inverse A-1 exists. So

Ax = b

can be written as

A-1Ax = A-1b.

Since A-1A = In , the above equation becomes

Inx = x = A-1b,

giving the solution vectorx = A-1b. To show the uniqueness, let y be a second solu-tion. Then

Ay = b and so y = A-1b = x.

The Gauss-elemination-method is illustrated with the help of the following example.

Example 9.3.2:

(i) Solve the systems of equations:

(9.3.6)

=++=++=+

1-xxx

12xx4x

02xx-x

321

321

321

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Solution:Here

A =

111

214

211

and Ab =

1111

1214

0211

.

We apply row operations on Ab to reduce it to an echelon form. Thus

Ab =

1111

1214

0211

R

~

1120

1650

0211

R2+(-4)R1 and R3 + (-1)R1

R

~

575700

1650

0211

//

R3 + (-2/5)R2

By back substitution we get:

x3 =- 1,

5x2 6x3 = 5x2 6(-1) =1 and

x1 - x2 + 2x3 =x1 -(-1) + 2(-1)= 0.

Therefore x1 = 1, x2 = 1, x3 = 1 is a solution of (9.3.6).

The following row operations transform the augmented matrix to a matrix into re-duced echelon form without fractions. These transformations are based on experi-ence and are not suitable for a computer program (but a computer has no problemswith fractions). The solution remains of course the same.

Ab =

1111

1214

0211

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R

~

1120

1650

0211

R2 +(-4)R1 and R3 + (-1)R1

R

~

1120

3410

0211

R2 +(-2)R3

R

~

7700

3410

0211

R3 +(-2)R2

R

~

1100

3410

0211

(1/7)R3

By back substitution we get:

x3 = -1,

x2 4x3 = x2 4(-1) =3 and

x1 -x2 + 2x3 =x1 -(-1) + 2(-1)= 0.

Therefore we get the same solution x1 = 1, x2 = 1, x3 = 1.

Definiti on 9.3.4:

If the equation Ax = b has a solution, then it is called a consis tencesystemother-wise it is termed as inconsistencesystem.A linear system Ax = b is called overdeterminedifm > n, that is, it has more equations than unknowns. A linear system

Ax = b is called determinedifm = n, that is, the number of equations equals to thenumber of unknowns. A linear system Ax = b is called under determined ifm < n, that is, it has fewer equations than unknowns.

Example 9.3.3:

Examine the following system for a non-trivial solution:

(9.3.9)

=+++=++=++

02x2xx4x

0x2x3x

0x2xx-x

4321

421

4321

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Solution:The matrix of coefficients is

A =

2214

1023

1211

.

We bring A into the echelon form by the following elementary row operations:

R

~

1211

1023

2214

R13

R

~

12111023

2121411

(1/4)R1

R

~

2123451

2123450

2121411

R2 - 3R1 and R3 - R1

R

~

2123450

525610

2121411

(4/5)R2

R

~

0000

525610

535401

R1 - (1/4)R2 and R3 + (5/4)R2

The rank of the matrix A is 2 which is less than number of unknowns 4. Hence thesystem has a non-trivial solution. The first two rows of the last matrix give the follow-ing relations. In equation 2 we have 2 degrees of freedom. Assigning arbitrary val-ues to x3 and x4, we find the corresponding values of x1 and x2. Therefore system(9.3.9) has infinitely many solutions.

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9.4 Determinants

Determinants were originally introduced for solving linear systems. Although imprac-tical in computations, they have important engineering applications in eigenvalueproblems, differential equations, vector algebra etc. They can be introduced in sev-eral equivalent ways. Our definition is particularly practical in connection with linearsystems.

An n-th-order determinant is an expression associated with an (nn)-matrix (squarematrix) A = [aij], as we now explain, beginning with n = 2.

Definiti on 9.4.1:A determinantofsecond orderis defined by

(9.4.1) ( ) 211222112221

1211aaaa

aa

aa==Adet

So here we have bars, whereas a matrix has brackets.

Remark 9.4.1:

The determinant det(A) of order 2 is a real number associated with a square matrix

A of order 2. So, in this case, we may regard det:R2x2R as a function whose do-main is the set of all square matrices of order 2 and whose range is the set of thereal numbers.

