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# Matrix Binomials IA

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Introduction

Tenzin Zomkey                Maths SL Type 1

Maths Portfolio Standard Level

International Baccalaureate

Matrix Binomials

The main aim of this portfolio is to investigate the matrix binomials and observe and determine a general expression from the patterns that we obtain through the workings. Throughout the project, I shall be using solely matrices of 2 x 2 formations, and investigate the patterns I find.

1. To begin with, we consider the matrices X = and Y = .

The values of these matrices, each raised to the power of 2, 3 and 4 are calculated, as shown below;

X2= X = Y2 = x = X3= x = and     Y3 = x = X4 = x = Y4 = x = It can be observed that all the matrices calculated above are in the form of 2 X 2, they are all square matrices. The corresponding diagonal elements are also observed to be the same. Since the matrices of each nth power can be seen to be the value of 1 less than the nth term, the general expression for the matrix Xn in terms of n is -

Xn = And the general expression for Yn is –

Yn = Middle B3= = A4= = B4= = Note: A GCD calculator (TI 83) has been used throughout this portfolio to calculate the matrices and other calculations.

Observing the above calculations, we can detect a certain pattern in determining the values, which gives us the general expression of A and B in terms of n as;

An = and                      Bn= Proof:

Taking n to be 4, we substitute the values in the expression of An

A4= A4 = A4 = A4 = And since A4 = is the correct value as calculated previously, this expression is proven true and consistent.

Likewise, to find the general expression of (A + B) n , the values of (A + B) raised to the powers 2,3 and 4 are calculated;

First, we find the value of (A + B)

(A + B)= + = (A + B) 2 = = (A + B) 3 = = (A + B) 4 = = Observing the repeating patterns in the calculations above, we can deduce the general expression of (A+B) in terms of n to be;

(A + B) n = 2 n-1 Proof:

Taking n as 3, and substituting it in the above expression –

(A + B) 3 = 2 3-1 (A + B) 3 = 2 2 (A + B) 3 = 4 (A + B) 3 = Conclusion

n = 2 n-1 whereby a=2, b= -2, and taking n=3, we calculate (A+B) raised to the third power –

(A+B)3 = 23-1 = 22 = 4 = So since we know from previous calculations that M3= , we can say that , M3= (A+B)3 .

Therefore, the general statement of Mnin terms of aX and bY is;

Mn= (aX+bY)n

Proof:

To check the validity of this general statement, we shall take different values for a, b and n. Suppose a= 3, b=4, and n= 2 –

M2= (3X+4Y)2

M2= M2= M2= M2= And since M = (A+B) = (aX+bY),

M = M = So then,

M2= = Therefore, the statement is proven true and consistent with all values of a, b and n.

5.  Using the Algebraic method, the general statement is to be verified and explained again.

Taking the expression, Mn =2 n-1 , we find:

M = M2= = 2 M3= = 2 = 2 = 2 =4 Proof:

Substituting the above with the initial values of a and b, we find M3 -

M3= 4 M3 = 4 M3= 4 M3= 4 M3= 4 M3= Therefore, since it has been shown earlier in our work that the value M3= is true and correct, it shows that the general statement of Mnin terms of aX and bY is true.

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

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