1,·", 1,"',

SEC. 7.4 LinearInrilonon/....or"lro Rank of a Matrix. Vector

These are vector a r n l n ....n.r\C\ for rows. To switch to corumns, we write (3) in terms of components asn such with k== 1,"',n,

and collectn.n.1MY'\~,.n."Y\a"Y\11-C\in columns. we can write as

== ^Vlk

+ ... +

^Vrk

wherek == 1,·",n. Now the vector on the left is the kth column vector of A. We see that each of thesencolumns is a linear combination of the same columns on the Hence A cannot have more columns than rows, whose number is rank Now rows of are columns of the . For our conclusion is that cannot have more columns than rows, so that cannot have more

mdependent rows than columns. the number of columns

of A must be r, the rank of A. This the

The matrix in (2) has rank 2. From Example 3 we see that the first two row vectors are linearly independent and by "working backward" we can verify that Row 3⁼ 6 Row 1 -

i

Row 2. Similarly, the first two columns are linearly independent, and by reducing the last matrix in Example 3 by columns we find that

Column 3=

i

Column 1+

i

Column 2

'.n.1"Y1lhll-nllYl,(1rTheorems 2 and 3 we obtain

and Column 4 =

i

^{Column 1}+~ Column 2.

Consider p vectors each n componerus. If n <p, then these vectors are

The matrix A with thosep vectors as row vectors hasp rows andn<p r-.n.lllrY1lnc· hence Theorem 3 it has rank A ~ n <p, which linear Theorem 2.

The related are of interest in linear In the

context a clarification of essential of matrices and their role in connection with linear I;;.'\r';;:tpnl I;;.

(見補充資料)

例題3變成行向量作運算

向量數目(可視為方程式的數目) 向量分量數目(可視為未知數的數目)

向量空間 (見補充資料)

7 Linear Matrices, Determinants. Linear ...."11""''1-,, ...

Consider a set V of vectors where each vector has the same number of for any two vectors a and inV,we have that all their linear combinations aa

+ f3

any real are also elements ofV,and if, and

the laws and in Sec. as well as any vectors a, c in V

then V is a vector space. Note that here we wrote laws and of Sec. 7.1 lowercase b,C, which is our notation for vectors. More on vector spaces in Sec. 7.9.

The maximum number of vectors inVis called the U1

Vand is denoted dimV.Here we assume the dimension to be infinite dimension will be defined in Sec. 7.9.

A set in V of a maximum number of vectors

V is called a basis for V. In other any set of vectors V forms basis for V. That means, if we add one or more vector to that set, the set will

be also the of Sec. 7.4 on linear and

rI,::u'''H::llnrL:::II1,,.,,a of the number of vectors of a basis forV dim V.

The set of all linear combinations of vectors aCl), ... , with the same number of is called the of these vectors. a span is a vector space. If

o.4 ....j.-"-v.-..L''U'..L..L~the vectors aCl), ... , are then form a basis

for that vector space.

This then leads to another definition of basis. A set of vectors is a basis for a vector spaceVif the vectors in the set are

Vcan be as a linear combination of the vectors also say that the set of vectors the vector space V.

a of a vector spaceVwe mean a..LJL'U'.LJL"-'JL..llJLIIJ'" Y

that forms a vector space with to the two ....JLj'- .. "-'U'JL ....-'""-''U'IIJ"-'JL ...JL,JJL·....J \U,'U...i.L.A.V..LA.

..L..L.i.IL,JI...LL.LIJ.L.L",-,U,I,.LVJ,A.Jdefined for the vectors of V.

The span of the three vectors in Example 1 is a vector space of dimension 2. A basis of this vector space consists of any two of those three vectors, for instance,a(l), a(2),ora(l), a(3),etc.

We further note the

The vector space has dimension n.

all vectors with components (n realVJIJ1VWI101l"C'

A basis of n vectors is aCl) == [1 0 1 0

aCn) ==

o

].

For a matrix we call the span of the row vectors the of the span of the column vectors of A is called thecotumn

Theorem 3 shows that a matrix A has as many rows as columns. the definition of their number is the dimension of the row space or the column space of A. This proves

The row space and the column space to rankA.

matrixA have the same aimension.

基底

維數

展延空間

次空間

行空間，以Col(A)表示 列空間，以Row(A)表示 n個向量所展延而成的空間

向量的分量數目 n個向量，每個向量有n個分

量，其展延的空間有n維

SEC. 7.4 LinearInrilonon,l""'Iov""ro Rank of a Matrix. Vector

for a matrix A the solution set of the nomogeneous vector space, called the space of and its dimension is called the the next section we motivate and prove the basic relation

Ax == 0 is a of In

Find the rank. Find a basis for the row space. Find a basis for the column space. Hint. Row-reduce the matrix and its transpose. (You may omit obvious factors from the vectors

of these rank = rank

= rank AB. (Note the

a counterexample.

IfAis not square, either the row vectors or the column vectors ofAare

If the row vectors of a square matrix are so are the column vectors, andr»A"'",,p>'rc p>II"

Give that the rank of a of

matrices cannot exceed the rank of either factor.

Show the-tAIIA'nrlno-'

o

4 -6 -6

o

2. [ 2a -b

o

4 -2

5

o

0 3

5

0 0 5 [

- 2 1.

CAS Show expenmentallv

that then X nmatrixA = [ajkJ withajk=j + k - 1 has rank2for anyn. 20shows n=4.) to prove it.

