3.0. A COLLECTION OF NUMBERS AS A MATHEMATICAL OBJECT

Exercise 1: Pick three other applications and propose natural groupings of numbers for those

applications.

i. Group of whole numbers e.g.{0,1,2,3,4,5,6,7,8,9} are applicable in of coordinates axes

for graphical representation.

ii. Group of rational number for instance { 31,

32,

74

} are applicable in precise proration

concerning probability and statistics.

iii. Group o or set {0,1} is natural grouping and its application is in computational

Mathematical analysis appertaining to group of numbers leads us to will check into Set

theory

Definition: a set is a group or collection of distinct elements. Set’s are represented by capital

letters and are bracketed with curly brackets. Example D= {a, f, m, o, e, r, f} and N= {4, 9, 16,

25, 36}, Set D and N is a set of distinct letters and number respectively. Numbers can be group

or collected according to specific property, for instance this is a set {1, 4, 9, 16} is of numbers

related because they are perfect squares.

Function of sets is used in algebraic analysis especially in analytical operation of grouping of

various elements in categories that appertain to a certain property.

Below are illustration of binary operation of union denoted by “ ” and operation of intersection

denoted by “ ”

a) Operation of union denoted by “ ”.Definition: Union operation of sets is collection or

grouping of all elements contained either set or in both concerning sets and avoiding double

listing of coexisting elements or repetitive existence of an element or elements in the

concerning sets.

Example

Set Z= {k, h, e, f, m} and another set X={c, y. f. m, r, s}.

Union of Z and X that is },,,,,,,,{srycmfehkXZ note that f and m are listed once

b) Operation of Intersection denoted by

Definition: intersection of two sets contains only elements in both sets in our case only

elements in set Z and X

Example:

Set Z= {k, h, e, f, m} and another set X={c, y. f. m, r, s}.Intersection of Z and X that is

},{mfXZ

Two Sets can be added even with aid of tuple which is systematic additional of elements in sets

according to order of arrangement.

Exercise 2

i. Go back to the example with two coordinate systems, and verify by hand that addition

produces the same result.

In coordinate system addition of sets that is a coordinates containing elements of x axis and y

axis produces same result meaning the result has same tuple as the constituents added because

tuple being the number of elements is a set. In addition of two points as tuples like (4, 3) + (5,

7) = (9, 10).We illustrate this by first considering 1 dimension and additional operation is on

number line. First on x axis point (4, 0) is moved to point (5, 0) which is then moved to point

(9, 0). And on still1D dimension on y axis point (0, 3) is moved to point (0, 7) which is the

moved to point (0.10). Verification by hand from point (4, 3) to point (9, 10) move 5 units by

hand from 4 and get 9 on x-axis and move 7 units from3 by hand and get 10 thus the resulting

coordinate (9, 10) has x coordinate and y coordinate and this has same tuple as in point (4, 3)

and point (5, 7).The result (9, 10) is same as the previous one.

ii. How would you prove that addition produces the same result regardless of coordinate

system?

This is proven with aid moving vectors. Vector heading to (4, 3) from 0 and vector from 0

heading to (5, 7) when added give results (9, 10) this operation is illustrated as

10

9

7

5

3

4

This is the mathematical legitimate additional operation of vectors.

iii. By the rules above, we aren't allowed to change the orientation of an arrow when it's

moved for addition. Why is that? What would go wrong with the theory if we allowed

the direction to change?

Changing orientation of the moving arrow or vector for purpose of additions is not

allowed because the arrow drawn between (5, 7) and (9, 10) doesn’t exist as a tuple and if

orientation is changed it will not give correct result

3.1 Vectors-

a. Exercise 3: Why is it true that the orientation (angle), except for flips, is unchanged

by scalar multiplication?

Generally orientation of the vector is unchanged my scalar multiplication, in additional.

