ia mathhl type 1 - lasse

Lasse Ø. Løvik 2IBA Side 1

Mathematical Investigation

Type 1 13/05 2011

PATTERNS WITHIN SYSTEMS OF LINEAR EQUATIONS

Part A Consider this 2x2 system of linear equations: �� 2� � 32� � � � �4

�

When examining the constants of the first equation x +2y = 3 there is a very clear arithmetic

pattern. 1 is the constant for variable x, 2 for the variable y and 3 is the third term. As the

terms increase by 1 for every term the common difference is 1.

This means that U1 = 1. U2 = 2 and U3 =3. As they make up an arithmetic sequence they

follow the general formula:

Un = U1 + (n1) d

n is number of term

U1 is the first term of the sequence and

d is the common difference

The formula for the arithmetic sequence given would be:

Un = 1+ (n1) (1)

If we put in: U1 we get 1. U2 we get 2. U3 we get 3.

The second equation is also an arithmetic sequence. However, in this situation the terms

decrease by 3. Hence the common difference is 3.

The constant of x is 2, hence U1 = 2. 1 is the constant of y, ergo U2 = 1. And the third term

U3 = 4

Using the general formula:

Un = U1 + (n1) d

This arithmetic sequence can be expressed:

Un = 2 + (n1) (3)

There are mainly two ways to solve this 2x2 system of equations: The method of substitution

or by looking at the coordinates of the point of intersection of the graphs.

Using the substitution method leads to:


�� 2� � 3

2� � � � �4�

� � 2� � 3 � � �3 � �

2

Inserting � � ��

� into 2x – y = -4 gives: y = 2.

Putting y = 2 into � � 2� � 3, gives x = 1

To solve this graphically one should first solve both equations for y:

� � 2� � 3 � � �3 � �

2

2� � � � �4 � � � 4 � 2�

The point of

intersection is the

solution:

x = 1, y = 2.

At the point (1,2)

the lines are equal

y = 4 +2x

y �3 � x

2

(1,2)

x

y


The graphs intersect in the point (1,2). We get the same answer when solving the graphs

graphically as when one use the substitution method.

In order to make similar graph systems, one must first define what is meant by similar. One

can assume that similar would refer to a system of two linear equations which have

arithmetic characteristics. In order of being similar to the previously examined system it

should have two linear equations where one has a positive common difference while the

other one is

negative.

�4x � 7y � 10

13x � 6y � �1�

� ��

�

� ��1 � 13�

6

x = 1

y = 2

�5� � 10� � 15

3� � 2� � 1�

� � ��

��

� �1 � 3�

2

x = 1

y = 2

x

y

(1,2)

� �10 � 4�

7

� ��1 � 13�

6

x

y

� �15 � 5�

10

� �1 � 3�

2

(1,2)


�3� � 4� � 56� � 5� � 4

�

� �5 � 3�

4

� �4 � 6�

5

x = 1

y = 2

�13� � 50� � 87

40� � 11� � �18�

� � ��

��

� �18 � 40�

11

x = 1

y = 2

y

� �5 � 3�

4

� �4 � 6�

5

x

y

� �87 � 13�

50

(1,2)

(1,2)

� ��18 � 40�

11

x


�101� � 201� � 301

15� � 12� � 9�

� �301 � 101�

201

� � ��

��

x = 1

y = 2

In these examples one equation increases while the other one declines. They all end up

intersecting each other in the point (1,2). In the following graph all graphs of my examples

are shown. The red lines are the ones with a negative common difference.

x

� �301 � 101�

201

(1,2)

x

y

� �9 � 15�

12

(1,2)

y


But what if we had a 2x2 system with two linear arithmetic equations which both have either

a positive or negative common differences?

�� 2� � 3

5� � 13� � 21�

� � ��

�

� �21 � 5�

13

x = 1

y = 2

��13� � 110� � 233

50� � 83� � 116�

� �233 � 13�

110

� � ��

��

x = 1

y = 2

x

y

(1,2)

� �3 � �

2

� �21 � 5�

13

x

(1,2)

� �233 � 13�

110

� �116 � 50�

83

y


Both systems that contained arithmetic function with positive common difference had the

same intersection point as the previous lines (1,2)

�16� � � � �18

�5� � 10� � �15�

� � 18 � 16�

� �15 � 5�

10

x = 1

y = 2

�110� � 20� � �70�10� � 33� � �56

�

� ��

��

� �56 � 10�

33

x = 1

y = 2

x

y

� �15 � 5�

10

� � 18 � 16�

(1,2)

(1,2)

x

y

� ��70 � 110�

20

� �56 � 10�

33


Also the two systems containing negative common differences have the same intersection

point (1,2). Seemingly every two linear equations on the form ax + by = c where the

coefficients a, b and c form an arithmetic will intersect in the coordinates x = 1 and y = 2.

