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Maths Portfolio Matrix Binomials

Extracts from this document...

Introduction

Maths Portfolio

SL Type 1

Matrix Binomials

In this mathematics portfolio we are instructed to investigate matrix binomials and algebraically find a general statement that combines perfectly with our matrices and equations given.

MATRIX BINOMIALS

Given that:

image01.png  and  image02.png

We calculated image13.png

Therefore:

image24.png

And

image35.png

We were then requested to find expressions for image46.png

by considering the integer powers of X and Y:

image56.png

These expressions were found by observing that the result of image67.pngwas always the matrix to the power of n multiplied by 2 to the power of n-1. The sequence of results gives us: 1, 2, 4 and 8, reaffirming our expressions are correct because image72.png.

Given that:

image73.png  and  image03.pngwhere a and b are constants

We were asked to find image04.pngusing different values of a and b and then find the expressions for image05.png.

Therefore:

image06.png

And:

image07.png

Assuming that a = 2

image08.png

Assuming that a = 3

image09.png

Assuming that a = 6

image10.png

Assuming that a = -4

image11.png

...read more.

Middle

As a result we can see that:

image19.png

And since we know image20.png

Hence:  image21.png

To find the expression for image22.png I did the same as for image23.png changing the value a to b and X to Y.

For example (assuming b = -1):

image25.png

As a result we can see that:

image26.png

And since we know image27.png

Hence:  image28.png

To find image29.pngI used the binomial theorem

image30.png

From the binomial theorem we can see that the values of A and B multiply by each other on every term except the first and the last, where we findimage31.png.

However if we multiply matrix A by B we will see that the product will be a zero matrix.

image32.png

This allows us to cancel every term in which A multiplies B or vice-versa, as the result will be zero.

...read more.

Conclusion

 = 1

image55.png

image57.png

Assuming: a = -5, b = 2, n = 1

image58.png

image59.png

Assuming: a = 1, b = 3, n = 2

image60.png

image61.png

Assuming: a = 1/2, b = 2, n = 1

image62.png

image63.png

Assuming: a = 2, b = 2, n = -2

image64.png

Assuming: a = 2, b = 2, n = 0

image65.png

image66.png

Not Compatible

After testing the validity of my general statement we can see that the results for both formulas were mostly compatible for all numbers of a, b and positive n, proving the validity of our statement. However when n is zero or a negative integer we find some problems with it. As we can’t power a matrix to a negative number n can’t be a negative number and when 0 we find out all matrices to the power of 0 form identity matrices that when added in the formula image68.png it differs from our other result which is a identity matrix as well.

To get to this formula algebraically I did:

image69.pngimage00.png

This can be concluded by:

image70.png

By knowing that AB =0:

image71.png

And finally I found the general expression:

image51.png

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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