MATEMATICS 1

Exercises, questions, applications

•

e-book

Vaclav NYDL, Renata KLUFOVA, Vivian WHITE

Department of Applied Mathematics and Informatics

Faculty of Economy, University of South Bohemia in Ceske Budejovice

1

Tato publikace vznikla v ramci projektu IP15 02/2/ -”Inovace vyuky matematiky na EF

JU se zamerenım na paralelnı vyuku v anglickem jazyce“

Zvlastnı ocenenı autoru si zaslouzı studentka oboru Ekonomicka informatika EF JU AnnaStepura, ktera se vyznamnym zpusobem podılela na kontrole vysledku vetsiny uloh v tetopublikaci.

c⃝ Vaclav Nydl, 2016

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PREDMLUVA

Tento studijnı text byl sestaven tymem pedagogu Katedry aplikovane matematiky a infor-matiky Ekonomicke fakulty JU na zaklade dlouholetych zkusenostı s vyukou predmetu mate-matika na Ekonomicke a Zemedelske fakulte JU. Ma slouzit studentum 1. rocnıku k dalsımuprocvicenı latky predmetu Matematika 1.

V sesti tematickych celcıch je pokryta latka predmetu v zimnım semestru. Kazdy tema-ticky celek zacına prehledem zakladnıch pojmu, vlastnostı a vzorcu. Mimochodem, prehledvzorcu ma student k dispozici pri kazdem psanı zapoctovych testu.

Oddıl ULOHY vyzaduje prevazne pocıtanı, ale tez analyticke cinnosti. Pro tuto situacije mozno si dosadit zname nemecke uslovı:

”Die Ubung macht den Meister“.

Oddıl OTAZKY rozvıjı logicke myslenı a prohlubuje porozumenı dane latce. Pri jehoprıprave pro kazde tema jsme meli na mysli, ze matematika je idealnı predmet pro povzbu-zovanı kreativity - ostatne jako motto nam vzdy poslouzila vseobecne znama fraze (snadnosi ji muzete i

”vygooglit“):

”The purpose of education is to replace an empty mind with an

open one.“

Zaverem tematu je pripraven oddıl APLIKACE, ktery zasazuje probırane tema do sirsıchsouvislostı. Je to i reakce na celkem casto se opakujıcı dotazy studentu typu

”k cemu to je

dobre?“ Chteli bychom zde studentovi vysvetlit, ze mozna bude jednou vykonavat povolanı,ktere dneska jeste vubec neexistuje. A pak je otazkou, jake predpoklady pro takove povolanıbudou klıcove. To, co prave studuje, nenı jen matematika pro budoucıho ekonoma. Matema-tika ma totiz presah do nejruznejsıch oblastı lidske cinnosti. Ukazme si jeden prıklad:

V geoinformatice se pracuje obvykle se dvema zakladnımi typy dat: vektorovymi arastrovymi. V podobe rastrovych dat se ukladajı letecke a druzicove snımky nebo nasnımaneobrazky. Rastr je jednoducha pravidelna sıt’ bunek, ve kterych je zaznamenan prevladajıcıjev v danem mıste (napr. dominantnı vegetacnı druh, teplota, mnozstvı srazek, nadmorskavyska apod.) Tyto jevy muzeme ukladat v pameti pocıtace jako zakladnı rastrove vrstvy.Rastr je mozno chapat jako matici.

Jednotlive rastrove vrstvy muzeme prekladat pres sebe a zıskavat tak nove informace.K tomu slouzı tzv. mapova algebra - souhrn pravidel pro matematicke operace s rastry.Pomocı mapove algebry dochazı matematickymi, ale i jinymi operacemi ke kombinaci mezivıce rastrovymi vrstvami a temito operacemi posleze k vypoctu hodnot rastru v podobe novevrstvy.

Ceske Budejovice, prosinec 2015autori

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TOPIC 1-2. Notations, Vectors, and Matrices

NotationsV1 ∨ V2 or V1 ∧ V2 . . . the logical disjunction or conjunction of propositions V1,V2

{a, b, c} . . . the set containing elements a, b, cx ∈ M , x /∈ M element x belongs to set M , does not belong to set MN and R . . . the set of all natural and the set of all real numbers(a, b) or ⟨a, b⟩ . . . an open or a closed interval⟨a, b), (a, b⟩ . . . two kinds of half-closed (half-open) intervals

Arithmetic vectorsVn . . . the space of all arithmetic n-component vectors ( n-dimensional vectors)vi . . . the i-th component of vector v = (v1, v2, . . . , vn)

Basic operations on vectors

Addition/subtraction u± v = (u1, . . . , un)± (v1, . . . , vn) = (u1 ± v1, . . . , un ± vn)

Multiplication by c c · v = c · (v1, . . . , vn) = (c · v1, . . . , c · vn)

Norm ∥v∥ = ∥(v1, v2, . . . , vn)∥ =√v21 + v22 + · · ·+ v2n

Dot product u · v = (u1, . . . , un) · (v1, . . . , vn) = u1 · v1 + u2 · v2 + . . .+ un · vnCross product (u1, u2, u3)× (v1, v2, v3) = (u2v3 − u3v2, u3v1 − u1v3, u1v2 − u2v1)

The formula for the angle of two non-zero vectors u and v : cosα =u · v

∥u∥ · ∥v∥Notes

• o = (0, 0, . . . , 0) is called the zero vector. Vector v is called a unit vector if ∥v∥ = 1.• Vectors u, v are called orthogonal (perpendicular) if their dot product is equal to 0.• Vectors u, v are called parallel if there exists a number c such that u = c · v.• The dot product is a number; the cross product is a vector - it is defined only in V3.

MatricesSize of matrix A . . . gives the number of rows and columns (write m× n, read ”m by n”)aij . . . the entry of matrix A which can be found in row i and column jPrincipal diagonal in A consists of all entries of A of the form aiiNull matrix O . . . has all its entries equal to zeroSquare matrix . . . any n× n matrix (we also talk about a matrix of order n)Unit matrix E . . . a square matrix with diagonal entries equal to 1, the others are 0s

Basic matrix operations

Transposing if Z = AT and A is an m× n matrix then Z is an n×m matrixand always zij = aji

Addition/subtraction if Z = A±B then A,B,Z have the same size and zij = aij ± bij

Multiplication by c if Z = c ·A then A,Z have the same size and zij = c · aijMatrix if Z = A ·B, where A is an m× s matrix, B an s× n matrix, then Zmultiplication is an m× n matrix and always zij = ai1 · b1j + ai2 · b2j + . . .+ ais · bsj

Notes

• A+O = O+A = A and A ·O = O ·A = O hold whenever the operations are defined.• For every matrix A there is A · E = E · A = A, if the operations can be performed.• A power of a square matrix can be obtained by means of repeated multiplication,e. g. A4 = A · A · A · A.

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EXERCISES

Exercise 1-2.1 [operations on intervals] For two open intervals I1 = (−1, 4), I2 = (3,+∞)and two closed intervals I3 = ⟨−2, 2⟩, I4 = ⟨−3, 3⟩, perform the specified operations. Writeyour result in the form of an interval (if possible).

(a) I1 ∩ I2 (b) I3 ∩ I4 (c) I1 ∩ I4 (d) I2 ∩ I3

(e) I1 ∪ I2 (f) I3 ∪ I4 (g) I2 ∪ I4 (h) I1 ∪ I4

Exercise 1-2.2. [vector operations] Four vectors u1, u2, v1, v2 are given as u1 =(−2, 0), u2 = (1, 2), v1 = (−1, 1, 3), v2 = (1, 2,−4).

(A) Find the resulting vector (if possible) and then determine its norm.

(a) 3u1 + 2u2 (b) (v1 · v1)·v1 (c) (v1 · v2)·u1 (d) (u1+v1)− (u2+v2)

(e) v1 × 2v2 (f) (v1+v2)× (v2−v1) (g) u2 × u1 (h) (v1×v2)− (v2×v1)

(B) Find the measure of the angle of two vectors (a) u1, u2, (b) v1, v2, (c) u1, v2.

Exercise 1-2.3 [matrix operations] Four matrices A,B,C,D are given as

A =

[1 −2 −24 0 −1

], B =

[0 3 −41 2 3

], C =

[10 −11 0

], D =

1 −14 20 3

.(A) Find the resulting matrix (if possible).

(a) 2A−B +DT (b) C3 (c) C · (A+B) (d) B ·D(e) (C · A)− 3B (f) (A ·B)− C (g) (D ·DT)T (h) D · AT

(B) Without knowing the result, determine the size of the resulting matrix (if it exists).

(a) AT · C8 ·B (b) (A+ 2B − 3D)3 (c) D · A ·D ·B

Exercise 1-2.4 [creative exercises](a) For vector v = (1,−1, 5, 3), find at least 3 vectors x such that v · x = −4.

(b) For matrix M =

[1 −2

−4 8

], find at least 2 matrices X such that M ·X =

[5

−20

].

(c) Create a 2× 2 matrix Q such that the entries of matrix Q ·Q are non-zero and amongthem there are: (c1) just 2 positive, (c2) just 1 positive, (c3) just 3 positive values.

Exercise 1-2.5 [exercises with an unknown]

(A) Find the value of an unknown real number y:

(a) ∥(1, 2, y)∥ = 4 (b) (y, 8,−y) and (−7, 2, 1) are orthogonal vectors,

(c) (35 , y) is an unit vector, (d) if u = (y, 4, 0), v = (y,−y, 2), then u · v < 3.

(B) Find an unknown vector x (the vectors used here come from Exercise 1.2):(a) v1 + x = v2 (b) 2u1 − 3x = 2x+ 4u2 (c) u1 · x = 5

(C) Find an unknown matrix X (the matrices used here come from Exercise 1.3):(a) A+X = B (b) (A ·BT)−X = X + 2C (c) 3D −XT = BT

(D) Determine the size of an unknown matrix - you do not have to find the matrix itself(the matrices used here come again from Exercise 1.3):

(a) Y ·B = D ·DT (b) A · Y = Y · C (c) Y · Y T = C

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QUESTIONS

Questions 1-2.1 [number intervals]

(a) Can the union of two open intervals be an open interval as well?

(b) Does the union of two closed intervals have to be a closed interval as well?

(c) Can the intersection of two open intervals be an open interval as well?

(d) Does the intersection of two closed intervals have to be a closed interval as well?

(e) Can the intersection of two half-open intervals be an open interval?

(f) Can the intersection of two half-open intervals be a closed interval?

(g) Can the union of two half-open intervals be an open interval?

(a) Can the union of two half-open intervals be a closed interval?

Questions 1-2.2 [vectors]

(a) Can a vector be treated as a matrix?

(b) A three dimensional vector has three components: 5,−1 and 0 in some order. Howmany such vectors do exist?

(c) Determine the minimal and the maximal norm of vector v = (4,−3, 0, x) where x isan arbitrary real number.

(d) Vector z = (1, 5, 7). How many vectors orthogonal and how many vectors parallel toz can you find?

(e) Vector u = (1, 2, 0). How many vectors w such that u× w = (−3, 0, 2) are there?

(f) On a sheet of paper, two vectors u and v are sketched; the angle between them isα = 35◦. How will this angle change if we rotate the sheet by exactly 10◦ counterclockwise?