I now want to check the above assumption:

If we let the first term of the first equation be U1 and the common difference d the first

equation will be:

U1 x + (U1 + d) y = (U1 + 2d)

If we set the first term of the second equation to be Q1 and the common difference to w the

second equation would become:

Q1 x + (Q1 + w) y = (Q1 + 2w)

Solving the two equation gives:

�� 2�� 2��

�

First of all we need to eliminate x. This can be done by multiplying the first equation with Q1

and the second equation with U1. This results in:

U1Q1 x + (U1Q1 + Q1 d)y = (U1Q1 + Q12d)

Q1U1 x + (Q1U1 + U1w)y = (Q1U1 + U12w)

Thereafter we subtract the first equation with the second equation to remove the similar

variables. This cancels out the x variable. This leads to:

U1Q1 y +Q1 d y = U1Q1 + Q1 2d

Q1U1y + U1wy = Q1U1 + U12w

Furthermore :

U1Q1 x – Q1U1 x + Q1dy – U1wy = U1Q1 – Q1U1 + Q1 d – U12w

U1Q1 x – Q1U1 x + Q1dy – U1wy = U1Q1 – Q1U1 + Q12d – U12w

Q1dy – U1wy = Q12d – U12w

y ( dQ1 – wU1) = 2 (dQ1 – wU1)

y = 2

As the constants on both side of the equation are equal, y must be 2 to solve the equation.


To solve for x we put y = 2 in the equations:

Equation 2:

Q1 x + (Q1 + w) y = (Q1 + 2w)

Q1x + (Q1 + w)2 = (Q1 + 2w)

Q1x + 2Q1 + 2w = Q1 + 2w

Q1x = Q1 – 2Q1

Q1x = -Q1

x = 1

Hence the solution / point of intersection is (1,2) for any 2 x 2 systems following this

pattern.

Now I want to investigate 3 x 3 systems with this same arithmetic pattern.

Starting out with three planes with arithmetic coefficients :

x + 2y + 3z = 4

x – 3y – 7z = -11

3x + 7y + 11z = 15

From the graph we can deduce that

these three planes intersect in a

straight line.

Using row reduction we can find the

equation of this line :

�


Observing that every number in the third row is 0 we can conclude that we have an infinite

number of solutions. Let z = t, where t is a parameter.

From row two : y +2t = 3. Hence y = 3 – 2t

Row one : x + 2(3 – 2t) + 3t = 4

x = t – 2

The parametric equation of the line of intersection between the three planes is then :

x = t – 2

y = 3 – 2 t

z = t

Looking at other examples :

x + 7y + 13z = 19

7x – y – 9z = 17

11x +21y +31z = 41

�

Here as well we see that the third row only contains zeroes. Therefore also this has an

infinite number of solutions. But do the planes meet in the same line as the planes of the

previous example ?

Let z = t, where t is a parameter.

From row two : y + 2t = 3. Hence y = 3 – 2t

Row one : x + 7(3 – 2t) + 13t = 19


x + 21 – 14t + 13t = 19

x = t – 2

The parametric equation of the line of intersection between the three planes is then again:

x = t – 2

y = 3 – 2 t

z = t

Another example :

x – 23y – 47z = 71

13x + 21y + 29z = 37

x + 3y + 5z = 7

�

Using above method we get the line:

x = t – 2

y = 3 – 2t

z = t


x – 9y – 19z = 29

2x +7y +12z = 17

x + 11y + 21z = 31

�

x = t – 2

y = 3 – 2t

z = t

We can observe that each of the 3x3 systems intersect in the same line. Assuming this is true

for all 3x3 systems with similar pattern.

Let u be the first term in the first equation, and d the common difference :

�� 2�� 3��

Let q be the first term in the second equation, and w the common difference:

�� 2�� 3��

In the third equation let the first term be t, and the common difference m

�� 2�� 3��

The general formula for the 3x3 system = �� 2�� 3�� 2�� 3�� 2�� 3��

�


�

Let z = t.

