整理方式：

1 1 1

2 2 2

3 1 2 1 2

3 3 4

5 4 3

0 0 0

0 0 0

2 2 0 0

0 0 0

0 0 0

x c c

x c c

x c c c c

c c x

x c c

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ = − − ⎥ ⎢ = − ⎥ ⎢ + − ⎥ ⎢ ⎥ ⎢ ⎥ + +

⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥ ⎢ ⎣ ⎥ ⎢ ⎦ ⎣ ⎥ ⎢ ⎥ ⎢ ⎥ ⎦ ⎣ ⎦ ⎣ ⎦

⎣ ⎦ ⎣ ⎦

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)

(-1)

^i+j

a

ij

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)

(-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

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

+ +

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

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.

｜A｜=｜A

^T

｜

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.

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

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

行列式常必須先進行列運算化簡，之後再展 開...可簡化問題的複雜度

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

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

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！

Gauss消去法

Gauss-Jordan 消去法

如何求出反矩陣，

是解題重點！

Gauss消去法

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

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

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！

在文檔中 data-think linear (頁 48-69)