# Math IA - Matrix Binomials

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Introduction

IBO INTERNAL ASSESSMENT |

Matrix Binomials |

Mathematics SL Type I |

Katie Xie Mrs. Cheng |

9/23/2008 |

Let X=and Y=. Calculate X2, X3, X4; Y2, Y3, Y4.

X2=X ● X

X2=

X2=

X2=

X3=X2 ● X

(Matrix multiplication is associative)

X3=

X3=

X3=

X4=X3 ● X

X4=

X4=

X4=

Y2=Y ● Y

Y2=

Y2=

Y2=

Y3=Y2 ● Y

(Matrix multiplication is associative)

Y3=

Y3=

Y3=

Y4=Y3 ● Y

Y4=

Y4=

Y4=

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:

, where Un=a specific term

a=first term

Middle

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

A2=A●A

=

=

=

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

A3=A2●A

=

=

=

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

A4=A3●A

=

=

=

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

A= -2

A=

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

A2=A●A

=

=

=

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

A3=A2●A

=

=

=

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

A4=A3●A

=

=

=

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

A=10

A=

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

A2=A●A

=

=

=

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

A3=A2●A

=

=

=

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

A4=A3●A

=

=

=

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= 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:

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

r=4

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

r=4

4=2●2

r=2a

We can also express in terms of a:

For An when a= -2:

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

Conclusion

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

(A+B)3=

=

= (A+B)

=

But AB and BA both equal 0, as:

A=aX

=a

=

B=bY

=b

=

So AB=

=

=

=0

BA=

=

=

=0

We re-write as

=

=

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=(A+B)

=

=

=

=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=, where M=A+B:

M=A+B

= aX+bY

=+

=

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

=(aX)2+(bY)2

=+

=+

=+

=

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

M=A+B=

M2=A2+B2=

M3=A3+B3=

M4=A4+B4=

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.

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

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