Questions 1-2.3 [matrices]

(a) Can a 3× 4 matrix be treated as a vector?

(b) D is a 2× 2 matrix and such that 2 of its entries equal 2 and 2 of its entries equal3. How many matrices of this kind are there?

(c) F is a 2×2 matrix and such that each of its entries equals 0 or 1. How many matricesof this kind are there?

(d) How many 3× 3 matrices M such that M = MT do exist?

(e) A is a 2 × 2 matrix and such that a11 = 2 and a12 = 4. How many matrices Asatisfy the identity AT = A?

(f) We know that for matrices A and B the sum A+B exists. Then also the differenceA−B is defined. Why?

(g) For what matrices M the sum M +MT is defined?

(h) For what matrices M the product M ·MT is defined?

(i) For what matrices can you get the sum A+O +B?

(j) If A ·B is a 2× 3 matrix, determine the size of matrix B · A?

(k) If C =

[0 11 0

], then the power C100 can be easily obtained. Why?

(l) For what matrices can you calculate the product A ·O ·B?

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APPLICATIONS

Application 1-2.1 [angles in the triangle] In △ABC calculate the measure of all itsangles, (a) in the plane for A = [−1, 3], B = [3, 10], C = [8,−2],

(b) in the space for A = [1,−1, 3], B = [3, 7, 10], C = [0, 8,−2].

Application 1-2.2 [Markov models of employment - Simon & Blume, 1994] In 1966, amacroeconomist R. Hall analyzed the data on employment in some regions of the USA. Ifx1, . . . , xn are the numbers of employed and y1, . . . , yn the numbers of unemployed men,respectively, in weeks 1, . . . , n, we can describe this phenomenon as: p is the probability thata man employed in a given week will remain employed next week and q is the probabilitythat a man unemployed in a given week will become employed next week; if probabilitiesp, q are stable during some time period, we form equations about weeks n and n + 1 in theform xn+1 = p · xn + q · yn (employed) and yn+1 = (1−p) · xn + (1−q) · yn (unemployed).

Example. In one of the models, x1 = 5220, y1 = 105, p = 0.998, q = 0.136.Then x2 = 0.998·5220+0.136·105 .

= 5224 (employed) and y2 = 0.002·5220+0.864·105 .= 101

(unemployed). We repeat the calculation once more and we get x3 = 0.998·5224+0.136·101 .=

5227 and y3 = 0.002 · 5224 + 0.864 · 101 .= 98 . . . , etc. Using matrices, we get:[

xn+1

yn+1

]=

[p q

1−p 1−q

]·[xnyn

]. Matrix P =

[p q

1−p 1−q

]is so called transition matrix - the

values after one week are obtained by means of multiplication by P . After k weeks:[xn+k

yn+k

]= P k ·

[xnyn

].

Example - continued. If (above) p = 0.998, q = 0.136, form the transition matrix P andcalculate its square P 2. Then, using matrix algebra, determine the values x3 and y3 if youknow that x1 = 5220 and y1 = 105. Try to find x6 and y6.

Application 1-2.3 [dominance matrices - Lay, 2003] In a voleyball tournament of 5 teamsA,B,C,D,E, each team plays the other 4 once. The results are represented by matrix K -each team is associated with a corresponding row and column; a 1 is placed at the (i, j)entry of the matrix if the team corresponding to row i defeated the team corresponding tocolumn j, otherwise a 0 is recorded. We also calculated matrices K2 and K +K2.

K =

0 0 1 1 01 0 1 0 10 0 0 1 00 1 0 0 01 0 1 1 0

, K2 =

0 1 0 1 01 0 2 3 00 1 0 0 01 0 1 0 10 1 1 2 0

, K +K2 =

0 1 1 2 02 0 3 3 10 1 0 1 01 1 1 0 11 1 2 3 0

.K is the matrix of one-step dominance and K2 the matrix of two-step dominance (forexample, in K2, the value 3 placed at the (2, 4) entry means three times confirmed two-stepdominance of B over D, namely, B → A → D, B → C → D, B → E → D. In matrix K+K2,the entries give the number of one- or two-step dominance. The sum of a row in K gives thenumber of wins - this sum is the same for teams B and E. Now, we calculate the sums ofrows of K +K2 and we get: A - 4, B - 9, C - 2, D - 4, E -7. The team B has the highestpower. Which team is the least powerful?

Note. Vectors are applied in geometry, physics or engineering. Another example is computergraphics in the CorelDRAW software. Spreadsheets like the Microsoft Excel exploit directlythe properties of matrices, their structure and operations on the entries.

7

TOPIC 3. Rank

Systems of vectors in Vn

S : v1, v2, . . . , vk a system of vectors; the order matters, the vectors can repeat

Linear combination . . . lin. combination of system S with coefficients c1, c2, . . . , ckis vector v = c1 · v1 + c2 · v2 + . . .+ ck · vk

Trivial combination . . . all its coefficients equal 0 (the result is the zero vector o)Non-trivial combination has at least one non-zero coefficient

≪ S ≫ . . . the linear span of S, i. e. the set of all lin. comb. of vectors of SDependent system S . . . o can be obtained as some non-trivial combination of SIndependent system S o can be obtained as only the trivial combination of SP =≪ G ≫ . . . P is a subspace of Vn, G its system of generators (P spanned by G)Basis of subspace P any independent sytem of generators of P

Coordinates of vector v the unique system of coefficients c1, . . . , ck such thatwith respect to basis B v = c1 · b1 + . . .+ ck · bk where B : b1, . . . , bk

h(S) . . . the rank of system S= the size of its maximal indep. subsystem

h(A) . . . the rank of matrix A= the rank of the system of all its row vectors

dim(P ) . . . the dimension of subspace P = the number of vectors in any basis

Notes

• The empty system is supposed to be independent with rank 0. In a dependent system, one ofthe vectors is a lin. combination of the others, in an indep. system, such a vector doesn’t exist.• S having k vectors is dependent if and only if h(S) < k; it is indep. if and only if h(S) = k.• Vector v belongs to the span of system S, if and only if it does not increase its rank.• The rank of a system of vectors is determined by means of the rank of a matrix. The dimensionof a subspace is the rank of any system of its generators. Especially, dim(Vn) = n.• A basis of subspace P can be obtained by removing dep. vectors from a system of generators.

• The problem of finding coordinates of a vector leads to a system of linear equations.

Determining the rank of a matrixA matrix is in the Gaussian form if it doesn’t have a zero row and any row contains more left

zeros than the previous one (e. g. E). The rank of a matrix in the Gaussian form equals the

number of rows. Each non-zero matrix can be converted into a matrix in the Gaussian form –

the rank is the same. In this way, we can determine the rank of any matrix.

Equivalent transformations of (row operations on) a matrix(operations (1) - (4) can be repeated)

(1) To change the order of rows arbitrarily. (4) To remove or to join a (new) zero-row.(2) To multiply any row by a non-zero number (thus also to divide).(3) To add to any row (thus also to subtract from) a lin. combination of the other rows.

Notes

• In a matrix in the Gaussian form, two consecutive rows can differ by more than oneleft zeros. Moreover, already the first row of such a matrix can have 1 or more left zeros.• Only zero matrices have their rank equal to zero.• Equivalent transformations are also possible for the columns of a matrix (we never do it!).

• For every matrix h(A) = h(AT) which is connected to the previous note.

8

EXERCISES

Exercise 3.1 [linear combination] S : v1 = (−1, 2), v2 = (2,−1), v3 = (3, 1) is a system ofvectors in ve V2. Calculate

(a) the trivial combination of the system S,(b) some linear combination of the system S with non-zero coefficients,(c) the linear combination of the system S with coefficients c1 = 2, c2 = 10, c3 = −3,(d) the lin. combination of the system S with coefficients c1 = v1·v2, c2 = v2·v3, c3 = v3·v1.

Exercise 3.2 [linear dependence] A system of two or more vectors is linearly dependent ifand only if some of these vectors is a linear combination of the others. All the systems beloware dependent. Your task is justify this fact the following way: you will find a vector that isa combination of the others finding by guessing coefficients c1, c2 for such combination.

(a) u1 = (1, 2, 3), u2 = (4, 2,−1), u3 = (0, 0, 0), (b) u1 = (2, 1), u2 = (4, 1), u3 = (2, 1),(c) u1 = (1, 2), u2 = (100, 100), u3 = (101, 102) (d) u1 = (−1,−1), u2 = (1, 1), u3 = (0, 1).

Exercise 3.3 [matrices in the Gaussian form] Which of the following matrices are in theGaussian form?

(a)

[1 10 −1

](b)

[0 10 0

](c)

1 0 00 3 00 1 3

(d)

0 0 10 1 01 0 0

(e)

[0 5 0 10 0 3 0

](f)

[0 0 −2 2 20 0 0 0 1

]

Exercise 3.4 [equivalent transformations (1) and (4)] Using only equivalent transformati-ons (1) or (4), but not (2) or (3), convert the given matrix into the Gaussian form.

(a)

[1 10 0

](b)

[0 12 0

](c)

1 0 20 0 01 1 3

(d)

5 5 00 0 30 3 3

(e)

[0 0 0 11 0 3 0

](f)

0 0 0 0 0 01 0 0 0 1 10 1 0 0 1 1

Exercise 3.5 [equivalent transformations (2) and (3)] Using only equivalent transformati-ons (2) or (3), but not (1) or (4), convert the given matrix into the Gaussian form.

(a)

[1 11 3

](b)

0 12 03 1

(c)

1 2 21 4 31 0 3

(d)

5 5 00 0 30 0 3

(e)

0 1 3 01 1 0 10 0 3 0

(f)

[0 0 0 −2 2 20 0 0 2 1 1

]

Exercise 3.6 [equivalent transformations - the rank] Using equivalent transformations(1),(2), (3) or (4), every matrix can be converted into the Gaussian form. Convert each of thegiven matrices into the Gaussian form, and thus determine its rank.

(a)

[1 11 3

](b)

0 11 13 12 1

(c)

1 2 22 4 31 2 1

(d)

0 0 00 0 30 0 30 0 4

(e)

0 1 3 01 1 0 10 2 6 0

(f)

0 0 0 −2 2 20 0 0 2 1 10 0 0 1 1 1

Exercise 3.7 [exercises with an unknown] Find all the values of the real number x so thatthe given matrix is in the Gaussian form:

(a)

1 2 20 1 30 x+2 1

(b)

1 2 20 4−x2 30 0 1

(c)

1 2 20 0 30 0 1+x2

(d)

1 2 3 2 40 0 2x2 1 50 0 0 0 1

9

QUESTIONS

Questions 3.1 [linear combination]

(a) Explain the difference between the concepts of trivial and non-trivial lin. combin.

(b) Explain the difference between a linear combination with non-zero coefficients anda combination with positive coefficients.

(c) How do you create a linear combination of one vector?

(d) Can you find a vector which is a linear combination of itself?

(e) S : a = (1,−2), b = (0, 0) is a system of two vectors in V2. How many different linearcombinations can you create? Why the vector c = (1, 2) never occurs?

Questions 3.2 [matrices in the Gaussian form]

(a) Is the zero matrix O a matrix in the Gaussian form or not? Why?