There is special vector0= {0, 0, 0} which is 0 additive. Orientation (angle) of flip changes

by negative scalar multiplication stretch and Negative scalar multiplication stretch

negate the values of coordinates and the direction linearly shifts to negative quadrant of

the Cartesian plane thus orientation is changed. Flips direction is likely to be an

orientation (angle) of 180 0 (angles on a straight line).

3.2 Linear Combination

4: Exercise:

Download LinCombExample.java and DrawTool.java. Then, examine the code

in LinCombExample to see how vectors and arrows are drawn. Then, draw the

remaining arrows to complete the parallelogram.

OEFG is the parallelogram

/ Instructions:

// 1. Read the code in main() to understand how DrawTool is used.

// 2. Compile and execute to see.

// 3. Draw the remaining arrows to complete the parallelogram.

public class LinCombExample {

public static void main (String[] argv)

{DrawTool.display ();

DrawTool.setXYRange (-10,10, -10,10);

DrawTool.drawMiddleAxes (true);

double[] u = {6,4};

double[] v = {1.5,6};

double[] z = {7.5,10};

DrawTool.drawVector (u);

DrawTool.drawVector (v);

// Alternatively: DrawTool.drawArrow (0,0, 6,4);

DrawTool.setArrowColor ("blue");

DrawTool.drawVector (z);

// INSERT YOUR CODE HERE: (Use DrawTool.drawArrow())

class DrawObject {

double x,y; // For Points, one end of lines,

// and top-left of ovals/rectangles.

double x2,y2; // For the other end of lines, arrows.

double width,height; // Forrectangles.

double a, b, c; // For the line ax+by+c=0

BasicStroke drawStroke = new BasicStroke (1.0f);}

// What we draw: points, lines, \\rectangle.

static java.util.List<DrawObject>lines, rectangles,

// Animated versions: these will clear between successive frames.

static java.util.List<DrawObject> animPoints, animLines, animOvals,

animRectangles, animLabels;

static double minX=0, maxX=10, minY=0, maxY=10; // Bounds.

static int numIntervals = 10; // # tick marks.

static int pointDiameter =7.5; // Size of dot.

// GUI stuff.

static JPanel drawArea = new DrawTool ();

static Dimension D; // Size of drawing area.

static int inset=60; // Inset of axes and bounding box.

// Static initializer.

static{rectangles = Collections.synchronizedList (new

ArrayList<DrawObject>());

public static void drawLine (double x1, double y1, double x2, double y2)

{drawLine (x1, y1, x2, y2, false)}

public static void drawArrow (double x1, double y1, double x2, double y2)

{drawLine (x1, y1, x2, y2, true);}

public static void drawVector (double x, double y)

{drawLine (0, 0, x, y, true);public static void drawVector (double[] x)

{drawVector (x[0], x[1]);}

public static void drawLineFromEquation (double a, double b, double c)

{// Draw the equation ax+by+c=0 in the available range.

DrawObject L = new DrawObject ();

L.color = lineEqnColor;

L.a = a; L.b = b; L.c = c;

L.sequenceNum = currentSequenceNum;

L.drawStroke = drawStroke;

synchronized(eqnLines) { eqnLines.add (L); }

drawArea.repaint ();}

public static void drawLin

public static void drawRectangle (double x1, double y1, double width, double

height)

{

DrawObject R = new DrawObject ();

R.color = rectangleColor;

R.x = x1; R.y = y1; R.width = width; R.height = height;

R.sequenceNum = currentSequenceNum;

R.drawStroke = drawStroke;

if (animationMode) {synchronized(animRectangles) { animRectangles.add

(R); }}else {synchronized(rectangles) { rectangles.add (R); }drawArea.repaint

();} public static void drawImage (int[][][] pixels, int startX, int startY)

{} );}cPane.add (drawArea, BorderLayout.CENTER);drawArea.addMouseListener

(new MouseAdapter(){public void mouseClicked (MouseEvent

e){handleMouseClick(e);

}

public void mouseReleased (MouseEvent e){handleMouseReleased (e);}});

void drawOvalOrRectangle (Graphics g, DrawObject R, boolean isRect)

