• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8

# math portfolio type 1

Extracts from this document...

Introduction

IB Mathematics SL

Portfolio Type I

Matrix Powers

Done by: Bassam Al-Nawaiseh

IB II

• Introduction:

Matrices are rectangular tables of numbers or any algebraic quantities that can be added or multiplied in a specific arrangement. A matrix is a block of numbers that consists of columns and rows used to represent raw data, store information and to perform certain mathematical operations. The aim of this portfolio is to find general formulas for matrices in the form    .        of

Each set of matrices will have a trend in which a general formula for each example is deduced.

• Method 1:

Consider the matrix M = when k = 1.

Table 1: Represents the trend in matrix M = as n is changed in each trial.

 Power Matrix n = 1 n = 2 n = 3 n = 4 n = 5 n = 10 n = 20

Matrix M is a 2 x 2 square matrix which have an identity. As n changes the zero patterns is not affected while the 2 is affected. 2n is raised to the power of n. When n =1, 21 = 2, when n = 2, 2² = 4 and when n = 3, 2³ = 8 and so on. So as a conclusion, Mn =

• Method 2

Consider the matrices P =

Middle

1 = 20 =

n = 2

2 = 21 =

n = 3

3= 22 =

n = 4

4= 23=

n = 5

5= 24=

As for two consecutive matrices, the trend is found so as the power n increases by a factor of one. The scalar is doubled in each trial and then a certain factor is added to the elements of matrix S. taking n = 2 as an example, the scalar is found by doubling the power n = 1 (1x2=2). The factor of addition is determined by multiplying the difference in x and y inside the matrix with a power less with one by 3. In matrix n = 1, the difference in the elements is 2 (4-2=2), the answer is multiplied by 3 (2x3=6). Finally 6 is added to the matrix of n = 2. The general formula for this trend is Sn = 2  where n = 1. Notice that this formula needs two consecutive matrix powers in order to be applied. Another general formula can be derived for this sequence. As the power n changes, the power in which the scalar is raised will change also.

Sn = 2n-1  where n is an integer. To check the validity of this formula n = 4 is used as an example.

Using GDC: 4 =

Using the General Formula:4=23=8=.

• Method 3:
1. Consider the matrices in the form Q = .

Table 3: Represents the trend in matrix Q as k is increased by one in each trial.

 Power Matrix General Formula k = 1 M = Mn = 2n-1 k = 2 P = Pn = 2n-1 k = 3 S = Sn = 2n-1 k = 4 D = Dn = 2n-1 k = 5 F = Fn = 2n-1 k = 6 N = Sn = 2n-1

Conclusion

n, so matrix B = . From rule A and the above examples, the following formula is deduced: Bn = 2n-1

Using GDC: 2 =

Using Rule A: 2 = 22-1  = 2  =

As a conclusion, rule A also can be applied when k is a negative value.

1. Fraction Values:

Matrix L represents the matrix where k is a fraction value raised to the power of n, k = 3/2 and n = 3 in a matrix of the form n, so matrix L =. From the above examples, Ln = 2n-1.

Using GDC:  3 =

Using Rule A: 3 = 23-1 = 4  = .

From the above results, it is shown that rule A is valid even for fraction numbers.

1. Irrational Values:

Matrix U represents the matrix where k is an irrational number which is raised to the power of n, k =  and n = 2 in a matrix of the form

n, so matrix L = . From the above examples,

Un = 2n-1.

Using GDC: 2 = .

Using Rule A: 2 = 22-1  =

2  =

From those final results, we can prove that this statement, Rule A, can also be applied to irrational integers.

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 IB HL math portfolio type II. Deduce the formula Sn = ...

2 (a3 + a4) = (a1 + a2) + (a5 + a6) So we get: 2 (an + an+1) = (an-2 + an-1) + (an+2 + an+3) Proof: 2[a1 + (n - 1)d + a1 + nd] = a1 + (n - 3)d + a1 + (n - 2)d] +

