矩陣乘法補充：齊

CHAP. 7 Linear Determinants. Linear ...^,/c...Oi""Y"1lC'

Now suppose further that the transformation, say,

<")-:"iV:"iICIII is related to a WIw2-system by another linear

(7) x= [::]

Then the is related to the we wish to express this relation

a linear too, say,

lAln·- ... \.f .... It-".111I indirectly via the xIx2-system, and

Substitution will show that direct relation is

y==

we obtain

+ +

YI == + + +

Y2==

+ +

+ + +

with we see that

+ +

+ +

This proves that C == AB with the defined as in For matrix sizes idea and result are the same. the number of variables cnanzes, We then have m variables Y and n variables x andp variables w. The matrices and C == then

have sizes m X n, n X p, and m X p, And that C be the

"f"....n'rllIlI.ni"AB leads to formula in its form.This motivates matrix muuuiucauon.

transpose of a matrix its rows as columns its

columns as This also to thet"r<:llnCll""\r\('Aof vectors. a row vector becomes a column vector and vice versa. In for square we can "reflect"

the elements the main entries that are svmmetncauv

"f"r\(:"lI1"lI,n.nc.hrlwith to the main to obtain the HenceaI2 becomes

a21, a3Ibecomes a13,and so forth. 7 illustrates these ideas. Also note if A is the then we denote itst"r<"lnC1l1'"'\r\('O

If

-8

o ^then

x1, x2系統與w1, w2

系統的關係

代入

y1, y2系統與w1, w2

系統的關係

y= Ax = ABw=Cw，即C=AB (矩陣相乘 可應用於不同系統之間的轉換)

(7)式代入(6)式中

x1 x2

c11 c12

c21 c22

x1 x2

整理成(8) 式做係數 亦即C=AB 比較

矩陣轉置

A → A

(行、列位置互換)

矩陣乘法補充：齊

次座標系統的轉換

SEC. 7.2 Matrix ^{1\.11 - ...}

A little more compactly, we can write

[:

-8

o

Furthermore, the transpose [6 2 3JTof the row vector [6 2 3] is the column vector

[:l

...LU..L..Lk)J-1'Vk)v of an m X n matrix A == [ajk] is the n X m matrix A

that has first rowofA as its first the second row of as its second\...-U£'VH'''lr,,~and so on. Thus the ofA in is == [akj], written out

Asa case, ....~n.'nCl..-..~Cl-... n ...converts row vectors to column vectors and^{0A111l ...}rt:::..rC't::l>.I...T

us a choice in that we can work either with the matrix or its

L.1.UJL.l0I--fV0I\,.1qwhichever is more convenient.

Rules for are

Note that in the matrices are reversed the proofs as an exercise in Probs. 9 and 10.

We leave

Certain kinds of matrices will occur

most ones of them.

....rO.n111£:..nt"ll ...rin our and we now list the

~"lnnl1r1il1ptrj~ and ~Ke~T..Svmmetrrc Matrices, rise to two useful

classes of matrices. matrices are square matrices whose transpose equals the

行向量 列向量

A

A → A

(行、列位置互換)

研究所考過證明 (交大機械) (見補充資料)

｜A｜=｜A

｜

特殊矩陣

對稱矩陣(R) 反對稱矩陣(S) 注意下標的差異

CHAP. 7 Linear Matrices, Vectors, Determinants. Linear "'\fC~T~lrnc:

matrix itself. matrices are square matrices whose the matrix. Both cases are defined in (11) and illustrated

[ 20 A= 120 200

120 10 150

200]

150 30

is symmetric, and

=[-; ~ ^=:]

For instance, if a company has three building supply centers C1,C2 ,C3 ,then A could show costs, say,^ajjfor handling 1000 bags of cement at center and^ajk(j =/=k)the cost of shipping 1000 bags from CjtoCi:Clearly,

ajk = akjif we assume shipping in the opposite direction will cost the same.

