Пресс-релиз популярных книг

Авторы: 111 А Б В Г Д Е Ж З И Й К Л М Н О П Р С Т У Ф Х Ц Ч Ш Щ Э Ю Я

Книги: 164 А Б В Г Д Е Ж З И Й К Л М Н О П Р С Т У Ф Х Ц Ч Ш Щ Э Ю Я

На сайте 111 авторов, 92 книг, 72 статей, 5913 глав.

Chapter 4 DETERMINANTS

Back

DEFINITION 4.0.1 If A =

a11 a12

a21 a22

, we de¯ne the determinant of

A, (also denoted by detA,) to be the scalar

detA = a11a22 ¡ a12a21:

The notation

¯¯¯¯

a11 a12

a21 a22

¯¯¯¯

is also used for the determinant of A.

If A is a real matrix, there is a geometrical interpretation of detA. If

P = (x1; y1) and Q = (x2; y2) are points in the plane, forming a triangle

with the origin O = (0; 0), then apart from sign, 1

¯¯¯¯

x1 y1

x2 y2

¯¯¯¯

is the area

of the triangle OPQ. For, using polar coordinates, let x1 = r1 cos µ1 and

y1 = r1 sin µ1, where r1 = OP and µ1 is the angle made by the ray

OP with

the positive x{axis. Then triangle OPQ has area 1

2OP ¢ OQsin ®, where

® = \POQ. If triangle OPQ has anti{clockwise orientation, then the ray -

OQ makes angle µ2 = µ1 + ® with the positive x{axis. (See Figure 4.1.)

Also x2 = r2 cos µ2 and y2 = r2 sin µ2. Hence

AreaOPQ =

OP ¢ OQsin ®

OP ¢ OQsin (µ2 ¡ µ1)

OP ¢ OQ(sin µ2 cos µ1 ¡ cos µ2 sin µ1)

(OQsin µ2 ¢ OP cos µ1 ¡ OQcos µ2 ¢ OP sin µ1)

72 CHAPTER 4. DETERMINANTS

©©©©©©©©

µ1

¢@

Figure 4.1: Area of triangle OPQ.

(y2x1 ¡ x2y1)

¯¯¯¯

x1 y1

x2 y2

¯¯¯¯

Similarly, if triangle OPQ has clockwise orientation, then its area equals

¡1

¯¯¯¯

x1 y1

x2 y2

¯¯¯¯

For a general triangle P1P2P3, with Pi = (xi; yi); i = 1; 2; 3, we can

take P1 as the origin. Then the above formula gives

¯¯¯¯

x2 ¡ x1 y2 ¡ y1

x3 ¡ x1 y3 ¡ y1

¯¯¯¯

or ¡

¯¯¯¯

x2 ¡ x1 y2 ¡ y1

x3 ¡ x1 y3 ¡ y1

¯¯¯¯

;

according as vertices P1P2P3 are anti{clockwise or clockwise oriented.

We now give a recursive de¯nition of the determinant of an n£n matrix

A = [aij ]; n ¸ 3.

DEFINITION 4.0.2 (Minor) Let Mij(A) (or simply Mij if there is no

ambiguity) denote the determinant of the (n ¡ 1) £ (n ¡ 1) submatrix of A

formed by deleting the i{th row and j{th column of A. (Mij(A) is called

the (i; j) minor of A.)

Assume that the determinant function has been de¯ned for matrices of

size (n¡1)£(n¡1). Then detA is de¯ned by the so{called ¯rst{row Laplace

expansion:

detA = a11M11(A) ¡ a12M12(A) + : : : + (¡1)1+nM1n(A)

j=1

(¡1)1+ja1jM1j(A):

For example, if A = [aij ] is a 3 £ 3 matrix, the Laplace expansion gives

detA = a11M11(A) ¡ a12M12(A) + a13M13(A)

= a11

¯¯¯¯

a22 a23

a32 a33

¯¯¯¯

¡ a12

¯¯¯¯

a21 a23

a31 a33

¯¯¯¯

+ a13

¯¯¯¯

a21 a22

a31 a32

¯¯¯¯

= a11(a22a33 ¡ a23a32) ¡ a12(a21a33 ¡ a23a31) + a13(a21a32 ¡ a22a31)