(b) Is the unit matrix E in the Gaussian form or not? Why?

(c) A matrix M having only one row is nearly always in the Gaussian form. Explain.

(d) Can a matrix in the Gaussian form contain some negative numbers?

(e) Is there a simple characteristic of all 2× 2 matrices that are in the Gaussian form?

(f) None 3× 2 matrix is in the Gaussian form. Why?

(g) A =

[1 2 4 −20 0 −1 2

], B =

[1 2 0 70 0 1 −2

]are two matrices in the Gaussian form. But

neither matrix A+B nor matrix A−B is in the Gaussian form. Is it always true?

(h) Let C be a matrix in the Gaussian form. Then its transposed matrix CT can be,

but not necessarily, in the Gaussian form. Show some examples.

Questions 3.3 [equivalent transformations]

(a) When transforming a matrix, can we insert a zero-row into the matrix?

(b) When transforming a matrix, can we divide a zero row by zero ?

(c) When transforming a matrix, can we insert a copy of any row into the matrix?

(d) When transforming a matrix, can we add all the other rows to the first one?

(e) Explain why the following is incorrect:

[3 2 4 27 0 −1 2

]∼[

21 14 28 14−21 0 3 −6

]∼[21 14 28 140 14 31 8

]

(f) Explain why the following is incorrect:

−3 2 4 71 2 5 21 2 5 2

∼

−3 2 4 70 0 0 00 0 0 0

∼[−3 2 4 7

](g) When transforming a matrix, we also can delete one of two identical rows. Why?

Questions 3.4 [matrix rank]

(a) What is the rank of the zero matrix O and what is the rank of the unit matrix E?

(b) What is the maximum rank of a 4× 2 matrix?

(c) Can you find a matrix M such that h(M) = h(MT)?

(d) I have a matrix. What can happen with the rank if I add to Q some matrix of thesame size? Can the change of rank be big?

10

APPLICATIONS

Application 3.1 [dependence and independence] Determine if the systems are dependentor independent:

(a) (3,−1), (−3, 1), (12,−4), (b) (1, 2, 3), (1, 1, 1), (2, 1, 8), (−2,−2,−2),

(c) (1, 1, 1, 1), (0, 1, 1, 0), (0, 1,−3, 1), (1, 1, 1,−4) (d) (1, 0, 2, 0, 1), (1, 2, 0, 0, 1), (1, 2, 1, 1, 1),

(e) (1, 0, 1, 1, 1, 1, 1), (0, 0, 1, 1, 0, 0, 0), (1, 1, 1, 0, 0, 1, 1), (0, 0, 1, 1, 0, 1, 1).

Application 3.2 [belonging to the linear span] In each case, decide which of vectors p, q, r, sbelong to the span of system S.

(a) S : (1,−2), (−1, 2), (0, 0); p = (8, 9), q = (8,−9), r = (1, 1), s = (−2, 4),

(b) S : (1,−2, 1), (−1, 2, 2), (0, 0, 3); p = (1, 1, 1), q = (2, 8, 0), r = (1, 0, 1), s = (0, 0, 4),

(c) S : (1,−1, 1, 1), (0, 1,−1, 2), (0, 0, 0, 0);

p = (−2, 2, 2, 2), q = (0,−2, 2,−4), r = (1, 0, 0, 3), s = (1, 1, 1, 1),

(d) S : (1, 1, 1, 0, 1), (1, 0, 0, 0, 0), (1, 0, 1, 0, 1);

p = (0, 0, 1, 8, 9), q = (2, 3, 3, 0, 0), r = (1, 1, 1, 1, 1), s = (2, 0, 0, 0, 0).

Application 3.3 [testing bases of Vn] Every basis of Vn must have just n vectors and itsrank must be n. In each case, a candidate to a basis of Vn is to be tested. Always decide ifit is a basis of Vn or not and justify why.

(a) V2; (1,−1), (−3, 1), (1,−4), (b) V4; (1, 1, 1, 1), (0, 1, 1, 0), (0, 1,−3, 1), (1, 1, 1, 0)

(c) V3; (1, 2, 3), (1, 1, 1), (2, 1, 8), (0, 2, 1), (d) V5; (1, 0, 2, 0, 1), (1, 2, 0, 0, 1), (1, 2, 1, 1, 1),

(e) V6; (1, 0, 1, 1, 1, 1), (0, 0, 1, 1, 0, 0), (1, 1, 1, 0, 0, 1), (0, 0, 1, 1, 0, 1), (1, 1, 1, 1, 1), (1, 1, 0, 0, 1).

Application 3.4 [determining the dimension of a subspace using its system of generators]If a subspace P is described by its system of generators G, i.e. P = ⟨⟨G⟩⟩, then its dimensiondimP equals h(G). In each of the following cases, determine the dimension of the subspacespanned by its generators.

(a) (3,−1), (−3, 1), (12,−4), (b) (1, 2, 3), (1, 1, 1), (2, 1, 8), (−2,−2,−2),

(c) (1, 1, 1, 1), (0, 1, 1, 0), (0, 1,−3, 1), (1, 1, 1,−4) (d) (1, 0, 2, 0, 1), (1, 2, 0, 0, 1), (1, 2, 1, 1, 1),

(e) (1, 0, 1, 1, 1, 1, 1), (0, 0, 1, 1, 0, 0, 0), (1, 1, 1, 0, 0, 1, 1), (0, 0, 1, 1, 0, 1, 1).

Application 3.5 [transforming a system of generators into a basis of a subspace] Whentransforming a system of generators G,

”redundant vectors disappear“ and finally we obtain

a basis of the subspace spanned by G (usually, there are many possibilities). Solve in thecases below.

(a) (3,−1), (−3, 1), (6,−2), (b) (1, 2, 3), (1, 1, 1), (2, 3, 4), (−2,−2,−2),

(c) (0, 1, 1, 1), (0, 1, 1, 0), (0, 1,−3, 1), (0, 1, 1,−4) (d) (1, 0, 2, 0, 1), (1, 2, 0, 0, 1), (2, 2, 2, 0, 2),

(e) (1, 0, 1, 1, 1, 1, 1), (0, 1, 1, 1, 0, 0, 0), (1, 1, 2, 2, 1, 1, 1), (0, 0, 1, 1, 0, 1, 1).

11

TOPIC 4. Square Matrices

Order of square is the number of its rows (columns), i. e. A is an n× n m.matrix A . . .

det(A) . . . determinant of A

Aij. . . the co-factor (or algebraic complement) of entry aij in A

A, adj(A) . . . the adjoint matrix of A

A−1 . . . the inverse matrix of A (there is A · A−1 = A−1 · A = E)

Non-singular matrix . . . the rank is equal to the order, the determinant is not equal(regular m.) to zero; only a non-singular matrix has its inverse

Singular matrix . . . the rank is less than the order, the determinant equals zero

Notes

• The co-factor Aij in A is calculated as Aij = (−1)i+j · det(SA,i,j), where SA,i,j is thesubmatrix of A obtained by deleting of the i-th row and the j-th column from it.• The adjoint matrix of A is first formed by the co-factors and then transposed. Everysquare matrix has its adjoint matrix.• Higher order determinants are calculated by the development (the Laplace expansion).

Calculation of the determinant

Order 2 (the ”cross”multiplication): det

[a bc d

]= a · d− c · b, order 3 (the Sarrus rule):

det

a11 a12 a13a21 a22 a23a31 a32 a33

= a11a22a33+ a12a23a31+ a13a21a32− (a13a22a31 + a11a23a32+ a12a21a33)

Development by row i: det(A) = ai1Ai1 + ai2Ai2 + . . .+ ainAin

Development by column j: det(A) = a1jA1j + a2jA2j + . . .+ anjAnj

Notes

• If the rows of A are permuted then the det(A) can change its sign. A factor common toall the entries of a row can be taken before the whole det(A). The value of the determinantdoes not change if to a chosen row a linear combination of the other rows is added.• For any matrices of order n there is det(A) = det(AT), det(A ·B) = det(A) · det(B).

Calculation of the inverse to a regular matrixBy means of the adjoint matrix: A−1 = 1

detA· A

Elimination:

[A | E

]∼ . . . ∼ . . . ∼

[E | A−1

]Notes

• For order 2 there is: A−1 =

[a bc d

]−1

= 1detA ·

[d −b

−c a

]• For large matrices the computation of the inverse matrix can be quite time-spending.

• If A is a regular matrix, the matrix equation A ·X = B can be solved like:

A ·X = B −→ A−1 · A ·X = A−1 ·B −→ E ·X = A−1 ·B −→ X = A−1 ·B.

• If A is a regular matrix, the matrix equation X · A = B can be solved like:

X · A = B −→ X · A · A−1 = B · A−1 −→ X · E = B · A−1 −→ X = B · A−1.

12

EXERCISES

Exercise 4.1 [the rank, singular and non-singular matrices] Determine the rank and thendecide if the matrix is singular or not:

(a)

[1 −22 −1

](b)

[−3 16 −2

](c)

2 1 14 2 50 0 3

(d)

1 0 50 1 01 1 0

(e)

0 5 0 11 0 3 20 1 3 30 0 3 0

(f)

1 −1 2 01 1 1 21 0 3 0

−1 3 3 0

Exercise 4.2 [co-factors] In a matrix M , we use the symbol Mij forthe co-factor of the entry mij. Calculate the values of the given expres-sions:

M =

1 −3 12 2 31 0 10

(a) M11 + 2M22 + 3M33 (b) m11M21 +m22M23 (c) m12M12 +m22M22 +m32M32

(d) M31 ·M32 +M13 (e) m13M12 −m21M21 (f) m31M21 +m32M22 +m33M23

Exercise 4.3 [determinants and singular and non-singular matrices] Evaluate the deter-minant of each matrix from Exercise 4.1 and then decide if the matrix is singular or not; thefollowing is required:

(i) In Ex. 4.1 (c), (d) use the Sarrus’ Rule.(ii) In Ex. 4.1 (c), (e) use the Laplace expansion along the last row.(iii) In Ex. 4.1 (d), (f) use the Laplace expansion along the last column.

Exercise 4.4 [the determinant: transforming and the Laplace expansion] Transform mat-rices of order 4 or 5 and then use the Laplace expansion to find their determinants. Whichof the matrices are singular?

(a)

1 2 0 11 0 −1 20 1 3 30 2 3 6

(b)

2 5 0 14 0 3 20 1 3 3

−1 4 3 2

(c)

2 5 0 1 −11 0 3 0 10 2 0 0 03 5 3 0 −14 −1 3 2 1

(d)

1 −1 2 0 11 1 1 1 11 0 3 −1 2

−1 3 3 −2 31 2 3 1 2

Exercise 4.5 [the inverse matrix of order 2 by means of the adjoint matrix]Matrix A of order 2 is shown at the right. Now, use the adjoint matrices tocalculate the inverse matrices of B,C,D, F,G, and H (if the inverses exist).

A =

[1 20 −1

]

B = AT, C = A+ A, D = A · A, F = A, G = A−1, H = det(A) · A.

Exercise 4.6 [the inverse matrix by means of elimination] Use the elimination method tocalculate the inverse matrices.