Symmetric matrices have several general properties which make them important. This will be seen as we proceed.

ipper trtangularmatricesare square matrices that can have nonzero diagonal, whereas any below the must be trtangutarll1nl':lIf"1"'U£ll.o~can have nonzero entries on and the

on the mainU-lU,F,VII..lU.lof a matrix may be zero not

-1 6

3 9 -3

a

9 2 3

lJUU!Onal Matrices.. These are square matrices that can have nonzero entries on

the main above or below the main must be zero.

If all the ^{rl1r:l1n A nr:II} entries of a matrix S are say, c. we call a because of any square matrix of the same size S has the same

effect as the a that

AS == SA == cA.

whose entries on the mainriI ...,~r..·rlI"'"^I are all 1, is called a

11"16:1I"tjt'7 ....c,.iI-,..,.." ...T\and is denoted or For formula becomes

== A.

注意特徵！

三角矩陣

上三角矩陣 下三角矩陣

對角矩陣

只有對角線 非全為0

純量矩陣 單位矩陣 S=c I

=c IA 交換律成立 A為任一矩陣，其為一對

稱矩陣與反對稱矩陣的和

(見補充資料)

SEC.7.2 MatrixI\JII.I ...i"'lil,.."'ll ... i"'...

Supercomp Ltd produces two computer models PC1086 and PC1186. The matrix A shows the cost per computer (in thousands of dollars) and B the production figures for the year 2010 (in multiples of 10,000 units.) Find a matrix C that shows the shareholders the cost per quarter (in millions of dollars) for raw material, labor, and miscellaneous.

Quarter

PC1086 PCl186 2 3 4

[1.2 1.6] Raw Components

B= [ : ^{8 6}

~]

A= 0.3 0.4 Labor

2 4 PCl186

0.5 0.6 Miscellaneous

1 2 3 4

[13.2

15.6]

C=AB= 3.3 3.2 3.4 3.9 Labor

5.1 5.2 5.4 6.3 Miscellaneous

Since cost is given in multiples of $1000 and production in multiples of 10,000 units, the entries of Care multiples of $10 millions; thusell = 13.2 means $132 million, etc.

Suppose that in a weight-watching program, a person of 185 lb burns 350 cal/hr in walking (3 mph), 500 in bicycling (13 mph), and 950 in jogging (5.5 mph). Bill, weighing 185 lb, plans to exercise according to the matrix shown. Verify the calculations (W=Walking, B= Bicycling, J=Jogging).

W B J

MON 1.0 0 0.5 825 MON

[350]

WED 1.0 1.0 0.5 1325 WED

500 =

FRI 1.5 0 0.5 1000 FRI

950

SAT 2.0 1.5 1.0 2400 SAT

Suppose that the 2004 state of land use in a city of 60 mi²of built-up area is

C: Commercially Used 25% I: Industrially Used 20% R: Residentially Used 55%.

Find the states in 2009, 2014, and 2019, assuming that the transition probabilities for 5-year intervals are given by the matrix A and remain practically the same over the time considered.

FromC From I FromR

[0.7

~.2]

A= 0.2 0.9 To I

0.1 0 0.8 ToR

CHAP. 7 Linear Matrices, Vectors, Determinants. Linear

(d) To visualize a three-

dimensional object with plane faces (e.g., a cube), we may store the position vectors of the vertices with respect to a suitable^{XIX 2X}3-coordinate system (and a list of the connecting edges) and then obtain a two- dimensional image on a video screen by projecting the object onto a coordinate plane, for instance, onto the Xlx2-plane by setting ^X3 == O. To change the appearance of the image, we can a linear transformation on the position vectors stored. Show that a diagonal matrix with main diagonal entries3, 1,~ from an x == [Xj] the new vector y == Dx, where Yl == (stretch in the

by a factor 3),Y2 == ^{X2 Y3} == (con- traction in the What effect would a scalar matrix have?

Rotations space. Explainy ==Ax geometrically whenA is one of the three matrices

[:

0 0

cos () -sin () sin () cos ()

cos<.p 0 -sin^<.p cosljJ -sinljJ 0

0 0 sinljJ cosljJ o .

sin<.p 0 cos<.p 0 0

What effect would these transformations have in situations such as that described in

We now come to one of the most use of that matrices to

solve of linear We showed in 1 of Sec. how

to the information contained in a of linear a called

the matrix. This matrix will then be used in the linear of

""""-'I •.-...""JL'-J'JUlU.Our to linear is called the Gauss elimination method.

