Mathematics Review for MS Finance Students Lecture Notes

Transcription

1 Mathematics Review for MS Fiace Studets Lecture Notes Athoy M. Mario Departmet of Fiace ad Busiess Ecoomics Marshall School of Busiess Uiversity of Souther Califoria Los Ageles, CA 1

2 Lecture 1.1: Basics I. Sets 1. A set is a list or collectio of objects. The objects which compose a set are termed the elemets or members of the set. Remark 1: A set may be writte i tabular form, that is, by listig all of its elemets. As a example, cosider the set A which cosists of the positive itegers 1, 2 ad 3. We would write this set as A = {1, 2, 3}. Note that whe we defie a set, the order of listig is ot importat. That is, i our above example, A = {1, 2, 3} = {3, 1, 2}. Aother otatio for writig sets is the set-builder form. I the above example, A = {x x is a positive iteger, 1 x 3}. or A = {x : x is a positive iteger, 1 x 3} The otatio : or reads such that. The etire otatio reads A is the set of all umbers such that x is a positive iteger ad 1 x 3. Remark 2: The symbol "" reads "is a elemet of". Thus, i the same example 1 A ad 1, 2, 3, A. 2. If every elemet of a set S 1 is also a elemet of a set S 2, the S 1 is a subset of S 2 ad we write S 1 S 2 or S 2 S 1. The symbol "" reads "is cotaied i" ad "" reads "cotais". Examples: 2

3 #1. If S 1 = {1, 2}, S 2 = {1, 2, 3}, the S 1 S 2 or S 2 S 1. #2 The set of all positive itegers is a subset of the set of real umbers. Def 1: Two sets S 1 ad S 2 are said to be equal if ad oly if S 1 S 2 ad S 2 S 1. Remark 3: The largest subset of a set S is the set S itself. The smallest subset of a set S is the set which cotais o elemets. The set which cotais o elemets is the ull set, deoted. Remark 4: I Remark 3 we oted that the ull set is the smallest subset of ay set S. To see that the ull set is cotaied i ay set S, cosider the followig proof by cotradictio. Proof: Assume S, the there exists at least oe x such that x S. However, has o elemets. Thus, we have a cotradictio ad it must be that S. Remark 5: The set S which cotais zero as its oly elemet, i.e. S = {0}, is ot the ull set. That is, if S = {0}, the S sice 0 S. As a example of the ull set, cosider the set A where A = {x : x is a livig perso 1000 years of age} Clearly, A =. Def 2: Two sets S 1 ad S 2 are disjoit if ad oly if there does ot exist a x such that x S 1 ad x S 2. Example: If S 1 = {0} ad S 2 = {1, 2, 4}, the S 1 ad S 2 are disjoit. Def 3: The operatios of uio, itersectio, differece (relative complemet), ad complemet are defied for two sets A ad B as follows: (i) (ii) (iii) (iv) A B {x : x A or x B}, A B {x : x A ad x B}, A - B {x : x A, x B}, A {x : x A}. 3

4 Remark 6: I the applicatio of set theory all sets uder ivestigatio are likely to be subsets of a fixed set. We call this set the uiversal set ad deote it as U. For example, i huma populatio studies U would be the set of all people i the world. Remark 7: We have the followig laws of the algebra of sets: 1. Idempotet laws 1. a. A A = A 1. b. A A = A 2. Associative laws 2. a. (A B) C = A ( B C) 2. b. (A B) C = A ( B C) 3. Commutative laws 3. a. A B = B A 3. b. A B = B A 4. Distributive laws 4. a. A (B C) = (A B) (A C) 4. b. A (B C) = (A B) (A C) 5. Idetity laws 5. a. A = A 5. b. A U = U 5. c. A U = A 5. d. A = 6. Complemet laws 6. a. A A = U 6. b. (A) = A 6. c. A A = 6. d. U =, = U 4

5 Practice Problems I 1. Give the sets S 1 = {2,3,5}, S 2 = {3,5,6} ad S 3 = {7}, fid: a. S 1 S 3 b. S 1 S 2 c. S 2 S 2 d. S 2 U e. S 2 2. Give A = {4,5,6}, B = {3,4,6,7}, ad C = {2,3,6}, verify the distributive law. II. The Real Number System 1. The real umbers ca be geometrically represeted by poits o a straight lie. We would choose a poit, called the origi, to represet 0 ad aother poit, to the right, to represet 1. The it is possible to pair off the poits o the lie ad the real umbers such that each poit will represet a uique real umber ad each real umber will be represeted by a poit. We would call this lie the real lie. Numbers to the right of zero are the positive umbers ad those to the left of zero are the egative umbers. Zero is either positive or egative. -1-1/2 0 +1/2 +1 We shall deote the set of real umbers ad the real lie by R. 2. The itegers are the whole real umbers. Let I be the set of itegers such that I = {..., -2, -1, 0, 1, 2,...}. 3. The ratioal umbers, Q, are those real umbers which ca be expressed as the ratio of two itegers. Hece, Def: Q = {x x = p/q, p I, q I, q 0}. Clearly, each iteger is also a ratioal, because ay x I may be expressed as (x/1) Q. So we have that the set of itegers is a subset of the set of ratioal umbers, i.e., I Q. 5