Example 9.4.1:

(i) 165235152

31det ===

(ii) 101424278382

73

===

Definiti on 9.4.2:

A determinantofth i rd ordercan be defined by

( )2322

1312

31

3332

1312

21

3332

2322

11

333231

232221

131211

aa

aaa

aa

aaa

aa

aa

aaa

aaa

aaa

+== aAdet .

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Definiti on 9.4.3:

(i) Let A be a square matrix of order n. A matrix obtained from A by deletingits i-th row and j-th column is again a matrix Aij of order (n -1). Aij is called the i j-thminor of A .

(ii) Let Aij be the ij-th minor of a square matrix A of ordern. Then Cij = (-1)i+j det(Aij) is

called the i j -th co factor of A .

Example 9.4.2:

Let det(A) =333231

232221

131211

aaaaaa

aaa

be a determinant of order 3. Then applies:

det(A11)=3332

2322

aa

aaand det(A23)=

3231

1211

aa

aa.

Definit ion 9.4.4 (Cofactor exp ansion for a determinant o f ord er n):

For the matrix

A = [ ija ] =

mn2m1m

n22221

n11211

a......aa

...

...

a......aa

a......aa

, i= 1,2,,n andj= 1,2,,n

of ordern, we define det(A) for a given i {1,,n} by

(9.4.2) det(A) = ai1Ci1 + ai2Ci2 +.+ ainCin.

This expression is called an expansion of det(A) by cofactors of ith row of A.We can also write (9.4.2) in terms of minors as

(9.4.3) det(A) = ( )=

+n

1j

ij

jiA1- ija , i {1,,n}.

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Remark 9.4.2:

(i) Although the above technique to evaluate determinant of a (nn)-matrix seemsquite straight forward, yet, in practice, it is very laborious to work with when n > 3because it involves a lot of calculations. We shall later discuss methods which sim-plify these calculations.

(ii) For some special type of matrices, their determinants can be easily evaluated.For instance, if a matrix A is triangular that is, all its entries above or below the maindiagonal are zero, then its determinant is the product of the elements on the maindiagonal.

(iii) Analog to expanding a determinant of a square matrix A across the i-th row it canbe expanded across thej-th column. In this case we get:

(9.4.4) det(A) = ( )=

+n

1i

ij

jiA1- ija , j {1,,n}.

Example 9.4.3:

(i) We compute the determinant of the matrix

=

1342

2360

0123

1021

A .

Expanding det(A) by cofactors of the first column leads to

=

134

236

012

A11 ,

=

134

236

102

A21 ,

=

134

012

102

A31 ,

=

236

012

102

A41 .

Therefore we get:

( ) ( ) ( ) ( ) ( )

( )

72

16224332

==

+= 41312111 Adet2Adet0Adet3Adet1Adet

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(ii) The determinant of the matrix

=

7000

300

010

1021

A7

2

is given by det(A) = 1237 = 42.

(9.4.5) General prop erties of determ inants :

(i) Interchange of any two rows (columns) multiplies the value of determinantby (-1).(ii) Addition of a multiple of a row (column) to another row does not alter the value

of the determinant.(iii) Multiplication of a row (column) by a scalarrmultiplies the value of the determi-

nant by r.(iv) If the two rows (columns) are identical then value of determinant will be zero.(v) A zero row (column) renders the value of a determinant zero.

(vi) For any (nn)-matrices A and B, det(AB) = det(BA) = det(A)det(B).

(9.4.6) An Alg orith m to evaluate det(A):

By an algorithm we mean a sequence of a finite number of steps to get a desired re-sult. The step by step evaluation of det(A) of ordern is obtained as follows:

Step 1:By an interchange of rows of A bring a non-zero entry to position (1,1).

Step 2:By adding suitable multiples of the first row to all the other rows, reduce the (n -1)entries, except (1,1) in the first column, to zero. Expand det(A) by its first column.Repeat this process for the minor A

11or continue the following steps.

Step 3:Repeat step 1 and step 2 with the last remaining rows concentrating on the secondcolumn.

Step 4:

Repeat step 1, step 2 and step 3 with the remaining (n -2) rows, (n -3) rows and soon, until a triangular matrix is obtained.

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Step 5:

Multiply all the diagonal entries of the triangular matrix and then multiply it by its signto get det(A).