{if (R == null) {return;}

if ((sequencingOn) && (R.sequenceNum!=currentSequenceNumDisplay)) {

return;}

Graphics2D g2 = (Graphics2D) g;

g2.setStroke (R.drawStroke);

int x1 = (int) ( (R.x-minX)/(maxX-minX) * (D.width-2*inset) );

int y1 = (int) ((R.y-minY) / (maxY-minY) * (D.height-2.0*inset) );

double x = R.x + R.width;

double y = R.y – R.height;

int x2 = (int) ( (x-minX)/(maxX-minX) * (D.width-2*inset) );

int y2 = (int) ((y-minY) / (maxY-minY) * (D.height-2.0*inset) );

if (isRect) {

g.drawRect (inset+x1, D.height-y1-inset, x2-x1, y1-y2);}else {

g.drawOval (inset+x1, D.height-y1-inset, x2-x1, y1-y2);}}

// Remove this line to leave arrow head unpainted:

g2d.fillPolygon(tmpPoly);}

private int yCor (int len, double dir) {return (int)(len * Math.cos(dir));}

private int xCor (int len, double dir)

{return (int)(len * Math.sin(dir));}

void drawImage (Graphics g, DrawObject I){

// Instructions:

// Compile and execute.

import java.util.*;

public class LinCombExample {

public static void main (String[] argv)

{Function G = new Function ("g(x) vs x");for (double x=0; x<=6.283; x+=0.1) {

G.add (x, g(x));}

// Draw the curve.

G.show ();}

static double g (double x)

{// Array of coefficients (scalars in the linear combination).

double[] alpha = {0, 1.0, 0, -0.16666666666666666};

// Here, f_1(x) = x, f_2(x) = 0, f_3(x) = x^3.

double g = alpha[1] * x + alpha[3] * Math.pow(x,3) ;

return g;}}

Exercise 5: Download LinCombExample2.java (you already have DrawTool). Write

code to implement vector addition and scalar multiplication, and test with the example

in main().

// Instructions:

// 1. Write code to implement vector addition and scalar multiplication

// 2. Then compile and execute.

public class LinCombExample2 {

public static void main (String[] argv) {

DrawTool.display ();

DrawTool.setXYRange (-12,12, -12,12);

DrawTool.drawMiddleAxes (true);

double[] u = {6,4};

double[] v = {1.5,6};

double alpha = 1.5, beta = 2;

double[] z = add (scalarMult(alpha, u), scalarMult(beta,v));

DrawTool.setArrowColor ("blue");

DrawTool.drawVector (z);}

static double[] add (double[] u, double[] v)

{/ INSERT YOUR CODE HERE}

static double[] scalarMult (double alpha, double[] v)

{// INSERT YOUR CODE HERE}}

Exercise 6: Download LinCombExample3.java. The double-for loop tries to systematically

search over possible values of α,β to see if some combination will work. Write code to see if

the linear combination αu+βv is approximately equal to z

// Instructions:

// Try values of n=3, n=5, n=7, n=13 in the function g().

// Do the same after changing the intervalEnd to Math.PI.

import java.util.*;public class LinCombExample3 {

public static void main (String[] argv){double totalError = 0;

double deltaX = 0.1;

double intervalEnd = 2*Math.PI;

for (double x=0; x<=intervalEnd; x+=deltaX) {

double errorAtX = Math.abs (g(x) – Math.sin(x));

totalError += errorAtX * deltaX;}

System.out.println ("Total error: " + totalError);}

static double g (double x)

{// Array of coefficients (scalars in the linear combination).

double[] alpha = {0, 1.5, 0, -0.1819178256658966614, 0, 0.007545573338983125,

0, -1.847412689841269839E-6, 0, 2.6857319228185884E-4, 0, –

1.69871083854125764E-8, 0, 2.69822585583682161E-10};

// How many terms to include: Start with n=1.5, then change to n=7.5,

// n=6, then n=10.

int n = 4;

double g = 0;

for (int k=0; k<=n; k++) {g = g + alpha[k] * Math.pow (x, k);}

return g;}}

Exercise 7: Solve the above problem by hand: are there values of α,β such

that αu+βv=z when u=(1,4), v=(3,2) and z=(7.5,10)?

