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

Math IB portfolio assignment - MATRICES

Extracts from this document...

Introduction

Ankit Shahi

Investigating Matrix Binomials

Introduction to Matrix Binomials

Matrix Binomials can be defined as a type of a 2 by 2 matrix. Generally speaking, matrix binomials come in the form . These matrix binomials can be defined as the sum of two component matrices. One component should be known as the positive matrix. All elements within the positive matrix have the same positive value. The other part should be called as the negative matrix. All elements within the negative matrix have the same magnitude but the top right and bottom left elements have a negative value.

The overall goal of this project is to investigate the properties of these matrix binomials in relation to its positive and negative matrix components.

The first step would be investigating the positive and negative matrix components separately as they are the simplest components. We shall begin by defining X and Y as the simplest positive and negative matrices respectively and finding their general expressions.

Let   From these experimentations with the matrix X, we notice a clear pattern. In each consecutive matrix, the values of all four elements are twice as great as the values within the previous matrix. Therefore, as the power of X is increased by one, the values of all the elements within the matrix are multiplied by two. This trend is understandable since the process of matrix multiplication is row by column.

Middle

Having explored two very basic 2 by 2 matrices that contained elements with values of 1 or -1 and found the very interesting property of (X+Y)n = Xn+ Yn, the next step would be explore more complex 2 by 2 matrices that are multiplied by a scalar multiple.

To investigate the positive and negative matrices, we shall suppose a matrix A where A=aX and a matrix B where B=bX and then let a = 2 and b =-3. Clearly, the values of the elements in each consecutive matrix are four times greater than the previous matrix. Therefore, with every increase in the power of A, the values of the elements within the product matrix are multiplied by a factor of 4. Once again, it should be noted that all elements stayed positive. The value of each element in the subsequent matrix is now multiplied by a power of -6. In other words, every time the power of B is increased by 1, the values of every element are multiplied by a factor of -6. It should be noted that multiplication by a factor of -6 does change the value of the elements from positive to negative and negative to positive.

Table 3: Matrices for An and Bn

 n An Bn 5  6  7  8  Once again, we have found the general patterns for matrices A and B when they are raised to a power under 5.

Conclusion

= An + Bn which can also be written as (A+B)n = An + Bn. This important property allows us to factor out matrix binomials and thus allows us to manipulate matrix binomials much more easily.

The general statement was also found through this project. Although a large formula, this general statement can help us quickly analyze and factor matrix binomials. Using this formula, we can factor out the base matrices X and Y rapidly in order to find the scalar multiples that were multiplied to them.

Nevertheless, despite the fact that many tests were done to test these properties and the general statement, they have yet to be proven true for all possible cases. Some cases that were missed include non-integer values for a, b, and n.

Another interesting property that was found was . This property could probably be extended to all matrices and not just matrix binomials, but insufficient time was spent investigating this property. More tests on different numbers and different base matrices must be done to verify this property further.

Overall, the research was a quick glimpse into a major property (Mn = An + Bn) of a matrix binomial. Many different tests were done to verify and to try to find a counter-example, but so far, all tests support this property. Further tests and an inductive proof is needed to verify this property. Of course, a general statement was also found based upon this property. The accuracy of this general statement is dependent on the accuracy of this property.