Since this method is so fundamental to linear the student should be alert.

A shorter term for of linear Linear ~VI~rp1TI~

model many and many other areas.

Electrical and markets may serve as

of<:l1"'\1nl1,0<:lt·1n.l'lC

A linear the form

of n ^rmunnwns Xl, ...^,X_n is a set of a.ntll"Jl"Jl ...·..~lI'.,.C' of

is called linear because each variable xj appears in the first power

a.ntlll"Jl"t·l~nof a line. all, ... ,a_{m n}are called the coefficients

of the b_{l , ... ,}b_m on the are also numbers. If all theb_j are zero, then (1) is called a If at least one bj is not zero, then is called a

高斯消去法

擴大矩陣

(1) n個未知數 (2) m個方程式

齊次系統 (bi=0) 非齊次系統

(bi≠0)

係數 必須先按照順序排

好，缺項補為0

SEC. 7.3 Linear''If:OT"O''''''''f:O of1-1"'1111 "'I>TI.,...I"\("' Gauss Elimination

A solution of (1) is a set of numbers Xl, ...,Xn that satisfies all the m equations.

A solution vector of (1) is a vector x whose components form a solution of (1). If the system (1) is homogeneous, it always has at least the trivial solutionxI == 0, ...,Xn == 0.

Matrix Form of Linear From the definition of matrix ~1"IIt-",,,,I^..,,,,..,.t-·.~,....

we see that the mt3r1111<:l-t1,n.nC' of (1) may be written as a single vector ....,.... '...IALJL'U'..I...I.

where the coefficient A == [ajkJ is them X n matrix

and and b ==

are column vectors. We assume that the coefficientsajk are not all zero, so that A is

not a zero matrix. Note that has n whereas The

matrix

A==

a_{m n}

is called the The vertical line could be

V..l..L..L.Ll-l-'...\.-D., as we shall do later. It is a reminder that the last column ofAdid not

come from matrix A but came from vector the matrix A.

Note that the A ^{because it}

contains all the

If^m = n= 2,we have two equations in two unknowns ^{Xl, X2}

If we interpretXl, X2as coordinates in the Xlx2-plane, then each of the two equations represents a straight line, and^{(Xl, X2)}is a solution if and only if the pointP with coordinates^{Xl, x2}lies on both lines. Hence there are three possible cases (see Fig. 158 on next page):

(a) Precisely one solutionifthe lines intersect (b) Infinitely many solutions if the lines coincide (c) No solutionifthe lines are parallel

恆零解→齊次系統獨有

變成AX=C矩陣形式 (一定要會的技巧！)

增廣矩陣 增大矩陣 擴大矩陣

三種解的狀況，及 其對應的幾何情況

Matrices, Vectors, Determinants. Linear " ...

For instance,

Xl+X2=1 2xI -x2=0 Case (a)

Xl+x2=1 2x I+2x2=2

Case (b)

Xl+x₂= 1

xl+x₂=0 Case (c) Unique solution

Infi nitely many solutions

p

the system is homogenous, Case (c) cannot happen, because then those two straight lines pass through the origin, whose coordinates (0, 0) constitute the trivial solution. Sirnilarly, our present discussion can be extended from two equations in two unknowns to three equations in three unknowns. We give the geometric interpretation of three possible cases concerning solutions in Fig. 158. Instead of straight lines we have planes and the solution depends on the positioning of these planes in space relative to each other. The student may wish to come with some specific examples.

5 3

Its -:lH4:TtTIpntpr! matrix is

2 -30.

+ +

2 -26

may have solution. This leads to such have solution? Under what conditions does it have

nr~:l>01lCP."uone solution? it has more than one how can we characterize the set

of all solutions ? We shall consider such in Sec. 7.5.

hr"""'1a"1Cl>1I'" let us discuss method for linear~V;"~I.LliI~.

