IB Math IA

02/06/12

Part A

Consider this 2× 2 system of linear equations: x + 2y = 3

2x - y = -4

We can see patterns in the constants of both equations. In the first equation, the constants are 1, 2, and 3. The common difference between the constants is 1: 3 – 2 = 1

2 – 1 = 1

Based on this, we can set up a general formula for the constant of this equation: Un = U1 + (n - 1)d Where:

n: The number of the series

d: the common difference in the series.

For the second equation, we also can see that the constants belong to a arithmetic series, which has the common difference of -3: -1 – 2 = -3

-4 – (-1) = -3

Un = U1 + (n+3)d

Solving the equation:

x + 2y = 3

2x – y = -4

* 2x + 4y = 6

2x – y = -4

* 5y = 10

* y = 2

* x + 4 = 3

* x = -1

Looking at the graph, we can see the intersection point A has the coordinate (-1, 2). At this point, the two lines are equal. Solving equation with similar formats:

x + 6y = 11Common difference is 5

6x – 2y = -10Common difference is -8

* 6x + 36y = 66

* 38y = 76

* y = 2

* x = -1

6x + 7y = 8Common difference is 1

2x + 4y = 6Common difference is 2

* 6x + 12y = 18

* 5y = 10

* y = 2

* x = -1

Looking at these systems of linear equations, we can conclude that any equations which have their constants follow a arithmetic series will have the answer for the variable y as 2 and variable x as -1. Because they all have the same solutions for x and y, their graph will all intersect at one point. In order to make a conjecture for these systems, we can generate a general formula for the general equations similar to those equations above: Let the first constant of the first equation be U, and the common difference be e. Let the first constant of the second equation be V, and the common difference be t. The first equation of the system will be:

Ux + (U + e)y = U + 2e

The second equation of the system will be:

Vx + (V + t)y = V + 2t

In order to solve these equations, we have to multiply the first equation by V to eliminate X, then multiply the second equation by U to eliminate x: Ux + (U + e)y = U + 2e

* UVx + (U + e)Vy = (U + 2e)V

Vx + (V + t)y = V + 2t

* UVx + (V + t)Uy = (V + 2t)U

Now, subtract the first equation to the second equation to eliminate UVx: (U + e)Vy – (V + t)Uy = (U + 2e)V – (V + 2t) U

* UVy + Vye – UVy – Uty = UV + 2eV – UV – 2tU.

* Vye – Uty = 2eV – 2tU

* y(Ve – Ut) = 2(eV – tU)

* y = 2

* Ux + 2(U + e) = U + 2e

* Ux + 2U + 2e = U + 2e

* Ux = -U

* x = -1

We can see the result is applicable to any equations which their constants follow arithmetic series. However, in a particular 2×2 system, the two common differences must be different from each other. Because if they are the same, the two equations will never touch each other, so there will be no possible answers for x and y.

Now, we are going to extend the system to 3×3 system. Consider the following equations: x + 3y + 5z = 7 (common difference is 2)

6x + 2y - 2z = -6 (common difference is -4)

2x + 9y + 16z = 23 (common difference is 7)

The system can be solved using matrix:

xyz 13562-22916 7-623

C A D

In order to find x, y, z, we need to find the inverse of the matrix A, then multiply it with D. Using technology to calculate detA, we get detA = 0. Therefore, this matrix is singular matrix. That means the system will either have many solutions or no solutions at all. Now, we are going to prove that any 3×3 system which has constants follow arithmetic series will either have many solutions or no solutions at all. Let the first term of the first equation be U1 and the common difference d Let the first term of the second equation be P1 and the common difference be c Let the first tem of the third equation of be Q1 and the common difference b The equations will look like:

U1 x + (U1 + d)y + (U1 + 2d)z = (U1 +3d)

P1...

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