Do the same whenajk = j +k+c,whereCis any positive integer.

(c) What is rankAifajk = to find other large matrices of low rank mdependent ofn.

7. 0 -1

-0.6 IJ [0 0

[9 7 5 3

[0 8 [1 3

subset with the

last of the vectors [3 0 1 [6 1 0 [12 1 2 4

J,

[6 0 2 and [9 0 1 2], omit one after another until you get a linearly

set.

Are the sets of vectors Show the details of your work.

25.

220 19.

200

[3 4 0

J,

[2 -1 3 [1 6 -8

18. [1 ~

i [! !

~ ~^J

[1 0 IJ, [1 OJ, [0 OJ

[ 2 3 4J, [2 3 4 5J, [3 4 5 6J,

[4 5 6 7J

[2 0 0 7J, [2 0 0 8J, [2 0 0 [2 0 1 OJ

-0.2 [9 8 7 6 [4 -1 [2 6 IJ

[6 0 -1 3], [2 2 5 [-4 -4 -4 -4J

o

o

o

2 4

o

o

-4

o -4

-4 -11 2

2 4 8 16

16 8 4 2

4 8 16 2

2 16 8 4

o

5 -2 -2

o

6. -1

o

80

10.

o o

o o

5

o

8 4

o

4

o o

-3

o

o

-1

o

o

-1 2

5

o

6

o 2 -2 5. 0

2

9.

齊次系統之解所形成之向量空間，稱為零空間，以N(A)表示

無效度

(r) (n-r) (n)

CHAP. 7 Linear Matrices, Vectors, Determinants. Linear ....\lc~,..onnC'

All vectors with negative components 32. All vectors inR³with3V1 - 2V2 +V3 =0,

4V1 +5V2 = 0

All vectors inRⁿ with

Iv

^j

I

= 1 with3V1 - V3 = 0,

=0

= 1,"',n

=3V3= 4V4

withV1 = 33. All vectors in

2V1 +3V2 -

35. All vectors in Is the given set of vectors a vector space? Give reasons. If

your answer is yes, determine the dimension and find a basis. (V1, V2, ... denote components.)

27. All vectors in withV2 - V1 +4V3= 0 28. All vectors inR³with3V2 +V3 = k 29. All vectors inR²such thatV1 :::; V2

All vectors in R" with the first n - 2 components zero

columns

f'r.·II"Y\-r~la1rainformation aboutL:l>""lT1'C1 ....":>."Y\f'.a Unl(lUE~ne~ss,and structure of the solution set of linear as follows.

A linear of in n unknowns has a

matrix and the matrix have the same rank n, and many solutions if that common rank is less than n. The has no solution if those two matrices have different rank.

To state thisn1"'.a,01 c.a hT and prove it, we shall use the of a

.:::llUIU'lI..II.JIUfiL,.II...IlAof A. this we mean any matrix obtained from A some rows or

definition this includes A itself the matrix obtained this isn1"'<:10 ....10<:11

Existence..A linear system eauauons in n unknowns^{Xl' ... ,}_Xn

is that is, has

if

and

if

the matrix and the

matrix

A

have the same rank.

all aln ^all aln

A==

and

A==

aml amn aml amn

has one solution

if

and only

if

this n.

(補充資料已教過...) 如何從高斯消去法所 得之列梯形矩陣的 Rank數目，來判斷滿 足唯一解、無窮解、

無解的三種條件！！

SEC. 7.5 Solutions of Linear ....\11:'· ..._ ... • I-vH~1"or"\,..o

If

solutions exist, they can all be obtained method will reveal whether or not Inttmtelv many solutions..If this common rank r is less than n, the system (1) has infinitely many solutions. All of these solutions are obtained by determining r suitable unknowns (whose submatrix of coefficients must have rank r) in terms of the remaining n - r unknowns, to which arbitrary values can be assignea.

Example 3 in Sec. 7.3.) Gauss elimination the Gauss elimination.

see Sec.

We can write the

CCI), ... , cCn)ofA:

in vector formAx == or in terms of column vectors

(2)

A

is obtained A a column b. Theorem 3 in Sec.

rank

A

rank A or rank A

+

Now if has a solution then shows that b must be a linear combination of those column vectors, so that and A have the same maximum number of column vectors and thus the same rank.

if rank

A

== then must be a combination of the column

(2*)

since otherwise rank

A

==rank A

+

1. But

Xl == al,· ..,xn == an, as can be seen If rank A ==n,thencolumn vectors in in Sec. 7.4. We claim that then the representanon

This would all terms to the with a minus

+ ... +

1nrla-n,an,rla"""l0a But this means that the scalars

Xl, ...,Xn that the solution of is

If rank A == rank == r< n, then Theorem 3 in Sec, 7.4 there is a

mdenendentsetKof rcolumn vectors of A such that the other n - r column vectors of A are linear combinations of those vectors. We renumber the columns andllnllTnr'...T.","C'

r l a ...r . ....·.nrTthe renumbered A, so that { } is that

setK.Then becomes

+

^{0 0 .}

+ + +

^{0 0 .}

+

_{== b,}

0 0 0 ,CCn) are linear combinations of the vectors of and so are the vectors

jjxpn:~SSJLngthese vectors in terms of the vectors of K and collect-

in the form

(3)

+

^{0 0 0}

+

參數解，有n-r個參數

CHAP. 7 Linear Matrices, Vectors, Determinants. Linear ...JV~]LC;III.::ll

... , CCn)xn ; (3). These fixes the and results from the n - r terms

has a there are Y1,"',Yr with Yj == Xj

+

f3j, where

j == 1,"',r. Since the scalars are sinceK is

....,'Lf..l.·..I...,'-'IJ'Lf.L..I.U.L..I..L~Xj ==Yj - wherej ==

This was discussed in Sec. 7.3 and is restated here as a reminder.