The Gauss elimination method can be motivated as follows. Consider a linear that

is upper such as

0A-r-rpc~n{"\nti1In^0- coefficient matrix lie

"' ... ""' ...~... ! Then we can solve the

J.ULlIl-\./\.-IU(..IlU.'-J'J.J.for the x2 ==

and then work uU"",,,JL'1l.. V'V UJL'u.c substitutmzx2 == -2 into the first and

Aht'-::nnlno-Xl ==!^{(2 - ==} !^{(2 - 5 . == 6. This} us the idea of first-rarilllllf~lI1Y".n-

to form. For instance,let the be

No solution

We leave the first as is. We eliminate x1 from the second to For this we add twice the first and we do the same

恰有一解 (交於一點)

無解 (平行線) 無限多解

(重合線) 三個空間中的平面，

可能的交會狀況

反代換

高斯消去法的目的：化簡成三角矩陣形式

反代換求解

注意步驟！

(1)寫出增廣矩陣

SEC. 7.3 Linear .... ',.-_... .-... ....'-! ....\A''-,_, ,....Gauss Elimination

operation on therowsof the augmented matrix. This -30 + 2 . 2, that is,

+

+ 2

13x2== -26 Row +

[~

where Row

is theGauss 2Orn1<:11"11"'1""0

which back substitution now Since a linear

eumtnauon can be done

We do this in the next exampte,0rlt1I1t"\h<:101'7'11"'!ICTthe matrices

the OrIl111<:1 ....1I"'....\1:'behind in order not to track.

Solve the linear system

80.

(KVL). In any closed loop, the sum of all voltage drops equals the impressed This is the system for the unknown currents

Xl = is-X2 = i2 , X3 =i3in the electrical network in Fig. 159. To obtain it, we label the currents as shown, choosing directions arbitrarily; if a current will come out negative, this will simply mean that the current flows against the direction of our arrow. The current entering each battery will be the same as the current leaving it.

The equations for the currents result from Kirchhoff's laws:

Kirchhoff's Current Law (KCL). At any point of a circuit, the sum of the inflowing currents equals the sum of the outflowing currents.

Kirchhoff's Voltage electromotive force.

Node Pgives the first equation, node Q the second, the right loop the third, and the left loop the fourth, as indicated in the figure.

NodeP: -,- i₂+ i₃= 0

~^{90 V NodeQ: -i1+ i2- i3= 0}

80

Right loop: + =90

Left loop: 20i₁+10i₂ =80

Network in 2and the

Solution Gauss This system could be solved rather quickly by noticing its particular form. But this is not the point. The point is that the Gauss elimination is systematic and will work in general,

(2)進行高斯消去法，

變成三角矩陣形式 (進行Row運算) (4)反代換求解 (3)寫出化簡後

之方程式

(1)注意對齊性 (2)缺項補0

利用高斯消去法求解！

CHAP. 7 Linear Matrices, Determinants. Linear""V.~""""III'"

also for large systems. We apply it to our system and then do back substitution. As indicated, let us write the augmented matrix of the system first and then the system itself:

Augmented Matrix A Equations

-1 0 X2+ X3 = 0

-1 0 X2 - X3 = 0

10 25 90 10x2 +25x3 = 90

10 0 80 = 80.

Step1. Elimination ofXl

Call the first row of A thepivot rowand the first equation thepivot equation. Call the coefficient 1 of its x I-term thepivotin this step. Use this equation to eliminatex1(get rid ofx1)in the other equations. For this, do:

Add 1 times the pivot equation to the second equation.

Add -20 times the pivot equation to the fourth equation.

This corresponds torow operationson the augmented matrix as indicated in (3). So the operations are performed on thepreceding matrix. The result is

behind thenew matrix in

-1 0 Xl - X2+ X3 = 0

0 0 0 0 + 0= 0

(3)

0 10 25 90 10x2+25x3 = 90

0 30 -20 80 30X2 - 20X3 = 80.

Step 2. ofx2

The first equation remains as it is. We want the new second equation to serve as the next pivot equation. But since it has nox_2-term(in fact, it is 0= 0), we must first change the order of the equations and the corresponding rows of the new matrix. We put 0= 0 at the end and move the third equation and the fourth equation one place up. This is called partial pivoting (as opposed to the rarely used total pivoting, in which the order of the unknowns is also changed).Itgives

~O

o

-1

o

25 -20

o o

90 80

o

25x3= 90 20X3==80

0= O.

