MA122 MidtermRosanna Lo - Coordinator

Shawn Lucas – TutorGreg Krikorian - Tutor

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About Us

• SOS started in 2004• Raise money for children in Latin

America• Many volunteer opportunities• Check out our website:http://www.studentsofferingsupport.ca/• Video!

Shawn Lucas

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Program: BBA w/Math & Comp Sci MinorYear: SecondHometown: WaterdownFavourite Team: Blue Bots (Obviously..)Qualifications: Taken all first year math courses; A+ in allLove: Corny Math Jokes

Greg Krikorian

• 2nd year BBA/Financial Math• From

Kingston/Pittsburgh/Vancouver/Ottawa• I.A. for MA129• Favourite Hockey Team is the Penguins

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About this Session

• Welcome to Linear AlgeBRO• Listen!• Participate!• Ask questions!• Yell at Us System

• What does little mermaid wear?– Algae-bra

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Agenda

• Chapter 1 – Systems of Linear Equations and Matrices– Linear Systems– Gaussian Elimination– Matrices & Matrix Operations– Algebraic Properties of Matrices– Elementary Matrices & Inverse– Matrix Inverses & Linear System

• Chapter 2 – Determinants– Cofactor Expansion of Determinants– Evaluating Determinants– Properties of Determinants

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• Chapter 3 – Euclidean Vector Spaces– Vectors in 2-, 3-, n-space– Norm, Dot Product, & Distance in

• Study Tips

nR

Chapter 1:Systems of Linear Equations

and Matrices

Linear Systems of Equations

• What is a Linear Equation?– of the form

• What is a Linear System?– two or more linear equations using the same

variables – solution must satisfy each equation in the system

bxaxaxa nn ...2211

How do we Solve?

• How do we solve a Linear System?– Can use substitution & elimination– Can draw a graph– Or..– Augmented Matrices!

11

Matrix Forms

• Row Echelon Form (REF)– Leading 1s– Leading 1 in an upper row must be to the

left of the leading 1 in the row below it• Once we get into this form, can use back

substitution

11

12

Matrix Forms (ctd.)

• Reduced Row Echelon Form (RREF)– In row echelon form and– All entries except leading 1s are zero

• Once we get into this form, can use Gauss-Jordan Elimination

12

Using Matrices to Solve

• Elementary Row Operations– Multiply a row by any nonzero constant– Interchange any two rows– Replace a row by itself plus a multiple (+ or -) of

another• Caution: DO NOT combine the first and last; it’s a shit

storm

• Gaussian Elimination– Step 1: Get leading 1 in first row

– Step 2: Get zeroes for all entries below the first leading 1

– Step 3: Repeat Process with second row, and so on

Using Matrices to Solve (ctd.)

• Gaussian Elimination– Step 1: Get leading 1 in first row

– Step 2: Get zeroes for all entries below the first leading 1

– Step 3: Repeat Process with second row, and so on

• Now we have REF, but we need RREF:– Use EROs to introduce zeroes above leading 1s

• ERO = Elementary Row Operations

Example

• Solve the system of equations

15

1523 321 xxx

0235 321 xxx

1133 321 xxx

30246 321 xxx

Solution

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Step 1: Get leading 1 in first row

13

1R

12 5RR 13 3RR 14 6RR

Step 2: Get zeroes for all entries below the first leading 1Step 3: Repeat Process with second row, and so on

23R

23 3RR

37

1R

Now we have REF, but we need RREF!!

21 3

2RR

32 11RR 31 3

21RR

Example

• Solve the system of equations

• Solution:

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1523 321 xxx

0235 321 xxx

1133 321 xxx

30246 321 xxx

41 x 22 x 73 x

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Matrix Operations

• Addition/Subtraction– Two matrices A and B can only be added if:

• They are the same size

– How does this work?• Add numbers in same position in each matrix

18

19

Example

• Find the matrix C, where C = A + B, given the following matrices:

19

461

232

035

342

624

754

A = B =

20

Solution

C = A + B

C = +

C =

20

461

232

035

342

624

754

789

456

789

123

456

789

21

Matrix Operations

• Scalar Multiplication– The matrix A can also be multiplied by a constant,

say c• Multiply each entry in the matrix by c

– Quick Example:• Find the matrix cA if A = and c = 2• Solution:

21

95

31

1810

62

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Matrix Operations

• Matrix Multiplication– Suppose we have two matrices, A and B, and we

wish to multiply them to form a new matrix AB..– Important: The number of columns in A must equal

the number of rows in B• e.g is 2 x 3, and so it can only be

multiplied by a matrix that is 3 x n:

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f

c

e

b

d

a

lk

ji

hg

“The entry of the jth row and kth column of the new matrix AB is the dot product of the jth row of A with the kth column of B.”

f

c

e

b

d

a

lk

ji

hg=

nm

o

nm

po

nm

m

23

Example

• Find the matrix C = AB if

23

100

312

211

342

624

754

A = B =

342

29246

7112

C =

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Matrix Operations

• Transpose– For a matrix A, the transpose is denoted AT

– How? Turn rows into columns and columns into rows

– E.g. Find the transpose of

– Solution: Rotate 45° clockwise and then swap columns

OR; Your R1 is now C1, R2 is now C2, etc. 24

253

461

42

65

13

24

56

31

24

56

31

- Solution:

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Matrix Operations

• Trace– tr(A) – sum of the entries on the main diagonal of A

25

ihg

fed

cba

26

Algebraic Properties

• Inverse– The matrix A has an inverse matrix B if:

• BA = AB = I• Recall, I is the identity matrix:

– How do we find the inverse?• Row reduce the matrix [ A | I ] until the left part is in

RREF• You will end up with a matrix of the form [ I | B ] where B

= A-1

– Note that if the RREF of A is not I, then A is not invertible

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100

010

001

Example

• Determine whether A = is

invertible, and if so, determine its inverse.

27

342

221

211

Solution

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100

010

001

342

221

211

102

011

001

120

010

211

120

011

012

100

010

201

120

011

252

100

010

001

R2-R1

R3-2R1

R1-R2

R3-2R2

R1+2R3

-R3

Example

• Determine whether A = is

invertible, and if so, determine its inverse.

• Solution:• Check:

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342

221

211

120

011

252

342

221

211

120

011

252

100

010

001

=

Algebraic Properties

• Using the Inverse to Solve a System– Suppose we have a system of equations

Ax=b– E.g.

– If we multiply both sides of the equation Ax=b by A-1, we get:A-1 Ax= A-1 b

Ix= A-1 b x= A-1 b

– So now we have another way to solve for x besides row reducing!

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l

k

j

ihg

fed

cba

Example

• Solve the system of equationsx1+x2+2x3 = 2

x1+2x2+2x3 = 7

2x1+4x2+3x3 = 5

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Solution

• Step 1: Put into a Matrix

• Step 2: Find the Inverse A-1

(we found this already in the last example)

• Step 3: Multiply the Inverse by b

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342

221

211

120

011

252

120

011

252

5

7

2

5

7

2

A=

b=

=

9

5

21

Example

• Solve the system of equationsx1+x2+2x3 = 2

x1+2x2+2x3 = 7

2x1+4x2+3x3 = 5

• Solution:

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9

5

21

x1 = - 21, x2 = 5, x3 = 9

Algebraic Properties

• Matrix Polynomials – For a given polynomial ,

we define the matrix polynomial in A to be

• In other words, we substitute A for x and replace a0 with a0I

• In better words; Change x to A (input matrix A), and any constants in your equation change to the identity matrix

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nnxaxaxaaxp ...)( 2

210

nnAaAaAaIaAp ...)( 2

210

Example

• Given and A = find p(A).

• Solution:

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12)( 2 xxxp

12

13

IAAAp 22)(

10

01

12

13

12

13

12

132)(Ap

10

01

12

13

38

4112)(Ap

10

01

12

13

616

822)(Ap

614

720)(Ap

Chapter 2:Determinants

The Determinant

• What is the determinant?

– For a 2x2 matrix:• det(A) = a11a22-a12a21

– For a 3x3 matrix:

• det(A) = ai1Ci1+ai2Ci2+ai3ci3 “Cofactor Expansion”

• Where C is the cofactor

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2221

1211

aa

aa

333231

232221

131211

aaa

aaa

aaa

Cofactor

• What is the cofactor?– First, some notation:

• Let A(j,k) be the 2x2 matrix obtained from A by deleting row j and column k

– Then the cofactor Cjk = (-1)(j+k)·det[A(j,k)]

– Checkerboard method on board

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Example

• Find the determinant of .