6 4. The irratioal umbers, Q', are those real umbers which caot be expressed as the ratio of two itegers. Hece, the irratioals are those real umbers which are ot ratioal. They are the o-repeatig ifiite decimals. The set of irratioals Q is just the complemet of the set of ratioales Q i the set of reals (Here we speak of R as our uiversal set). Some examples of irratioal umbers are 5, 3 ad The atural umbers N are the positive itegers: N = {x x > 0, x I}, or N = {1, 2, 3,...}. 6. The prime umbers are those atural umbers say p, excludig 1, which are divisible oly by 1 ad p itself. A few examples are 2, 3, 5, 7, 11, 13, 17, 19, ad We may illustrate the real umber system with the followig lie diagram. Real Ratioal Irratioal Itegers Negative Itegers Zero Natural Prime 8. The exteded real umber system. The set of real umbers R may be exteded to iclude - ad +. Accordigly we would add to the real lie the poits - ad +. The result would be the exteded real umber system or the augmeted real lie. We deote these by R. The followig operatioal rules usually apply. (i) If "a" is a real umber, the - < a < + (ii) a + = + a =, if a - (iii) a + (-) = (-) + a = -, if a + 6

7 (iv) If 0 < a +, the a = a = a (-) = (-) a = - (v) If - a < 0, the a = a = - a (-) = (-) a = + (vi) If a is a real umber, the a a 0 9. Absolute Value Def 1: The absolute value of ay real umber x, deoted x, is defied as follows: x x if x 0 x if x 0 Remark 1: If x is a real umber, its absolute value x geometrically represets the distace betwee the poit x ad the poit 0 o the real lie. If a, b are real umbers, the a-b = b-a would represet the distace betwee a ad b o the real lie. Remark 2: The followig properties characterize absolute values: (i) a 0, (ii) a b a b, (iii) a b a b, (iv) a b a b, b 0, where a, b R. 10. Itervals o the real lie. Let a, b R where a < b, the we have the followig termiology: 7

8 (i) The set A = {x a x b}, deoted A = [a, b], is termed a closed iterval o R. (ote that a, b A) (ii) The set B = {x a < x b}, deoted B = (a, b], is termed a ope-closed iterval o R. (ote a b, b B) (iii) The set C = {x a x < b}, deoted C = [a, b), is termed a closed-ope iterval o R. (ote a C, b c) (iv) The set D = {x a < x < b}, deoted D = (a, b), is termed a ope-iterval o the real lie. (ote a, b D) Practice Problems II 1. Write the followig i set otatio ad i iterval otatio: a. The set of real umbers betwee 2 ad 10, iclusive. b. The set of real umbers less tha Determie the followig: a. 5-6 b. 6/ c. 10(-) d (-) e. 12/- III. Fuctios, Ordered Tuples ad Product Sets Ordered Pairs ad Ordered Tuples 1. If we are give a set {a, b}, we kow that {a, b} = {b, a}. As we oted above, order does ot matter. We could call {a, b} a uordered pair. 2. However, if we were to desigate the elemet a as the first listig of the set ad the elemet b as the secod listig of the set, the we would have a ordered pair. We deote this orderig by (a, b). Similarly it would be possible to defie a ordered triple say 8

9 (x 1, x 2, x 3 ), where the ordered triple is a set with three elemets such that x 1, is the first, x 2 the secod, ad x 3 the third elemet of the set. Likewise, a ordered N-tuple (x 1, x 2,..., x ) could be defied aalogously. A ordered N-tuple of umbers may be give various iterpretatios. For example, it could represet a vector whose compoets are the umbers, or it could represet a poit i -space, the umbers beig the coordiates of the poit. (Note (a, b) = (b, a) iff b = a, (a, b) = (c, d) iff a = c, d = b) The product set 1. Let X ad Y be two sets. The product set of X ad Y or the Cartesia product of X ad Y cosists of all of the possible ordered pairs (x, y), where x X ad y Y. That is, we costruct every possible ordered pair (x, y) such that the first elemet comes from X ad the secod from Y. The product set of X ad Y is deoted X Y, which reads X cross Y. More formally we have Def 1: The product set of two sets X ad Y is defied as follows: X Y = {(x, y) x X, y Y} Examples: #1. If A = {a, b}, B = {c, d}, the A B = {(a, c), (a, d), (b, c), (b, d)} #2. If A = {a, b}, B = {c, d e}, the A B = {(a, c), (a, d),(a, e), (b, c), (b, d), (b, e)}. #3 The Cartesia plae or Euclidea two-space, R 2, is formed by 9