Remark 9.4.3:

For (3x3)-matrices A there is a simple rule called Sarrus Ruleto compute thenumber det(A).

(9.4.7)

.aaaaaaaaaaaaaaaaaa

a

a

a

a

a

a

aaa

aaa

aaa

det(A)

122133112332132231322113312312332211

32

22

12

31

21

11

333231

232221

131211

++=

=

Example 9.4.4:

Suppose

=

563

212

320

A , then we get by Sarrus Rule:

det(A) = 53200936120

6

1

2

3

2

0

563

212

320

=+=

.

Definiti on 9.4.5:

Let A be an (nxn)-matrix and Cij be the cofactor ofaij in A. We set

B =

nnn2n1

2n2221

1n1211

C...CC

...

...

C...CC

C...CC

.

Then the adjointof A, written ad j(A ),is the transpose of B. Thus

+++

---

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9-48

adj(A) = BT =

nn2n1n

n22212

n12111

C...CC

...

...

C...CC

C...CC

.


Let A be a (nn)-matrix. Then applies:

(i) A is non-singular if and only if det(A) 0

(ii) A is singular if and only if det(A) = 0

(iii) Aadj(A) = det(A)In

(iv) ( ) ( )TAdetAdet =

(v) If det(A) 0, then det(A-1) =( )Adet

1.

Proof:We prove only (iv). By the product of determinants, we have

det(A-1A) = det (A-1)det (A).

But det(A-1A) = det (I) = 1. Hence det(A

-1)det(A) = 1. Therefore

det(A-1) 0 and det(A) 0.

So det(A-1) =( )Adet

1.

Cor ollary 9.4.1:

(i) Let A be an (nn)-matrix.Then A is invertible if and only if det(A) 0.

(ii) If det(A) 0, then A-1 =( )Adet

1adj(A).

Proof:If A is invertible, then A is non-singular and therefore det(A) 0.Next ifdet(A) 0, then

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A( )Adet

1adj(A) =

( )Adet

1A adj(A) =

( )

( )Adet

AdetI= I.

Hence A is invertible and

A-1

=( )Adet

1adj(A).

Example 9.4.5:Find, by adjoint method, the inverse of

A =

725

812

543

.

Solution:Here, the cofactors are

C11 = (-1)1+1

72

81

= 9, C12 = (-1)

1+2

75

82= 26,

C13 = (-1)1+3

25

12

= 1, C21 = (-1)2+1

72

54

= -38,

C22 = (-1)2+2

75

53= -4, C23 = (-1)

2+3

25

43

= 26,

C31 = (-1)3+1

81

54

= 37, C32 = (-1)

3+282

53= -14,

C33 = (-1)3+3

12

43

= -11.

Now

det(A) = a11 C11 + a12 C12 + a13 C13 = 3 9 + 4 26 + 5 1 = 136.

Therefore

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A-1 =( )Adet

1adj(A) =

136

1

11261

14426

37389

.

In the following we discuss a classical method of solving a system ofn equations inn unknowns using determinants. Consider the system of equations

(9.4.8)

=+++

=+++=+++

nnnn22n11n

2nn2222121

1nn1212111

bxa...xaxa.

.

.bxa...xaxa

bxa...xaxa

which can be written as Ax = b, where

A = [ija ] is a (nn)-matrix, x = [x1,x2,...,xn]

T and b = [b1,b2,,bn]T.

Suppose that det(A) 0. Then (9.4.8) has the unique solution x = A-1b where the in-verse may be calculated by any of the methods discussed earlier. However, in the

following we describe a method of finding a solution with row (column) reduction.

(9.4.9) Cramer s Rule:

Let [ ]n1 a,...,a=A Rnxn be a (nn)-matrix, where [ ]Tnjj1 a,...,a=ja is thej-th column

of A, with det(A) 0. Then the unique solution of the linear system Ax = b is given by

(9.4.9)( ) ( )

( )n1j1j1jj ,...,,,,...,detAdet

1D

Adet

1x aabaa +==

where Dj = n1j1j1 ,...,,,,...,det aabaa + ,j= 1,2,,n, is the determinant obtained from

A by replacing in A thejth column by the column with entries b = [b1,b2,,bn]T.