To eliminatex2, do:

Add - 3 times the pivot equation to the third equation.

The result is

o -95 -1

o 10

- 95x3= -190

o.

o

90

0=

3 3

o

90 -190

o o

25

o o o

(4)

Back Substitution. Determination of X3,X2,Xl(in this order)

Working backward from the last to the first equation of this "triangular" system (4), we can now readily find X3, then X2, and then Xl:

- 95x3=-190 90

o

X2 = 10(90 - 25x3) = iz = 4 [A]

Xl = X2 - X3 = il = 2 [A]

where A stands for "amperes." This is the answer to our problem. The solution is unique.

元素開頭為1的列，移至最上方 注意列運算的步驟！

注意兩邊的 對應關係

R14(-20) = R1 × (-20) + R4

R12(1) = R1 × (1) + R2

元素全為0的列

，移至最下方

R23(-3) = R2 × (-3) + R3

形成上△矩陣的原則：

(1)每一列中，第一個不為0 之項，所在之行皆不相同 (2)愈下方之列，第一個非0 項愈靠右

列出化簡後 之方程式

反代換求解

SEC. 7.3 Linear .... ,,._...,. of ...."-1...\,,4 . . . ..;p.Gauss Elimination

Example 2 illustrates the operations of the Gauss elimination. These are the first two of three operations, which are called

Interchange of two rows

Addition of a constant of one row to another row row anonzero constantc

These A-nt::l>'t",=,1rl A1''' CI are for rows,not columns!

ttarot-r-tar tv»r ra of twoeQurau~ons

Addition one eauation to another eauauon

anonzero constantc

addition because we can undo it muitmncalJO][1, which we can undo

ClAh,1lt-lIr.. 1I1Iset. Neither does their

"l1'"'nll'=''t''I'{T for their

c*

if can be elimination and

Because of this the same solution sets are often called

But note well that we are withrow No column

" ' ...L:l>~n1rll"'1t"CI on the matrix are in this context because would

the solution set.

A linear is than 1I,,""'IIT1I1Ir.."T7"''''''''

as in 2,determinedifm == n, as inJI_IL'l..IUl...L..L..LV.l.V

fewer than unknowns.

is calledconsistentif it has at least one solution one

solution or many but if has no solutions at as

Xl + ^X2 == 1,Xl +^X2 == 0 in 1, Case

We have seen, in 2, that Gauss elimination can solve linear that have a unique solution. This leaves us to Gauss elimination to a system with l f l ....1 I f l l f " a h T

many solutions (in Example 3) and one with no solution 4).

基本列運算 列等效系統

矩陣列運算

(聯立)方程式運算 (高中時代的做法) 請了解兩者的相似性！

過定(可能無解, m>n)

靜不定(無限多解, m<n) 靜定(唯一解, m=n)

唯一解、無窮多解、無解

CHAP. 7 Matrices, Vectors, Determinants. Linear ....\lC~TOrnC

Solve the following linear system of three equations in four unknowns whose augmented matrix is

,30O ^{2.0 2.0 -5.0 I} SOOJ ^+2.0X2+ 2.0X3 - 5.0X4= 8.0

I

(5) 0.6 1.5 1.5 -5.4 ^I 2.7 . Thus, + 1.5x2 + 1.5x3 - 5.4x4 = 2.7

I

L1.2 -0.3 -0.3 2.4 ^I 2.1 - 0.3X2 - 0.3X3+ 2Ax4=2,L

Solution. As in the previous example, we circle pivots and box terms of equations and corresponding entries to be eliminated. We indicate the operations in terms of equations and operate on both equations and matrices.

Step Elimination ofXl from the second and third equations by adding

= -0.2 times the first equation to the second equation, -1.2/3.0 = -0.4 times the first equation to the third equation.

This gives the following, in which the pivot of the next step is circled.

l:OO

2.0 2.0 -500: SOOJ ^3.0X1+2.0X2+2.0X3 - 5.0X4= 8.0

(6) 1.1 1.1 -4.4 ^I 1.1 Row 2 - 0.2 Row 1 1.1x3 - 4.4x4= 1.1

I

-1.1 -1.1 4.4 ^I-1.1 3 - 0.4 Row 1 - 1.1x3+4.4x4= -1.1.