(a)

[1 −20 −1

](b)

[3 11 1

](c)

1 1 10 1 10 0 1

(d)

0 1 11 0 11 1 0

(e)

0 1 0 01 0 0 00 0 1 10 0 1 0

(f)

1 0 0 01 1 0 01 0 1 0

−1 3 0 −2

Exercise 4.7 [exercises with an unknown] Find all the real numbers x such that:

(a) matrix

[−1 x0 10

]does not have its inverse, (b) matrix

[2 1

4−x2 0

]is non-singular,

(c) the det. of matrix

−2 2 x1 0 −31 x −7

is negative, (d) the det. of matrix

1 2 3x 1 x1 1 4

is not 5.

13

QUESTIONS

Question 4.1 [singular or non-singular matrices]

(a) Is the zero matrix O singular or non-singular?

(b) Is the unit matrix E singular or non-singular?

(c) A matrix M of order 3 is in the Gaussian form. Is M singular or non-singular?

(d) A matrix U is non-singular and a matrix V is singular. Which of them has itsdeterminant greater than the other one?

(e) Are there more singular matrices or non-singular matrices?

Question 4.2 [the determinant]

(a) In a unit matrix E, the equality Eij = eij holds for every i and j. For order 4,investigate the details.

(b) We know that for a matrix M of order 4 there is detM = 0. Can we say somethingabout its rank h (M)?

(c) M is a square matrix, detM = 54. Can we say something about its rank h (M)?

(d) In less than 1 minute, are you able create two different matrices B1, B2 of order 2such that detB1 = detB2 = 120 ?

(e) At the right, matrix Q is given. It is easy to confirm thatdet(Q+Q) = detQ+ detQ. Does it hold for any matrix Q?

[−1 23 −6

](f) The formula det(A · B) = det(A) · det(B) holds for every two matricesA,B of the same order. In 1 minute, can you determine the value of det(W 19)where W is the matrix at the right?

[1 11 0

]

(g) For the given matrix, find the number x such that the matrix has itsdeterminant (i) as small as possible, (ii) as big as possible.

[x −14 x

]

Question 4.3 [the inverse matrix]

(a) For the matrix at the right, find the number x such that the matrix(i) has its adjoint matrix, (ii) does not have its adjoint matrix.

[x −x4 x

](b) For the matrix at the right, find the number x such that the matrix

(i) has its inverse matrix, (ii) does not have its inverse matrix.

[2 −14 x

]

(c) If E is a unit matrix then E · E = E. In means that E−1 = E. A matrix M , forwhich M−1 = M holds, is called a selfinverse matrix. Find a selfinverse matrix of order 2,that is different from E.

(d) It is easy to create the adjoint matrix of any matrix of order 2. So, we can see thatthe unit matrix E of order 2 is selfadjoint, i.e. E = E. Also the zero matrix O of order 2 isselfadjoint, but it is not selfinverse. Why? Now, find a matrix F of order 2 such that F

(i) is selfinverse, but not selfadjoint, (ii) is not selfinverse,F = O, but F is selfadjoint.

(e) The formula det(A · B) = det(A) · det(B) is correct for every two matrices A,B ofthe same order. We can also use the formula for C and C−1. It follows that if detC = 4 thenwe know the value of detC−1. Find it.

(f) If G is a square matrix in the Gaussian form then detG equals the product of alldiagonal entries. Give an example of order 4.

14

APPLICATIONSApplication 4.1 [the cross product] u = (u1, u2, u3), v = (v1, v2, v3)are two vectors in V3. We create matrix Z of order 3 (at the right)so that its first row is vector u, its second row is vector v and thesymbols z31, z32, z33 create its third row. We get (using the co-factors)u× v = (Z31, Z32, Z33). Check for u = (1, 2,−1), v = (3, 4, 1).

Z =

u1 u2 u3v1 v2 v3z31 z32 z33

Application 4.2 [determinants in geometry] A = [a1, a2], B =[b1, b2], C = [c1, c2] are three points in the plane. We create matrix S

of order 3 (see at the right). Then the area of △ABC equals∣∣∣12det(S)

∣∣∣. S =

a1 a2 1b1 b2 1c1 c2 1

(a) Use the formula on △A1B1C1 : [1, 1], [4, 3], [−2,−5], △A2B2C2 : [1, 0], [−4, 1], [−5, 3].

(b) Find the value of x so that the area of triangle with vertices [x, 4], [2, 1], [−1, 2] is 5.

Application 4.3 [bases of Vn] The vectors of any basis of Vn must create a non-singularmatrix of order n. Use the determinant to decide if the system of vectors is a basis of Vn:

(a) (1,−1), (−3, 1), (b) (1, 0, 2, 0, 1), (1, 2, 0, 0, 1), (1, 2, 1, 1, 1), (1, 2, 1, 1, 0), (1, 0, 1, 1, 1),

(c) (1, 2, 3), (1, 1, 1), (2, 1, 8), (d) (1, 1, 1, 1), (0, 1, 1, 0), (0, 1,−3, 1), (1, 1, 1, 0),

(e) (1, 0, 1, 1, 1, 1), (0, 0, 1, 1, 0, 0), (1, 1, 1, 0, 0, 1), (0, 0, 1, 1, 0, 1), (1, 1, 1, 1, 1, 1), (1, 1, 0, 0, 0, 1).

Application 4.4 [using inverse matrices to solve matrix equations] P=

[1 21 1

], Q=

[1 2−1 0

]are non-singular matrices appearing in the equations below. Solve (a) through (f).

Help: In part (d), start as: X · Q + X · P = X · (Q + P ) = . . .; in part (f) , start as:Q ·X −X = Q ·X − E ·X = (Q− E) ·X = . . ..

(a) P ·X =

[1 2 1 2−1 0 −3 0

](b) X ·Q =

[10 −2

](c) P ·X + 2Q =

[3 −612 1

]

(d) X ·Q+X · P =[20 −12

](e) Q ·X · P =

[2 −10 4

](f) Q ·X −X =

[3132

]

Application 4.5 [input-output macroeconomic analysis, Simon & Blume, 1994] Consider

the economy of an organic farm that produces two goods: corn and fertilizer. Corn is produced

using corn (to plant) and fertilizer. Fertilizer is made from old corn stalks (and perhaps by feeding

he corn to cows, who then produce useful end products). The production of 1 ton of corn requires

as inputs 0.1 ton of corn and 0.8 ton of fertilizer. The production of 1 ton of fertilizer requires as

inputs no fertilizer and 0.5 ton of corn. Let d1 and d2 be the amounts of corn and fertilizer the

farm wants to sell. The total production will be x1 tons of corn (the consumption at the farm +

selling) and x2 tons of fertilizer (the consumption at the farm + selling). We have (in tons):

corn: x1 = 0.1x1 + 0.5x2 + d1fertilizer: x2 = 0.8x1 + 0.0x2 + d2

i.e. X =

[0.1 0.50.8 0.0

]·X +D, where X=

[x1x2

], D=

[d1d2

].

Further, X =

[0.1 0.50.8 0.0

]·X+D −→ E ·X−

[0.1 0.50.8 0.0

]·X = D −→

([1 00 1

]−[0.1 0.50.8 0.0

])·X = D

−→[

0.9 −0.5−0.8 1.0

]·X = D −→ X =

[0.9 −0.5

−0.8 1.0

]−1

·D. It remains to find the inverse matrix.

If the farmer wants to sell 4 tons of corn and 2 tons of fertilizer (i.e. d1=4, d2=2) whatamounts of these two products (i.e. x1, x2) are to be produced? So, matrix D is known(given) and it is easy to determine matrix X. Finish the calculation.

15

TOPIC 5. Systems of Linear Equations

a11 x1 + a12 x2 + . . .+ a1n xn = b1a21 x1 + a22 x2 + . . .+ a2n xn = b2. . . . . . . . .

am1 x1 + am2 x2 + . . .+ amn xn = bm

A =

a11 a12 . . . a1n. . . . . . . . . . . .am1 am2 . . . amn

, A∗ =

a11 a12 . . . a1n b1. . . . . . . . . . . . . . .am1 am2 . . . amn bm

A and A∗ . . . the system matrix and the augmented matrixm, n . . . . . . the number of equations, the number of unknowns (variables)

b. . . = (b1, b2, . . . , bm), the right side vectorx . . . = (x1, x2, . . . , xn), the variable vectorSolution . . . any vector x satisfying all equationsh(A) = h(A∗) the Frobenius condition; the system is solvable ⇔ the F.c. is satisfiedRegular system i. e. h(A) = h(A∗) = n = m; such system has a unique solution

Notes• If h(A) = h(A∗) < n then the system has infinitely many solutions.• If h(A) = h(A∗) = n then the system has a unique solution.

Methods of solving regular systems of equationsThe Cramer Rule. Denote by Aj the matrix obtained by subtituting the j’s columnin A with the right-side column. Then xj = det(Aj)/ det(A).

The Gauss Elimination. The augmented matrix is transformed into an echelon form.Then, from the new system of equations, we gradually get the values of xn, xn−1 to x1.[

A | bT]∼ . . . ∼ . . . ∼

[A′ | cT

]The Jordan Elimination. We transform the augmented matrix until we have the unitmatrix E were A was originally. Then the solution vector xT appears on the place of bT.[

A | bT]∼ . . . ∼ . . . ∼

[E | xT

]The use of the inverse matrix. We solve the matrix equation A·xT = bT by multiplyingboth its sides by A−1 from the left.

A · xT = bT −→ A−1 · A · xT = A−1 · bT −→ E · xT = A−1 · bT −→ xT = A−1 · bT.

The Homogeneous System, i.e. b = oThe set of all solutions is a subspaceH of Vn and dim (H) = k = n−h(A). After the Gauss’elimination we distinguish the leading and the free variables. Using an appropriate choiceof the values of free variables we get the vectors h1, . . . , hk forming a basis of H.

The general solution is h = c1 · h1+ . . .+ ck · hk, where c1, . . . , ck are arbitrary constants.

The Inhomogeneous System, i.e. b = oWe must obtain one (particular) solution p = (p1, . . . , pn) of the system (e.g. usingGauss’ elimination and then choosing the values of all free variables equal to zero).Then it is necessary to solve the associated homogeneous system - to give its generalsolution h. We use the scheme General = Partial + Homogeneous which means that thegeneral solution is x = p+ h = p+ c1 · h1 + . . .+ ck · hk (c1, . . . , ck arbitrary constants).

16

EXERCISES

Exercise 5.1 [regular system of linear equations]

(a) Which of the four systems in unknowns x1, x2 are regular?

2x1 − x2 = 16x1 + 3x2 = 2

2x1 + 2x2 = 23x1 + 3x2 = 6

2x1 − x2 = 26x1 − 3x2 = 4

2x1 − 5x2 = 01x1 − 3x2 = 0

(b) Use the Cramer’s Rule to solve the systems from 5.1 (a), if it is possible.

(c) Use the inverse matrix to solve the systems from z 5.1 (a), if it is possible.

(d) Use the Gauss-Jordan elimination to solve the systems from 5.1 (a), if it is possible.

Exercise 5.2 [the Frobenius’ condition] For every given system of equations in unknownsz1, z2, z3, z4 check the Frobenius’ condition and determine the number of solutions.

