m horizontal rows and n vertical columns is called an m n ... ongoing... · d pro 5 ma 3 ma × 3 m...

Ch 06 matrices v29 of 54 www.utstat.utoronto.ca/sharp

Data Projector Frameworks: MATA33 W11 Sharp

(partly stolen from Haeussler et al 13e Ch 6)

Matrix ‘size’

• A matrix consisting of m horizontal rows and n vertical columns is called an m×n matrix or a matrix of size m×n.

• For the entry aij, we call i the row subscript

and j the column subscript.

mnmm

n

n

aaa

aaa

aaa

...

......

......

......

...

...

21

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11211

A

b

c

d

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Equality

Matrices A = [aij

] and B = [bij] are equal if

they have the same size and aij

= bij for each i

and j.


Transpose of a Matrix

A transpose matrix is denoted by AT

Roughly, rotate it around its diagonal if it had one. Note that (AT)T = A

654

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Properties of Scalar Multiplication:

k(A+B) =kA + kB

(k+l)A=kA+lA k(lA)=(kl)A

0A=0 k0=0 (A+B)T = AT +BT

(kA)T= k AT


Matrix Subtraction:

(a)

(b)

13

08

84

3203

1144

2662

30

14

26

23

14

62

52

80

42

66

10

262BAT


Matrix of 'size' 2 by 4 3 2 5 6

4 3 7 1

Matrix of 'size' 2 by 3 3 2 6

4 3 1

Maybe this 2 by 3 matrix indicates games of soccer and of chess played by

Rob, Sal and Ted last week:

Rob Sal Ted

Soccer 3 2 6

Chess 4 3 1

Call the matrix A. If they play the same number of games 4 weeks in a row,

then total for the short month is:

12 8 24

16 12 4

We have multiplied the matrix A by the scalar 4 in the obvous way. Note 4A=A4 (commutative

Let's do a matrix addition of a fifth week's data, B, also in the obvious way:

2 2 3 12 8 24

3 1 3 16 12 4

14 10 27

19 13 7

Working element by element, since x+y=y+x for scalars, then matrix addition is commutative:

Also, since( x +y)+z=x+ (y+z) for scalars, then matrix addition also is associative

)

( )

( )

( )

+

=A

(

)

( )

( )

B+A= A+B

(A+B)+C= A+(B+C)

4A=

B+4A=

= (

M

•

Siz

A

B

AB C D CD

ijc

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Matrix

• AB

give

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= 3 ×

= 5 ×

B = 3

= 3 ×= 7 ×D = u

n

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atrix

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atrix

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Matrix by matrix multiplication examples

2 3 3 2 5 6 18 13 31 15

0 1 4 3 7 1 4 3 7 1

2*6+3*1=15

2 2 ‐2 3 3 2 5 2 13 0 36 11

0 1 ‐3 1 4 3 7 4 ‐1 2 18 ‐1

2 ‐1 ‐3 2

1 ‐4 2 1

5 0 0 0 ‐1 0.2 0.0 0.0 0.0

0 2 0 0 = 0.0 0.5 0.0 0.0

0 0 ‐1 0 0.0 0.0 ‐1.0 0.0

0 0 0 1 0.0 0.0 0.0 1.0)

( )( )

( ) (

( )* =

( )) )( (* =


More matrix products

a)

b)

c)

d)

32

6

5

4

321

183

122

61

61

3

2

1

1047

0110

11316

212

315

201

401

122

031

2222122121221121

2212121121121111

2221

1211

2221

1211

babababa

babababa

bb

bb

aa

aa


Example: Cost Vector

Given the price and the quantities, calculate the total cost.

Solution:

The cost vector is

432P

C of units

B of units

A of units

11

5

7

Q

73

11

5

7

432

PQ


Example: Matrix Mult is Associative

compute ABC in two ways

Associative: A(BC)=(AB)C

11

20

01

211

103

43

21CBA

1318

94

43

12

43

21

11

20

01

211

103

43

21BCA

1318

94

11

20

01

5413

521

11

20

01

211

103

43

21CAB


Example: Raw materials and cost:

Quantities of each of 3 house types:

Use by each type of 5 raw materials:

Price of products:

Materials cost for 1 copy:

Materials cost for the quantities produced:

1275Q

1500

150

800

1200

2500

C

71650

81550

75850

1500

150

800

1200

2500

1358256

21912187

17716205

RC

900,809,1

71650

81550

75850

1275

RCQQRC

1358256

21912187

17716205

R


Examples using I and 0 matrices:

00

00

10

01

41

23

103

101

51

52

OIBA

31

22

41

23

10

01 a.

