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MATH1231 Maple TA Explanations for Week 6, SEM 2 Christy Oh∗ Anonymous† August 29, 2018

Contents 1 Introduction To Linear Maps

3

1.1

Question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

1.2

Question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4

1.3

Question 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5

1.4

Question 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6

1.5

Question 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7

2 Linear Maps

8

2.1

Question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8

2.2

Question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11

2.3

Question 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

12

2.4

Question 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14

2.5

Question 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

16

∗ Explanations † Explanations

written for Sections 2 and 3. written for Sections 1 and 4.

1

3 Linear Transformations Through Pictures

19

3.1

Question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

20

3.2

Question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21

3.3

Question 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

22

3.4

Question 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

23

3.5

Question 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

25

4 Reflections

28

4.1

Question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

28

4.2

Question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

29

4.3

Question 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

30

4.4

Question 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

31

4.5

Question 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

32

5 Further Linear Algebra II

34

2

1 1.1

Introduction To Linear Maps Question 1

Dot product distributive law Integration sum rule Triangle inequality (|x + y| 6 |x| + |y|)

√ Consider x = y = 1 ⇒ 2 = 2 Cross product distributive law

Let λ = 2, x =

π 2

⇒0=2 Constant factor rule Let λ = −1 ⇒ |x| = −|x|

Consider x = y = 1 ⇒ 4 = 2 Consider x = y = π2 ⇒ 0 = 2 Consider x = y = 0 ⇒ 1 = 2

Consider x = y = 1 ⇒ ln 2 = 0 Differentiation sum rule

Constant multiple rule Consider λ = 0 ⇒ 1 = 0

Scalar multiplication property

3

1.2

Question 2

For the functions T , we determine if the function value of 0 is also 0.       0 0 0×0×0 0 For example, T (0) = T ( )= = =0 0 0 0 0

For the functions T , we determine if the function value of −v is the negative of the function value of v.     a b −a −b For example, T (−v) = T (− ) ) = T( −c −d c d   −a × −b × −c (Definition of T ) = −d   abc =− d   a b ) (Definition of T ) = −T ( c d = −T (v) 4

For the functions T that satisfied both properties, we determine if the addition and scalar multiplication conditions of linear maps are also satisfied.     a b λa λb For example, T (λv) = T (λ ) ) = T( λc λd c d   λa × λb × λc (Definition of T ) = λd  3  λ abc = λd       abc λabc a b ) = λ = . However, λT (v) = λT ( c d d λd Since T (λv) 6= λT (v), the scalar multiplication condition is not satisfied. Hence, the function T is not an example of a linear map.

1.3

Question 3

      1 0 2 + 6 ) ) = T (2 T( 0 1 6     1 0 = T (2 ) ) + T (6 1 0     1 0 = 2T ( ) ) + 6T ( 1 0

(Addition condition) (Scalar multiplication condition)

5

     7 1 20 = 2 −4 + 6 −8  = −56 −2 −4 −28 

1.4

Question 4

v3 = 2v1 + v2 T (v3 ) = T (2v1 + v2 ) = T (2v1 ) + T (v2 )

(Addition condition)

= 2T (v1 ) + T (v2 )       −2 6 4 + = =2 6 2 14

(Scalar multiplication condition)

6

1.5

Question 5

7

2 2.1

Linear Maps Question 1

Since T is a linear map, we know that it satisfies two conditions: 1. Addition Condition T (v1 + v2 ) = T (v1 ) + T (v2 ) 2. Scalar Multiplication Condition T (λv) = λT (v) We can use these conditions to find:

8

      0 2 2 + ) ) = T( T( 0 5 5     0 2 ) ) + T( = T( 5 0     0 1 ) ) + T (5 = T (2 1 0     0 1 ) ) + 5T ( = 2T ( 1 0     −1 7 +5 =2 −8 3     −5 14 + = −16 15   9 = −1

(Addition condition)

(Scalar multiplication condition) (Definition of the linear map T)

We use the same method to find the general vector:       0 x x + ) ) = T( T( 0 y y     x 0 = T( ) ) + T( y 0    1 0 = T (x ) ) + T (y 1 0     1 0 = xT ( ) ) + yT ( 1 0     −1 7 +y =x 3 −8     −y 7x + = −8x 3y   7x − y = 3y − 8x

(Addition condition)

(Scalar multiplication condition) (Definition of the linear map T)

Using the linear map S is slightly more difficult since the linear combination of the function values of S may not be as obvious.