(a)z1 + 2 z2 + z3 + z4 = 1

3 z1 − z2 − 2 z3 + z4 = 02z1 + 4z2 + 2 z3 + z3 = 2

(b)

z1 − 2 z2 + 2 z3 + z4 = 2−3 z1 + 6 z2 − 2 z3 + z4 = −62 z1 − z2 + 3 z3 + z4 = 3

−2 z1 + 4 z2 + z3 + z4 = −4

(c)z1 − 2 z2 + 2 z3 + z4 = 2

3 z1 + 6 z2 − 2 z3 + z4 = −6(d) −4z1 − 2 z2 + 2 z3 + z4 = 2

Exercise 5.3 [homogeneous systems] Find one possible basis of the space of all solutionsof the given homogeneous system in unknowns t1, t2, t3, t4:

(a)t1 +2t2+ t3 + t4 =02t1 +5t2+2t3− t4 =0t1 +3t2+ t3− 2t4 =0

(b)t1 + t2 + t3 + t4=02t1 +2t2 + t3 + t4=03t1 +3t2 +2t4=0

(c)t1 + t2 − t3− t4=0

−t1− t2 + t3 + t4=02t1 +2t2 − 2t3− 2t4=0

Exercise 5.4 [inhomogeneous systems] Give the general solution of the inhomogeneoussystem in unknowns x1, x2, x3:

(a)x1 + 2x2 + x3 = 62x1 + x3 = 1

(b)x1 + 2x2 − x3 = −62x1 + 4x2 − 2x3 = 0

(c)x1 + 2x2 + x3 = 62x1 + 4x2 − x3 = 6

Exercise 5.5 [exercises with a parameter] Find all real numbers p such that

(a) the systempx1 + 3x2 = 5

−6x1 − 2px2 = 1in unknowns x1, x2 can not be solved by means of

the Cramer’s Rule;

(b) the system2y1 +4y2 − 2y3 =3y1 +2y2 +5y3 = p

in unknowns y1, y2, y3 satisfies the Frobenius’ condi-

tion;

(c) the space of all solutions of the homogeneous systemy1 +2y2 − 2y3 =0

2y1 +4y2 + py3 =0in unknowns

y1, y2, y3 has dimension 2;

(d) the systempy1 +2y2 − 2y3 =1y1 − 4y2 + y3 =2

2y1 +4y2 =3in unknowns y1, y2, y3 has just one solution.

17

QUESTIONS NOTE: all the equations in these questions are linear.

Questions 5.1 [the coordinates of a vector] (a) Peter and Paul solved the same problem:determine the coordinates c1, c2 of vector (−4, 8) with respect to basis (1, 2), (1, 3). Boththe students formed a system of equations in unknowns c1, c2, namely:

Peter:c1 + 2c2 = −4c1 + 3c2 = 8

Paul:c1 + c2 = −42c1 + 3x2 = 8

.Peter’s system is incorrect while

Paul’s system is correct. Why?

(b) Linda is solving the problem: determine the coordinates c1, c2 of vector (4,−8)

with respect to basis (−1, 2), (3,−6). She forms a system of 2 equations in unknowns c1, c2and applies the Cramer’s Rule. But it does not work. It is really weird - you can easily seethat (4,−8) = (−1)·(−1, 2)+1·(3,−6) and it means that c1=−1, c2=1. Give an explanation.

(c) A basis B : b1=(−1, 10), b2=(3, 5) of V2 is given. Which of the problems (i) and (ii)below is easier to solve? Why?

(i) Determine the coordinates of vector (2, 1) with respect to B.(ii) The coordinates of vector v with respect to basis B are 2, 1. Find the vector v.

Questions 5.2 [the Cramer’s Rule]

(a) What homogeneous systems can be solved by means of the Cramer’s Rule?

(b) The calculation of the coordinates of a vector with respect to a given basis of Vn canalways be realized by means of the Cramer’s Rule. Why?

(c) We are given a system of equations that can not be solved by means of the inversematrix. Then, necessarily, also the Cramer’s Rule can not be applied. Why?

(d) We know that a given system of equations can not be solved by means of the Cramer’sRule. How many solutions does the system have?

Questions 5.3 [the Frobenius’ condition, the system of linear equations]

(a) Every homogeneous system satisfies the Frobenius’ condition. Why?

(b) At the right, a system in unknowns x1, x2, x3 is given. Peter hasnot noticed, that, in the second equation, two unknowns (x2, x3) had

x1 +2x2 +5x3 =14x1 +2x3 +5x2 =15

been swapped. His incorrect solution is given below. Show a correct calculation.

x1 +2x2 +5x3 =14x1 +2x3 +5x2 =15

→[1 2 5 141 2 5 25

]∼[1 2 5 140 0 0 11

]→ h(A)=1, h(A∗)=2 → no solution.

(c) How many solutions can a system of 2 equations in 3 unknowns have? Show examples.

(d) How many solutions can a system of 3 equations in 2 unknowns have? Show examples.

(e) Show an example of a system of 1 equation in 4 unknowns that does not satisfy theFrobenius’ condition.

(f) A system of 2 equations in unknowns t1, t2, t3 and parameter p isgiven. Why the value of p does not influence the number of solutions?

t1 − 2t2 +2t3 =32t1 − 4t2 +5t3 = p

(g) A system of 2 equations in unknowns u1, u2, u3 and parameter pis given. Why the value of p does not influence the number of solutions?

u1 − 2u2 +2u3 =32u1 − 4u2 + pu3 =6

(h) A system of 2 equations in unknowns v1, v2, v3 and parameter pis given. Why the value of p does not influence the number of solutions?

v1 − 2v2 +2v3 =32v1 − 4v2 + pv3 =5

(i) A system of 2 homogeneous equations in unknowns w1, w2, w3 isgiven. The dimension of the space P of all its solutions equals 1, i.e. itsbasis is always formed by one vector. Explain why the system of onevector v = (1, 1, 1) can not be a basis of P .

w1 − 2w2 +2w3 =02w1 − 4w2 +5w3 =0

18

APPLICATIONSApplication 5.1 [the coordinates of a vector with respect to a basis]A regular system of equations is to be solved. For example, to deter-mine the coordinates c1, c2, c3 of vector (−4, 8, 2) with respect to basis(1, 2, 1), (1, 3, 2), (1,−3, 2) we form the system at the right.

c1 + c2 + c3 =−42c1 +3c2 − 3c3 = 8c1 +2c2 +2c3 = 2

Solve it using (i) the Cramer’s Rule, (ii) the Gauss’ Elimination. What was quicker?

Application 5.2 [the regression line] The data are N pairs of coordinates [xi, yi] of Npoints in the plane. For example, N = 5 and the points are [1, 2], [0, 0], [−1,−1], [4, 2], [4, 3].For this data, we need the regression line y = k · x+ q, where k, q the unknown parameters.

The Least Squares Method gives the system of equations inunknowns k, q at the right.

N · q+∑

xi · k=∑

xi∑xi · q+

∑x2i · k=

∑xiyi∑

xi = 1+0+(−1)+4+4 = 8,∑

yi = 2+0+(−1)+2+3 = 6,∑x2i = 1+0+1+16+16 = 34,

∑xiyi = 2+0+1+8+12 = 23. Now, put the values into the

equations and use the Cramer’s Rule to find the values of q, k.

Application 5.3 [decision making] The International Red Cross is making plans to aircraft

emergency food and medical supplies into a large South American city which has recently suffered

from extensive flooding. The amount of money available for the purchase of supplies is 155,000

USD, the plane has the volume capacity of 6,000 cubic feet and the weight capacity 40,000 pounds.

Four items will be airlifted in containers with their following parameters: a container of blood

(volume 20 cubic feet, weight 150 pounds, price 1,000 USD per container), a container of medical

supply kits (volume 30, weight 100, price 300), a container of food (volume 8, weight 55, price

400) and a container of water (volume 6, weight 70, price 200). Let x1, x2, x3, x4 be the numbers of

containers of blood, medical supply, food, and water respectively; we can form a system of three

equations (the equation for the volume, the equation for the weight, the equation for the price) in

4 unknowns. Then we create the augmented matrix and, moreover, one free variable (we chose x4)

will be moved to the right sides. Now, we use the Gauss-Jordan elimination:

20x1 + 30x2 + 8x3 + 6x4 =6000150x1 +100x2 + 55x3 + 70x4 =40 000

1 000x1 +300x2 +400x3 +200x4 =155 000−→

20 30 8 6 6 000150 100 55 70 40 000

1 000 300 400 200 155 000

−→

20 30 8 6 000− 6x4150 100 55 40 000− 70x4

1 000 300 400 155 000− 200x4

−→

1 0 0 927.1− 3.008x40 1 0 120.8− 0.083x40 0 1 −2021 + 7.083x4

−→x1 = 927.1− 3.008x4x2 = 120.8− 0.083x4x3 = −2021 + 7.083x4

If x4 = 290 then x1 = 54.8, x2 = 96.7, x3 = 33, i.e. the numbers of containers can be 54,96, 33, and 290, respectively. Solve the problem if 30 containers of blood are required.

Application 5.4 [optimization of production] A company produces 3 products A, B, andC with the following requirements per unit: A - 5 labour-hours, 15 pounds of material,B - 2 labour-hours, 10 pounds of material, C - 4 labour-hours, 12 pounds of material.

Each month there are 1,300 labour-hours and 4,700 pounds of raw material available. Ifcombined monthly production for the three products should equal 400 units, we are to findthe number units of each product. Thus, we form 3 equations in the unknowns n1, n2, n3 -the numbers of products A , B , and C respectively - below. Solve it.

5n1 + 2n2 + 4n3 =130015n1 +10n2 +12n3 =4700

n1 + n2 + n3 = 400

19

TOPIC 6. Relations, Mappings, Functions

For a finite set X, we denote |X| the number of its elements; for the empty set |∅| = 0.

r : X −→ Y a relation with domain D(r) = X and range Y ; 2|X|·|Y | possible relations; the

number of k-arrow relations is(|X|·|Y |

k

); for P ⊆ X, Q ⊆ Y we call r(P ) the

image, and r−1(Q) the preimage in r; Ar is the adjacency matrix of r

r|P,Q . . . the restriction of r on P ⊆ X, Q ⊆ Y ; from Ar some rows and somecolumns are deleted; the number of such restrictions is 2|X|+|Y |

r−1, s ◦ r the inverse and the composed relation; for the inversion, Ar is transposed

we write s ◦ r in the reverse order, but not the boolean multiplication Arb· As

f : D −→ H a mapping; D = ∅ and every element x∈D has just one image f(x)∈H;|H||D| of possible mappings; if D,H ⊆ R we talk about a function

injection a mapping such that for every y ∈ H there is |f−1(y)| ≤ 1; the number

of injections is |Y |!(|Y |−|X|)! if |X| ≤ |Y |, or 0 (for bijections =|Y |! or 0)

surjection a mapping for which f(D) = H; on the contrary, for a constant |f(D)| = 1

bijection injection ∧ surjection; f is a bijection iff both f , f−1 are mappings

y = f(x) . . . functional notation; f(x) is the value of function at point x ∈ D(f); alsowe write y = y(x); x, y . . . independent and dependent variable

the graph all points [x, f(x)] in the plane for x ∈ D(f)

y = |x| the ”absolute value”function; |x|=x for x ≥ 0, . . .=− x for x ≤ 0

y = sign(x) the ”signum”function; sign(0)=0, sign(x) = 1 for x > 0, . . . = −1 for x < 0

y = chM(x) the characteristic function of set M ; chM (x)=1 for x∈M , . . .=0 for x/∈My = ex . . . the exponential function (e

.= 2.71... the base of natural logarithms),

D = R, H = (0,+∞); e0 = 1; the function is increasing on D;for x < 0 there is ex ∈ (0, 1) and for x > 0 there is ex ∈ (1,+∞)

y = ln x . . . the natural logarithm; D = (0,+∞), H=R; ln 1=0; increasing on D;lnx ∈ (−∞, 0) for x ∈ (0, 1), and ln x ∈ (0,+∞) for x > 1

Notes

• Inverse functions to trigonometric functions sin, cos, tg or tan, and cotan denoted byarcsin or sin−1, arccos or cos−1 etc. are called circular or inverse trigonometric functions.