2. ## Extended Essay- Math

ï¿½'-ï¿½I~3ï¿½ï¿½]'ï¿½eDwï¿½ï¿½ï¿½ï¿½ï¿½A"Iï¿½ï¿½ï¿½"ï¿½gï¿½-"(tm)ï¿½uï¿½"@ï¿½"ï¿½vï¿½3/4ï¿½*Zï¿½ ï¿½-ï¿½`/ï¿½ï¿½|ï¿½ï¿½Kï¿½"ï¿½ï¿½ ï¿½S7 ï¿½<c~ï¿½ ?ï¿½ï¿½ND\\\NNNï¿½]ï¿½nÂï¿½ï¿½Fï¿½-ï¿½nï¿½ï¿½Nï¿½ï¿½ï¿½`3/4ï¿½ï¿½3/4ï¿½ ï¿½ï¿½ï¿½ï¿½? ï¿½]ï¿½ï¿½(r)ï¿½ï¿½/Dï¿½ï¿½ï¿½3/4ï¿½Âï¿½"ï¿½? ï¿½ï¿½&ï¿½ï¿½an ï¿½ï¿½c?--ï¿½Zqï¿½"ï¿½...ï¿½Eï¿½'ï¿½.0ï¿½ï¿½3fï¿½@ï¿½ßodï¿½0ï¿½Ç>'!Rï¿½)ï¿½/"q1ï¿½ï¿½ï¿½ï¿½ï¿½cï¿½ï¿½ï¿½ï¿½8ï¿½Cv(ï¿½ ï¿½ï¿½6ï¿½ji-8i;ï¿½RRR,ï¿½ï¿½ï¿½ï¿½ï¿½1/2pï¿½#sï¿½)9ï¿½ï¿½'ï¿½ï¿½ï¿½Íï¿½ï¿½;ï¿½'mZ..ï¿½0ï¿½k8ï¿½ï¿½`ï¿½ï¿½`ï¿½/-s1/2%"Dï¿½x-,Ybï¿½ÞÚµkgeeéï¿½Bï¿½ï¿½-Ôï¿½ï¿½Ò¥ ï¿½ï¿½Z"ï¿½ï¿½ï¿½ï¿½ï¿½3/4ï¿½;ï¿½#@ï¿½Kjb"ï¿½ï¿½-=zï¿½--ï¿½ï¿½FXï¿½9ï¿½-ï¿½4pï¿½ï¿½3g ï¿½x-_ï¿½ ï¿½ï¿½ï¿½20ï¿½Eï¿½Sï¿½Qï¿½ï¿½ï¿½k{pï¿½KOOwÒ¦ï¿½ï¿½ï¿½RWdHï¿½Kg&8ï¿½wï¿½Ü¹U"Vï¿½\$ï¿½j?=ï¿½...ï¿½-OA\$ï¿½<ï¿½ï¿½Wï¿½ï¿½ g{lï¿½)'ï¿½Eï¿½-OY\$ï¿½<ï¿½ï¿½ï¿½ï¿½k{ï¿½vï¿½ï¿½"-ï¿½aï¿½\ï¿½"ï¿½ï¿½ï¿½b>ï¿½ï¿½ï¿½;ï¿½1/4ï¿½Mï¿½|ï¿½3`?ï¿½ï¿½ï¿½ocnn(r)ï¿½ï¿½Eï¿½P^8ï¿½[~ï¿½Mï¿½Ø·ï¿½ï¿½sï¿½ï¿½+ï¿½ï¿½ï¿½ï¿½x~ï¿½È¨}ï¿½ï¿½-ï¿½3/48igfÛ±oOï¿½zï¿½ï¿½ï¿½'ï¿½ubï¿½L2(r)ï¿½;oJï¿½ï¿½ï¿½Vï¿½^m"od=kÖ¬ï¿½`ï¿½'!Rï¿½)[~ï¿½mï¿½;imï¿½ï¿½ï¿½`(c)ï¿½-ï¿½ï¿½ï¿½;R4ï¿½ï¿½ï¿½ .../ï¿½ï¿½8iï¿½VV\aï¿½Kï¿½ï¿½ï¿½ï¿½"3ï¿½ï¿½ H~'ï¿½"ï¿½'m~ï¿½ï¿½'ï¿½Oï¿½#Rbï¿½rcï¿½"ï¿½Axï¿½\$ï¿½pï¿½W ï¿½ï¿½;v|ë­·ï¿½ï¿½ï¿½ï¿½ï¿½l,ï¿½[ï¿½ï¿½ï¿½24Ýï¿½ïRSS"''"ï¿½ynï¿½ï¿½xï¿½ï¿½ï¿½ï¿½wï¿½Æ³gï¿½ï¿½F:3'mZ.ï¿½}ï¿½ï¿½Ç¨--ï¿½*sï¿½\$ï¿½ï¿½Aï¿½zØ°aï¿½ vã¬*Z~Nï¿½Gï¿½Hï¿½ï¿½\$ï¿½hï¿½bF~lï¿½ï¿½ï¿½#-ï¿½9ï¿½ï¿½gï¿½2'oDï¿½"zHï¿½6'ï¿½x1ï¿½nï¿½Iï¿½Öï¿½nï¿½ï¿½[ï¿½ï¿½Ì¹'ï¿½Bï¿½7^ï¿½aßï¿½k{!ï¿½FÏï¿½!'Qï¿½&ï¿½/Î¬ï¿½3hï¿½ 8ï¿½;ï¿½ï¿½Y[ï¿½ï¿½Úµ+r>U'_ï¿½)ï¿½"Wï¿½^vGï¿½5ï¿½\ï¿½ï¿½ï¿½ï¿½]ï¿½'ï¿½ï¿½ï¿½...