  0 ) To find S( 1 9

  0 1     1 1 and into a linear combination of 0 1

we need to first manipulate the vector

      −1 1 0 + ) ) = S( S( 0 1 1     1 −1 ) ) + S( = S( 1 0     1 1 ) ) + S( = S(− 1 0     1 1 ) ) + S( = −S( 1 0    1 8 =− + 1 4    −1 8 = + −1 4   7 = 3

(Addition condition)

(Scalar multiplication condition) (Definition of the linear map S)

  x ): Similarly, we can find S( y       x y x−y S( + ) ) = S( 0 y y     x−y y = S( ) ) + S( y 0     1 1 = S((x − y) ) ) + S(y 1 0     1 1 ) + yS( ) = (x − y)S( 0 1     8 1 +y = (x − y) 4 1     8y x−y + = x−y 4y   x + 7y = x + 3y

10

(Addition condition)

(Scalar multiplication condition) (Definition of the linear map S)

2.2

Question 2

    2 1 in the original graph should map to in the new graph The vector 0 0     0 −1 and the vector in the original graph should map to in the new graph 1 −2 The first graph satisfies these conditions. We can use the graphs for the rest of this question. We observe in the new graph that the tip of the roof has been mapped to the point [− 21 , −3]. The point [1, −2] in the new graph is the point [1, 1] in the original graph.

And the point [1, 0] in the new graph is the point [ 12 , 0] in the original graph.

11

2.3

Question 3

12

        0 0 −9 −9 T (−7) = T ( 0  + −7 + 0) 0 0 2 −2       0 0 −9 (Addition condition) = T ( 0 ) + T ( −7) + T (0) 0 0 2       1 0 0 = T (−9 0) + T (−7 1) + T (2 0) 0 0 1       0 0 1 (Scalar multiplication condition) = −9T ( 0) − 7T (1) + 2T (0) 1 0 0       0 5 1 = −9  1  − 7 −2 + 2 2 (Definition of the linear map T) 2 −4 −4       2 0 −35 = −9  +  14  + 4 4 28 36   −33 = 9  68

13

2.4

Question 4

14

15

2.5

Question 5

Use Maple to solve. Make sure to put the Original Coke vector in the right hand column since we are trying to write it as a linear combination of the others.

16

4 1 1 . So we have α = , β = , γ = 3 2 15

17

18

3

Linear Transformations Through Pictures A linear transformation can be represented by a matrix.

If we are given the vectors which are mapped by the standard basis vectors, then these vectors become the column vectors of the matrix A which satisfies T (x)=Ax, where T is a linear map. This is called the Matrix Representation Theorem.

For example, let T : R2 → R2 be a linear map.     1 0 The standard basis vectors for R2 are ei = . and ej = 1 0     3 2 If T(ei ) = , and T(ej ) = −1 2   3 2 satisfies T (x)=Ax. then the matrix A = 2 −1 This allows us to find the function value (i.e. T (x)) of any vector x by matrix multiplying it with the matrix A.

The next three questions use this theorem.

19

3.1

Question 1

Under the transformation S(x)=Nx, we use matrix multiplication to find: −−→ −−→ −−→ −−→ −−→ −−→ S(OE) = N OE S( OD) = N OD S(OC) = N OC   7  1      −4 0 1 −4 0 −4 0 8 2 = = = 3 3 0 3 0 3 0 3 1 2 2    7    −2 + 0 −4 + 0 −2 + 0 = = = 0+3 0 + 29 0 + 29      7 −4 −2 −2 = = 9 = 9 3 2 2

20

3.2

Question 2

Under the transformation T (x)=M x, we use matrix multiplication to find: −−→ −−→ −−→ −−→ −−→ −−→ T(OE) = M OE T( OD) = MOD T(OC) = MOC    7   1     4 −1 4 −1 1 4 −1 8 2 = = = 3 3 0 2 0 2 0 2 1 2 2       7 3 4−1 2 − 23 − 2 = = 2 = 0+2 0+3 0+3     1  3 2 = = = 2 2 3 3

21

3.3

Question 3

Under the transformation R(x)=Lx, we use matrix multiplication to find:

22

−−→ −−→ −−→ −−→ −−→ −−→ R(OE) = L OE R( OD) = LOD R(OC) = LOC  7     1     1 3 1 3 1 3 1 8 2 = = = 3 3 −1 2 −1 2 −1 2 1 2   7 9 2   1 9  + + 1+3 = 87 2 = = 21 2 −1 + 2 −8 + 3 −2 + 3    43    4 5 8 = = = 5 17 1 2

3.4

Question 4

23

8

By moving the vectors of the original image (normal house) to the corresponding location on the transformed image (transformed house), the app used shows that the matrix   −2 −1 was used in the transformation. K= −1 1 U(v) = Kv    4 −2 −1 = −1 1 −5   −8 + 5 = −4 − 5   −3 = −9