10 R R = R 2. Each poit x i the plae is a ordered pair x = (x 1, x 2 ), where x 1 represets the coordiate o the axis of abscissas ad x 2 represets the coordiate o the axis of ordiates. The -fold Cartesia product of R is Euclidea -space. Fuctios Def 2: A fuctio from a set X ito a set Y is a rule f which assigs to every member x of the set X is sigle member y = f(x) of the set Y. The set X is said to be the domai of the fuctio f ad the set Y will be referred to as the codomai of the fuctio f. Remark 1: If f is a fuctio from X ito Y, we write f : X Y, or f X Y. This otatio reads f is a fuctio from X ito Y or f maps from X ito Y. Def. 2. a: The elemet i Y assiged by f to a x X is the value of f at x or the image of x uder f. Remark 2: We deote the value of f at x or the image of x uder f by f(x). Hece, the fuctio may be writte y = f(x), which reads y is a fuctio of x, where y Y, x X. Def. 2. b: The graph Gr(f) of the fuctio f : X Y is defied as follows: Gr(f) {(x, f(x)) : x X}, where Gr(f) X Y. Def 2. c: The rage f [X] of the fuctio f : X Y is the set of images of x X uder f or f[x] { f(x): x X}. 10

11 Remark 3: Note that the rage of a fuctio f is a subset of the codomai of f, that is f[x] Y. Def 3: A fuctio f: X Y is said to be oto if ad oly if f[x] = Y. Def 4: A fuctio f: X Y is said to be oe-to-oe if ad oly if images of distict members of the domai of f are always distict; i other words, if ad oly if, for ay two members x, x X, f(x) = f(x) implies x = x. Oe-to-oe fuctios have a useful property i that they are capable of beig iverted. That is, it is possible to reverse the mappig ad write x as a fuctio of y. The poit is that if y = f(x) is oe-to-oe, the each y has a associated uique x i the origial mappig. Thus, there is a reverse fuctioal relatioship. Result: If y = f(x) is a oe-to-oe fuctio, the a iverse fuctio x = f -1 (y) exists ad f -1 : f[x] X. That is the iverse fuctio assigs to each image value f(x ) the value x, for all x. Examples: Give the fuctio y = 2x, the iverse fuctio exists ad is give by x = y/2. Give the fuctio y = 3 + 4x, a iverse fuctio exists ad is give by x = -3/4 + y/4. Practice Problems III 1. Give S 1 = {1,3,4}, S 2 = {a,b} ad S 3 = {m,}, fid the Cartesia products: a. S 1 S 2 b. S 2 S 3 c. S 1 S 3 d. S 1 S 2 S Is it true that S 1 S 2 = S 2 S 1? Explai. 3. If the domai of the fuctio y = 5 + 3x is the set {x 1 x4}, fid the rage of this fuctio ad express it as a set. 4. Does a circle draw i a rectagular coordiate plae represet a fuctio? Explai. 5. Is a hill shaped fuctio oe-to-oe? Explai. 6. Does y = 2 + 4x have a iverse? If so what is it? 11

12 IV. Commo Fuctios 1. Whe we write y = f(x), we mea that a fuctioal relatioship betwee y ad x exits (each x maps ito oe y), however, we have ot made the rule of the mappig explicit. I this sectio, we cosider several specific fuctioal types. Each is used i busiess applicatios. 2. The first example is a specific example of a liear fuctio. Let the fuctio f assig to every real umber its double. Hece, for every real umber x, f(x) = 2x, or y = 2x. Both the domai ad the codomai of f are the set of real umbers, R. Hece, f: R R. The image of the real umber 2 is f(2) = 4. The rage of f is give by f[r] = {2x : x R}. The Gr(f) is give by Gr(f) = {(x, 2x): x R}. We illustrate a portio of G r (f) R R = R 2. f(x) 2 1 x Clearly, the fuctio f(x) = 2x is oe-to-oe. Moreover, f(x) = 2x is oto, that is, f[r] = R. 3. Example #1 is a specific example of a geeral class of liear fuctios. Let a ad b be two real umbers. The fuctio y = f(x) = a + bx is a liear fuctio of x. The graph of this fuctio is show below for the case where both a ad b are positive. The costat a is called the itercept 12

13 of the fuctio, because, for x = 0, y = a. That is the fuctio crosses the y axis at the poit a. The costat b is called the slope coefficiet, because the slope of the graph of this fuctio is b at each poit. By slope, we mea the rate of chage of y per chage i x. Let x chage by some arbitrary amout x = x - x. This chage i x geerates a chage i y through the fuctioal relatioship. The iduced chage i y is give by y = (a + bx ) - (a + bx ) = b(x - x ). Thus, the rate of chage of y per chage i x is just y b(x'' x') = = b. x (x'' x' ) f(x) = a + bx slope = b y y a x y 0 x x x If i the above example, b = 0, the f is said to be a costat fuctio. For every value of x, y is equal to the costat a. I this case, the graph of f is a flat lie (i.e., the slope is zero). 13

14 y a x 3. Next, we cosider a geeral class of fuctios termed polyomial. The term polyomial meas multi-term. A polyomial fuctio has the geeral form y = a o + a 1 x + a 2 x a x. If = 0, the y = a o ad we have a costat fuctio. If = 1, the y = a o + a 1 x ad we have a liear fuctio. If = 2, the y = a o + a 1 x + a 2 x 2 ad we have a quadratic fuctio. If = 3, the y = a o + a 1 x + a 2 x 2 + a 3 x 3 ad we have a cubic fuctio. The superscript idicators of the powers of x are called expoets ad the highest power ivolved i the fuctio is called the degree of the polyomial fuctio. A cubic fuctio is therefore called a third-degree polyomial. We have already discussed the case of a liear fuctio. Whe a quadratic fuctio is plotted, it appears as a parabola. This is a curve with a sigle bump. A example is give i the diagram below. 14