= a11a22a33 ¡ a11a23a32 ¡ a12a21a33 + a12a23a31 + a13a21a32 ¡ a13a22a31:

(The recursive de¯nition also works for 2 £ 2 determinants, if we de¯ne the

determinant of a 1 £ 1 matrix [t] to be the scalar t:

detA = a11M11(A) ¡ a12M12(A) = a11a22 ¡ a12a21:)

EXAMPLE 4.0.1 If P1P2P3 is a triangle with Pi = (xi; yi); i = 1; 2; 3,

then the area of triangle P1P2P3 is

¯¯¯¯¯¯

x1 y1 1

x2 y2 1

x3 y3 1

¯¯¯¯¯¯

or ¡

¯¯¯¯¯¯

x1 y1 1

x2 y2 1

x3 y3 1

¯¯¯¯¯¯

;

according as the orientation of P1P2P3 is anti{clockwise or clockwise.

For from the de¯nition of 3 £ 3 determinants, we have

¯¯¯¯¯¯

x1 y1 1

x2 y2 1

x3 y3 1

¯¯¯¯¯¯

¯¯¯¯

y2 1

y3 1

¯¯¯¯

¡ y1

¯¯¯¯

x2 1

x3 1

¯¯¯¯

x2 y2

x3 y3

¯¯¯¯

x2 ¡ x1 y2 ¡ y1

x3 ¡ x1 y3 ¡ y1

¯¯¯¯

One property of determinants that follows immediately from the de¯ni-

tion is the following:

THEOREM 4.0.1 If a row of a matrix is zero, then the value of the de-

terminant is zero.

74 CHAPTER 4. DETERMINANTS

(The corresponding result for columns also holds, but here a simple proof

by induction is needed.)

One of the simplest determinants to evaluate is that of a lower triangular

matrix.

THEOREM 4.0.2 Let A = [aij ], where aij = 0 if i < j. Then

detA = a11a22 : : : ann: (4.1)

An important special case is when A is a diagonal matrix.

If A =diag (a1; : : : ; an) then detA = a1 : : : an. In particular, for a scalar

matrix tIn, we have det (tIn) = tn.

Proof. Use induction on the size n of the matrix.

The result is true for n = 2. Now let n > 2 and assume the result true

for matrices of size n ¡ 1. If A is n £ n, then expanding detA along row 1

gives

detA = a11

¯¯¯¯¯¯¯¯¯

a22 0 : : : 0

a32 a33 : : : 0

...

an1 an2 : : : ann

¯¯¯¯¯¯¯¯¯

= a11(a22 : : : ann)

by the induction hypothesis.

If A is upper triangular, equation 4.1 remains true and the proof is again

an exercise in induction, with the slight di®erence that the column version

of theorem 4.0.1 is needed.

REMARK 4.0.1 It can be shown that the expanded form of the determi-

nant of an n £ n matrix A consists of n! signed products §a1i1a2i2 : : : anin,

where (i1; i2; : : : ; in) is a permutation of (1; 2; : : : ; n), the sign being 1 or

¡1, according as the number of inversions of (i1; i2; : : : ; in) is even or odd.

An inversion occurs when ir > is but r < s. (The proof is not easy and is

omitted.)

The de¯nition of the determinant of an n £ n matrix was given in terms

of the ¯rst{row expansion. The next theorem says that we can expand

the determinant along any row or column. (The proof is not easy and is

omitted.)

THEOREM 4.0.3

detA =

j=1

(¡1)i+jaijMij(A)

for i = 1; : : : ; n (the so{called i{th row expansion) and

detA =

i=1

(¡1)i+jaijMij(A)

for j = 1; : : : ; n (the so{called j{th column expansion).

REMARK 4.0.2 The expression (¡1)i+j obeys the chess{board pattern

of signs: 2

6664

+ ¡ + : : :

¡ + ¡ : : :

+ ¡ + : : :

...

7775

The following theorems can be proved by straightforward inductions on

the size of the matrix:

THEOREM 4.0.4 A matrix and its transpose have equal determinants;

that is

detAt = det A:

THEOREM 4.0.5 If two rows of a matrix are equal, the determinant is

zero. Similarly for columns.