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. Extended Essay- Math

vï¿½ï¿½ï¿½='"`"X<ï¿½<ï¿½ï¿½Æï¿½ï¿½ï¿½-!y#Y_Ç-Äï¿½Å°I"ï¿½ï¿½'ï¿½vÖï¿½>*-ï¿½6Õ¶ï¿½mï¿½ï¿½;6:H+Iï¿½d(r)ï¿½Nï¿½Jï¿½Jï¿½ï¿½ï¿½ï¿½\ï¿½ï¿½ï¿½yï¿½ï¿½s(tm)ï¿½(r)ï¿½hï¿½ï¿½ï¿½Ý£ï¿½Cï¿½3/4 ï¿½)ï¿½*1/2KSï¿½1/2ï¿½l(sï¿½iï¿½ï¿½ Q0'ï¿½ï¿½ï¿½46Jï¿½ï¿½!ï¿½f1/4Kï¿½Zjï¿½"(c)ï¿½Lï¿½ï¿½+ 0[ ï¿½Xï¿½N*lï¿½Ò°IVBiï¿½O11/4Kï¿½FJNb]]ï¿½ï¿½ ï¿½2ï¿½Lï¿½ï¿½(tm)ï¿½ï¿½VYmd -ï¿½ï¿½' ï¿½ï¿½T!(sï¿½ï¿½VEgï¿½0ï¿½ï¿½W#-2ï¿½ï¿½ hR5bï¿½_(ï¿½ï¿½;ï¿½ï¿½cNï¿½ï¿½ï¿½Cï¿½ï¿½Jï¿½(tm)ï¿½[ï¿½Ó­+ -Ð¬bï¿½ï¿½ï¿½\J)ï¿½ 3/4`6"ï¿½(tm)ï¿½q4ï¿½_ï¿½?ï¿½ï¿½wiï¿½\$ï¿½dï¿½Gï¿½3/4Kï¿½FJJQ9ï¿½ëï¿½ ï¿½ï¿½ï¿½qï¿½ï¿½ï¿½ï¿½ï¿½ ï¿½ï¿½ï¿½X6t"ï¿½ï¿½Q]\*ï¿½"kÓ(tm)ï¿½eï¿½]~ï¿½,-ï¿½Jï¿½ï¿½K3cï¿½%86;ï¿½ï¿½ï¿½ï¿½Ôï¿½Swï¿½_ï¿½ï¿½ï¿½ï¿½ ï¿½4P'(c)ï¿½eï¿½ï¿½Wï¿½wï¿½djï¿½ï¿½ï¿½X(ï¿½*ï¿½)j6lï¿½ï¿½ï¿½ b- ï¿½z1ï¿½fÓ¯ï¿½ï¿½w'sï¿½ï¿½%(c)...ï¿½?ï¿½Æï¿½ï¿½1/43ï¿½ï¿½"ï¿½ï¿½ï¿½" ï¿½È¸1/4ï¿½"3TBï¿½/ï¿½ÛJR(c)'1/42ï¿½]ï¿½ï¿½hï¿½ï¿½ï¿½ï¿½psG(c)ï¿½PQ ï¿½Åï¿½ï¿½8ï¿½;((tm)+ XØ/lW7ï¿½rqi Gj\$"(tm)wYï¿½}ï¿½ï¿½ï¿½nï¿½eï¿½!%%M}ï¿½ï¿½`ï¿½ï¿½|?ï¿½ï¿½}8ï¿½U,ï¿½qï¿½ï¿½×²ï¿½ï¿½ï¿½[iëºX+ï¿½\$xï¿½ï¿½3ï¿½=~ï¿½?|bï¿½ï¿½.ï¿½Ò¥wï¿½7yï¿½ï¿½ï¿½yï¿½I|ï¿½`sï¿½Ø¸ï¿½\g'ï¿½ ' ï¿½ï¿½ ï¿½"Kï¿½jSï¿½ï¿½Aï¿½ï¿½Wï¿½ueï¿½ï¿½Bï¿½ï¿½4IDATï¿½yg3...(tm)'ï¿½?ï¿½ï¿½ï¿½ï¿½ï¿½J-ï¿½]ï¿½H\$R:ï¿½ï¿½(tm)ï¿½0e-ï¿½VIï¿½'-&Sï¿½sï¿½ï¿½\ï¿½ï¿½ï¿½tdZo[ï¿½"Mï¿½{ï¿½ï¿½:?ï¿½ï¿½"ï¿½ï¿½Gï¿½â¯ G&q#!ï¿½ï¿½&ï¿½JYï¿½Xï¿½ï¿½tÐï¿½-f< "ï¿½ï¿½ï¿½ ï¿½^ï¿½ï¿½ Nï¿½Vï¿½ï¿½8ï¿½Kï¿½ï¿½Xï¿½ï¿½ï¿½ï¿½ï¿½<ï¿½ ï¿½Qï¿½ï¿½ï¿½Mï¿½ï¿½V9ï¿½c"0ï¿½6ï¿½O×.ï¿½RkTï¿½T(1ï¿½ ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½tï¿½Zï¿½wNï¿½ï¿½mï¿½ï¿½ï¿½ï¿½tï¿½qlr8ï¿½ 3ï¿½ï¿½ï¿½"\$ï¿½7ï¿½3ï¿½ï¿½Þï¿½iP(ï¿½d2ï¿½ï¿½b1*/ï¿½ï¿½ï¿½ï¿½ï¿½JºÎªï¿½' /ï¿½ï¿½ï¿½0ï¿½bï¿½+ï¿½ ï¿½=ï¿½3/4ï¿½ï¿½ï¿½ Bï¿½ï¿½mï¿½Uï¿½"'N&(r)%ï¿½ï¿½F3Pï¿½ï¿½"4-ï¿½Yixï¿½a~~~ï¿½Mï¿½"(c)ï¿½b`*ï¿½1/2ï¿½1/2"ï¿½Lï¿½> Q (c)ï¿½9ï¿½ï¿½%ï¿½<ï¿½`2ï¿½ï¿½ï¿½"ï¿½(c)Ï¦ï¿½7b<T>ï¿½D"ï¿½*&...{6ï¿½mï¿½ï¿½ï¿½ï¿½eï¿½ï¿½ï¿½ ï¿½"ï¿½bï¿½ï¿½×ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½1/2Kï¿½jï¿½x&ï¿½(2*ï¿½ï¿½ï¿½-Oï¿½ï¿½ï¿½P1qï¿½z6\$-fï¿½9ï¿½ySï¿½ï¿½Iï¿½3ï¿½9ï¿½ï¿½ï¿½ ï¿½.-ï¿½uVvFï¿½ï¿½(r)Þ¦1\M.ï¿½N<ï¿½ ï¿½ï¿½ï¿½%ï¿½>B ï¿½ï¿½aPF"ï¿½\:ï¿½8Ý"ï¿½ï¿½Pï¿½"ï¿½\ï¿½3/4ï¿½ï¿½ :R-rï¿½1/4"3ï¿½ï¿½R"Nï¿½Wï¿½^ï¿½ï¿½ï¿½@oKrï¿½L&1ï¿½ #e#- /ï¿½ï¿½ï¿½ï¿½@.@ï¿½ï¿½ï¿½ï¿½ï¿½m"ï¿½w ï¿½ï¿½ï¿½ï¿½"E@.@"(r)^1/2zï¿½ï¿½Jnï¿½zï¿½gï¿½ï¿½ï¿½ï¿½-ï¿½ï¿½;'ï¿½ï¿½Ý²ï¿½Ú*...ï¿½qÞµDwï¿½Gxï¿½ï¿½ï¿½ï¿½!ï¿½ï¿½ï¿½sf_Tï¿½\$ï¿½ï¿½ï¿½!ï¿½u1/4,å;Phï¿½Ð°ï¿½...