ï¿½=Lï¿½ï¿½9RPyï¿½ï¿½ï¿½"/ï¿½yBΕ/qWï¿½ï¿½amï¿½Ø±7n"sï¿½ï¿½Aï¿½ï¿½ï¿½ ï¿½''v`ï¿½ï¿½ï¿½ ï¿½ï¿½ï¿½7ï¿½a"ï¿½xï¿½"6ï¿½ï¿½ï¿½#ï¿½ï¿½ï¿½ï¿½QTï¿½c% ï¿½m7#ï¿½ï¿½bï¿½ï¿½Hï¿½-ï¿½5lï¿½ï¿½^ï¿½+(r)Cï¿½ï¿½~{Mï¿½tï¿½R"OÚ´VtOV(r)\imï¿½ï¿½MNï¿½ï¿½ï¿½hxï¿½cmï¿½ï¿½"6ï¿½U~ï¿½ï¿½ï¿½k-ï¿½ï¿½ï¿½ï¿½ï¿½1/2v(tm)n1/4ï¿½7~ï¿½(tm)gï¿½(tm)ï¿½1/27ï¿½"Eï¿½ï¿½9snï¿½ï¿½ï¿½S1ï¿½ï¿½ï¿½eï¿½ï¿½ï¿½ï¿½ï¿½ï¿½"ï¿½ï¿½4" ï¿½-_ï¿½ï¿½Û(r)Dsï¿½oï¿½10aï¿½ï¿½Ç9iï¿½r|ï¿½\ï¿½6aï¿½ï¿½Ê-ï¿½ï¿½ï¿½ï¿½ï¿½pï¿½>}ï¿½KWï¿½kÔ¨ï¿½ï¿½ï¿½ï¿½W ï¿½r\$ï¿½ï¿½ï¿½ &ï¿½ï¿½wï¿½.ï¿½Mku`\$_kï¿½"67vï¿½-6ï¿½ï¿½(c)Sï¿½χfï¿½ï¿½|f '_ [0a)ï¿½?ï¿½@Hï¿½,ï¿½-ï¿½ï¿½vEï¿½ï¿½ &,\ï¿½ï¿½ï¿½Sï¿½pï¿½ï¿½Ni%4ï¿½K>H~ï¿½l7ï¿½.]ï¿½ Dï¿½Sehï¿½ï¿½ï¿½gYÍ5#'ï¿½2Ë5 ï¿½qï¿½ï¿½ Kï¿½?2*ï¿½Yï¿½ï¿½ï¿½(tm)pï¿½oï¿½uï¿½gï¿½ï¿½ï¿½(tm)(tm)(tm)jï¿½""T(c)£ï¿½ï¿½(c)ï¿½ï¿½-ï¿½ï¿½`xï¿½ï¿½ï¿½ï¿½fï¿½'ï¿½ ï¿½*!ï¿½ ï¿½ï¿½(tm)ï¿½Qï¿½ï¿½ï¿½kï¿½ï¿½Zï¿½nÝ¤Iï¿½ï¿½ï¿½tï¿½ï¿½léï¿½ï¿½"ï¿½ lï¿½blï¿½ï¿½Zï¿½wï¿½ï¿½Ð¡Cï¿½f\$ï¿½pO*^1/48/dï¿½ï¿½Xï¿½ï¿½o PD1/4ï¿½Iï¿½Xï¿½Sï¿½...ÄWï¿½LU ï¿½_&ï¿½ï¿½ï¿½ï¿½ï¿½ ï¿½<ï¿½ ï¿½ï¿½ï¿½ï¿½ï¿½'ï¿½ï¿½Qï¿½ï¿½7nSï¿½ï¿½ï¿½1ï¿½ï¿½ï¿½Fï¿½1/4fÍï¿½ï¿½ï¿½'ï¿½ï¿½(c)" ï¿½ ï¿½ï¿½x×­['ï¿½Iï¿½5 ï¿½Zï¿½Zï¿½xzOlï¿½ï¿½ï¿½~#ï¿½ï¿½Luï¿½]"Vï¿½ ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½"ï¿½íï¿½ï¿½ï¿½ï¿½Uo:#oï¿½aË³ï¿½ ï¿½ï¿½...ï¿½"ï¿½"ï¿½ï¿½<ï¿½ï¿½9ï¿½+n.JOï¿½ï¿½l,iOÚY"h/`Fï¿½ï¿½'ï¿½ï¿½ï¿½1/276ï¿½edd ï¿½ï¿½Í1/2ï¿½Fï¿½ï¿½ï¿½=ï¿½ï¿½Wï¿½fËï¿½pï¿½fï¿½3/4}ï¿½ï¿½ Vï¿½ï¿½k""ï¿½ï¿½ï¿½;\+>ï¿½ï¿½ï¿½ ï¿½ï¿½`>Bï¿½\$&&ï¿½ï¿½Tï¿½ï¿½8Fï¿½Tï¿½3/4\$ï¿½^ï¿½Λ7Oï¿½"6ï¿½*ï¿½/Vï¿½ï¿½ï¿½ï¿½3/4ï¿½Lï¿½28ï¿½ï¿½e04ï¿½ï¿½(r)ï¿½eï¿½/ ï¿½5D ,ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½nï¿½Bï¿½ï¿½G7%ï¿½ï¿½"DÏ¸qï¿½<2ï¿½G"N3/4ï¿½h(vMï¿½G7(ï¿½ï¿½)))ï¿½"*-ï¿½N3/4ï¿½(tm)ï¿½-B\$ï¿½-,B^aï¿½ï¿½ï¿½=WGï¿½ (c)-Λtï¿½ï¿½ ""qï¿½ï¿½jï¿½ 6lï¿½ï¿½ï¿½.=~#'nï¿½ï¿½ï¿½ï¿½Ø¶ï¿½/-Sï¿½ ï¿½ï¿½a 7ï¿½ï¿½ï¿½ï¿½>}ï¿½\$\$\$"Mï¿½ï¿½Lï¿½aï¿½ï¿½ï¿½Jï¿½(tm)ï¿½kV"ï¿½ï¿½Gï¿½-6...