Step 2.Ettmuuuion of X2 from the third equation of (6) by adding

= 1 times the second equation to the third equation.

This gives

l:oo

2.0 2.0 -5.0 ^I

SOOJ ^{3.0X1+2.0X2+2.0X3 - = 8.0}

I

(7) 1.1 1.1 -4.4 ^I 1.1 1.1x 2+ 1.1x3 4.4x4 = 1.1

I

0 0 0 ^I 0 Row 3+Row 2 0=0.

From the second equation, X2= 1 - X3+4X4'From this and the first equation, Xl = 2 - X4. Since X3 and X4 remain arbitrary, we have infinitely many solutions. If we choose a value of X3 and a value of x4, then the corresponding values of x1and x2are uniquely determined.

Notation. If unknowns remain arbitrary, it is also customary to denote them by other letters t1, t2, ....

In this example we may thus write Xl 2 - X4 = 2 - t2, X2 = 1 - X3+4X4= 1 - t1+4t2, X3 t1(first arbitrary unknown), X4= tz (second arbitrary unknown).

What will happen if we apply the Gauss elimination to a linear system that has no solution? The answer is that in this case the method will show this fact by producing a contradiction. For instance, consider

2

2 4

Step Elimination ofXlfrom the second and third equations by adding

- itimes the first equation to the second equation, - ~= -2 times the first equation to the third equation.

無窮多解

無解 x3、x4可為任意值

，故為無窮多解 (形成參數解) 令

x3=t1 x4=t2 故

x2=1-t1+4t2 x1=2-t2

代入

得x1=2-x4

R12(-0.2) = R1 × (-0.2) + R2

R13(-0.4) = R1 × (-0.4) + R3

R23(1) = R2 × (1) + R3

R12(-2/3) = R1 × (-2/3) + R2

^{R13(-2) = R1 × (-2) + R3}

SEC. 7.3 Linear ...~/..."iI""...ofB.."'I'''''1l,AI. 1'-'1 II...Gauss Elimination

This gives

[:

2

-:]

3Xl +^2X2+ ^X3 3

1 1

Row 2 - _~Row 1 lX3 = -2

-3 3

-2 2 Row3 - 2 Row 1 ^2X3 = O.

Step 2. of^X2from the third equation gives

[:

2^{1 1} -2

3]

-3 3

0 0 12 Row3 - 6 Row 2 0= 12.

The false statement 0= 12 shows that the system has no solution.

At the end of the Gauss elimination the form of the coefficient the augmented and the itself are called the row echelon In rows of zeros, if are the last rows, in each nonzero row, the leftmost nonzero is farther

to the than in the row. For in 4 the coefficient matrix

and itsauamenteu in row echelon form are 2

o

and

2 1

o ⁰

that the leftmost nonzero entries 1 since this would so-called reduced echelon in which those Note that we do not

no theoretic or numeric ~rI'\J~nt~^(JfP

entries are 1, will be discussed in Sec.

The of m in n unknowns has matrix

is to be row reduced to matrix I The two Ax == and == arepnlll1"'U"t;:lilp·nt"·

if either one has a so does the and the solutions are identical.

At the end of the Gauss elimination the back the row echelon form of the matrix will be

r¹¹ r₁₂ t,

[2

I, fr+l

fm

r ~ m, r11 -=I=- 0, and all entries in the blue and blue are zero.

The number of nonzero rows, r, in the row-reduced coefficient matrix is called the of and also the of A. Here is the method for whether Ax == b has solutions and what are:

No ris less than m l~on1n1no

all and at least one of thenrl1l'Y'l norC'

r>l't'~~r>II'l:Thas at least one row of

... ,fm is not zero, then the R23(-6) = R2 × (-6) + R3

矛盾，故無解 列階梯型式

如何從高斯消去法所得之列梯 形矩陣形式，來判斷滿足三種 解之條件！！

(見補充資料)

CHAP. 7 Linear .-...L l I L A .Matrices, Determinants. Linear ....\Ic:',.on"l~

Rx == f is inconsistent: No solution is possible. Therefore the system Ax == b is inconsistent as well. See Example 4, where r == 2< m == 3 and ==13 == 12.