THEOREM 4.0.6 If two rows of a matrix are interchanged, the determi-

nant changes sign.

EXAMPLE 4.0.2 If P1 = (x1; y1) and P2 = (x2; y2) are distinct points,

then the line through P1 and P2 has equation

¯¯¯¯¯¯

x y 1

x1 y1 1

x2 y2 1

¯¯¯¯¯¯

= 0:

76 CHAPTER 4. DETERMINANTS

For, expanding the determinant along row 1, the equation becomes

ax + by + c = 0;

where

a =

¯¯¯¯

y1 1

y2 1

¯¯¯¯

= y1 ¡ y2 and b = ¡

¯¯¯¯

x1 1

x2 1

¯¯¯¯

= x2 ¡ x1:

This represents a line, as not both a and b can be zero. Also this line passes

through Pi; i = 1; 2. For the determinant has its ¯rst and i{th rows equal

if x = xi and y = yi and is consequently zero.

There is a corresponding formula in three{dimensional geometry. If

P1; P2; P3 are non{collinear points in three{dimensional space, with Pi =

(xi; yi; zi); i = 1; 2; 3, then the equation

¯¯¯¯¯¯¯¯

x y z 1

x1 y1 z1 1

x2 y2 z2 1

x3 y3 z3 1

¯¯¯¯¯¯¯¯

= 0

represents the plane through P1; P2; P3. For, expanding the determinant

along row 1, the equation becomes ax + by + cz + d = 0, where

a =

¯¯¯¯¯¯

y1 z1 1

y2 z2 1

y3 z3 1

¯¯¯¯¯¯

; b = ¡

¯¯¯¯¯¯

x1 z1 1

x2 z2 1

x3 z3 1

¯¯¯¯¯¯

; c =

¯¯¯¯¯¯

x1 y1 1

x2 y2 1

x3 y3 1

¯¯¯¯¯¯

As we shall see in chapter 6, this represents a plane if at least one of a; b; c

is non{zero. However, apart from sign and a factor 1

2 , the determinant

expressions for a; b; c give the values of the areas of projections of triangle

P1P2P3 on the (y; z); (x; z) and (x; y) planes, respectively. Geometrically,

it is then clear that at least one of a; b; c is non{zero. It is also possible to

give an algebraic proof of this fact.

Finally, the plane passes through Pi; i = 1; 2; 3 as the determinant has

its ¯rst and i{th rows equal if x = xi; y = yi; z = zi and is consequently

zero. We now work towards proving that a matrix is non{singular if its

determinant is non{zero.

DEFINITION 4.0.3 (Cofactor) The (i; j) cofactor of A, denoted by

Cij(A) (or Cij if there is no ambiguity) is de¯ned by

Cij(A) = (¡1)i+jMij(A):

REMARK 4.0.3 It is important to notice that Cij(A), like Mij(A), does

not depend on aij . Use will be made of this observation presently.

In terms of the cofactor notation, Theorem 3:0:2 takes the form

THEOREM 4.0.7

detA =

j=1

aijCij(A)

for i = 1; : : : ; n and

detA =

i=1

aijCij(A)

for j = 1; : : : ; n.

Another result involving cofactors is

THEOREM 4.0.8 Let A be an n £ n matrix. Then

(a)

j=1

aijCkj(A) = 0 if i 6= k:

Also

(b)

i=1

aijCik(A) = 0 if j 6= k:

Proof.

If A is n£n and i 6= k, let B be the matrix obtained from A by replacing

row k by row i. Then detB = 0 as B has two identical rows.

Now expand detB along row k. We get

0 = detB =

j=1

bkjCkj(B)

j=1

aijCkj(A);

in view of Remark 4.0.3.

78 CHAPTER 4. DETERMINANTS

DEFINITION 4.0.4 (Adjoint) If A = [aij ] is an n £ n matrix, the ad-

joint of A, denoted by adjA, is the transpose of the matrix of cofactors.

Hence

adjA =

6664

C11 C21 ¢ ¢ ¢ Cn1

C12 C22 ¢ ¢ ¢ Cn2

...