The theorem is illustrated in Sec. 7.3. In 2 there is a

A

== rankA == n == 3 can be seen from the last matrix in the 3 we have rank

A

== rankA == 2 < ⁿ == 4 and can choose X3 and X4 In

..."7,.-..,...,.,.,.·-.,'" 4 there is no solution because rank A == 2< rank

A

== 3.

Recall from Sec. 7.3 that a linear is called if all the are zero, and if one or several are not zero. For the hAlrnn,O-A·nt::l>r'l"1"

we obtain from the Fundamental Theorem the results.

A nomogeneous linear system

has the ^Xl == 0, ... ,Xn == O. Nontrivial solutions exist

A< n. A == r < n, these with

Sec. n - called the spaceof

In and^X(2)are solution vectors then ^x == ^C1x(1)

+

^C2x(2)

with any scalars C1 and C2 is a solution vector does not for the term solution space is

1l'"'\1I"'r"1l'"'\n,CllIi-·.r-,.""", can be seen from the

that rank

A

== rank so that ahAlrnno-t::l>lnt::l>r"'l"1"

rank A == n, the trivial solution is the

If rankA

<

n,there are nontrivial solutions r)r-r~n^....,r'lll1'1lo-

~dilis

+ +

where c is If rank A == r < n,Theorem 1 suitable call themx_r+1, ... ,Xn ,in an

obtained in this way. Hence a basis for the solution space, solutions of is Y(1), . . . ,YCn-r), where the basis vector

SEC. 7.6 For Reference: Second- and Third-Order Determinants

The solution space of is also called the the solution space of (4). Its dimension is called the

of becauseAx == for every x in of Hence Theorem 2 states that

where nis the number of unknowns of columns of the definition of rank we have rank

Theorem 2 this the ...",/"'...11,,-,," 111"'11:7llrln ...n ...n1n ....

Hence if m< n,

A linear system with

nontrivial solutions.

The characterization of solutions of the

eauations than unknowns

is now

has

as follows.

a nonnomogeneous linear system ontatnea as

is C011Slj.terU, then all its solutions are

solution andXh runsrM1I",nU(;tMall the solutions the

COjrreSD,r:Jn(1Jn~J!nomogeneous system

The difference Xh== x - Xo of any two solutions of - b == Since the solutions of we take any solutionXo of solution space of

is a solution of because is any solution of all and let Xhvary t"n-r'f'i>1I11o-n',",,1lIt"the

This covers a main of our discussion of the solutions of of linear next main is determinants and their role in lineara.ri1ll11l<:1l ....1r11"YlC'

We created this section as a reference section on second- and third-order determinants. It is compteterv 1l1l1lriCl>1t"'l.Cl>1l1IriCl>1n .... of the Sec. 7.7 and suffices as a

reference for many of our Since this section is for go

next

a121== alla22 - a12a21·

a22 So here we have bars uTI1IPn:3l~1;;. a matrix hasnrackets

高中時代已學過....請自行複習！

高職生？聽說有些學校沒教....

交叉相乘且相減 矩陣:

(1)可為mxn或nxn型式 (2)不可展開成一數值

行列式:

(1)一定是nxn型式 (2)可展開成一數值

可利用第四章與第八章矩 陣運算求解(特徵值與特 徵向量)！！

無效度 (n-r) (r)

CHAP.7 Linear

Cramer's for

Matrices,

linear

Determinants. Linear ....'If·''1"'on'''lC'

of two c.rill-.ra-r..." ' ....' oin two unknowns

(2)

is

with D as

a121

a22 Xl == - - - - ==

D

D

*

^0.

The valueD ==

°

appears for h"'1rY\AcC..o·nO/~-'-'CI svsremswith nontrivial solutions.

We prove To eliminateX21rY\rllt"1l1nl"\:r -a12and

to eliminateXI

Assummzthat D == alla22 - al2a21

*

^0,

two O r l l l l l t : l t " l l A ....'CI as we obtain

and the sides of these

1

12

~I I:

¹²¹

4Xl+^3X2= 12 -8 84 -8 -56

If then Xl ^- 6,

X2=

I:

^{= - = -4.}

2Xl+^5X2= -8

I

4²

~I

¹⁴

~I

¹⁴

A can be defined

all al2 al3

la 22 a231 laI2 aI31 laI2 a131·

D== _{a21 a22 a23} == all - a21

+

a31

a32 a33 a32 a33 a22 a23

a31 a32 a33

三階行列式展開的原則：

(1) 降階(降成二階)展開(可對任何一列或任一行展開) (2) 三階直接展開(交叉相乘且相減)(建議)

對第一行降階展開

= a

11

a

22

a

33

+a

21

a

32

a

13

+a

12

a

23

a

31

-a

31

a

22

a

13

-a

32

a

23

a

11

-a

21

a

12

a

33

三階可直接展開(四階 以上不可直接展開)

SEC. 7.7 Determinants. Cramer's Rule

Note the following. The signs on the right are

+ - +.

Each of the three terms on the right is an entry in the first column of D times its that the second-order determinant obtained from D deleting the row and column of that for ^all delete the first row and first and so on.