AI

63

63

20

02

41

233

10

012

41

23323 b. IA

OAO

00

00

41

23 c.

IAB

10

01

41

23 d.

103

101

51

52


7

4

38

52

2

1

x

x

BAX

Matrix form of linear equations:

Can be written:

738

452

21

21

xx

xx

1. 2. 3.

a

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Properties of a reduced matrix

• All zero-rows at the bottom. • For each nonzero-row, leading entry is 1 and

the rest in that column are zeros. • Leading entry in each row is to the right of the

leading entry in any row above it.


Reduced matrices: Not red (3 sb 1) Reduced Not reduced

Reduced Not reduced Reduced

0000

2100

3010

f.

010

000

001

e. 000

000d.

01

10c.

010

001b.

30

01 a.


Evil Nerd 1: he likes to make things complex

He knows that x=3 y=5 but he likes to baffle classmates so instead he says: x =3 2x +y =11 or even better 3x + y = 14 4x+2y=22

Classmates say ‘go away to Evil Nerd’ and disentangle it:

3 1 | 144 2 | 22 (start, now R2 /2 to give:)

Typeequationhere.

3 1 | 142 1 | 11 (now R1 – R2 to give:)


1 0 | 32 1 | 11 (now R2 – 2R1 to give)

1 0 | 30 1 | 5 (hence x=3, y=5, ha ha nerd)

So

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Evil Nerd 2 x=3 y=2 z=4 Above is what he knows, but he makes it complex: x+2y=7 (adding 2*(y=2)) y=2 z=4 x+2y+z=11 (adding (z=4)) x+3y+0z=9 x+3y+z=13 x+2y+z=1 2x+5y+z=20 x+3y+z=13


Decode Evil Nerd

1 2 1 112 5 1 201 3 1 13

R1 swap R22 5 1 201 2 1 111 3 1 13

R1-R21 3 0 91 2 1 111 3 1 13

R2-R11 3 0 90 -1 1 21 3 1 13

R2*(-1)1 3 0 90 1 -1 -21 3 1 13

R3-R11 3 0 90 1 -1 -20 0 1 4

R2+R31 3 0 90 1 0 20 0 1 4

R1-3R21 0 0 30 1 0 20 0 1 4

x=3, y=1, z=4: Nerd foiled


Section 6.4, Example 3 Matrix Reduction

Example 3

2 3 -12 1 51 1 1

R1-3*R3-1 0 -42 1 51 1 1

(-1)*R11 0 42 1 51 1 1

R2-2*R11 0 40 1 -31 1 1

R3-R1-R21 0 40 1 -30 0 0

x=4y=-3

1

52

132

yx

yx

yx


Section 6.4, Example 5 Parametric Form of Solution

1 3/2 1 3 50 1 2 1 20 -9/2 -6 -3 -6

-(3/2)R2+R11 0 -2 3/2 20 1 2 1 20 -9/2 -6 -3 -6

(9/2)R2+R31 0 -2 3/2 20 1 2 1 20 0 3 3/2 3

R3/31 0 -2 3/2 20 1 2 1 20 0 1 1/2 1

-2R3+R21 0 -2 3/2 20 1 0 0 00 0 1 1/2 1

2R3+R11 0 0 5/2 40 1 0 0 00 0 1 1/2 1

9633

22

06232

431

432

4321

xxx

xxx

xxxx


Section 6.4, Example 5 Parametric Form (Cont)

Where x4 (or r= x4 looks neater) is any real value x1 = 4 – (5/2) r x2 = 0 x3 = 1 – (1/2) r x4 = 0 + r and let r vary.

x1 4 -5/2

x2 = 0 +r 0

x3 1 -1/2

x4 0 1

This is a line in 4-space, going through (4,0,1,0)T

421

3

2

425

1

1

0

4

xx

x

xx


Two Parameter Families of Solutions (Ch 6.5)

These are found when e.g. we use matrix reduction and end up with two free variables, and can express the other variables in terms of them. It makes more sense with an example:


Probs06.05Q06: Solve the set: w + x+ y +2z =4 2w+ x +2y +2z =7 w+ 2x+ y+ 4z = 5 3w -2x + 3y -4z = 7 4w- 3x + 4y -6z = 9