15 y a o, a 1 > 0 ad a 2 < 0. y a o, a 2 > 0 ad a 1 < 0. a o a o x x y = a o + a 1 x + a 2 x 2 y = a o + a 1 x + a 2 x 2 Whe a cubic fuctio is plotted, it exhibits two bumps as show i the example below. y a o y = a o + a 1 x + a 2 x 2 + a 3 x 3 (a o, a 1, a 2 > 0 ad a 3 < 0) x 4. Ratioal fuctios are fuctios i which y is expressed as the ratio of two polyomial fuctios i x. A example is y = x 1. x 2 4x Uder this defiitio, every polyomial fuctio is also a ratioal fuctio. Oe commo ratioal fuctio ecoutered i busiess applicatios is the rectagular hyperbola. This fuctio has the form y = a/x. Give a > 0 ad x 0, the graph of this fuctio is as i the diagram below. 15

16 y y = a/x, a > 0 x 5. All of the fuctios discussed thus far are termed algebraic. Geerally, such fuctios are expressed i terms of polyomials ad/or roots of polyomials (e.g., square root of a polyomial.). For example, y = square root {(x 3 + x 2 )} is ot ratioal, but it is algebraic. Noalgebraic fuctios iclude two types that are used i busiess applicatios. The first is the expoetial fuctio, y = b x, where the idepedet variable appears i the expoet. The secod is the closely related logarithmic fuctio, y = log b x. We will discuss both of these fuctios, but first let us digress o the topic of expoets. A Digressio o Expoets Above, i the itroductio to polyomial fuctios, we cosidered the expoet as the idicator of the power to which a variable or umber is to be raised. For example, 3 2 meas that the umber 3 is to be raised to the secod power or that 3 is to be multiplied by itself. We have that 3 2 = 3 3 = 9. Geerally, x x x x (-terms) = x. As a special case, x 1 = x. Expoets obey the followig rules. Rule 1. x x m = x +m. Rule 2. x /x m = x -m. 16

17 The proofs of Rules 1 ad 2 are obvious. However, if < m, the the power of x becomes egative from Rule 2. What does this mea? Actually, Rule 2 tells us the aswer. If = 4 ad m = 6, the x x 4 6 xxxx xxxxxx 1 xx 1. 2 x Thus x -2 = 1/x 2 ad this ca be geeralized ito aother rule: Rule 3. x - = 1/x. Aother special case of Rule 2 is where = m, i which case x /x = x 0. This must be oe. Thus, i accordace with Rule 2, ay ozero umber raised to the zero power is oe. Rule 4. x 0 = 1. ( x 0) So far we have thought of expoets as itegers. How do we iterpret fractioal expoets? Usig Rule 1, we ca iterpret a umber such as x 1/2 : x 1/2 x 1/2 = x 1 = x. That is, because x 1/2 multiplied by itself is x, x 1/2 is the square root of x. Likewise, x 1/3 is the cube root of x. Geerally, Rule 5. x 1/ = x. Two other rules obeyed by expoets are as follows. Rule 6. (x m ) = x m. Rule 7. x m y m = (xy) m. Expoetial ad Logarithmic Fuctios 1. A expoetial fuctio is a fuctio i which the idepedet variable appears as a expoet: y = b x, where b > 1. 1 A logarithmic fuctio is the iverse fuctio of b x. That is, 1 We take b > 0 to avoid complex umbers. b > 1 is ot restrictive because we ca take (b -1 ) x = b -x for this case. 17

18 x = log b y. The expoetial fuctio is graphed below. b x y 1 x 2. Rules of logarithms. Rule 1. If x, y > 0, the log (yx) = log y + log x. Rule 2. If x, y > 0, the log (y/x) = log y - log x. Rule 3. If x > 0, the log x a = a log x. 3. A preferred base is the umber e (More formally, e = lim [1 + 1/].) e is called the atural logarithmic base. The correspodig log fuctio is writte x = l y, meaig log e y. 4. Coversio ad iversio of bases. a. coversio log b u = (log b c)(log c u) (log c u is kow) Proof: Let u = c p. The log c u = p. We kow that log b u = log b c p = plog b c. By defiitio, p = log c u, so that log b u = (log c u)(log b c). b. iversio log b c = 1/(log c b). Proof: Usig the coversio rule, 1 = log b b = (log b c)(log c b). Thus, log b c = 1/(log c b). Practice Problems IV 1. Graph the fuctios a. y = 2 + 3x b. y = 2-3x. I each case, let x 0. 18

