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Math IA - Matrix Binomials

Extracts from this document...

Introduction

IBO INTERNAL ASSESSMENT

Matrix Binomials

Mathematics SL Type I

Katie Xie

Mrs. Cheng

9/23/2008


Let X=image07.pngand Y=image08.png. Calculate X2, X3, X4; Y2, Y3, Y4.

X2=X ● X

X2=image07.pngimage07.png

X2=image104.png

X2=image24.png

X3=X2 ● X

(Matrix multiplication is associative)

X3=image24.pngimage07.png

X3=image122.png

X3=image09.png

X4=X3 ● X

X4=image09.pngimage07.png

X4=image19.png

X4=image12.png

Y2=Y ● Y

Y2=image08.pngimage08.png

Y2=image40.png

Y2=image37.png

Y3=Y2 ● Y

(Matrix multiplication is associative)

Y3=image37.pngimage08.png

Y3=image56.png

Y3=image58.png

Y4=Y3 ● Y

Y4=image58.pngimage08.png

Y4=image73.png

Y4=image39.png

As we can see, a general trend emerges as we increase the power of the matrix. There is a definite relationship between the power of the matrix and the end product (entries in the matrix). We observe that when X is to the power of 2, i.e. X2, the matrix’s entries are all 2’s; when X3, the entries are all 4’s; when X 4, the entries are all 8’s.

For the Y matrix, a similar pattern emerges, except in this case, we must note the negative signs. However, these two negative numbers always occupy the same position in the matrix when the power is increased.

Through consideration of the integer powers of X and Y, we can now find expressions for Xn, Yn, (X+Y)n.

We observe that the elements of the matrices appear to form a geometric sequence; thus we can use the general equation for a geometric sequence to determine the expressions for Xn, Yn, (X+Y)n:

image87.png, where Un=a specific term

                                    a=first term

...read more.

Middle

When a=2, i.e. A2=(2X)2:

A2=A●A

=image24.pngimage24.png

=image123.png

=image12.png

When a=2, i.e. A3=(2X)3:

A3=A2●A

=image12.pngimage24.png

=image124.png

=image125.png

When a=2, i.e. A4=(2X)4:

A4=A3●A

=image125.pngimage24.png

=image126.png

=image15.png

When a=-2, i.e. A= -2X:

A= -2image07.pngimage05.png

A=image10.png

When a= -2, i.e. A2=(-2X)2:

A2=A●A

=image10.pngimage10.png

=image11.png

=image12.png

When a= -2, i.e. A3=(-2X)3:

A3=A2●A

=image12.pngimage10.png

=

=image13.png

When a= -2, i.e. A4=(-2X)4:

A4=A3●A

=image13.pngimage10.png

=image14.png

=image15.png

When a=10, i.e. A=10X:

A=10image07.pngimage05.png

A=image16.png

When a=10, i.e. A2=(10X)2:

A2=A●A

=image16.pngimage16.png

=image17.png

=image18.png

When a=10, i.e. A3=(10X)3:

A3=A2●A

=image18.pngimage16.png

=image20.png

=image21.png

When a=10, i.e. A4=(10X)4:

A4=A3●A

=image21.pngimage16.png

=image22.png

=image23.png

Through consideration of the integer powers of A (2, -2, and 10), we can observe a pattern and find an expression for An.

For An when a=2:

When n=1, 2, 3, 4, … (integer powers increase), then the corresponding elements of each matrix are:

             1, 4, 16, 64, … These terms represent the pattern between the scalar values multiplied to

                                      A=aX where a=2 and hence A=image24.png to achieve an end product of An.

Thus, we can now deduce the geometric sequence of these scalar values using the general equation listed above:

image03.pngimage06.png

In the sequence {1, 4, 16, 64}, f=1

                                                   r=4

image25.png

image26.png

image27.png

Here, we can express r in terms of a (a=2):

r=4

4=2●2

image28.pngr=2a

We can also express image27.pngin terms of a:

image27.png

image29.png

For An when a= -2:

When n=1, 2, 3, 4, … (integer powers increase)

...read more.

Conclusion

(A+B)3=(A+B)2(A+B)

(A+B)3=image68.png

=image67.pngimage69.png

= image67.png(A+B)

=image70.png

But AB and BA both equal 0, as:

A=aX

=aimage07.png

=image71.png

B=bY

=bimage08.png

=image72.png

So AB=image71.pngimage72.png

=image74.png

=image75.png

=0

BA=image72.pngimage71.png

=image76.png

=image75.png

=0

We re-write image70.pngas image77.png

=image78.png

=image79.png

Where n=4, i.e. (A+B)4:

(A+B)4=( A+B)( A+B)( A+B)( A+B)

(A+B)4=( A+B)3(A+B)

(A+B) 4=image80.png(A+B)

=image81.png

=image82.png

=image83.png

=A4+B4

We can now find an expression for (A+B)n through consideration of the integer powers of (A+B) and observing the pattern that has emerged.

(A+B)=A+B

(A+B)2=A2+B2

(A+B)3=A3+B3

(A+B)4=A4+B4, and so on.

Thus, we can conclude that (A+B)n=An+Bn, which can be rewritten as follows (to be expressed in terms of X and Y):

(A+B)n=An+Bn

=(aX)n+(bY)n

=anXn+bn+Yn

Now we consider M=image84.png, where M=A+B:

M=A+B

image85.png= aX+bY

image85.png=image86.png+image88.png

image85.png=image85.png

image89.png

We can also prove that M2=A2+B2 (which we can use to later calculate M3=A3+B3, M4=A4+B4 to find a general statement for Mn, in terms of aX and bY):

M2=A2+B2

image84.pngimage84.png=(aX)2+(bY)2

image90.png=image86.pngimage86.png+image88.pngimage88.png

image91.png=image92.png+image93.png

image94.png=image95.png+image96.png

image94.png=image94.png

image97.png

Now, tracing the pattern in order to find the expression for Mn, we see that:

M=A+B=image85.png

M2=A2+B2=image94.png

M3=A3+B3= image98.png

M4=A4+B4=image99.png

Expressing Mn in terms of aX and bY (=A and B respectively), we have arrived at the general statement:

Mn=(aX)n + (bY)n, which we can also express as

Mn=anXn + bnYn.

...read more.

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