1 1 1 2 4

2 1 2 2 7

1 2 1 4 5

3 ‐2 3 ‐4 7

4 ‐3 4 ‐6 9

1 1 1 2 4

0 ‐1 0 ‐2 ‐1

0 1 0 2 1

0 ‐5 0 ‐10 ‐5

0 ‐7 0 ‐14 ‐7

1 1 1 2 4

0 1 0 2 1

0 1 0 2 1

0 ‐5 0 ‐10 ‐5

0 ‐7 0 ‐14 ‐7

1 0 1 0 3

0 1 0 2 1

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

Last matrix above is the unique reduced form


Probs06.05Q06 (cont)

The rows of zeros mean that y, z can be regarded as taking all real values-it is traditional to say y=r, z=s: w=-r+3 x=-2s +1 y=r z=s

w 3 -1 0

x = 1 +r 0 +s -2y 0 1 0z 0 0 1

Letting r and s vary, this is a plane in 4-space.


Homogeneous Systems

Is called homogeneous if c1 =c2 =….=cm =0 (all the c=0). Reduce it, say that k non-zero rows survive.

If homogeneous then If k<n there’s infinite solutions If k=n there’s a unique solution Unique but trivial: xi =0 for all i

o Example of trivial: x+y=0 3x+y=0 has unique trivial solution x=0, y=0

o Example of infinite: x+y=0 3x+3y=0 has infinite solutions: line y=-x/3

mnmnmm

nn

cxaxaxa

cxaxaxa

...

.

.

.

.

...

2211

11212111


Homogenous: Infinite Solutions or all-zero solution? Probs06.05Q20: Solve the homogeneous set: 2x+ y +z =0 x -y +2z =0 x+ y+ z = 0


2 1 1 0

1 ‐1 2 0

1 1 1 0

Drop last column: all zeros

2 1 1

1 ‐1 2

1 1 1

1 0 0

1 ‐1 21 1 1

1 0 0

0 ‐1 2

0 1 1

1 0 0

0 1 ‐2

0 1 1

1 0 0

0 1 ‐2

0 0 3

1 0 0

0 1 ‐2

0 0 1

1 0 0

0 1 0

0 0 1 Last matrix above is the unique reduced form. k=n=3.It gives us x=0, y=0, z=0: the unique trivial solution.


Homogenous: Infinite Solutions or all-zero solution? Probs06.05Q20 changed a bit: Solve the homogeneous set: 2x+ y +z =0 x -y +2z =0 x + z= 0 (was x+ y+ z = 0 previously) You can maybe see that adding the first and second equations gives the third (time 3). That is, the third equation gives no more info, so we have two equations, three unknowns. But let’s check by reducing it:


2 1 1 0

1 ‐1 2 0

1 0 1 0

Drop last column: all zeros

2 1 1

1 ‐1 2

1 0 1

1 1 0

1 ‐1 21 0 1

1 1 0

0 ‐1 1

1 0 1

1 0 1

0 1 ‐1

1 0 1

1 0 1

0 1 ‐1

0 0 0

1 0 1

0 1 ‐1

0 0 0

Last matrix above is the unique reduced form. k=2, n=3. Setting r=z x=-r y=r z=r


(6.6) Matrix Inverse to solve equation

Say we have a system of equations Ax = B

Here’ some logic:

Matrix inverse A-1 exists =>

A-1 exists such that A-1 A =I =>

x=A-1 B , a unique solution But we already know that in trying to solve for x, there might be 0, 1 or an infinite number of solutions. So the logic above says that a matrix inverse can only exist if there is just 1 solution. It turns out that this can only happen if A is a square matrix.


Matrices A has an inverse?

So only square matrices can have inverses, and not all of them do. But if a square matrix A does have an inverse A-1 then:

A-1 also is square A-1 is unique A-1 is the same size as A A-1 commutes with A: A-1 A = I = AA-1


Example 1: Inverse of Matrix

So C is the inverse of A, A is ‘invertible’.

Which was expected since a matrix commutes with its inverse. AC=CA=I

73

21A

13

27C

ICA

10

01

73

21

13

27

IAC

10

01

13

27

73

21


Finding Inverse by Reduction

[A|I] => => => [R|B]

Where R is the reduced form of A

If R=I, then A is invertible and A-1 =B (we did it‼) If R≠I, then A is not invertible (no-one can do it!)