From row two : y + 2t = 3, hence y = 3 – 2t

Row one : � � ��

��

��

�

� � � � 3�

��

�� 2��

� ��

�

� � � � 3�

��

�� 2��

� �3 � 2��

�

� � 1 �3��

� � �2��

�� 3 �

3��

� 2� �2��

�

� � � � 2

So the line of intersection for every 3x3 system where the coefficients form an arithmetic

sequence, is

x = t – 2

y = 3 – 2t

z = t

To conclude we can say that all 2x2 systems with arithmetic properties as seen above will

intersect in the same point (1,2). The restrictions are that the first term and the common

difference can not be zero. Of course the coefficients a, b and c of the equations on the form

ax + by = c must form an arithmetic sequence.

The 3x3 systems that have the same arithmetic properties as examined above will all

intersect in the same line as shown above. The restrictions are that it must have the same

arithmetic pattern as observed above . The first term and the common difference can not be

zero.


Part B Consider this 2x2 system �� 2� � 4

5� � � � �

�

�

If we start by looking at the first equation one can observe that the constants behave in a

geometric manner. x has the constant 1, y has 2 and then we have 4. The common ratio

between them is 2. If we look at the general equation :

��

Where n is the number of term

u1 is the first term

And r the common ratio

Using the general equation the first equation would look like this

�� 1 � 2��

And the terms could be found by changing n :

�� 1 � 2��

�� 1 � 2��

�� 1 � 2��

u1 = 1, u2 = 2, u3 = 4.

Examining the second equation 5� � � � �

� it is not as clear what the common ratio is. But

the constant for x = 5, for y = 1, and the last one is �

�

Using the formula : � � ��

�� we find � � � �

�

Transforming this equation into the general one would give:

�� 5 � ��15

��

So also the second equation act as a geometric sequence .

To look at the two equation from a different angle it might be smart to transform them from

the �� format to y = ax +b, (the slope – intercept form).

The equation would become :

� � 2� � 4 � � � ��

� � � � �

�

�� 2


5� � � ��

� � � � 5� �

�

�

If we now examine the relationship between the constants in each equation we get :

� � ��

�� 2 , a = �

�

�, b = 2

� � 5� ��

�, a = 5, b = �

�

�

It is very clear that both equation have something in common as in both cases b is the

negative reciprocal of a. This means that � � � in equation one = � � � in equation two.

�12

� 2 � 5 � �15

And we find that

� � � � �1

Also we see that b = r, that is the yintercept is the same as the common ratio. This means

that a = � �

� and the slope – intercept form then can be written as � � � �

��

When plotting these two lines and other families following the same patterns into GeoGebra

we find that the straight lines form an outline of a parabola which the lines never intersect:

a beautiful parabola symmetric about the xaxis.

x

Y


We can create a general expression for 2x2 systems incorporating the pattern seen in the

previous graph:

For the first equation:

��

Let u1 be the first term.

r is the common ratio.

For the second equation:

��

Let q1 be the first term.

w is the common ratio.

The easiest way of solving this is to first eliminate x by multiplying with q1 in the first

equation and u1 in the second equation and subtracting.

��

��

��

��

� ��

� � �

� � � � �

Putting the value of y into the second equation gives:

��

� ��

��

� � ��

� � ��

So for lines � � � �

�� and � � � �

�� the point of intersection is (wr, r + w).


Using the general equation ��

This can be rearranged: �� 0

Treating as a quadratic equation with r as the variable we get that

� �� 4��

2��

Seeing as all the lines are tangents to the parabola, the discriminant equals zero

�� 4�� 0

�� 4��

�� 0

�� 4� � 0

Drawing this parabola shows that this is the parabola outlined by my straight lines on the

form �� :

Same graph as the previous except the colours.

Restrictions: r and w cannot be 0.

��

x

Y


To sum up part B; 2x2 systems of linear equations with geometric coefficients, I found that

the point of intersection only depended on the common ratio of the geometric equations.

The investigation only covered geometric linear equations, which means that first term and

common ratio must be different from zero. When plotting many similar equations they

formed a region with a parabolic shape, where there are no solutions to the 2x2 systems.

Technology used:

For row reduction and matrices I used wxMaxima.

For 2d graphs I used GeoGebra and Graphic Calculator 3D for 3D graphs.

Due to a maximum allowed amount of graphs in GeoGebra it was impossible to add more

straight lines to better show the parabola.

Texas TI84 plus

��

ia mathhl type 1 - lasse

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