19 2. Codese the followig expressios: a. x 6 x 4 b. x 3 /x -2 c. (x 1/2 x 1/3 )/x 2/3 m 3. Show that x m/ = x ( x). 4. Simplify the followig: a. l(x 6 y z 5 ) b. log 10 1 c. log 10 2 = log 2? d. l(x/yz) 5. Give log e 2, how do we covert this to log 10? Explai. m V. Fuctios of Two or More Idepedet Variables 1. So far we have oly cosidered fuctios of a sigle idepedet variable. However, the cocept ca be exteded to the case of two or more idepedet variables. If y is the depedet variable ad x 1 ad x 2 are two idepedet variables, the a fuctio of the form y = f(x 1,x 2 ) expresses a relatioship where each ordered pair (x 1, x 2 ) is assiged oe image value y. For example, y = 6x 1 + 4x 2 ad y = x x 2 exemplify such fuctios. I this case the domai of the fuctio is a set of ordered pairs (x 1, x 2 ) ad the co-domai is the real lie. The fuctio maps from a poit i a two dimesioal space to a poit i oe dimesioal space. 2. Whe graphig a fuctio of two idepedet variables, we plot (x 1, x 2 ) i a horizotal plae ad let values of y be show as verticals perpedicular to that plae. The graph of the fuctio cosists of a set of ordered triples (x 1, x 2, f(x 1, x 2 )) ad thus is a surface. I the diagram below, we show oe poit i the graph of such a fuctio. 19

20 y (x 1, x 2,f(x 1, x 2 )) x 1 x 2 3. It is obvious that the cocept of a multivariable fuctio ca be exteded easily to the case of idepedet variables. Such a fuctio is a mappig from poits i -dimesioal space to the real lie. The domai cosists of ordered -tuples (x 1,,x ) ad the fuctio is writte as y = f(x 1,,x ). The graph of the fuctio is a surface i (+1)-dimesioal space. For greater tha 2, it is of course impossible to graph the fuctio. 20

21 Appedix to Lecture 1.1 Some Commo Notatio 1. Summatio Notatio: We have x x x. 2. Product Notatio: We have x i1 Notes o Logical Reasoig i1 i x 1. Notatio for logical reasoig: a. meas "for all" b. ~ meas "ot" c. meas "there exists" 2. A Coditioal Let A ad B be two statemets. AB meas all of the followig: 1 x i 1. If A, the B A implies B A is sufficiet for B B is ecessary for A 3. Provig a Coditioal: Methods of Proof a. Direct: Show that B follows from A. b. Idirect: Fid a statemet C where CB. Show that AC. c. Cotrapositive: Show that (~B) (~ A). d. Cotradictio: Show that (~ B ad A) (false statemet). 3. A Bicoditioal Let A ad B be two statemets. 21

22 AB meas all of the followig: A if ad oly if B ( A iff B) A is ecessary ad sufficiet for B A ad B are equivalet A implies B ad B implies A 4. Provig a Bicoditioal BA. Use ay of the above methods for provig a coditioal ad show that AB ad that 22

23 Practice Problems I 1. Give the sets S 1 = {2,3,5}, S 2 = {3,5,6} ad S 3 = {7}, fid: a. S 1 S 3 b. S 1 S 2 c. S 2 S 2 d. S 2 U e. S 2 2. Give A = {4,5,6}, B = {3,4,6,7}, ad C = {2,3,6}, verify the distributive law. Aswers: 1. a. S 1 S 3 =, b. {2,3,5,6}, c. {3,5,6}, d. S 2, e.. 2. A (B C) = {4,5,6} {3,6} = {3,4,5,6} (A B) (A C) = {3,4,5,6,7}{2,3,4,5,6}= {3,4,5,6}. Practice Problems II 1. Write the followig i set otatio ad i iterval otatio: a. The set of real umbers betwee 2 ad 10, iclusive. b. The set of real umbers less tha Determie the followig: a. 5-6 b. 6/ c. 10(-) d (-) e. 12/- Aswers: 1a. [2,10] = {x x is real, 2 x 10}. 1b. (, 15) = {x x is real,15 > x}. 2. a. 1, b. 0, c. -, d. -, e. 0. Practice Problems III 1. Give S 1 = {1,3,4}, S 2 = {a,b} ad S 3 = {m,}, fid the Cartesia products: a. S 1 S 2 b. S 2 S 3 c. S 1 S 3 d. S 1 S 2 S Is it true that S 1 S 2 = S 2 S 1? Explai. 3. If the domai of the fuctio y = 5 + 3x is the set {x 1 x4}, fid the rage of this fuctio ad express it as a set. 4. Does a circle draw i a rectagular coordiate plae represet a fuctio? Explai. 23

24 5. Is a hill shaped fuctio oe-to-oe? Explai. 6. Does y = 2 + 4x have a iverse? If so what is it? Aswers 1. a. S 1 S 2 = {1,3,4} {a,b} = {(1,a),(1,b),(3,a),(3,b),(4,a),(4,b)}. b. S 2 S 3 = {a,b}{m,}= {(a,m),(a,),(b,m),(b,)}. c. S 1 S 3 = {(1,m),(1,),(3,m),(3,),(4,m),(4,)}. d. S 1 S 2 S 3 = {1,3,4} {a,b} {m,} = {(1,a,m), (1,a,), (1,b,m), (1,b,), (3,a,m), (3,a,), (3,b,m), (3,b,),(4,a,m), (4,a,), (4,b,m), (4,b,) }. 2. This is true if the two sets are equal. I discrete case, if Si have uequal # elemets but all the same elemet, this also is sufficiet. 3. Rage is [8,17]. 4. No, a x ca be assiged two y's. 5. No, there are y's which are commo to two x's. 6. Yes, x = y. Practice Problems IV 1. Graph the fuctios a. y = 2 + 3x b. y = 2-3x. I each case, let x Codese the followig expressios: a. x 6 x 4 b. x 3 /x -2 c. (x 1/2 x 1/3 )/x 2/3 m 3. Show that x m/ = x ( x). 4. Simplify the followig: a. l(x 6 y z 5 ) b. log 10 1 c. log 10 2 = log 2? d. l(x/yz) 5. Give log e 2, how do we covert this to log 10? Explai. m Aswers 1. This ca be doe o your ow. 24