• the functions y = ex and y = ln x are inverse to each other, i.e. elnx = x, ln(ex) = x;

ea+b = ea · eb; ea−b = ea/eb; (ea)b = ea·b, ln(a · b) = ln(a)+ ln(b); ln(a/b) = ln(a)− ln(b);

ln(ab) = b · ln(a); the definition of the general power: AB = eB·lnA pro A > 0.

Getting the formulas for inverse and composite functionsThe inverse function y = f−1(x): if f : D −→ H is a bijection then f−1 : H −→ Dis a function satisfying f−1(b) = a iff f(a) = b. To get the formula for f−1 we first swapthe symbols x and y in the formula of y = f(x), i. e. we will have x = f(y), and then weexpress y in terms of x.

The composite function (compositum) y = g(f(x)): if a ∈ D(f) and if b = f(a) ∈ D(g)then we define g(f(a)) = g(b); f is called the inner and g is called the outer function. Toget the formula, we can first introduce a helper notation like w = f(x), y = g(w); now,in the formula of g(w), all the symbols of w are replaced with the formula of f(x).

20

EXERCISES

Exercise 6.1 [operations on relations] U = {α, β, γ}, V = {a, b, c},W = {1, 2, 3, 4, 5} are3 sets and p : U −→ V, q : V −→ W are 2 relations given as

Ap =

[1 0 11 1 00 1 1

], Aq =

[1 0 1 1 10 1 0 1 01 0 0 1 1

]. Find the adjacency matrices of the following relations:

(i) q|{b,c},{1,3,5} (ii) p−1 (iii) q ◦ p (iv) p−1 ◦ q−1 (v) q−1 ◦ q.

Exercise 6.2 [counting relations and mappings] Use the sets from Exercise 6.1 and deter-mine the number of all:(i) relations from W to V (ii) relations from V to W (iii) 3-arrow rel. from W to V(iv) mappings from W to V (v) mappings from U to W (vi) injections from W to V(vii) inject. from U to W (viii) constants from W to U (ix) constants from U to V .

Exercise 6.3 [the functions | |, ’sign’ and ’ch’](a) Evaluate the expressions:

(i) (−2) · sign (4) (ii) | − 8| − |5| (iii) 5− ch(−2,3⟩ (3)

(iv) sign (4− |5|) (v) 1− sign (2)− | − 2| (vi) ch(−4,3) (| − 3|)− 3 · ch(−2,3) (1).

(b Evaluate the expressions if x = −3:

(i) (−2) · sign (x+ 3) (ii) |x− 8| − |5 + 2x| (iii) 5− ch⟨−3,3) (x− 3)

(iv) sign (x− |5|) (v) x · sign (2x)−|x+4| (vi) ch(−4,3) (|x−3|)− ch{−1} (x+2).

Exercise 6.4 [the exponential and the logarithmic function] With a help of a calculator,evaluate the expressions if x = −1 and round your results to 2 decimal places:

(i) e2x − e−x (ii) ln 2x− ln(2− x) (iii) ex2−3x−2

(iv) ln(x2 − 3x+ 2) (v) e + sign(ln(−x

2))

(vi) esign (x) + ch(−2,3) (ln(−x)).

Exercise 6.5 [domains of definition] Determine the domains of the given functions:

(i) y = 2x2−3x−4

(ii) y =√x2 + 2 (iii) y = ln(2− x)

(iv) y = ln(x2 − 3x+ 2) (v) y = 31−sign (x)

− 4x

(vi) y =√x+ 2 +

√4− 2x

Exercise 6.6 [graphing] Sketch the graphs:

(i) y = −x2 + 3x+ 4 (ii) y = 1 + sign (x) (iii) y = 1− ch⟨−3,3) (x)

(iv) y = |x2 − 3x| (v) y = sign (x2 − 1) (vi) y = ch(−3,3) (x) + ch⟨−1,1⟩ (x)

(vii) y = sign ex (viii) y = |x2 + 1| (ix) y = 3ch(−3,+∞) (x2 + 1)

Exercise 6.7 [the composite and the inverse function](a) Find an inverse function formula:(i) y = 2

x+2(ii) y = 4− e2x (iii) y = 1 + ln(2− x)

(iv) y =√x− 3 (v) y = x+1

x+2(vi) y = 3

ex+2

(b) If f(x) = x+ 2, g(x) = x2 + 2, h(x) = e2x form composite function formulas:

(i) y = f(g(x)) = . . . (ii) y = g(f(x)) = . . . (iii) y = h(g(x)) = . . .

(iv) y = f(h(x)) = . . . (v) y = g(h(x)) = . . . (vi) y = f(f(x)) = . . ..

21

QUESTIONS

Questions 6.1 [operations on relations and mappings; please, pay attention to our require-ment that a mapping must always have a non-empty domain]

(a) Show a relation which is not a mapping. (b) Show a mapping which is not a relation.(c) Find relations r, s such that both Ar, As are 3× 3 matrices but s ◦ r is not defined.(d) Find a relation r which is not a mapping but r−1 is a mapping.(e) Find a relation r which is not a mapping and r−1 is not a mapping, too.(f) Find 2 relations r and s, such that s ◦ r is a mapping and, moreover,(i) r is, but s isn’t a mapping, (ii) r is not, but s is a mapping, (iii) neither r nor s is a mapping.

Questions 6.2 [counting relations and mappings]

(a) Determine the number of all relations from P to Q knowing that:(i) |P | = 2, |Q| = 5 , (ii) |P | = 0, |Q| = 5, (iii) |P | < |Q| < 2.

(b) Determine the number of all mappings from P to Q knowing that:(i) |P | = 2, |Q| = 5 , (ii) |P | = 0, |Q| = 5, (iii) |P | < |Q| < 2.

(c) If you know that the number of all relations from A to B equals 16, determine thenumber of mappings from A to B (there are 3 solutions).

Questions 6.3 [the functions | |, ’sign’ and ’ch’]

(a) Evaluate the functions if x = −3:

(i) y = |3− x| (ii) y = sign (2x+ 3) (iii) y = ch⟨−4,0⟩(x+ 4).

(b) Given the function y = ch(−6,7)(x2 − 4) + ch{−6,7}(x + 1) determine its values for 4

choices of x: (i) x = 2 (ii) x = 6 (iii) x = −7 (iv) x = 1.

Questions 6.4 [the functions y = ex and y = ln x] Evaluate the expressions without anyuse of a calculator:

(i) 5e0 − 4 ln 1 (ii) 5eln 4 + ln e5 (iii) ln(ln e) + ee0 − e

(iv) ln(sign 5) + sign (ln 5) (v) ln(sign 12) + sign (ln 0.5) (vi) esign 5 · esign (− 1

2)

Questions 6.5 [determining domains of definition]

(a) Give a formula for a function with the empty domain.

(b) Give a formula for a function with the one-element domain.

(c) Give a formula for a function with the domain equal to R.

(d) Which function has larger domain y = 3 ln(x− 1) or y = 2√x− 1?

(e) Which function has larger domain y = 2 ln(x− 1) or y =3√x− 1

?

Questions 6.6 [the composite and the inverse function]

(a) Evaluate the function if x = 5:

(i) y =|x− 6||6− x|

(ii) y = sign (sign (sign (x))) (iii) y = ch(−1,0⟩(ch(−1,0⟩(ch(−1,0⟩(x))).

(b) Function f is called self-inverse if f and f−1 can have the same formula. Which ofthe following functions are self-inverse?

f1(x) = x− 2, f2(x) = 2− x, f3(x) =2

x, f4(x) =

2

x2.

22

APPLICATIONS

Applications 6.1 [measuring temperature](i) The formula f = 32 + 9

5c is a function which converts temperature c in degrees of

Celsius onto degrees of Fahrenheit. Example: f(20) = 32+95·20 = 68 converts 200C onto 680F.

Now, we create the inverse function formula: f = 32+ 95c → f−32 = 9

5c → c = 5

9(f−32).

It converts the temperature f in 0F onto 0C; for example, c(68) = 59(68− 32) = 200C.

(ii) An entomologist described a behaviour of a species of cricket. The number of cricketchirps per minute, n, is related the temperature, f (in 0F), by the function n = 4 ·f−40. Forexample, n(59) = 4 · 59 − 40 = 196 converts the temperature 590F to 196 chirps/min. Theinverse function formula will be: n = 4·f−40 → n+40 = 4·f → f = 10+ n

4; this function

converts n chirps/min to the temperature f in 0 F; for example, f(196) = 10 + 1964

= 590F.

(iii) Using the functions from (i) and (ii) we can create the composite function c(f(n)) =c(10 + n

4) = . . . which will convert n chirps/min onto the temperature in 0C. Finish.

Applications 6.2 [mathematics of finance](a) The model of continuous compounding is used in financial mathematics. For exam-

ple, if the interest rate is 4% p.a. the formula is y = 800 · e0.04x, where x is time in yearsand y is the future value of an investment of 800 EUR after x years from now. Find:

(i) the future value y of the investment of 800 EUR after 6 and 3/4 years,(ii) the time necessary to double the investment of 800 EUR,(iii) the inverse function formula and explain what it calculates.

(b) [Swokowski, 1992] A company purchased a machine. The machine’s resale value y(in USD) will decrease according to the function y = 4400 · e−0.2·x + 900 where x is time inyears. Create an inverse function formula and explain what it calculates.

Applications 6.3 [health care](a) A mathematical model relating the age and the systolic blood pressure of an adult

patient has the form of function y = 40 + 25 ln(x+ 1) where x is the age in years and y thepressure in mm Hg. Create an inverse function formula and explain what it calculates.

(b) Public health records in a city indicate that x weeks after the outbreak of a certainform of influenza, approximately y = 80

4+76e−1.2x thousand people had caught the disease.

Create an inverse function formula and explain what it calculates.

Applications 6.4 [models in economy](a) It is estimated that if x thousand EUR are spent on advertising, approximately

y = 50−41e−0.15x thousand units of a certain commodity will be sold. Express x as functionof y. How much should be spent on advertising to generate sales of 40 000 units?