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601

532

114

Solution

• Recall two important formulae:– det(A) = ai1Ci1+ai2Ci2+ai3ci3

– Cjk = (-1)(j+k)·det[A(j,k)]

• Choose to expand along the last row:

• det(A)= 1·C31+6·C33

• det(A)= 1·(-1)4·det[A(3,1)] +6·(-1)6·det[A(3,3)]• det(A)=

• det(A)= (-1)(5)-(3)(1) + 6[(4)(3)-(2)(-1)]• det(A)= 76 40

601

532

114

32

146

53

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Properties of Determinants

• Adjoint– Create a matrix of the cofactors for each

number in the original matrix– Find the transpose of this matrix

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333231

232221

131211

aaa

aaa

aaa

333231

232221

131211

CCC

CCC

CCC

332313

322212

312111

CCC

CCC

CCC

Original Matrix A

Matrix of Cofactors

Transpose

Example

• Find the adj(A) for A =

• Solution:– The cofactors are

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021

386

481

C11 = -6 C12 = -3 C13 = 20C21 = 8 C22 = 4

C23 = -10C31 = -8 C32 = 21 C33 = -40

Solution

• Solution:– The cofactors are

Transpose

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C11 = -6 C12 = -3 C13 = 20C21 = 8 C22 = 4

C23 = -10C31 = -8 C32 = 21 C33 = -40

40218

1048

2036

401020

2143

886

Properties of Determinants

• We can use the adjoint to find the inverse of a matrix:

44

)()det(

11 AadjA

A

Example

• Find the inverse of A =

• We know adj(A) from the last example…what’s det(A)? – Recall, we found C21 = 8, C22 = 4, C23 = -10

– Using this we know (Cofactor expansion middle)det(A) = (6)(8) + (8)(4) + (3)(-10)det(A) = 50

– And so… 45

021

386

481

Solution

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5

4

5

1

5

250

21

25

2

50

325

4

25

4

25

3

1A

401020

2143

886

50

11A

Cramer’s Rule

• Suppose we have a system of linear equations Ax = b…

• Also suppose det(A) is not zero…• Then, we know the system has a

unique solution, and that solution is:

• What is An? It’s the matrix A, except that we substitute the entries in the nth column with the entries in b 47

)det(

)det(

A

Ax nn

Example

• Use Cramer’s rule to solve the system of equations:x – 4y + z = 64x – y + 2z = -12x + 2y – 3z = -20

• Step 1: Put into a matrix

48

322

214

141

20

1

6

A =

B =

Solution

49

322

214

141

22

141

32

244

32

211)det(

A

)10(1)16(4)1(1)det( A

55)det( A

3220

211

146

3202

214

161

2022

114

641

A1 =

A2 =

A3 =

144)det( 1 A 61)det( 2 A 230)det( 3 A

Solution

50

55

144

)det(

)det( 1 A

Ax

55

61

)det(

)det( 2 A

Ay

11

46

)det(

)det( 1 A

Az

Chapter 3:Vectors in 2, 3, N-Spaces

Vectors

• What is a vector?– A great way to start your day– A quantity that involves both a number and a

direction– denoted

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Initial Point (A)

Terminal Point (B)

ABv

v

Operations on Vectors

• Vector Addition– Parallelogram Rule

– Triangle Rule

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v

w

v + w

v

w

v + w

Operations on Vectors

• Vector Subtraction– The negative of a vector v, denoted -v, is a vector

with the same length as v but in the opposite direction

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v

w-w

v - w

v

w

v - w

Operations on Vectors

• Scalar Multiplication– The scalar product of a vector v by a

constant nonzero scalar k is that same vector but,• with length k times the original length• with direction the same if k is positive and

opposite if k is negative

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v2

1v v2

A Vector and a Scalar…

• The vector was walking down Cartesian drive when he bumped into a confused Scalar.

• The vector asked him what was wrong and he replied “Help, I have no direction”

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Vectors in Coordinate Systems

• If we place the initial point of a vector at the origin, then the vector is determined by the coordinates of its terminal point– We call these coordinates the components

of v

x

y

),( 21 vv

),( 21 vvv

vy

x

z

v ),,( 321 vvv

),,( 321 vvvv

Vectors in Coordinate Systems

• What if the vector doesn’t begin at the origin?– Then we have where is the

initial point and is the terminal point

– Also,

ABv ,...),( 21 aaA,...),( 21 bbB

OAOBABv

ABv

A

B

OA

OB

OA

ABv

OAOBABv),(),( 2121 aabbv ),( 2211 ababv

x

y

Example

• Find the components of the vector from point to point . Sketch.