Example 3: Inversion by Reduction

Can we invert A? Can we reverse the changes that Evil Nerd made? Turns out we can:

So we have

22

01A

10

01

22

01IA

12

01

20

01

2/11

01

10

01

2/11

011A


Example 5 – Use Inverse to Solve a System

1 0 ‐2 1 0 0

4 ‐2 1 0 1 0

1 2 ‐10 0 0 1

1 0 ‐2 1 0 0

0 ‐2 9 ‐4 1 0

0 2 ‐8 ‐1 0 1

1 0 ‐2 1 0 0

0 ‐2 9 ‐4 1 0

0 0 1 ‐5 1 1

1 0 0 ‐9 2 2

0 ‐2 9 ‐4 1 0

0 0 1 ‐5 1 1

1 0 0 ‐9 2 2

0 ‐2 0 41 ‐8 ‐9

0 0 1 ‐5 1 1

1 0 0 ‐9 2 2

0 1 0 ‐20.5 4 4.5

0 0 1 ‐5 1 1

1 3

1 2 3

1 2 3

2 1

4 2 2

2 10 1

x x

x x x

x x x

1021

124

201

A


Example 5 –Use Inverse to Solve a System(cont)

X=A-1 B

115

4

229

29

2411A

4

17

7

1

2

1

115

4

229

29

241

3

2

1

x

x

x


Two easy inverses:

10 1

10 1

c

b

a

D

00

00

00

c

b

a

D

/100

0/10

00/11


Remember about matrices

A is symmetric if AT = A o needs A T ij = Aij i,j o hence needs Akl = Alk k, l.

Homogeneous systems Ax = 0,

n unknowns m equations k=number of nonzero rows in reduced coeff. mat If kn: (always true if m n)

o Infinity solutions, including trivial xi =0 i If k=n: one unique solution, the trivial xi =0 i

Matrix Inverse

A-1 is unique if it exists at all A-1 commutes with A: that is A-1 A= I= A A-1 If A-1 and B-1 exist, then (AB)-1 = B-1 A-1 If A-1 exists then (AT)-1 = (A-1) T

Ch 06 matrices v2

(d) S

Com-9 = 5k2 =

29

MA

Solutio

mmute i6-3k2

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ATA3

on

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o

33 Fin

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MATA33 Final Summer 2009 LAQ3a Let k be an arbitrary real constant and let x and y be variables. Find all solutions to the system [6 points] kx + 2y = x x - ky = y

Solution (k-1) x + 2y =0 x -(k+1)y=0 Fast: Can e.g. substitute second equation in first and get (k-1)(k+1)y + 2y=0=(k2+1)y, hence y=0, x=0 Slow: Reduce:

1 21 1

1 11 2

1 10 2 1 1

1 10 1

1 00 1

So only the trivial solution x=0, y=0 for all k Hence homogeneous, with trivial solution x=0, y=0


MATA33 Final Summer 2009 LAQ3b

What equation must real constants a, b, and c satisfy in order that the system of linear equations x + 3y + z = a 2x - y - z = b 4x + 5y + z = c has infinitely many solutions ? [6 points] Solution Can only have infinitely many solutions if number of non-zero rows in reduced matrix, k <3=n, number of variables.

1 3 12 1 14 5 1

1 3 10 7 30 7 3

2 4

So if the second and third row are the same, we have two equations in three unknowns, infinite solutions, which needs: b-2a=c-4a or 2a +b-c=0

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MATA33 Final Summer 2009 LAQ8

In all of this question let P be an arbitrary n × n matrix where n ≥ 2. (a) Assume P is invertible and prove mathematically that the matrix equation PX = Q has a solution for every n × 1 matrix Q. [3 points] Solution: P invertible => P-1 exists => P-1PX= P-1 Q => X= P-1 Q (b) Assume the matrix equation PX = B has a solution for each n×1 matrix B and prove mathematically that P is invertible. (Note: parts (a) and (b) are independent of each other (they are not related.)) [6 points] Let Mi be the n×1 matrix with all zeros except a 1 in row i. Let Si be the n×1 matrix which solves PX = Mi, so P Si= Mi

Let Q be the n×n matrix with columns the Si

Solution: Q= (S1, S2, …..... Sn) PQ = P(S1, S2, …..... Sn) = (M1, M2, ……..Mn) = I Hence Q is an inverse to P and P is invertible.

m horizontal rows and n vertical columns is called an m n ... ongoing... · d pro 5 ma 3 ma × 3 m...

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Transcript of m horizontal rows and n vertical columns is called an m n ... ongoing... · d pro 5 ma 3 ma × 3 m...