If we write out the minors in (4), we obtain

(5)

is

*

⁰⁾

with the determinant D the system and

Note that column of the

Cramer's rule follows from the

are obtained sides of

can be derived case

eliminations similar to those for in the next section.

the

but it also

differential vector be introduced in severalp.rnl1U'':\ 1p.1nt

linear ^.'1'". ' ' ' , " , 1 1Ie,.

A nis a scalar associated with an n X n A == [ajk

J,

and is denoted

D == det A ==

matrix

行列式→均以nxn（方陣）形式出現

=值

CHAP.7 Linear ..._c:"~L. . n . Matrices, Vectors, Determinants. Linear,JV';';)I,,'I;,;;I 1 I.;)

Forn == 1, this determinant is defined

(2) Forn ~ 2

or

D==

+

+

+ ... +

+ ... +

(j == 1, 2, ... ,orn)

(k == 1, 2, ... , or

== (-

and is a determinant of order n - 1, the determinant of the submatrix ofA obtained fromA the row and column of the ajk,that row and the kth column.

In this way, is defined in terms ofndeterminants of order n - 1, each of which in turn, defined in terms of n - 1 determinants of order n - 2, and so on-until we

arrive at second-order in which those submatrices consist of entries whose determinant is defined to be the itself.

From the definition it follows thatwemay in the entries in any row or""''U'JlU-..LJl..L..L..L.C'111r1l1I-::1lrhT

aettnuton isunammeuous,

columns or rows we choose in A is in 4.

Terms used in connection with determinants are taken from matrices. InD we

ajk,also n a n d n and a on which^all,a22, ... , ann

stand. Two terms are new:

is called the

For later use we note that

ajk in and the coractor may also be written

n D==

D==

n

(-

(-

(j == 2, ... , orn)

(k== 1,2,"',

In (4) of the previous section the minors and cofactors of the entries in the first column can be seen directly.

For the entries in the second row the minors are

and the cofactors areC_{2 1}= -M_{2 1 ,}C_{2 2} = +M_{2 2 ,}andC_{2 3} = -M_{2 3 "}Similarly for the third row-write these down yourself. And verify that the signs in form acheckerboard

+ +

+

+ +

行列式可對任何 一列(或行)展開

對列展開

對行展開

副式 餘因子

選擇展開行(或列)的原則：

1.該行(或列)越多○越好 2.可先行(或列)運算後，再 選擇越多○之行(或列)展 開

不要死記公式，注意計 算原則最重要！

(見補充資料)

SEC. 7.7 Determinants. Cramer's Rule

D= 2

-1

3 0

6

o ~I

= 1(12 - 0) - 3(4+4)+0(0+6)= -12.

This is the expansion by the first row. The expansion by the third column is

D =01 2

-1 :1=0-12+0=-12.

Verify that the other four expansions also give the value - 12.

-3 0 0

6 4 0 =

-31: ^~I

⁼

-3 .

^{4 . 5= -60.}

-1 2 5

Inspired by this, can you formulate a little theorem on determinants of triangular matrices? Of diagonal matrices?

There is an attractive way of

row to so we obtain an

Sec. for definition with "matrix"^{raor\ I}~{-'aorl

easy to the of its entries. This ~n1l'""IlrA'':l0h

not the to what we did to matrices in Sec. 7.3. In nnlV';1f>'11Iff1JO

tntercnanetng two rows in a determinant introduces a muuioucauve the determinant! Details are as follows.

muitunies the value of the determinant -1.

row to another row does not alter the value

a row a nonzero constantc muuuiues holds also whenc == 0, but no

induction. The statement holds forn == 2 because

an elementarv

bl

^{= ad - be,}

dl

but

~I ⁼

^{be - ad.}

DI

(-1)

¹⁺²

× 3

(見補充資料)

對第一列展開

對第三行展開

結 果 一 樣

CHAP. 7 Linear Matrices, Determinants. Linear ....'IC""I"'OI"'ll...C"

We now make the induction that (a) holds for determinants of ordern - 1 ~ 2 and show that it then holds for determinants of ordern. Let D be of order n. Let E be obtained from D by the interchange of two rows. Expand D andE a row that is not one of those call it the jth row. Then

n

(- E==

n

(- ^kajk jk^NT

whereNjk is obtained from the minor of ajk in the of those two rows which have been in D which N_{j k} must both contain because we another Now these minors are of order n - 1. Hence the induction

and N_{j k} == Thus E == - D

Add c times Rowi to Row Let be the new determinant. Its entries in Rowj are ajk +caik. If we this Row j, we see that we can write it as

i5

== + where == D has in Rowj the ajk,whereas has in that Rowj the

ajk from the addition. Hence has ajkin both Row i and Rowj. these

two rows but on the other hand it

== 0, so thatD ==

the determinant the row that has been -rv'i~lll1"lIl1"hc.rI

det == en det A e det

Because of Theorem 1 we may evaluate determinants by reduction to triangular form, as in the Gauss elimination for a matrix. For instance (with the blue explanations always referring to thepreceding determinant)

2 0 -4 6

4 5 0

D=

0 2 6 -1

-3 8 9

2 0 -4 6

0 5 9 -12

0 2 6 -1

0 8 3 10 +

2 0 -4 6

0 5 9 -12

0 0 2.4 3.8 3- 2

0 0 -11.4 29.2

2 0 -4 ₆

0 5 9 -12

0 0 2.4 3.8

0 0 -0 47.25 + 3

= 2 . 5 . 2.4 . 47.25 = 1134.