(b) A cosmetic company is planning the introduction and promotion of a new lipstickline. The marketing research department, after test marketing the new line in a large carefullyselected city, found that the demand in that city is given approximatelly by p(x) = 10e−1.1x,where x thousand lipsticks were sold per week at a price of p dollars each. Give the inversefunction formula and interpret it.

(c) [Bartlett & Ziegler, 1990] In a national tour of a rock band, the demand of T-shirtsis given by p = 15−4 ln x where x is the number of T-shirts (in thousand) that can be soldduring a single concert at a price of p EUR per item. Create an inverse function formulaand explain what it calculates.

23

TOPIC 7. Sequences and Series

{an}∞n=1 . . . or simply {an} is a sequence, an its nth term, n the index of the term

limn→∞

an = L the limit of {an} as n approaches infinity equals L;

for L ∈ R the sequence is said to be convergent; for L = ±∞ thesequence is said to be divergent to ±∞; otherwise {an} is divergent

sn =n∑

i=1an = a1 + a2 + . . .+ an is called the nth partial sum of the sequence {an}

∞∑i=1

an = a1 + a2 + . . .+ an + . . . is called an infinite series; if there exists thelimit of the partial sums sequence s∞ = lim

n→∞sn = S ∈ R, we say that

the series∑

an converges and its sum is S; otherwise∑

an is divergent

If in sequence {an}∞n=1 for every n ∈ N there is

(1) an < an+1, then the sequence is increasing, (2) an > an+1, then the seq. is decreasing,(3) an ≤ an+1, then the seq. is nondecreasing, (4) an ≥ an+1, then the seq. is nonincreasing.If there exists M ∈ R such that for every n there is an ≤ M , we call {an} bounded from upward.If there existsm ∈ R such that for every n there is an ≥ m, we call {an} bounded from downward.

A sequence having both an upper bound and a lower bound is called bounded.

Arithmetic sequence - recurrence relation: an+1 = an + d, d is the difference

an = a1 + (n− 1) · d, or an = am + (n−m) · d; sn = a1+an2

· n = n · a1 + d · n(n−1)2

,

Geometric sequence - recurrence relation: an+1 = an · r, r is the common ratio,

an = a1 · rn−1 or an = am · rn−m; sn = an+1−a1r−1

= a1 · rn−1r−1

, s∞ = a11−r

for |r| < 1.

Important limits and series; formal ”calculus”on numbers (r ∈ R) and symbols of ∞

limn→∞

n = +∞, limn→∞

1n= 0; lim

n→∞rn = 0 for |r| < 1 and = +∞ for r > 1; lim

n→∞n√n = 1,

limn→∞

n√A = 1 for A > 0; lim

n→∞

(1 + 1

n

)n= e =

∞∑i=0

1i!= 1 + 1

1!+ 1

2!+ 1

3!+ . . .+ 1

n!+ . . .

∞∑i=1

1ip=1+ 1

1p+ 1

2p+ 1

3p+ . . . converges for p > 1; harmonic s.:

∞∑i=1

1i=1

1+1

2+1

3+ . . . = +∞,

r+∞ = ∞+ r = ∞; r−∞ = −∞+ r = −∞; +∞+∞ = +∞; −∞−∞ = −∞;

(±∞) · (∓∞) = −∞; (±∞) · (±∞) = +∞;√+∞ = ln(+∞) = log(+∞) = +∞.

r · (±∞) = ±∞ for r>0∓∞ for r<0

; 1±∞ = 0; 1

0=

+∞ if all the terms are positive

−∞ if all the terms are negative

Indeterminate forms - the value cannot be determined in general

∥∞−∞∥, ∥0 · (±∞)∥,∥∥∥10

∥∥∥, ∥∥∥00

∥∥∥, ∥∥∥∞∞∥∥∥, ∥1±∞∥, ∥0±∞∥, ∥(±∞)0∥.

Limit convergence tests - for positive-term series

The Comparison Test: Let∑

an and∑

bn be such series that limn→∞

anbn=L>0, L∈R.

Then either both the series converge or both diverge.

The Ratio and Root test: We find limn→∞

an+1

an= L or lim

n→∞

√an = L. Then

(1) L < 1 ⇒ ∑an is convergent, (2) L > 1 ⇒ ∑

an is divergent.

24

EXERCISES

Exercise 7.1 [recurrence relations](a) For each of the sequences described below, calculate the values a1, a2, a3, a4.

(i) a1 = 1, an+1 = (−1)n · an (ii) a6 = 43, an+1 = 2an − 1

(iii) a1 = 5, a2 = 6, an+2 = an+1 + an + 3 (iv) a5 = 120, an−1 =ann.

(b) For each of the pair of sequences described below, determine the values b4, c4.

(i) b1 = 1, c1 = 2, bn+1 = bn + cn, cn+1 = bn · cn,(ii) b5 = c5 = 2, bn+1 = bn + 2cn − 1, cn+1 = 3bn − cn + 2.

Exercise 7.2 [arithmetic and geometric sequences] Find the sums of the finite sequences:

(i) 2+4+6+8+ . . .+600, (ii) 400+401+402+403+ . . .+600,

(iii) 2+4+8+16+ . . .+2048+4096, (iv) 10+10

3+10

9+10

27+ . . .+

10

6561.

Exercise 7.3 [evaluating the limits](a) Without the use of the calculus on infinities, evaluate the limits:

(i) limm→∞

(3

m√2 + 2

m√3)

(ii) limn→∞

(2− n

√n)

(iii) limk→∞

[(1 + 1

k

)k+ 99 · 0.999k

].

(b) Evaluate the limits of indeterminate forms of the type ∥∞−∞∥; here is a Sample:

limn→∞

(3n2−4n+7) = limn→∞

n2

n2 (3n2−4n+7) = lim

n→∞n2 ·

(3n2−4n+7

n2

)= lim

n→∞n2 ·

(3n2

n2 −4nn2+

7n2

)=

= limn→∞

n2 ·(3− 4

n + 7n2

)=∥∥∥(+∞)2 ·

(3− 4

+∞ + 7(+∞)2

)= (+∞) · (3− 0 + 0)

∥∥∥ = +∞.

(i) limn→∞

(3n−4n2+5) (ii) limm→∞

(2 · 3m − 3 · 2m) (iii) limk→∞

(√4k + 1−

√k + 4

).

(c) Evaluate the limits of indeterminate forms of the type ∥∞∞∥; here is a Sample:

limn→∞

3n−4n2+2

= limn→∞

(3n−4)/n(n2+2)/n

= limn→∞

3nn− 4

nn2

n+ 2

n

= limn→∞

3− 4n

n+ 2n

=

∥∥∥∥ 3− 4+∞

+∞+ 2+∞

= 3−0+∞+0 = 3

+∞

∥∥∥∥ = 0.

(i) limn→∞

3n2 − 4n+ 2

n2 + n+ 2(ii) lim

n→∞

n4 + 2

n2 + 8(iii) lim

m→∞

6 · 2m + 4

2m + 5(iv) lim

k→∞

6 · 3k + 4k

2k + 5k.

(d) Determine: (i) limp→∞

p+2q+3

(ii) limq→∞

p+2q+3

(iii) limp→∞

(√6p−√

4q

)(iv) lim

q→∞

(√6p−√

q4

).

Exercise 7.4 [infinite geometric series] Determine the sums of infinite geometric series:

(i) 9+0.09+0.0009+ . . ., (ii) 5− 52+5

4−5

8+ 5

16− . . ., (iii) 51

50+(5150

)2+(5150

)3+(5150

)4+ . . ..

Exercise 7.5 [convergence tests] Investigate the convergence of the given series, using the

(a) Limit Ratio Test: (i)∞∑n=1

1.1n, (ii)∞∑n=1

1

n+ 8, (iii)

∞∑m=1

1

2m + 3, (iv)

∞∑m=1

1

m!,

(b) Limit Root Test: (i)∞∑n=1

0.8n, (ii)∞∑n=1

1

(n+ 8)n, (iii)

∞∑m=1

1√m, (iv)

∞∑m=1

1

0.5m.

25

QUESTIONS

Questions 7.1 [recurrence relations]

(a) The formula an+1 = 2 · an is not sufficient to describe the sequence {an}. Why?

(b) The formulas b1 = 14, bn+2 = bn+1 + bn are not sufficient to describe {bn}. Why?

(c) In a sequence, if we know that c1 = 5, c2 = 7, can we determine the value of c3?

(d) Can you simply describe the sequence given as d1 = 23, dn+1 = dn?

Questions 7.2 [arithmetic and geometric sequences]

(a) Can you give an example of a sequence that is both arithmetic and geometric?

(b) The sequence described as b2 = 3, bnbn+1

= 3 is geometric. Explain why?

(c) In a geometric sequence, there is c1 = 5, c5 = 80. Then the value of c3 is unique,but the value of c4 is not. Explain in more details.

(d) For the sum S = 200+201+202+ . . .+498+499+500, two formulas were proposed.

Which is correct and why? (i) S = 200+5002

× 300, (ii) S = 200+5002

× 301.

(e) By January 1, 2015, the price of a painting was 45 000 EUR. If the price will increaseby 2% yearly we are to calculate the price after 5 years. Which of the three formulas iscorrect and why? (i) a5 = a0 · 1.025, (ii) a6 = a1 · 1.025, (iii) a5 = a1 · 1.024.

Questions 7.3 [evaluating limits]

(a) Petr and Jan evaluated the same limit and both are wrong. Why? How is it correct?

Petr: limn→∞

2n3n=∥∥∥∥2·(+∞)3·(+∞)

=+∞+∞=

+∞/+∞/ = 1

1

∥∥∥∥ = 1, Jan: limn→∞

2n3n=∥∥∥23· (+∞) · 1

+∞=(+∞) · 0∥∥∥ = 0.

(b) Petra and Jana evaluated the same limit and both are wrong. Why? How is it correct?

Petra: limk→∞

[2k

3k+(1+ 1

k

)k]=∥∥∥∥2+∞

3+∞ +(1+ 1

+∞

)+∞= +∞

+∞ + 1+∞ = 1 + 1∥∥∥∥ = 2.

Jana: limk→∞

[2k

3k+(1+ 1

k

)k]=∥∥∥∥2+∞· 1

3+∞ +(+∞+1+∞

)+∞= (+∞)·0 + (+∞)+∞

(+∞)+∞ = 0++∞+∞ = 0+1

∥∥∥∥=1.

(c) Calculations of two limits were spoiled. Why? Give the correct results.

The 1st limit: limn→∞

2n+13k+1

= limn→∞

(2n+1)/n(3k+1)/k

= limn→∞

2+ 1n

3+ 1k

=∥∥∥∥2+ 1

+∞3+ 1

+∞= 2+0

3+0

∥∥∥∥ = 23.

The 2nd limit: limk→∞

2n+13k+1

=∥∥∥2·∞+13·∞+1

= +∞+∞ = +∞/

+∞/ = 11

∥∥∥ = 1.