25 2. a. x 10, b. x 5, c. x 1/6 3. By defiitio true. x m/ = (x m ) 1/ = (x 1/ ) m. 4. a. l(x 6 y z 5 ) = 6lx + ly + 5lz, b. 0, c. log 10 2 = 1/log 2 10 (log b c = 1/(log c b).), d. l(x/yz) = lx - (ly + lz). 5. log 10 2 = (log 10 e) ( log e 2). Use log b u = (log b c)(log c u) (log c u is kow). 25

26 Lecture 1.2: Itroductio to Matrix Algebra Geeral 1. A matrix, for our purpose, is a rectagular array of objects or elemets. We will take these elemets as beig real umbers ad idicate a elemet by its row ad colum positio. A matrix is the a ordered set. 2. Let a ij R deote the elemet of a matrix which occupies the positio of the i th row ad j th colum. The dimesio of the matrix is defied or stated by idicatig first the umber of rows ad the the umber of colums. We will adopt the covetio of idicatig a matrix by a capital letter ad its elemets by the correspodig lower case letter. Example 1. a A 22 a a a Example 2. A a11 11 Example 3. Example 4. A a a a a a a a A 23 a a a A matrix is said to be (i) square if # rows = # colums ad a square matrix is said to be (ii) symmetric if a ij = a ji i, j, i j. Example. The matrix is square but ot symmetric, sice a 21 = 2 3 = a 12. The square matrix is symmetric sice a 12 = a 21 = 2, a 31 = a 13 = 4, ad a 32 = a 23 =

27 4. The priciple diagoal elemets of a square matrix A are give by the elemets a ij, i = j. The priciple diagoal is the ordered -tuple (a 11,..., a ). The trace of a square matrix is defied as the sum of the pricipal diagoal elemets. It is deoted tr(a) = i 5. A diagoal matrix is a square matrix whose oly ozero elemets appear o the pricipal diagoal. 6. A scalar matrix is a diagoal matrix with the same value i all of the diagoal elemets. 7. A idetity matrix is a scalar matrix with oes o the diagoal. 8. A triagular matrix is a square matrix that has oly zeros either above or below the pricipal diagoal. If the zeros are above the diagoal, the the matrix is lower triagular ad coversely for upper triagular. Remark: The followig otatios for idicatig a m matrix A are equivalet a ij i1,..., j1,..., m a11 a1m,,, or. a a 1 m 9. If a matrix A is of dimesio 1, the it is termed a row vector, a a a. ii A there is oly oe row, the row idex is sometimes dropped ad A is writte. 1 a. Sice a a A a11 matrix A of dimesio 1 is termed a colum vector, A. Likewise, sice there is oly a 1 a1 oe colum, this is sometimes writte as a. a 27

28 Algebraic Operatios o Matrices 1. Equality. Two matrices say A ad B, aiji1,, biji j1,, m 1,, j1,, m are said to be equal iff a ij = b ij i, j. 2. Additio ad Subtractio. Take A ad B as above with the same dimesios we have A B m m a b a b a b a b m 1m 1 1 m m 3. Scalar Multiplicatio. Let k R. k A kaij m. a b a b A B a 1 b 1 a b m 1m m m. 4. Multiplicatio. Two matrices A ad B ca be multiplied to form AB, oly if the colum dimesio of A = row dimesio of B. If this coformability requiremet is met, the it is possible to defie the product AB. I words, the colum dimesio of the lead matrix must equal the row dimesio of the lag matrix, for coformability. Example If A 23 ad B, the AB caot be defied, but B A ca be defied. I order to precisely preset the mechaics of matrix multiplicatio, let us itroduce the idea of a ier (dot) product of two -tuples of real umbers. Suppose x, y R. The the ier product of x ad y is defied by i i i1 x y x y x y x y x y Note that x x 2 ad x y y x. That is, the dot product is commutative. Give a i x i m matrix A, let us associate the k th colum of A with the ordered -tuple a a a Moreover associate the j th row of A with the ordered m-tuple a a jo j a jm Example. A ok 1,,. 1k,,. k 28