Remark 9.4.4:

If the system (9.4.8) is homogeneousand det(A) 0, then it has only the trivial solu-tionx1 =x2 = = xn = 0. If det(A) = 0, the homogeneous system also has non-trivialsolutions.

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Example 9.4.6:

Solve, by Cramers Rule, the system of equations:

13xx2x 321 =+ 2x2xx 321 =+

32x2x3x 321 =++ .

Solution:Here

A =

223

121

312

and therefore det(A) =

223

121

312

= 5,

Furthermore:

D1 =

223

122

311

= 7, D2 =

233

121

312

= 0 and D3 =323

221

112 = -3.

Hencex1 =( )Adet

D1 =5

7, x2=

( )Adet

D2 = 0 andx3 =( )Adet

D3 =5

3.

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9.5 Eigenvalues and Eigenvectors

From the standpoint of engineering applications, eigenvalue problems are amongthe most important problems in connection with matrices. Let A = [aij] be a given(nn)-matrix and consider the vector equation

(9.5.1) A x = x.

Here, x = [x1,x2,...,xn]Tis an unknown vector and is an unknown scalar and we want

to determine both. Clearly, the zero vectorx = 0 is a solution of (9.5.1) for any valueof. This is of no practical interest.

Definiti on 9.5.1:

(i) A value ofC for which Ax = x has a non-trivial solution x0 is called an ei -genvalueorcharacter ist ic valueof the matrix A.

(ii) The corresponding solutions x0 of Ax = x are called eigenvectorsorcharac-terist icvectorsof A corresponding to that eigenvalue .

(iii) The set of eigenvalues is called the spectrumof the matrix A.

(iv) The largest of the absolute values of the eigenvalues of A is called spectra lra-d iusof A.

(v) The set of all eigenvectors corresponding to an eigenvalue of A, together with 0,forms a vector space, called the eigenspaceof A corresponding to this eigenvalue.

The problem of determining the eigenvalues and eigenvectors of a matrix is calledan eigenvalue problem. Problems of this type occur in connection with physical,technical, geometrical, and other application.

The equation Ax = x written in components is

(9.5.2)

=+++

=+++=+++

nnnn22n11n

2nn2222121

1nn1212111

xxa...xaxa.

.

.xxa...xaxa

xxa...xaxa

Transferring the terms on the right side to the left side, we have

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(9.5.3)

( )

( )

( )

=+++

=+++=+++

0xa...xaxa.

.

.

0xa...xaxa

0xa...xaxa

nnn22n11n

nn2222121

nn1212111

.

In matrix notation:

(1.5.4) (A I)x = 0.

By Cramers Rule, this homogeneous linear system of equations (9.5.4) has a non-

trivial solution if and only if the corresponding determinant det(A I) of the coeffi-cients is zero; i.e.:

(9.5.5) det(AI) =

( )

( )

( )a......aa

...

...

a......aa

a......aa

nn2n1n

n22221

n11211

= 0.

Definiti on 9.5.2:

The determinant det(AI) is called the characterist ic determinant. Equation(9.5.4) is called the characterist ic equation of the matrix A. By developingdet(AI) we obtain a polynomial ofn-th degree in . This is called the characteris-t ic polynom ialof A.

This proves the following important theorem.

Theor em 9.5.1:

The eigenvalues of a square matrix A are the roots of the corresponding characteris-tic equation (9.5.4). Hence a (nxn)-matrix has at least one eigenvalue and at most nnumerically different eigenvalues.

Remark 9.5.1:The eigenvalues must be determined first. Once these are known, correspondingeigenvectors are obtained from the system (9.5.3), for instance, by the Gauss elimi-nation, where is the eigenvalue for which an eigenvector is wanted.

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Example 9.5.1:

Find the eigenvalues and the corresponding eigenvectors of the matrix

A =

22

31.

Solution:The eigenvalues of A are given by its characteristic equation:

det(AI2) =

22

31= 0

(1 )(2 ) 6 = 0

2 - 3 - 4 = 0

( 4)( + 1) = 0.

Hence the eigenvalues of A are 1 = -1 and 2 = 4.

Let the corresponding eigenvectors be v1 =

2

1

x

xand v2 =

2

1

y

y.

Therefore

(A1I2) v1= 0

+

+122

311

2x

1x

=

0

0

=+=+

03x2x

03x2x

21

21.