1. ## matrix power

For example, if, and In order to solve the matrix power above, we multiply the matrix by "n" number of times. Thus, if, and, then to solve we times the matrix "M" by two times. To multiply matrices, we take the rows of the left hand matrix and pair it with the column of the right hand matrix.

2. ## Math IA - Matrix Binomials

B3=(2Y)3: B3=B2?B = = = When b=2, i.e. A4=(2Y)4: B4=B3?B = = = When b=-2, i.e. B= -2Y: B= -2 B= When b= -2, i.e. B2=(-2Y)2: B2=B?B = = = When b= -2, i.e. B3=(-2Y)3: B3=B2?B = = = When b= -2, i.e. B4=(-2Y)4: B4=B3?B = = = When b=10, i.e.

1. ## IB Mathematics Portfolio - Modeling the amount of a drug in the bloodstream

Part B A patient is instructed to take 10 microgram of this drug every six hours. 1. Sketch... over a 24 hour period Model B a. I assume that the amount of drug in the slope of the curve after the initial dose would be steeper.

2. ## Math Portfolio: trigonometry investigation (circle trig)

equals a positive number on the positive x axis and the value of r equals a positive number as mentioned beforehand. Therefore, when the value of x is divided by the value of r, a positive number is divided by a positive number resulting to a positive number.

1. ## Math IA- Type 1 The Segments of a Polygon

Ratios of sides of triangles = 1:n Ratios of areas of equilateral triangles( Ratio of bigger triangle to smaller triangle) 1:2 7:1 1:3 13:4 1:4 21:9 Now if one were to look at the values of ratios above, then one can see relationships both in the numerator and denominator of the ratios.

2. ## Math Portfolio Type II

1.5 un = X2 = 60,000 rn = Y2 = 1 Substituting these values in the formula for slope we get = = = -1 x 10-5 Also, we have one pair of coordinates, that is, (u0, r0) = (10000, 1.5)

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to