29 a 02 = (2, 4) a 20 = (0, 4, 5) With this otatio i had, cosider two matrices A B m mk which are coformable for multiplicatio i the order AB. The product AB is the give by A B m mk That is if AB = C, the c m a b a b a b a b a b. jl ji il i1 Note it must be that AB. Example 1 a A B a k k k a a b b b b m i1 m a b 1i i1 a b m i1 m i i1 i1 i1 a b 1i a b i ik ik a b a b a b a b a b a b a b a b i1 2 i1 a b 1i i1 a b 2i i1 2 i1 2 i1 a b 1i i2 a b 2i i2 a b a b a b a b Example 2 A B AB

30 Example 3. Suppose that A is a l row vector A = a = (a 11 a 12 a 1 ) ad B a 1 col vector b11 B b. Hece we have b 1 b11 a b a 11 a1 1 1 b1 i a b 1i i1 11 This is a scalar ad the operatio is termed a scalar product. Note that ab = a b. (The scalar product is same as the ier product of 2 row vectors.) Moreover suppose that A a a,, a 1 b b b11 while B b m. The product 1 b 11 1 ba is well-defied ad give by ba a a 11 m1 m m1 b a b a bm1a bm1a 5. The operatio of additio is both commutative ad associative. We have m. (Com. Law) A + B = B + A (Associative) (A + B) + C = A + (B + C) The operatio of multiplicatio is ot commutative but it does satisfy the associative ad distributive laws. (Associative) (Distributive) (AB)C = A(BC) A(B + C) = AB + AC (B + C) A = BA + CA To see that AB BA cosider the example A 1 2 B 0 1, AB

31 BA Geerally, whe we take the product of a matrix ad a vector, we ca write the result as c = Ab. I this example, the matrix A is by ad the colum vectors c ad b are by 1. This product ca be iterpreted i two differet ways. Takig the case of a 22 matrix A, we have a. 2 b First, this ca be a compact way of writig the two equatios 1 = a + 3b 4 = 3a + 2b. Alteratively, we ca write the relatioship as a liear combiatio of the colums of A a b I the geeral case where A is, we have c = Ab = b 1 a b a, where a i is the ith colum of A. Further, i the product C = AB, each colum of the matrix C is a liear combiatio of the colums of A where the coefficiets are the elemets i the correspodig colums of B. That is, C = AB if ad oly if c i = Ab i. 7. Traspose of a Matrix. The traspose of a matrix A, deoted A, is the matrix formed by iterchagig the rows ad colums of the origial matrix A. Example 1. Let A = (1 2) the 1 A 2 31

32 1 2 Example 2. Let A , the A Properties : (i) (A) = A (obvious) (ii) (A + B) = A + B Proof: Let A + B = C, the c ij = a ij + b ij. Let c ij deote a elemet of C. Clearly, cij c ji a ji b. ji Let a, b be elemets of A ad B respectively such that aij a ji ad b b ij cij a ji b ji aij b. ij ji Thus, the elemets of C ad A + B are idetical. ij ij (iii) (AB) = BA Proof: Let A, B the m mk a b a b AB a b a b k k a b a b AB k a b a b k 0 0k b01 1 m B km b0 k 1 m A a10 a20 a 0 m m 1 32

33 b01a10 b01a 0 BA. b0 ka10 b0 ka 0 7. The Idetity ad Null Matrices. a. A idetity matrix is a square matrix with oes i its priciple diagoal ad zeros elsewhere. A idetity matrix is deoted I. Properties: (i) Let A be p. The we have I A = AI P = A. Proof: Exercise (ii) Let A be p ad B be p m. The we have A Ip B AI B AB. p p p pm p (iii) IIII I. I geeral, a matrix is termed idempotet, p terms whe it satisfies the property AA = A. b. The ull matrix, deoted [0] is a matrix whose elemets are all zero. Subject to dimesioal coformability we have Properties: (i) A + [0] = [0] + A = A (ii) [0]A = A[0] = [0]. Proofs: Exercise Remark : If AB = [0], it eed ot be true that A = [0] or B [0]. Example. A

34 B It is easy to show that AB = [0]. 8. Determiats ad Related Cocepts. a. A determiat is defied oly for square matrices. Whe takig the determiat of a matrix we attach a sig + or - to each elemet: sig attached to a ij = sig (-1) i+j. Thus, for example, sig a 12 = -, sig of a 43 = -, ad sig a 13 = +. b. The determiat of a scalar x, x,is the matrix itself. The determiat of a 2 2 matrix A, deoted A or det A, is defied as follows: A a a a a a 11a 22 1 a 21 a12. Example 3 6 A A c. The determiat of a arbitrary 2 matrix A ca be foud via the Laplace Expasio process. I order to itroduce this process, let us cosider some prelimiary defiitios. Let A. Defiitio.The mior of the elemet a ij, deoted M ij, is the determiat of the submatrix formed by deletig the i th row ad j th colum. Example 1. a Let A be 2 2, A a a a M11 a22 a22. Moreover M a, M a ad M a Example 2. Let A be