1. 可先行(或列)運算化簡後，再 選擇越多○之行(或列)展開。

2.或化簡變成△矩陣，直接對角 線乘開即可。

R

12

(-2) R

14

(1.5)

R

34

(4.75) R

23

(-0.4)

R

24

(-1.6)

SEC. 7.7 Determinants. Cramer's Rule

in Theorem 1hold also for columns.

1ransnosuton leaves the value of a determinant unaltered.

A zero row or column renders the value of a determinant zero.

from the fact that a ri~t-~rl1r"l1nl"}nt- can be 'O-V1".,.-s-'rfCJ,rf

t-rl"}-nC'r\AC'lt-"tr,n is defined as for that the

follow column. In

column of thetrl"}nC'1''''AC'~

If Rowj == c times

anlIn-hor"hl"}1''1lo~ of these rows r~-r'\-rA.rlrl/"",oC'

Hence == 0 and D == ==

o.

"-.:"t1r'1l"'l"tllllrh;T

It is of the rank of a matrix which is the

maximum number of row or column vectors of A Sec. can

be related to determinants. Here we may assume that rank A

>

0 because the matrices with rank 0 are the zero matrices Sec.

Consider an m X n matrix == [ajk]:

Ahas rank r ~ 1 has an rX r submatrix with a nonzero determinant.

The determinant of any square submatrix with more than r rows, contained

in a has a value to zero.

== n, we have:

An n X n square matrix has rank n

if

and

if

detA

*-

O.

The idea is that row alter neither rank Theorem

1 in Sec. nor the of a determinant nonzero Theorem 1 in this

u ...,,''-'-'-'U'-'--'-/. The echelon form

A

of A Sec. has r nonzero row vectors are

the first r row if and if rank A == r. Without loss of O~1''1l~-r11l1"t1-",{T

assume that r ~ 1. Let be the rX r submatrix in the left corner of

the entries of are in both the first r rows and r columns of Now is 1"rll~lnllnll/Jlr

with all entries rjj nonzero. det == r11···Trr

*-

O. Also det

*-

0 for

the rX rsubmatrix of A because results from row

VI-/",,,,.LUl-.LV.lJLeJ.This proves part (1).

det S 0 for any square submatrix S of r

+

1 or more rows

contained in A because the submatrix

S

^of

A

must contain a row of zeros

Ath~r'"lTlI C'~we would have rank A _~ r

+

so that det

S

== 0 Theorem 2. This proves

(2). we have proven the theorem for an m X n matrix.

CHAP. 7 linearI \ . U : : : C U I U . Matrices, Determinants. linear' \ / e ' t - o l " V ' \ e '

For ann X n square matrix we as follows. To prove (3), we apply (1) (already proven!). This us that rankA == n ~ 1if and ifA contains an n X n submatrix with nonzero determinant. But the only such submatrix contained in our square matrix isA hence detA

"*

^0.This proves

Theorem 3 opens the way to the classical solution formula for linear known as Cramer's ²which solutions as of determinants.Cramer's rule is

comnutauons for which the methods in Sees. 7.3 and 20.1-20.3 are suitable.

I-IA"''I10'(101'" Cramer's rule is oftheoretical interest in differential 2.10 and

and in other theoretical work that has pn(T1nl::J>pr"lna 1"Jl~1nl-g'''1"Jlt"1Ar,C'

linear system eauauonsin the same number Xl, ...,Xn

has a nonzero coetticient dOt01l'VJI1I 1 1 / J / ' 1 1 / J t

solution. This solution is

the system has one

remactnu inD the kth column D

where is the determinant ontatnea the column with the entries

Hence

if

the system isnomogeneous and D

"*

^{0, it has} the trivial solution^XI == 0,^X2 == 0, ... ,^Xn == 0. == 0, the system also has nontrivial solutions.

The matrix

A

of the

at most n. Now if

is of sizen X (n

+

Hence its rank can be

D == det A ==

"*

^0,

2GABRIEL CRAMER(1704-1752), Swiss mathematician.

舉範例說明 (見補充資料)

何謂D, D1, D2... Dn? 要弄清楚！

SEC. 7.7 Determinants. Cramer's Rule

then rank A == n by Theorem 3. Thus rank

A

⁼⁼ rank A.

Theorem in Sec. 7.5, the (6)has a unique solution.

Let us now prove(7). D its kth column, we obtain

the Fundamental

(9)

where is the cofactor of aikin D. If we the entries in the kth column of

D any other we obtain a new say, its expansion by

the kth column will be of the form withalk, ... , ank by those new numbers and the cofactors as before. In if we choose as new numbers the entries

all, ... , anl of thelthcolumn of D I

*-

we have a new determinant

fJ

which has the column [all once as itslth once as its kth because

of the Hence Theorem If we now

fJ

the column

that has been we thus obtain

+ .

^{e .}

+

_{anlCn k} ⁼⁼ 0 (I

*-

the second on both

.... 0C11 .. 1 ..-I-nrr\.I"-I\..lULJLV.L.L0e This We now1'Y'Il"uU....,nl ....Tthe firsto r n .. n"-l.n.-n

the last and add

(11)

+

.e.

+

+

.e.

