• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Matrix Binomials IA

Extracts from this document...

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.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related International Baccalaureate Maths essays

1. ## Math Studies I.A

Validity The choice of sampling is systemic sampling because it avoids human nature of emotion and bias. Other form of sampling such as simple random sampling, stratified sampling, convenience sampling and so on are not used. When simple random sampling is used human emotions can interfere while choosing countries, although

2. ## Math IA - Logan's Logo

Going back to the definition of a period, how long it takes for the curve to repeat itself, it makes sense then that by finding the difference between the x-values of the maximum and minimum points on the curve, we would find half the period.

1. ## matrix power

Also we also found that the end result of the new matrix forms an identity matrix. The above statements will perhaps be limited only to matrices in the form of, for other different types of matrices it is not known for it was not investigated by the student.

2. ## Math IA - Matrix Binomials

(We can also check this on a calculator). Now, using the expression, we find the value of X5 to be the same (proving the accuracy of our expression): X5=25-1 =24 =16 = We conclude that is indeed a valid expression.

1. ## MATH IA- Filling up the petrol tank ARWA and BAO

∴ 20km+d/10≤39.2\$km/(2×0.80\$) ∴20km+d/10≤24.5km ∴d/10≤2.5km ∴d≤25km ∴25km is the maximum distance of detour Bao should drive in order to get a 2% saving. Let’s investigate the relationship between d and p2 when E2 is kept constant! Let’s assume that f, r and w are constant at 20km/liter, 20km and 10days respectively.

2. ## Math IA Type 1 Circles. The aim of this task is to investigate ...

Part V Now that we have completed testing the validity, the scope and limitations of the general statement can be discussed. It is important to not only consider what is valid in relation to the general statement, but what is also valid in relation to the triangles and to the real world.

1. ## SL Math IA: Fishing Rods

Therefore we have determined that the quadratic equations given the points {(1,10), (6,96), (8,149)} is . Averaging of the Two Equations The next step in finding our quadratic function is to average out our established a, b, and c values from the two sets data.

2. ## Gold Medal heights IB IA- score 15

In order to make this function more accurate a few adjustments need to be made. The restrictions on the function as established in the beginning remains the same. Domain of the function in Figure 6 is {tÏµâ |1932 ≤ x ≤ 1980} and the range for this function is {hÏµâ |197 ≤ y ≤ 236}. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 