If the system is consistent (eitherr == m,orr< m andall the numberslr+l,lr+2, ....i«

are zero), then there are solutions.

solution. If the system is consistent and r == n, there is exactly one

u'U' ... ..,... ''U'.I..I..which can be found back substitution. See 2, wherer == n == 3

andm == 4.

many solutions. To obtain any of these choose values of Then solve the rth for xr terms of those then the (r - forXr-l, and so on up the line. See

Gauss elimination is reasonable in time and demand.

We shall consider those in Sec. 20.1 in the on numeric linear _ . Section 7.4 fundamental of linear such as linear moenenuence and rank of a matrix. These in turn be used in Sec. 7.5 to characterize the

hAI'1It:l'{71Arof linear in terms of and of solutions.

o

15

o

5 -'-2 4

3 -6

-1 2 0 0

13.. lOx + 2z = 14

- 15x + y+ 2z = 0

w+ x+ Y = 6

8w - 5x + - 10z = 26

14. 2 3 -11 1

5 -2 5 -4 5

-1 3 ^~3 3

3 4 -7 2 -7

15. relation. definition, an eauivatence relationon a set is a relation three conditions:

(named asmdtcatcd)

(i) Each element A of the set is to itself or its

[ 3.0 -0.5 1.5 4.5

4 0 4

5 -3 2

-9 2 ^~1 5

6.. 4 -8 3

-1 2 -5

3 -6

8.. + 3z = 8

2x z=2

3x + =5

Solve the linear system matrix. Show details.

-3x + 8y = 5 8x ~ = -11

8y + 6z = -4 -2x + - 6z = 18

x+ y - z= 2

5.. 13 12 6

-4 7 73

4 5 11

7.. 4 0 6

-1 -1

2 -4

9.. - 2z = 8 1O..

[-1:

^17]

3x + 4y - 5z = 8 21 -9 50

y+ z - 2w = 0 2x - 3y - 3z + 6w = 2 4x + y+ z - 2w = 4

IfA is equivalent toB,then Bis toA (Symmetry).

IfA is equivalent to BandBis to C, then A is equivalent to C (Transitivity).

Show that row equivalence of matrices satisfies these three conditions. Hint. Show that for each of the three elementary row these conditions hold.

CHAP. 7 Linear Matrices, Determinants. LinearSvsterns

Apply E_{1 ,}E_{2 ,} to a vector and to a 4X 3 matrix of your choice. Find B = lI...:J:-{JiLJ")'liU I.cA.

the general 4 X2 matrix. Is

(b) Conclude that are obtained doing the corresponding eiementarv operations on the 4 X 4

unit matrix. Prove that is obtained from by an elementary row operation, then

M=EA,

where is obtained from the n X n unit matrix by the same rowoneratt.on.

Since our next characterize the behavior of linear in terms

of existence and of solutions we have to introduce new

fundamental linear <:ll llrAh,r'<:l10 0j~nr'Ant"that will aid us in so. Foremost among

these are and the mind that these

lIr1l1-·11!'Y1l'""lf"all"'{Tlinked with Gauss elimination method and how

Given any set ofm vectorsa(1), ... , aCm) the same number of0.n.lnB1'""Ir.nonTCI

COlmtnnatJ.onof these vectors is an.o'V1n1l""Cl.0 011r.n of form

a

where C1, C2, ... , Cm are any scalars. Now consider the v'4'....UI,.,Jl'J.I..1.

this vectorannllnt"-.r.rIl

If this is the

a(1), ... , aCm) are said to form a

holds if we choose all zero, because then it becomes of scalars for which then our vectors tnaenenaent set or, more we call them if also holds with scalars not all zero, we call these vectors This means that we can express at least one of the vectors as a linear combination of the other vectors. For if holds say,

C1 "* 0, we can solve fora(1):

a(1) == + ... + where kj ==

may be zero. Or even all of linear

uenenuent, then we can

ifa(1) ==

if a set of vectors is

rid of at least one or more of the vectors until we

a IlIrIlt::lln1l""Iumdenenuent set. This set is then the smallest essential" set with

which we can work. we cannot express any of the vectors, of this set, in terms of the others.