+

+ ... +

'.n.II.of1>1--.-nrrterms with the same we can write the left side as

+ + ... + + ... + +

From this we see thatxk is multrpneo

shows that this X1is.L.L.LUl..LL..L~J.-L.-L\,./~

+ +

shows that this is zero when I

*-

^k.

so that (11) becomes

the left side of (11)

+ +

as defined in the its kth ^{V V ..}LU.LJL.L.L.L~

This proves Cramer's rule.

hA1rY"1n,rrO-nO£'\11C' and D

*-

0, then each has a column of zeros, so that ==

°

the trivial solution.

hA-rn.n.l-Y01""lOAllC' and D == 0, then rank A < n Theorem 3, so that

Theorem 2 in Sec. 7.5.

For n= 2, see Example 1 of Sec. 7.6. Also, at the end of that section, we give Cramer's rule for a general linear system of three equations.

CHAP. 7 Linear Determinants. Linear ....\lC~,.OrY'liC

an application for Cramer's rule be given in the next section.

with inverse matrices will

1.. General of Determinants..Illustrate each statement in Theorems 1 and 2 with an of your choice.

o o

o

-1 -1 -1

o o

0

16.. CAS EXPERIMENT.. Determinant Zeros and Find the value of the determinant of the nX n a

ways and Second-Order

second-order determinant in four show that the results agree.

3. Third-OrderDeterminant.Do the task indicated in Theorem 2. Also evaluate D tot r l <:11'"\ 0"111 <:1r

form.

EX1paIlsioln Nn11l1lg:lo1"'l(bgRII'iT ...[t8.,... ,"'...""'... Show that the computauon of an nth-order determinant

involves n! which if a muttmncanon takes sec would take these times:

4

o

Find the rank Theorem 3 is not very 1-' ... ...,,"'...,..,.../

and check row reduction. Show details.

years

that det years

,"-,,'VJLJLLIJJl"-''''....,the list in HV'r:ln1nlao 1.

sec

5. IVIU!tlpJHca1Uon kⁿ det 6..

-1 2 4

2 0 4

sina

I

1°04 0.61

^{-8 8 8}

ICOSa

8. 20. Curves

sinf3 cosf3 1.5 -0.5 The idea is to

I

^{cosni) sinnO}

I I

cosht sinht

I

^{from the} of the determinant

9.. 10. of a linear system as the condition for a

-sin^nf) cos^nO sinh t cosh t nontrivial solution in Cramer's theorem. We

6 -1 8 a b c the trick for such a system for the case of

a line L two PI: (Xl, and

II. 0 -2 9 c a b _{P2: (X2,} The unknown line is ax + by = r:c,

0 0 -4 b c a say. We write it as ax +by +c .1= O. To get a

nontrivial solution a, b, c, the determinant of the 0 4 -1 5 4 7 0 0 "coefficients"x, y, 1must be zero. The system is

-4 ₀ 3 -2 2 8 0 0

ax+ by +c·1=O L)

13.

-3 0 0 0 5 (12) aXI+ bYI+^{C •}1= 0 (Plan L)

-2 -1 0 0 -2 2 aX2+ +c·1=O onL).

SEC. 7.8 Inverse of a Matrix. ^~~IIICC__ If"'\rt'"'l=lnElimination

°

- 5z= -31 + z= 13

x+

3x -

2x - 4y = -24 5x +2y =

-2x + y +4z= 11

+ z= -7 + z= -5

= -1

4z= - 3z= 2

w w - 2x

x+ Y - 2z= 7 -2w + ^{X -} Y

-X+

21 .. 2x - Y= 5.15 3x +9y = 6.15 23.. 3x+

Solve by Cramer's rule. Check by Gauss elimination and back substitution. Show details.

this through are Derive fromD =

°

ⁱⁿ

(a)

(12) the familiar formula

x - Xl Y - YI

Xl - X2 YI-

(b) Find the analog of (12) for a three given points. it when the (1,1,1),(3,2,6),(5,0,5).

(c) Circle.. Find a similar formula for a circle in the plane three given Find and sketch the circle (2,6), (6,4), (7,1).

Find the analog of the formula in (c) for

four Find the

(0,0,5), (4, 0, 1), (0,4, 1), (0,0, -3)

== ^{[ajk ]} is denoted and is ann X nmatrix

so that we obtain and CA ==

whereI is then X nunit matrix Sec.

has an A is called a nonsmautar ..."..."'...,... If A has no then A is called aSII1J:!U.lar ....JilJlil ....,,.,...."r,c"••

has an

if both and then

the from

== CI ==

We prove next that has an inverse

~f'CIC'lI^{h l a}rankn. The will also show that x ==

and will thus a motivation for the inverse as well as a relation to linear,"'Iv,"IL\../III,'.

this will a method of Ax == because the Gauss

elimination in Sec. 7.3rArnl1·fOAC' fewer r>Aln1-nil1"t':l"t1f",nc

== n, The inverse an n X n matrix A exists

if

and

Theorem 3, Sec. 7.7) =I=- O.Hence A is nOJ1S111fZu:tar and issineutar

反矩陣

非奇異

(見補充資料)

detA=0即為奇異矩陣，

A

^-1不存在

(與Gauss消去法

的差異何在？

)

CHAP. 7 Linear Determinants. Linear ....\/C~"I"'O~C':"

Let A be a n X nmatrix and consider the linear

(2) Ax ==

If the inverse then.Ll..U.!U.L.A.lfJ.A..A.'-"IULJLV.A..L from the left on both sides and use of (1)

==x==

1lJ...., .."I ...·u....,~for another solution we

must have rank n the Fundamental This shows that has a solutionx,which is

have Au == b, so that == x. Hence Theorem in Sec. 7.5.