C1n C2n ¢ ¢ ¢ Cnn

7775

Theorems 4.0.7 and 4.0.8 may be combined to give

THEOREM 4.0.9 Let A be an n £ n matrix. Then

A(adjA) = (detA)In = (adjA)A:

Proof.

(AadjA)ik =

j=1

aij(adjA)jk

j=1

aijCkj(A)

= ±ikdetA

= ((detA)In)ik:

Hence A(adjA) = (detA)In. The other equation is proved similarly.

COROLLARY 4.0.1 (Formula for the inverse) If detA 6= 0, then A

is non{singular and

A¡1 =

detA

adj A:

EXAMPLE 4.0.3 The matrix

A =

1 2 3

4 5 6

8 8 9

is non{singular. For

detA =

¯¯¯¯

5 6

8 9

¯¯¯¯

¡ 2

¯¯¯¯

4 6

8 9

¯¯¯¯

+ 3

¯¯¯¯

4 5

8 8

¯¯¯¯

= ¡3 + 24 ¡ 24

= ¡3 6= 0:

Also

A¡1 =

¡3

C11 C21 C31

C12 C22 C32

C13 C23 C33

= ¡

66666666664

¯¯¯¯

5 6

8 9

¯¯¯¯

2 3

8 9

¯¯¯¯

2 3

5 6

¯¯¯¯

4 6

8 9

¯¯¯¯

1 3

8 9

¯¯¯¯

1 3

4 6

¯¯¯¯

4 5

8 8

¯¯¯¯

1 2

8 8

¯¯¯¯

1 2

4 5

¯¯¯¯

77777777775

= ¡

4 ¡3 6 ¡3

12 ¡15 6

¡8 8 ¡3

5 :

The following theorem is useful for simplifying and numerically evaluating

a determinant. Proofs are obtained by expanding along the corresponding

row or column.

THEOREM 4.0.10 The determinant is a linear function of each row and

column.

For example

(a)

¯¯¯¯¯¯

a11 + a011 a12 + a012 a13 + a013

a21 a22 a23

a31 a32 a33

¯¯¯¯¯¯

a11 a12 a13

a21 a22 a23

a31 a32 a33

¯¯¯¯¯¯

a011 a012 a013

a21 a22 a23

a31 a32 a33

¯¯¯¯¯¯

(b)

¯¯¯¯¯¯

ta11 ta12 ta13

a21 a22 a23

a31 a32 a33

¯¯¯¯¯¯

= t

¯¯¯¯¯¯ a

a21 a22 a23

a31 a32 a33

¯¯¯¯¯¯

COROLLARY 4.0.2 If a multiple of a row is added to another row, the

value of the determinant is unchanged. Similarly for columns.

Proof. We illustrate with a 3 £ 3 example, but the proof is really quite

general.

¯¯¯¯¯¯

a11 + ta21 a12 + ta22 a13 + ta23

a21 a22 a23

a31 a32 a33

¯¯¯¯¯¯

a11 a12 a13

a21 a22 a23

a31 a32 a33

¯¯¯¯¯¯

ta21 ta22 ta23

a21 a22 a23

a31 a32 a33

¯¯¯¯¯¯

80 CHAPTER 4. DETERMINANTS

¯¯¯¯¯¯

a11 a12 a13

a21 a22 a23

a31 a32 a33

¯¯¯¯¯¯

+ t

¯¯¯¯¯¯

a21 a22 a23

a31 a32 a33

¯¯¯¯¯¯

a11 a12 a13

a21 a22 a23

a31 a32 a33

¯¯¯¯¯¯

+ t £ 0

¯¯¯¯¯¯

a11 a12 a13

a21 a22 a23

a31 a32 a33

¯¯¯¯¯¯

To evaluate a determinant numerically, it is advisable to reduce the matrix

to row{echelon form, recording any sign changes caused by row interchanges,

together with any factors taken out of a row, as in the following examples.