The eigenvectors of the eigenvalue 1 = -1 are v1 =

3

2 for any scalar. If we

set = 3, then an eigenvector is v1 =

23

.

Similarly (A2I2)v2 = 0 becomes

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422

341

2

1

y

y=

0

0

==+

02y2y

03y3y

21

21

A solution is easy in this case and can be expressed as y1= y2= for anynon-zero scalar . Therefore the eigenvectors of the eigenvalue 2 = 4 are

v2 =

. If we put = 1, then an eigenvector is v2 =

1

1.

Remark 9.5.2:

(i) The eigenvalues of a Hermitian matrix (and thus of a symmetric matrix) are real.

(ii) The eigenvalues of a skew-Hermitian matrix (and thus of a skew-symmetric ma-trix) are pure imaginary or zero.

(iii) The eigenvalues of a unitary matrix (and thus of an orthogonal matrix) have abso-lute value 1.


Ifv is an eigenvector of a matrix A corresponding to an eigenvalue , so is rv with anr 0.

Proof:

Av = v implies r(Av) = A(rv) = r(v) = (rv).

Example 9.5.1:

Tank T1 in Figure 9.5.1 contains initially 100 gal ofpure water. Tank T2 contains initially 100 gal ofwater in which 150 lb of fertilizer are dissolved.Liquid circulates through the tanks at a constantrate of 2 gal/min, and the mixture is kept uniform bystirring. Find the amounts of fertilizery1(t) and y2(t)in T1 and T2, respectively, where tis the time.

T2

2 gal/min

2 gal/min

T1

Fig. 9.5.1 Fertilizer content in Tanks

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Solution:

(i) The time rate of change ( )ty1 ofy1(t) (amount of fertilizer in T1) equals inflow

minus outflow. Similar for tank T2. Thus we get:

1y = infolw/min outflow/min = 12 y100

2y

100

2 (Tank T1)

2y = infolw/min outflow/min = 21 y100

2y

100

2 (Tank T2).

The mathematical model of the mixture problem is the following system of first-orderdifferential equations:

(9.5.6)

=+=

212

211

y0.02y0.02y

y0.02y0.02y

As a vector equation with vector

=

2

1

y

yy and matrix A this becomes

(9.5.7) Ay = where

=

020020

020020

..

..A

(ii) We try an exponential function oft:

(9.5.8) texy = texy =

With (9.5.7) we get:

(9.5.9) texAy = tt ee xAx =

Dividing by te we obtain

(9.5.10) xAx = .

We need nontrivial solutions and hence we have to look for eigenvalues andeigenvectors of the matrix A. The eigenvalues are the solutions of the characteristicequation

(9.5.11) ( ) ( ) 00.040.020.02

0.020.02det =+=

= IA .

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We get 01 = and 0.042 = and therefore

(9.5.12) = =+ 0x0.02x0.02

0x0.02x0.0221

21 and ( )( )

=++ =++ 0x0.040.02x0.020x0.02x0.040.02

21

21 .

As easily seen we need only the first equations of the systems in (9.5.12)

(9.5.13) 0x0.02x0.02 21 =+ and ( ) 0x0.02x0.040.02 21 =++ .

Hence21 xx = and 21 xx = and we can take the eigenvectors

(9.5.14) ( )

= 1

11x and ( )

= 1

12x .

From (9.5.8) and the superposition principle we thus obtain a solution

(9.5.15) ( ) ( ) ( ) t.tt eccecect 04021211

1

1

121

+

=+= 21 xxy

where c1 and c2 are arbitrary constants. Later this will be called a general solution.

(iii) The initial conditions are y1(0) = 0 (no fertilizer in tank T1) and y2(0) = 150. From

this and (9.5.15) we get

(9.5.16) ( )

=

+

=

+

=

150

0

cc

cc

1

1c

1

1c0

21

21

21y

In components 0cc 21 =+ and 150cc 21 = . Thesolution is 75c75,c 21 == . This gives theanswer of our problem

(9.5.17) ( )t0.04

e1

1

751

1

75t

=y .

In components:

( ) t0.041 e7575ty= (Tank T1, lower curve in Fig. 9.5.2)

( ) t0.04e7575ty +=2 (Tank T2, upper curve in Fig. 9.5.2)

Fig. 9.5.2 Solution of the mixing problem

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