let rank A == n. Then the same1"ha,~-ra'YY1l

solution x for any Now the back substitution the elimination

shows that the Xj ofxare linear combinations of those of b. Hence we can write

x==

with to be determined. Substitution into

Ax ==

for any b. Hence C == the unit matrix...^lInnllR~;lrl"l;' if we substitute into we

x == Bb ==

for any x Hence BA == exists.

we can use a

llJ(i.uS~~-J^'UClU:UI elimination.'The

To determine the inverse variant of the Gauss elimination idea of the method is as follows.

we form nlinear

== eCn)

where the vectors ...,eCn) are the columns of the n X n unit matrix

eCI) == [1 0 ,e(2) == [0 1 0 , etc. These arenvectort:llnn-Ir}1"1I.~nCl

in the unknown vectors XCI), ... ,xCn)' We combine them into a matrix ....,"-1 ...~L...,^...

JORDAN (1842-1899), German geodesist and mathematician. He did important geodesic work in Africa, where he surveyed oases. [See Althoen, S.C. and R. McLaughlin, Gauss-Jordan reduction: A brief history. American Mathematical Monthly, VoL 94, No.2 (1987), pp. 130-142.]

We donot recommendit as a method for solving systems of linear equations, since the number of operations in addition to those of the Gauss elimination is larger than that for back substitution, which the Gauss-Jordan elimination avoids. See also Sec. 20.1.

(見補充資料)

注意與

Gauss消去法的差異

以反矩陣求解線性 系統，以實例補充 說明！

SEC. 7.8 Inverse of a Matrix. ^_~::lIICC__I"i"'I'"I~lnElimination

AX == I, with the unknown matrix X having the columns xCI), ... , xCn). Correspondingly, we combine the n augmented matrices eCI)], ... , eCn)] into one wide n X 2n

"augmented matrix"

A

== Now multiplication of AX == I from the left

gives X == to solve AX == for we can apply the Gauss

elimination to

A

== This gives a matrix of the form with upper triangular U because the Gauss elimination triangularizes systems. The Gauss-Jordan method reduces U by further elementary row operations to diagonal form, in fact to the unit matrix This is done the entries of U above the main diagonal and making the diagonal entries all 1 multiplication (see 1). Of course, the method operates

on the entire matrix some the entire

to [I This is the matrix" of IX == K. Now IX == X == ,as shown

before. K == ,so that we can read from [I

The illustrates the details of method.

Determine the inverse A-1of

A=[-~

_-1

^-1

_{3 4}

~J.

We apply the Gauss elimination (Sec. 7.3) to the followingnX2n= 3X6 matrix, where always refers to the previous matrix.

Row 3 - Row 1

Row 2

:J

001J ^{Row 2+ 3 Row}

:J

^{Row 3}

°

° °

⁰

o

3

-1 0

o

3 -4 -1

2 7

2

°

^-5

-1

3 4

2

2 7

2 2

2

This is as produced by the Gauss elimination. Now follow the additional Gauss-Jordan steps, reducing U to that is, to diagonal form with entries 1 on the main diagonal.

-Row 0.5 Row 2 -0.2 Row 3

Row 2 - 3.5 Row 3 Row 1+ 2 Row 3

-:J -~:~J

-0.2

0.3J Row 1+Row 2 0.7

-0.2

o

0.5 0.2 0.4 -0.2 0.2 0.2 -0.2 0.2 -1

1.5 0.8 0.6 -1.3 0.8 -0.7 -1.3 0.8

o o

o o

-2 3.5

o

o o o

-1

變成

I

^即為

A

^-1

I

^矩陣

Gauss- Jordan

消去法

Gauss

消去法 以實例說明較

容易瞭解！

AX=I → X=A

^-1

I=A

^-1

...(1) Ã=[A ︳I] → [U ︳H] → [I ︳K]

IX=K → X=K...(2)

(1)與(2)式比較 K=A^-1

Gauss消去法

Gauss-Jordan

文字內容的主要意思

CHAP. 7 Linear Matrices, Vectors, Determinants. Linear- ' y ..;"...I ....

The last three columns constitute A-¹.Check:

[

-

~

^-1

~] [=~:: _~:~ ~:~]

⁼

[~

⁰

~].

-1 3 4 0.8 0.2 -0.2 0 0 1

Hence AA- 1= Similarly, A-1A=

the inverse of a matrix is really a problem of of linear

not that Cramer's rule 4, Sec. come into

as Cramer's rule was useful for theoretical but not for

so too is the formula in the theorem useful for

theoretical considerations but not recommended for inverse1YYl 'llt"r...,0aC1

for the 2 X 2case as

The inverse nonstnguiar n X n matrixA== is

- _l_[C.

JT __

1_

- detA Jk - detA

where is the cotactor in , the cofactor

In the inverse

Note well that does in

is

We denote the side of and show that BA == We first write

(5)

and show that G == Now the form of B in we obtain

BA ==

the definition of matrix1YYlrllt"lI,nl",,"'I1lt"-If'"y\ and because of notC_{k s)}

(6)

n

c.;

1

gkl == det Aasl == det A

s=l

+ ... +

"""YU'.'-''YI.I<"/·

進行驗證！

A

^-1

I A

利用公式求解

A

^-1

1. 求det A

2. 求出Cij (計算注意！) 3. 取轉置

detA=0即為奇異矩陣，A

^-1不存在 以實例說明較

容易瞭解！

SEC. 7.8 Inverse of a Matrix. .<..-<..- __.r .....-r-J"".r~Elimination

Now (9) and (10) in Sec. 7.7 show that the sum ( ... ) on the right is D == det A when I == k,and is zero whenI ^-=/=- k.Hence

1

gkk == det A det == 1,

In particular, forn in the second row,

2 we have in in the first row,

== ^-a21, == ^all. This

The special casen == 2 occurs frequently in geometric and other applications. You

may want to memorize formula 2 an illustration of

a

= [ :

~l

Using (4), find the inverse of

1 [ 4 A-I ~-

10 -2

-1]

^{= [}

0.4

3 -0.2

-0.1]

0.3

A=

[-~

^-1

~].