EXAMPLE 4.0.4 Evaluate the determinant

¯¯¯¯¯¯

1 2 3

4 5 6

8 8 9

¯¯¯¯¯¯

Solution. Using row operations R2 ! R2 ¡ 4R1 and R3 ! R3 ¡ 8R1 and

then expanding along the ¯rst column, gives

¯¯¯¯¯¯

1 2 3

4 5 6

8 8 9

¯¯¯¯¯¯

1 2 3

0 ¡3 ¡6

0 ¡8 ¡15

¯¯¯¯¯¯

¯¯¯¯

¡3 ¡6

¡8 ¡15

¯¯¯¯

= ¡3

¯¯¯¯

1 2

¡8 ¡15

¯¯¯¯

= ¡3

¯¯¯¯

1 2

0 1

¯¯¯¯

= ¡3:

EXAMPLE 4.0.5 Evaluate the determinant

¯¯¯¯¯¯¯¯

1 1 2 1

3 1 4 5

7 6 1 2

1 1 3 4

¯¯¯¯¯¯¯¯

Solution.

¯¯¯¯¯¯¯¯

1 1 2 1

3 1 4 5

7 6 1 2

1 1 3 4

¯¯¯¯¯¯¯¯

1 1 2 1

0 ¡2 ¡2 2

0 ¡1 ¡13 ¡5

0 0 1 3

¯¯¯¯¯¯¯¯

= ¡2

¯¯¯¯¯¯¯¯

1 1 2 1

0 1 1 ¡1

0 ¡1 ¡13 ¡5

0 0 1 3

¯¯¯¯¯¯¯¯

= ¡2

¯¯¯¯¯¯¯¯

1 1 2 1

0 1 1 ¡1

0 0 ¡12 ¡6

0 0 1 3

¯¯¯¯¯¯¯¯

= 2

¯¯¯¯¯¯¯¯

1 1 2 1

0 1 1 ¡1

0 0 1 3

0 0 ¡12 ¡6

¯¯¯¯¯¯¯¯

= 2

¯¯¯¯¯¯¯¯

1 1 2 1

0 1 1 ¡1

0 0 1 3

0 0 0 30

¯¯¯¯¯¯¯¯

= 60:

EXAMPLE 4.0.6 (Vandermonde determinant) Prove that

¯¯¯¯¯¯

1 1 1

a b c

a2 b2 c2

¯¯¯¯¯¯

= (b ¡ a)(c ¡ a)(c ¡ b):

Solution. Subtracting column 1 from columns 2 and 3 , then expanding

along row 1, gives

¯¯¯¯¯¯

1 1 1

a b c

a2 b2 c2

¯¯¯¯¯¯

1 0 0

a b ¡ a c ¡ a

a2 b2 ¡ a2 c2 ¡ a2

¯¯¯¯¯¯

¯¯¯¯

b ¡ a c ¡ a

b2 ¡ a2 c2 ¡ a2

¯¯¯¯

= (b ¡ a)(c ¡ a)

¯¯¯¯

1 1

b + a c + a

¯¯¯¯

= (b ¡ a)(c ¡ a)(c ¡ b):

REMARK 4.0.4 From theorems 4.0.6, 4.0.10 and corollary 4.0.2, we de-

duce

(a) det (EijA) = ¡detA,

(b) det (Ei(t)A) = t detA, if t 6= 0,

82 CHAPTER 4. DETERMINANTS

It follows that if A is row{equivalent to B, then detB = c detA, where c 6= 0.

Hence detB 6= 0 , detA 6= 0 and detB = 0 , detA = 0. Consequently

from theorem 2.5.8 and remark 2.5.7, we have the following important result:

THEOREM 4.0.11 Let A be an n £ n matrix. Then

(i) A is non{singular if and only if detA 6= 0;

(ii) A is singular if and only if detA = 0;

(iii) the homogeneous system AX = 0 has a non{trivial solution if and

only if detA = 0.