• Solution:

)2,3,1( A )1,1,3( B

ABv)2,3,1()1,1,3( ABv

)3,4,2( v

x

z

y2

4

3v

Example

• Find the terminal point of the vector that is equivalent to and whose initial point is .

• Solution:

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)3,5,1(u)1,1,1(

ABv)1,1,1(),,()3,5,1( 321 bbb

),,()1,1,1()3,5,1( 321 bbbB)4,6,2(

Norm

• The length or magnitude of a vector

• Example: Find the norm (length) of the vector .

Solution:

61

222

21 ... nvvvv

)1,2,2,4(v

2222 1224 v25v5v

LAWL

• Why did the vector cross the road?• It wanted to be normal

62

LAWL

Normalization

• Multiplying a nonzero vector by the reciprocal of its length to get a unit vector

• What is a unit vector?– A vector with norm/length/magnitude of 1–

63

vv

u1

Example

• Find the unit vector u that has the same direction as .

• Solution:

)1,2,2,4(v

5v

)1,2,2,4(5

1u

5

1,5

2,5

2,5

4u

(from last example)

Check:2222

5

1

5

2

5

2

5

4

u

25

1

25

4

25

4

25

16u

25

25u

1u

A Few Notes

• Standard Unit Vectors– , – , , – , , … , – Every vector can be described as a linear

combination of these ( )

• Distance in Rn

– The distance between two points is the norm of the vector connecting the two

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)0,1(i )1,0(j

)0,0,1(i )0,1,0(j )1,0,0(k)0,...,0,0,1(1 e )0,...,0,1,0(2 e )1,...,0,0,0(ne

nnevevevv ...2211

Dot Product

• For two vectors u and v where θ is the angle between them:

– is the dot product of the components in u and v

–

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vu

vucos

vu

nnvuvuvuvu ...2211

Dot Product

• We know that– is acute if and only if – is obtuse if and only if – if and only if

• Example: Prove that the two vectors

and are orthogonal (perpendicular).

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2

0vu 0vu

0vu

)4,1,6(u)3,0,2( v

Solution

• If two vectors are orthogonal then their dot product is zero, so we need to show that their dot product is zero:

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)3,0,2()4,1,6( vu)3)(4()0)(1()2)(6( vu

01212 vu

Orthogonal Projections

• If we have two vectors u and a, then u can be expressed in only one way in the form , where– is a scalar multiple of the vector a – is orthogonal to the vector a

21 wwu 1w

2w

a1w

2wu

2w

Orthogonal Projections

projection of u onto a

vector component of u, orthogonal to a

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1w

2w

aa

auuproja 2

uproju a

cosua

auuproja

Distance Problems

• Distance between a point P and a line (R2) or a point and a plane (R3) – – Where n is the normal (a vector orthogonal to the

line) and Q is a point on the line

• Distance between two parallel planes (R3)– Find a point in the first plane and then find the

distance between that point and the other plane

71

QPprojD n

Example

• Find the distance between the lines -8x-6y-8=0 and 4x+3y+4=0.

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Solution

• Find a point on the first line: – P(1,0)

• Now we need to find the distance between this point and the line:– Find the normal:

• n=(4,3)

– Plug it all into the equation:

n

QPnQPprojn

)3,4(

)]0,1()1,3[()3,4( QPprojn

22 34

)1,4()3,4(

QPprojn

5

13QPprojn

Geometry of Linear Systems

• Vector and Parametric Equations of Lines– The equation of the line through a point x0 that is

parallel to the vector v is:

• Vector and Parametric Equations of Planes– The equation of the plane through a point x0 that

is parallel to v1 and v2 is:

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tvxx 0

22110 vtvtxx

Example

• Find the vector equation and parametric equations of the line that passes through the point P(1,2,-4) and is parallel to the vector v=(1,1,1).

• Solution:– Vector equation:

– Parametric equations:

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)1,1,1()4,2,1(),,( tzyx

tx 1 ty 2 tz 4

Example

• Find vector and parametric equations for the plane 2x-3y+z=4.

• Solution:– Let x and y be t1 and t2 respectively. Then

rearrange the equation for z and substitute in:

yxz 324 21 324 ttz

Solution

• So our parametric equations are:

• Rewrite for the vector equation:

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21 324 ttz 1tx 2ty

)324,,(),,( 2121 ttttzyx

)3,1,0()2,0,1()4,0,0(),,( 21 ttzyx

Conclusion..

Good Luck!

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