Solution

10

5.7

2

3

4

1

1

4

1024

5.731

2

20100

1024

30124

5.1

64

1044

10224

It’s confirmed there are values for the said scalars

Exercise 8: Suppose u= (1, 2), v=(3,6) and z=(7.5,10).

Download LinCombExample4.java and draw all three vectors. Then: Use the drawing to

explain why no linear combination of u and v will work.

Solve by hand to confirm the explanation algebraically.

Public static void main (String[] argv)

{DrawTool.display ();

DrawTool.setXYRange (-12,12, -12,12);

DrawTool.drawMiddleAxes (true);

double[] u = {1, 2};

double[] v = {6, 4};

double[] z = {7.5,10};

// INSERT YOUR CODE HERE.}}

The linear combination of vectors u and v gives a vector. Vector (1.5, 6) achieved from

stretching (1,) by scalar 1.5 and (6, 4) is achieved from stretching (3,2) by multiplying by scalar

2 and that is 2(3,2).This linear combination work since the result is a vector

Exercise 9: Now consider u=(1,2),v=(3,6) and z=(4,8). Write code

in LinCombExample5.java to draw these. Is there a unique solution? Explain both

geometrically and algebraically.

public class LinCombExample5 {public static void main (String[] argv) {

DrawTool.display ();

DrawTool.setXYRange (-10,10, -10,10);

DrawTool.drawMiddleAxes (true);

double[] u = {1, 2};

double[] v = {3, 6};

double[] z = {4,8};

DrawTool.setArrowColor ("blue");

// INSERT YOUR CODE HERE.}}

Exercise 10: Finally, consider u=(1,4),v=(3,2),w=(−1,1) and z=(7.5,10). We will ask whether

there is a linear combination αu+βv+γw such that αu+βv+γw=z. Write out the equations

and explain algebraically whether there is a solution. The code

in LinCombExample6.java draws all four vectors. Can a linear combination of any two

of u,v,w suffice to produce z?

public class LinCombExample6 {

public static void main (String[] argv){

DrawTool.display ();

DrawTool.setXYRange (-10,10, -10,10);

DrawTool.drawMiddleAxes (true);

double[] u = {1, 4};

double[] v = {3, 2};

double[] w = {-1, 1};

double[] z = {7.5, 10};

// INSERT YOUR CODE HERE.

DrawTool.setArrowColor ("blue");

DrawTool.drawVector (u);

DrawTool.drawVector (v);

DrawTool.drawVector (w);

DrawTool.setArrowColor ("green");

DrawTool.drawVector (z);}}

A linear combination of any two of u,v,w suffice to produce z?.Yes linear combination of u and

v when stretched by a given scalar suffice to produce z

Exercise 11: The code in LinComb3DExample.java displays the vectors in the above

example. (You need to have the Draw3D jar in your CLASSPATH). Write down the

equations for α,β,γ and solve by hand

public class LinCombExample6 {

public static void main (String[] argv){

DrawTool.display ();

DrawTool.setXYRange (-10,10, -10,10);

DrawTool.drawMiddleAxes (true);

double[] u = {1, 4};

double[] v = {3, 2};

double[] w = {-1, 1};

double[] z = {7.5, 10};

// INSERT YOUR CODE HERE.