EXAMPLE 4.0.7 Find the rational numbers a for which the following

homogeneous system has a non{trivial solution and solve the system for

these values of a:

x ¡ 2y + 3z = 0

ax + 3y + 2z = 0

6x + y + az = 0:

Solution. The coe±cient determinant of the system is

¢ =

¯¯¯¯¯¯

1 ¡2 3

a 3 2

6 1 a

¯¯¯¯¯¯

1 ¡2 3

0 3 + 2a 2 ¡ 3a

0 13 a ¡ 18

¯¯¯¯¯¯

¯¯¯¯

3 + 2a 2 ¡ 3a

13 a ¡ 18

¯¯¯¯

= (3 + 2a)(a ¡ 18) ¡ 13(2 ¡ 3a)

= 2a2 + 6a ¡ 80 = 2(a + 8)(a ¡ 5):

So ¢ = 0 , a = ¡8 or a = 5 and these values of a are the only values for

which the given homogeneous system has a non{trivial solution.

If a = ¡8, the coe±cient matrix has reduced row{echelon form equal to

1 0 ¡1

0 1 ¡2

0 0 0

and so the complete solution is x = z; y = 2z, with z arbitrary. If a = 5,

the coe±cient matrix has reduced row{echelon form equal to

1 0 1

0 1 ¡1

0 0 0

and so the complete solution is x = ¡z; y = z, with z arbitrary.

EXAMPLE 4.0.8 Find the values of t for which the following system is

consistent and solve the system in each case:

x + y = 1

tx + y = t

(1 + t)x + 2y = 3:

Solution. Suppose that the given system has a solution (x0; y0). Then the

following homogeneous system

x + y + z = 0

tx + y + tz = 0

(1 + t)x + 2y + 3z = 0

will have a non{trivial solution

x = x0; y = y0; z = ¡1:

Hence the coe±cient determinant ¢ is zero. However

¢ =

¯¯¯¯¯¯

1 1 1

t 1 t

1 + t 2 3

¯¯¯¯¯¯

1 0 0

t 1 ¡ t 0

1 + t 1 ¡ t 2 ¡ t

¯¯¯¯¯¯

¯¯¯¯1

1 ¡ t 2 ¡ t

¯¯¯¯

= (1¡t)(2¡t):

Hence t = 1 or t = 2. If t = 1, the given system becomes

x + y = 1

2x + 2y = 3

which is clearly inconsistent. If t = 2, the given system becomes

x + y = 1

2x + y = 2

3x + 2y = 3

84 CHAPTER 4. DETERMINANTS

which has the unique solution x = 1; y = 0.

To ¯nish this section, we present an old (1750) method of solving a

system of n equations in n unknowns called Cramer's rule . The method is

not used in practice. However it has a theoretical use as it reveals explicitly

how the solution depends on the coe±cients of the augmented matrix.

THEOREM 4.0.12 (Cramer's rule) The system of n linear equations

in n unknowns x1; : : : ; xn

a11x1 + a12x2 + ¢ ¢ ¢ + a1nxn = b1

a21x1 + a22x2 + ¢ ¢ ¢ + a2nxn = b2

...

an1x1 + an2x2 + ¢ ¢ ¢ + annxn = bn

has a unique solution if ¢ = det [aij ] 6= 0, namely

x1 =

¢1

; x2 =

¢2

; : : : ; xn =

¢n

;

where ¢i is the determinant of the matrix formed by replacing the i{th

column of the coe±cient matrix A by the entries b1; b2; : : : ; bn.

Proof. Suppose the coe±cient determinant ¢ 6= 0. Then by corollary 4.0.1,

A¡1 exists and is given by A¡1 = 1

¢ adjA and the system has the unique

solution

6664

...

7775

= A¡1

6664

...

7775

6664

C11 C21 ¢ ¢ ¢ Cn1

C12 C22 ¢ ¢ ¢ Cn2

...

C1n C2n ¢ ¢ ¢ Cnn

7775 2 6664

...

7775

6664

b1C11 + b2C21 + : : : + bnCn1

b2C12 + b2C22 + : : : + bnCn2

...

bnC1n + b2C2n + : : : + bnCnn

7775

However the i{th component of the last vector is the expansion of ¢i along

column i. Hence

6664

...

7775

6664

¢1

¢2

...

¢n

7775

6664

¢1=¢

¢2=¢

...

¢n=¢

7775

4.1. PROBLEMS 85

Пресс-релиз популярных книг

Chapter 4 DETERMINANTS

Популярные книги

Популярные статьи

Навигация