-1 3 4

Solution. We obtain det A= -1(-7) - 1 . 13+ 2 . 8= 10,and in (4),

1

-1 11

Cl l = 3 4 = -7,

Cl 2=

-I

_-1³ ₄^{11= -13,}

I

3 -11

C_{l 3}= =8,

-1 3

C^{2 1}=

-I ~ :I

^{= 2,}

1

- 1

21

C_{2 2} = = -2, -1 4

1

-1 11

C_{2 3} = - = 2, -1 3

C3 l = 1 1 -1

1

- 1 C3 2= - 3

1

- 1 C3 3= 3

~I

^{= 3,}

~I

^{= 7,}

11

^{= -2,}

-1 so that by (4), in agreement with Example 1,

[ - 0.7 A-^l = -1.3 0.8

0.2 -0.2 0.2

0.3]

0.7 . -0.2

rnasonar ^{J1 ....}J1U~"'.......,u A == [ajk], ajk == 0 when j ^-=/=- k,have an inverse if and if all is too, with entries ... , 1/ann.

For a diagonal matrix we have in

Cl l a22···ann

D alla22·· .ann all' etc.

步驟1

步驟2

步驟3：

步驟4

CHAP. 7 Linear Matrices, Vectors, Determinants. Linear....IV.:JL~III..;»

Let

Then we obtain the inverse A-1by inverting each individual diagonal element of A, that is, by taking 1/( -0.5),!, and

f

as the diagonal entries of A-1,that is,

o

0.25

o

Products can be inverted

inverses reverse

the inverse of each factor and YY1lllllli-lIr\h.TlInr.- these

Hence for more than two

The idea is to start from it on both sides from the

instead of that , which because of

and r'lr"'lllllli-lIl1nl'l:T

and then r'lr"'l1l111i-1I11"\I'l:T-snr.- this on both sides from the this time and

This proves and from it, follows induction.

We also note that the inverse the inverse is the as you may prove,

Section 7.2 contains that some r\1I"'r"r\Q.1I"'i-lIt:J>C\

those for and we are now able to cancenanon laws [2]and

rYlll-llh-nl1,r'':li-lIA-n deviate from

of the so-called

f'A1t"lf'O.-t"'ll-tC'that were not

等號兩邊從左邊乘上

A

^-1

AA

^-1

=I

等號兩邊從左邊乘上

C

^-1

取倒數

SEC. 7.8 Inverse of a Matrix. _~IIe'e'_I",",fI',l""I~nElimination

available in Sec. 7.2. The deviations from the usual are of must be carefully observed. are as follows.

Matrix multiplication is not commutative that in we have

A == or BA ==

[1 1] [-1 1] 2 2 1-1

AC == c== when A ^-=1=

"'-rY1l-nIa.·ta. answers to and are contained in the T",II""''II:TlI11,O"theorem.

Let Cbe nX n matrices. Then:

but -=1=

then == C.

A == nand AB ==

A== n, then

as well as ^-=1= then rank

is so are BA and

from the left

The determinant of a matrix ~-r£"',.ri~~,n1" AB or BA can be written as the of the

determinants of the and it is that det AB == det -=1=

in The formula is needed and can be obtained

Gauss-Jordan elimination 1) and from the theorem

For any n X n matrices

以前就教過的性質！

奇異，表示det(A)=0

CHAP. 7 Linearr-\LC:::C:UICIL.I\JI-:lI1"'Ii"'lrA!~C' Determinants. Linear ...,"""'1-...

and reduces to 0 == 0

~.1.U.f=.,'V.L.1.U.1.matrix

A

^{== [ajk]}

Theorem 1 in

reversal in row when

with the same effect on det Theorem 3(c), and

If or is so are

Theorem 3 in Sec. 7.7.

Now let and be nonsmzutar.

Gauss-Jordan Sec. 7.7, (a) and

But the same n.r\t::lI1"''=lir1Al''lC

Hence it remains to prove

o o o

⁰

We now take the determinant det the first row, from the

det

A

because for det and the

On the we can take out a factor from the nth. But this ~..,..n.ri" ....n1-

remammgdeterminant is det the same idea.

from

This our discussion of linear Section 7.9 on vector

spaces and linear transformations is Numeric metnoas are discussed in Sees, which are of other sections on numerics.

0 2 3

Gauss-Jordan (or if

0 5 6

[-: =;J

^{[ cos2()}

sin 8 9

2. 1 1 2

-sin20 cos '2 3 3

0 -0.2 0.75 0 0 0 5 0 ²_{3' 3}¹

3. 0.4 2 0 0.25 0 0 0 ²3'

0 0 8 0.80 0 0 0

0 0 -4 0 0

for

-2 0 0 8 13 in Probe1.

5 -4 0 3 5 Prove the formula in Probe 1.