DrawTool.setArrowColor ("blue");

DrawTool.drawVector (u);

DrawTool.drawVector (v);

DrawTool.drawVector (w);

DrawTool.setArrowColor ("green");

DrawTool.drawVector (z);}}

Solve α(1,3,1)+β(4,1,0)+γ(3,2,6)=(1,7,8)

Solution

1

44

1342

2

68

816

1

183183

18717611

4711

1716

241803

7213

4711

723

39123

8601

2

3

723

134

8

7

1

6

2

3

0

1

4

1

3

1

nsubtractio

Exercise 12: Add code to LinComb3DExample2.java to display the 3D version of the 2D

parallelogram. That is, draw arrows that show the stretched vectors when scaling

by α,β,γ respectively. Then draw arrows from the tips of the stretched vectors to the resultant

vector z.

import org.edisonwj.draw3d.*;

import javafx.application.*;

import javafx.scene.*;

import javafx.scene.paint.*;

import javafx.stage.*;

public class LinComb3DExample2 extends Application {

String title = "Vector example";

void drawingCommands () {

// Type all drawing commands in here.

// The three vectors u,v,w:

d3.setDrawColor (Color.BLUE);

d3.drawVector (1,3,1);

d3.drawVector (4,1,0);

d3.drawVector (3,2,6);

// Example of using drawArrow(). The vector z:

d3.setDrawColor (Color.BLACK);

d3.drawArrow (0,0,0, 1,7,8);

// Stretched vectors:

// Use drawArrow to draw the added stretch.

d3.setDrawColor (Color.GREEN);

// Write here …

// Use drawArrow to draw arrows from the tips of

// the stretched vectors to the final vector z=(1,7,8).

d3.setDrawColor (Color.RED);

// Write here …}

// No need to read further

//////////////////////////////////////////////////////////

Draw3D d3;

void preambleCommands (){

d3.setAmbientLight(false);

d3.setPointLight(true);

d3.setCumulate(false);

d3.setSequencingOn();

d3.setVectorRadius(1);

d3.setArrowRadius(1);}

public void start (Stage primaryStage) {

d3 = new Draw3D ();

Scene scene = d3.buildScene ();

preambleCommands ();

drawingCommands ();

d3.setStart ();

primaryStage.setScene (scene);

primaryStage.setTitle (title);

primaryStage.show (); }

public static void main (String[] argv){

launch (argv);}}

Arrows showing stretch vectors scales by and,

Exercise 13: Suppose now that u=(1,3,1),v=(4,1,0),w=(9,5,1) and z=(1,7,9). We seek α, β,

γ such that αu+βv + γw=z. Write down the equations for α,β,γ and solve by hand. The code

in LinComb3DExample3.java displays the four vectors. Explain geometrically why this is

consistent with your solution.

Geometrically the resultant values of scalars consistent with four vectors since they are scale by

the scalar to get the final values

0,0,9

4404422

84422

202

2733

753

42211

753

327123

9

753

1943

9

7

1

1

5

9

0

1

4

1

3

1

3.5 The matrix vector equation

Exercise 14: Apply the matrix A= [a11a12a21a22] to the vector x=[x1x2].

What are the dimensions of the resulting product?

2

1

222121

212111

2

1

2221

1211

c

c

xaxa

xaxa

x

x

aa

aa

The resulting product has got two dimensions (2D)

Exercise 15: Apply the matrix A= [2−301] to the vector x= [23] to get the

vector y. Then apply B= [120−3] to y to get z. What are y and z? Show your

calculations

9

1

,

3

5

9

1

3

5

30

21

3

5

3

2

10

32

10

32

3012

zy

ZyB

y

A

Exercise 16: Download MatrixVectorExample.java and write code to perform the matrix-vector

product. Then confirm your calculations above from the displayed vectors

Multiplication: (u1,u2,…,un)⋅(v1,v2,…,vn)=u1v1+u2v2+…+u n v n

(u1,u2,…,un)+(v1,v2,…,vn)=(u1+v1,u2+v2,…,un+vn)

Example: (1,2,0,3)+(-,5,6)=(1×-5,0×-5+,2×-5+3×6)=(-1,9)

Multiplication: (u 1 , u 2 ,…, un)⋅(v1,v2,…,vn)=u1v1+u2v2+…+unv=

n

i

iivu

1

Exercise 17: Compute the vector obtained by multiplying x above by the

matrix C= [2−10−3]. Compare with the value of z calculated earlier

9

1

3

2

30

12

30

12

3012

xC

CC

This value is same as the value of z and all of them are similar because they are column matrix

Exercise 18: Compute the vector obtained by multiplying z above by the matrix C2=

[1.0/2−1.0/60−1.0/3]. Compare with the value of x calculated earlier.