35 a a a a a a a a a M 13 a21 a22 a21a32 a31a22. a a Defiitio. The cofactor of the elemet a ij deoted C ij is give by 1 i j ij M. a12 a13 Example Let A be 3 3. The C21 1 1a 12a33 a32a13. a a Defiitio. The priciple mior of the priciple diagoal elemet a ii, deoted PM i is the determiat of the submatrix formed by retaiig oly the first i rows ad first i colums. The order of PM i is its row = col. dimesio. Example a a a a a a a a a PM a PM2 a11a22 a21a12 PM 3 A Laplace Expasio: Let A be. The A i1 a ij C ij (expasio by j th col) A a C j1 ij ij (expasio by i th row) Note that evetually cofactors degeerate to the 2 2 case. Example Expasio by 2d col. a a a A a a a a a a A 3 a C i 2 i 2 i1 a a a a a a a a a a a a a a a

36 Next cosider expasio via the 3rd row a a a a a A a c a a a a a a a a a a j1 3j 3j Let s check the two terms to see if they are equal. The middle term of the secod expressio is the same as the last term of the first expressio. Checkig the remaiig two terms, we have the followig. I the first case -a 12 a 21 a 33 + a 12 a 31 a 23 + a 22 a 11 a 33 - a 22 a 31 a 13. I the secod case a 31 a 12 a 23 - a 31 a 22 a 13 + a 33 a 11 a 22 - a 33 a 21 a 12. Thus, they are the same. Example 2. A is 3 3 ad give by I this case it is easiest to expad via the first col Properties of Determiats. A (i) A A (ii) The iterchage of ay two rows (or two col.) will chage the sig of the determiat, but will ot chage its absolute value. Examples of (i) ad (i) #1 A A A 2 4 A 2 #2 A , B A 2 B 2 (iii) The multiplicatio of ay p rows (or col) of a matrix by a scalar k will chage the value of the determiat to k p A. 36

37 (iv) The additio (subtractio) of ay multiple of ay row to (from) aother row will leave the value of the determiat ualtered, if the liear combiatio is placed i the iitial (the trasformed) row slot. The same holds true if we replace the word row by colum. (v) If oe row (col) is a multiple of aother row (col), the value of the determiat will be zero. (vi) If A ad B are square, the AB = A B. Iverse Matrix 1. Def. Give a square matrix A, the iverse matrix of A, deoted A -1, is that matrix which satisfies A -1 A = AA -1 = I. Whe such a matrix exists, A is said to be osigular. If A -1 exists it is uique. 2. Computig the Iverse. a. Let us begi by assumig that the matrix we wish to ivert is a matrix A with A 0. b. Def. The cofactor matrix of A is give by C. C ij Def. The adjoit matrix of A is give by adj A = C. c. Computatio of Iverse: A adj A A 1. Example: Let A A , C C ij , 2 3 adj A

38 A / 25 3/ / / AA / 25 3/ 25 I / 25 1/ Key Properties of the Iverse Operatio a. (AB) -1 = B -1 A -1. Proof: B -1 A -1 AB = B -1 IB = I ad ABB -1 A -1 = I. It follows that B -1 A -1 is the iverse of AB. b. (A -1 ) -1 = A. Proof: AA -1 = I ad A -1 A = I. Thus, the result holds. c. (A') -1 = (A -1 )'. Proof: AA -1 = A -1 A = I. Trasposig ad otig that I' = I, we have (A -1 )'A' = I = A'(A -1 )'. d. I -1 = I. Proof: II = I. 4. a. Note that AB = 0 does ot imply that A = 0 or that B = 0. If either A or B is osigular ad AB = 0, the the other matrix is the ull matrix. That is, the product of two o-sigular matrices caot be ull. Proof: Let A 0 ad AB = 0. The A -1 AB = B = 0. b. For sqaure matrices, it ca be show that AB = A B, so that, i this case, AB = 0 if ad oly if A = 0, B = 0, or both. Liear Equatio Systems, the Iverse Matrix ad Cramer s Rule. 1. Cosider a equatio system with ukows x i, i = 1,,. a x a x a x d a x a x a x d a x a x a x d

39 I matrix otatio this system ca be writte as Ax = d, x1 d1 A a, x, d. If A 0, A -1 exists ad we ca write x d where ij 1 1 A Ax A d I x A d 1 1 x1 1 A d. x Thus, if there is o liear depedece i the rows or colums of the coefficiet matrix we ca obtai a solutio to the equatio system. Sice A -1 is uique if it exists, this solutio is uique. Hece, a easy way to test for existece ad uiqueess of a solutio to a set of liear equatios is to determie whether the coefficiet matrix has a ovaishig determiat. 2. This solutio gives us values of the solutio variables, i terms of A -1, i vector form. A formula kow as Cramer s Rule gives explicit solutios for each x i. If A 0, we have x 1 1 A x Thus, C x A d adj Ad A A C1 d1 C11 d C1 1 x A d11 C1 d C i1 i1 Ci 1 di Ci di 11 C C 1 d1 d 39

40 x j 1 A i1 C di. ij Cosider the term i1 C di. Recall A a C ij i1 ij ij (expas. by jth col.). Thus i1 di C ij a d a1d a a A j Usig this otatio, we obtai for j = 1,,, x j = A j A. (Cramer s Rule) Remark. This method does ot ivolve computatio of A -1. Example 3x 4x x 1x x x 2 20 A x1 A x 2 A 2 A Let's check this by computig A -1 Check 1 C 4 A C adja / / / 21 3/ 21 40

41 1 A A x x x I 41