24

3.5

Question 5

25

Solving the simultaneous equations: a+b=0

a−b=2

c+d =3

c−d =1

c=2

d=1

We find: a=1

b = −1

        (1)x + (−1)y x−y x X )= = = R( So y (2)x + (1)y 2x + y Y X =x−y

(1)

Y = 2x + y

(2)

(1)+(2):

(2)–2×(1):

X + Y = 3x ⇒ x =

X +Y 3

Y – 2X = 3y ⇒ y =

Y − 2X 3

x2 + y 2 = 1

(

X +Y 2 Y − 2X 2 ) =1 ) +( 3 3

26

So

  X satisfies the first equation. Y (X + Y )2 + (Y − 2X )2 = 9

X 2 + 2XY + Y 2 + Y 2 − 4X Y + 4X 2 = 9

5X 2 − 2XY + 2Y 2 = 9

Hence

  X also satisfies the second equation. Y

27

4 4.1

Reflections Question 1

By the addition condition, T (v + w) = T (v) + T (w).

  −2 0     2 −2 )= T (T (v)) = T ( 0 0     −2 2 )= T (T (T (v))) = T ( 0 0 T (v) =

(Substituting T (v)) (Substituting T (T (v)))

We note that reflection in the y-axis involves changing the x component to −x. Therefore, the reflection of the line y = 6x + 1 in the y-axis is given by y = 6(−x) + 1 = −6x + 1 28

4.2

Question 2

    1x + 0y x )= (Similar to the LHS of a system of linear equations) R( y 0x − 1y     x 1 0 (Representing as a matrix) = 0 −1 y   1 0 represents the map R (such that R(x) = Ax). Hence, the matrix A = 0 −1

We note that reflection in the x-axis involves changing the y component to −y. Therefore, the reflection of the line y = 9x + 3 in the x-axis is given by −y = 9x + 3

y = −9x − 3

29

4.3

Question 3

       −1x + 0y −1 0 x x ) = = S( y 0x − 1y 0 −1 y   −1 0 represents the matrix S (S(x) = Ax). Hence, the matrix A = 0 −1

We note that reflection in the origin involves changing the x component to −x, and the y component to −y. Hence, the reflection of y = 9x + 3 in the origin is −y = 9(−x) + 3 = −9x + 3

y = 9x − 3

30

4.4

Question 4

We observe that the x and y components interchange when reflected.

    0x + 1y x )= (Similar to the LHS of a system of linear equations) U( y 1x + 0y    0 1 x = (Representing as a matrix) 1 0 y   0 1 Hence, A = has the required property that U (x) = Ax. 1 0

We observe that reflection in the line y = x involves changing the x component to y, and the y component to x. Therefore, the reflection of y = 7x + 9 is x = 7y + 9 x − 9 = 7y 9 1 x− =y 7 7

31

4.5

Question 5

    1 −3 × 1 + 4 × 0 1 )= Using the given result, V ( 0 5 4×1+3×0  −3 = 54 5

  0 )= Similarly, V ( 1   1 )= and V ( 1

 4   1 −3 × 0 + 4 × 1 = 53 5 4×0+3×1 5   1  1 −3 × 1 + 4 × 1 = 57 5 4×1+3×1 5

1. We realise that reflecting any point that already lies on the line of reflection will result in the same point. So to find the this line, we try solving     x x V( )= y y     1 −3x + 4y x = Using the given result, y 5 4x + 3y 1 Equating x components, (−3x + 4y) = x 5 −3x + 4y = 5x 4y = 8x y = 2x For the x component to remain unchanged after reflection, the condition y = 2x must be satisfied (similarly for the y component - equate y components instead). Hence, we deduce that V is a reflection about this line.

32

2. We observe that reflection using V involves changing the x component to 51 (−3x + 4y), and the y component to

1 (4x + 3y). 5

Therefore, the reflection of y = 8x + 7 using V is given by 1 1 (4x + 3y) = 8( (−3x + 4y)) + 7 5 5 4x + 3y = 8(−3x + 4y) + 35 4x + 3y = −24x + 32y + 35

28x − 35 = 29y 35 28 x− =y 29 29 35 28 x− y= 29 29

33

(Substituting) (Multiplying by 5)

5

Further Linear Algebra II

No explanations given. All of the information you need can be found in the Maple notes, lessons, and assessments, which can be found below: Maple Notes, Lesson and Assessments: https://moodle.telt.unsw.edu. au/file.php/34371/Maple/MapleSelfPaced/index.html – END OF MAPLE –

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