3494.7

15.0

9

1

3333.0

0167.0

5.0

3333.01670.05.03/0.160/0.12/0.12C

The values of product achieved are in column same as the values of x which is also in column

matrix.

Exercise 19: Earlier, you computed the product of A=[a11a12a21a22] with the vector x=[x1x2],

which produces a result vector. Multiply this result vector by B=[b11b12b21b22]. (The result is

admittedly not pretty.)

222121

2121111

2221

1211

222121

212111

2cbcb

cbcb

c

c

bb

bb

xaxa

xaxa

It true the result is long expression and complex and admittedly not pretty

3.6 Matrix-Matrix Multiplication

Exercise 20: Download MatrixVectorExample2.java and write matrix multiplication code to

make the example work. Confirm that you obtained the same matrix in Exercise 17, and the same

resulting vector.

(U 1 , U 2 ,,,,,,,,,,,,,,,U n )+(V 1 ,V 2 ,…,V n ))=(U 1 +V 1 ,U 2 +V 2 ,…,U n +V n )

Example: (1,2,3)+(4,5,6)=(1+4,2+5,3+6)=(5,7,9)

Multiplication: (U 1 ,U 2 ,…,U n )⋅(V 1 ,V 2 ,…,V n )=U 1 V 1 +U 2 V 2 +…+U n V n = n∑U i V i

The result is a single number.

Example: (1, 2, 3) ⋅ (3, 5, 6) =1×3+1×5+3×5=4+10+17=28

Exercise 21: Consider the matrices

A= [321−235003] and B=[−4101013−21]Let x=(1,−1,2) and

define y=Ax and z=By. Calculate each of y,z and the matrix product C=BA. Confirm

by calculating Cx.

7

9

7

2

1

1

10013

423

1514

10013

423

1514

300

532

123

123

101

014

12106

12106

18159

653

123

101

014

653

2

1

1

300

532

123

CxConfirm

BAC

ByZ

Axy

Exercise 22: Download MatrixVector3DExample.java, then fill in the code (from previous

exercises) to perform matrix multiplication and matrix-vector multiplication. Confirm that the

results match the calculations in the previous exercise.

(U 1 , U 2 ,…,U n )+(V 1 ,V 2 ,…,V n )=(U 1 +V 1 ,U 2 +V 2 ,…,U n +V n )

Example: (1,2,3)+(4,5,6)=(1+4,2+5,3+6)=(5,7,9)

Multiplication: (u1,u2,…,un)⋅(v1,v2,…,vn)=u1v1+u2v2+…+unvn=n∑i=1uivi

The result is a single number.

Exercise 23: Suppose A and B are two n×n matrices. Prove or disprove: AB=BA

10013

423

1514

369

81226

317

BAAB

I disapprove that AB is not equal to BA

3.7 Non-Square Matrices

Exercise 24: What is the result of

multiplying A=[010−120] and B=[11220−3]? What is Ax when x=(2,3,4)?

4

3

4

3

2

021

010

33

22

3

2

1

0

2

1

021

010

Ax

3.8 A review of three cases involving equations

Parallelogram edge that is parallel to the (1, 4) line in intersect with line lie on (3,2) which is

possible stretch and where they intersect is a unique point .

References

1. Stuart, J. (1988). Inflation Matrices and ZME-Matrices that Commute with a Permutation

Matrix. SIAM Journal on Matrix Analysis and Applications, 9(3), 408-418. doi:

10.1137/0609036.

2. Cambridge Univ Pr. (2016). Matices Intermediate + Eleteca.

3. Zhang, S., & Holzapfel, W. Orthogonal polarization in lasers.