2. Math IA - Matrix Binomials

For Yn: When n=1, 2, 3, 4, ... (integer powers increase), then the corresponding elements of each matrix are: 1, 2, 4, 8, ... These terms represent the pattern between the scalar values multiplied to Y= to achieve an end product of Yn. Thus, we can now deduce the geometric sequence of these scalar values using the general

1. matrix power

The first is multiplying the matrix by "n" times. The second method is simply bringing the individual digits in the matrix by the power of "n". Therefore for n = 10, 20, and 50 we can simply use the second method to determine the power of the matrix at the

2. Artificial Intelligence &amp;amp; Math

However, negative aspects can also be raised, for example, that laptops prevent students from concentrating on their school work, and degrade learning (Borja, 2000). Not all students can afford laptops to buy a laptop for school, so their introduction has also raised the issue of equality and financial discrimination (Corcoran, 2002).

1. Math IB HL math portfolio type I - polynomials

Solution: P(x) = a3x3 + a2x2 + a1x + a0 P(k) = a3k3 + a2k2 + a1k + a0 Suppose P(k) = 0 then a3k3 + a2k2 + a1k + a0 = 0 a0 = -(a3k3 + a2k2 + a1k)

2. Math Portfolio Type II

the following logistic function graph: - Again, the table and graph show how an initial growth rate of more than 2 makes the population grow rapidly beyond the sustainable limit (which is 60,000) to 62577 within 3 years which is more rapid and higher than when the growth rate is 2.

1. Circles Portfolio. The purpose of this assignment is to investigate several positions of ...

aâ=OP’= aâ= OP’= aâ= OP’=0.33 ________________ Diagram 3: r=1, OP=4 In diagram 3: Triangle OPA Triangle OP’A aâ=OP=4 aâ=OP’=? o=PA=4 oâ= P’A=1 p=p’=AO=1 p’=p=AO=1 Finding â¢OAP (I will use the procedure as in Diagram 1 and 2) : aâ²=o²+p²-2·o·p·cos (A)

2. Gold Medal heights IB IA- score 15

An acknowledgement of the specific point (1980, 236) is made as it shows a significant increase in height which may provide limitation of the sinusoidal function however, it still models the correlation better in comparison with the other function. Therefore, the sinusoidal function will best represent the data in figure 1 and will be used in this investigation. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to
improve your own work 