向量空間 線性獨立

線性獨立：

c1, c2, c3...cm全為0 線性相依：

c1, c2, c3...cm不全為0 線性組合

相依性：向量彼此間有特定的關係存在

a1,a2,a3...am可當 作基底(basis)

SEC. 7.4 Linear tnoeoenoence. Rank of a Matrix. Vector

The three vectors

a(1)= [3 0

a(2)= [-6 42

a(3)= [21 -21

2 2J

24 54J

o -15J are linearly dependent because

6a(1) -

Although this is easily checked by vector arithmetic (do it!), it is not so easy to discover. However, a systematic method for finding out about linear independence and dependence follows below.

The first two of the three vectors are linearly independent because^cla(1)+^c2a(2)= 0 implies^C2= 0 (from the second components) and then^Cl= 0 (from any other component of^a(1).

further discussion will show that the rank of a matrix is an11'Yl1·nn.1I~1"1JI1I'"\1"

1""\1l""",,1""\'::Jl1l"'i"1 OCl of matrices and linear C',,,~rpnlC'

The matrix

=

l~:

(2) 42 24 54

-21 0 -15

has rank 2, because Example 1 shows that the first two row vectors are linearly independent, whereas all three row vectors are linearly dependent.

Note further that rank A=0 if and only if A= O. This follows directly from the definition.

can be obtained from if

row-euutvatent to a matrix ctemcntarv row 'U'!-""""..L ... "..L'U'Juu-.>.

Now the maximum number of row vectors of a matrix does not if we the order of rows or a row a nonzero c or take a linear combination of a row to another row. This shows that rank is

in1liTQlri€1lntunder row /"'<1""\01l""""lf"lI,n.n,C'·

Hence we can determine the rank of a matrix the matrix to row-echelon form, as was done in Sec. 7.3. Once the matrix is in row-echelon we count the number of nonzero rows, which is the rank of the matrix.

如何找出係數呢？

到底a1,a2,a3誰是基底？

寫成矩陣型式

列運算後所得之列梯形矩陣，

非全為0的列數即為Rank

經過列運算後之矩陣，與原矩陣形成列等效矩 陣，彼此間有相同的Rank數

CHAP. 7 Linear Vectors, Determinants. Linear svstems

For the matrix in Example 2 we obtain successively

A=

1-:

L 21 -21 0 - 15....J

l

l

^5:

The last matrix is in row-echelon form and has two nonzero rows. Hence rank A= 2, as before.

... '<Tr'"""""'"f'...I""''' 1-3 illustrate the^{"tAlIAlUl1110"}useful theorem

the matrix ==

p == 3,n == 3, and the rank of

Consider p vectors that each have n components. Then these vectors are

inaepenaeniifthe matrix with these vectors as row vectors, has rank p.

these vectors are ifthat matrix has rank less than p.

Further111('1'nAlr1"r::ll"11"..,...·"..,,11"'.0.,'...1-.. ,.-,." will result from the basic

The rank r a matrix A column vectors

HenceAand its transpose

the maximum number

have the same rank.

In this we write "rows" and "columns" for row and column vectors. Let A be an m X nmatrix of rank A == r.Then definition A has r

rows which we denote v(1), ... , vCr) of their in and all the rows

3(1), ... , 3Cm) of are linear combinations of say,

(3)

3Cm) == Cm1V(1) +^Cm2v(2) +

進行列運算

R12(2) = R1 × (2) + R2

R13(-7) = R1 × (-7) + R3

R23(1/2) = R2 × (1/2) + R3

不全為0的列數=2

，故Rank A = 2 (1) 第1、2列向量線性獨立，可當 作基底

(2) 兩個基底，故可組成2維向量空間

Rank (A) = Rank (A^T) 行運算後所得之行梯形矩陣，

非全為0的行數即為Rank Rank (A)=p，則x(1), x(2),…x(p)線性獨立 Rank (A)<p，則x(1), x(2),…x(p)線性相依