Questions 7.4 [infinite series]

(a) The sequence 1, 12, 13, 14, . . . , 1

n, . . . can be described as a1 = 1, an+1 = an · n

n+1. We use

the formula for the sum of an infinite series (the common ratio equals r = nn+1

) and we get:

1+12+1

3+1

4+ . . . = s∞ = a1

1−r= 1

1− nn+1

= 1(n+1)−n

n+1

= 11

n+1

= n+1 = ∥(+∞) + 1∥ = +∞, which

is the correct sum od the harmonic series. But the calculation is incorrect. Why?

(b) Chose a real number x. Is the series∞∑n=1

(1

x+5

)nconvergent?

(c) Petr and Jirı tried to find a sum of the infinite series 1− 1 + 1− 1 + 1− 1 + . . ..

Petr: 1− 1 + 1− 1 + 1− 1 + . . . = (1− 1) + (1− 1) + (1− 1) + . . . = 0 + 0 + 0 + . . . = 0,Jirı: 1−1+1−1+1−1+ . . . = 1+(−1+1)+(−1+1)+(−1+1)+ . . . = 1+0+0+0+ . . . = 1.Who is right?

26

APPLICATIONS

Application 7.1 [finance] The compound interest is frequently used in mathematics offinance. For example, if the interest rate is 4% yearly, we use the formula An = 800 · 1.04n,where n the number of years and An is the future value of an investment of 800 EUR aftern years (a geometric sequence). Determine(i) the future value after 8 years, (ii) the time necessary for the doubling of the investment.

Application 7.2 [wildlife] In an area of Canada, the population of bobcat was investiga-ted. It was classified by age as kittens (less than one year old) and adults (at least one yearold). We know that: the ratio of males to females is 1; all adult females have a litter each June,with an average litter size of 3 kittens; the survival rate of kittens is 50% while that of adultsis 70% per year. Let An be the number adults and Kn be the number of newborn kittens bothin June of the year n. We have recurrent relations: Kn = An

2· 3, An+1 = 0.7An +0.5Kn.

If A1987 = 400 then determine (a) the number of females that had a litter in June 1986,(b) the number of animals adult already in June 1988 and still alive in June 1989,(c) the number of animals born in June 1988 but not alive in June 1989.

Application 7.3 [market equilibrium] In the study of market equilibrium, the cobweb

model relates the price sequence p1, p2, p3, . . . of a chosen product in the individual timeperiods to supply values S1, S2, S3, . . . and demand values D1, D2, D3, . . .. Suppose that in asimple linear model Dn = 4800−50pn (the demand equation) and Sn+1 = 4000+25pn (thesupply always reflects the price in the previous time period). The equilibrium condition isDn+1 = Sn+1, which yields 4800− 50pn+1 = 4000 + 25pn −→ pn+1 = 16− 0.5pn.Now, determine the values of p2, p3, p4, S2, S3, S4, D2, D3, D4 if p1 = 8 EUR.

Let us continue: after some time, the price reaches its equilibrium value p = limn→∞

pn. From

the equation pn+1 = 16 − 0.5pn, the limit process yields limn→∞

pn+1 = limn→∞

(16 − 0.5pn) −→p = 16 − 0.5p −→ 1.5p = 16 −→ p

.= 10.67 EUR, which is the equilibrium price.

Analogously, we get S = limn→∞

Sn+1 = limn→∞

(4000 + 25pn) = . . . and, further, D = limn→∞

Dn =

limn→∞

(4800− 50pn) = . . ., the values of the equilibrium supply and demand. Find S,D.

Application 7.4 [ecology - Swokowski, 1992] A stable population of 40,000 birds lives onthree islands. Each year, 10% of the population on island P migrates to island Q, 20% of thepopulation on island Q migrates to island R, and 5% of the population on island R migratesto island P. Let pn, qn, and rn denote the number of birds in year n on islands P,Q, and Rrespectively, before the migration takes place. We can write the recurrence relations:

pn+1 = 0.05rn + 0.9pn qn+1 = 0.1pn + 0.8qn, rn+1 = 0.2qn + 0.95rn.

If p1986 = 8200, q1986 = 10400, r1986 = 21400 estimate the sizes of the populations in 1988.

Application 7.5 [mathematics of finance] A deposit of 4,000 EUR is made on the firstday of each quarter year for five years in an account that pays 0.8%, compounded quarteryearly. Determine the balance S at the end of five years, if, besides 20 deposits, one more ismade at the end.

The analysis: the 1st deposit will be compounded 20-times, the 2nd one 19-times, . . . , the20th one only once, and the 21st one not at all; i.e. S = 4000 ·1.00820+4000 ·1.00819+ . . .+4000 · 1.0081 + 4000. We work with a geometric sequence (in the opposite order) startingwith the last deposit a1 = 4000, r = 1.008, n = 21 and we get S = s21 = . . .. Finish thecalculation and decide if the amount of 90,000 EUR will be reached.

27

RESULTSTOPIC 1-2 Exercise 1-2.1(a) (3, 4), (b) ⟨−2, 2⟩, (c) (−1, 3⟩, (d) ∅, (e) (−1,+∞), (f) ⟨−3, 3⟩, (g) ⟨−3,+∞), (h) ⟨−3, 4).

Ex. 1-2.2 (A) (a) (−4, 4), 4√2, (b) (−11, 11, 33), 11

√11, (c) (22, 0), 22, (d) not def. (e)

(−20,−2,−6), 2√110, (f) (−20,−2,−6), 2

√110, (g) not def. (h) (−20,−2,−6), 2

√110.

(B) (a) 116.56◦, (b) 136.36◦, (c) not def.

Ex. 1-2.3 (A) (a)[3 −3 06 0 −2

], (b)

[980 −9999 −10

], (c)

[5 8 −621 1 −6

], (d)

[12 −69 12

], (e)

[6 −29 −7

−2 −8 −11

],

(f) impossible, (g)

[2 2 −32 20 6

−3 6 9

], (h) impossible. (B) (a) 3× 3, (b) non-ex., (c) 3× 3.

Ex. 1-2.4 (a) e.g. (0, 4, 0, 0), (1, 13, 1, 1), (−2, 2, 0, 0), (b) e.g.[

1−2

], or

[50

],

(c1) e.g.[−1 −12 2

], (c2) e.g..

[1 1

−2 1

], (c3) e.g.

[2 1

−1 2

],

Ex. 1-2.5 (A) (a) y = ±√11, (b) y = 2, (c) y = ±0.8, (d) y ∈ (2−

√7, 2 +

√7).

(B) (a) (2, 1,−7), (b) (−8/5,−8/5), (c) (−5/2, x2), x2 arbitrary.

(C) (a)[−1 5 −2−3 2 4

], (b)

[−9 −7/21 1/2

], (c)

[3 9 4

−4 4 6

], (D) (a) 3×2, (b) non-ex., (c) 2×n, any n.

App. 1-2.1 (a) α.= 89.3◦, β

.= 52.4◦, γ

.= 38.3◦, (a) α

.= 71.8◦, β

.= 52.3◦, γ

.= 55.9◦.

App. 1-2.2 P 2=[0.996 0.2530.004 0.747

], x3=5226, y3=99, for x6, y6 you need P 5. App. 1-2.3 C.

TOPIC 3 Ex. 3.1 (a) (0, 0), (b) c1=c2=c3=1 → (4, 2), (c) (9,−9), (d) (11,−14).

Ex. 3.2 (a) u3 = 0 · u1 + 0 · u2, (b) u1 = 0 · u2 + 1 · u3, (c) u2 = (−1) · u1 + 1 · u3,(d) u1 = (−1) · u2 + 0 · u3. Ex. 3.3 a,e,f. Ex. 3.4 (a) delete r. 2, (b) swap r. 1 and r.2, (c) imposs., (d) swap r. 2 and r. 3, (e) swap r. 1 and r. 2, (f) delete r. 1.

Ex. 3.5 (a) subtract r. 1 from r. 2, (b) imposs., (c) subtract r. 1 from r. 2, subtract r. 1from r. 3,, add r. 2 to r. 3, (d) imposs., (e) add r. 2 to r. 1, subtract r. 1 from r. 2, (f) addr. 1 to r. 2.

Ex. 3.6 (a) 2, (b) 2, (c) 2, (d) 1, (e) 2, (f) 2. Ex. 3.7 (a) −2, (b) R−{±2}, (c) ∅, (d) R.

App. 3.1 (a)(b) dep., (c)(d)(e) indep. App. 3.2 (a) s, (b) s, (c) q, r, (d) s.

App. 3.3 (a) no - too many vec., (b) yes - n=k=h=4, (c) no - too many vec., (d) no -missing vec., (e) no - some have 5 components. App. 3.4 (a) 1, (b) 3, (c) 4, (d) 3, (e) 4.

App. 3.5 (a) e.g. (3,−1), (b) e.g. (1, 2, 3), (0, 1, 2), (c) e.g. (0, 1, 1, 1), (0, 0, 1, 1), (0, 0, 0,−1),(d) e.g. (1, 0, 2, 0, 1), (0, 2,−2, 0, 0), (e) e.g. (1, 0, 1, 1, 1, 1, 1), (0, 1, 1, 1, 0, 0, 0), (0, 0, 1, 1, 0, 1, 1).

TOPIC 4 Ex. 4.1 (a) non-s., (b) sing., (c) sing., (d) non-s., (e) non-s., (f) non-s.

Ex. 4.2 (a) 68, (b) 24, (c) 69, (d) 9, (e) −77, (f) 0.

Ex. 4.3 (i) 0, sing.; −5, non-sing., (ii) 0, sing.; 42, non-sing. (iii) −5, non-sing.; 6 non-sing.

Ex. 4.4 (a) −21, non-sing. (b) 132, non-sing. (c) 12, non-sing. (d) −6, non-sing.

Ex. 4.5 B−1=

[1 02 −1

], C−1=

[1/2 1

0 −1/2

], D−1=

[1 00 1

], F−1=

[−1 −20 1

], G−1=

[1 20 −1

], H−1=

[−1 −20 1

].

Ex. 4.6 (a)[1 −20 −1

], (b)

[1/2 −1/2

−1/2 3/2

], (c)

[1 −1 00 1 −10 0 1

], (d)

[−1/2 1/2 1/21/2 −1/2 1/21/2 1/2 −1/2

], (e)

0 1 0 01 0 0 00 0 0 10 0 1 −1

,(f)

1 0 0 0−1 1 0 0−1 0 1 0−2 1.5 0 −0.5

. Ex. 4.7 (a) x ∈ R, (b) x = ±2, (c) x ∈ (2, 4), (d) x = −1.

App. 4.1 result (6,−4,−2), App. 4.2 (a) the first 6, the second 2, (b) x1 = −17, x2 = 3.App. 4.3 all are bases of some Vn. App. 4.4 (a)

[−3 −2 −7 −22 2 4 2

], (b)

[−1 −11

],

(c)[

27 12−13 −11

], (d)

[10 −52

], (e)

[−4 41/2 1/2

], (f)

[−95/231/2

], App. 4.5 x1 = 10, x2 = 10

28

TOPIC 5Ex. 5.1 (a) yes, no, no, yes, (b, c, d) the first: x1=5/12, x2=−1/6; the fourth: x1=0, x2=0.

Ex. 5.2 (a) F.C. yes; ∞ sol., (b) F.C. yes; 1 sol., (c) F.C. yes; ∞ sol., (d) F.C. yes; ∞ sol.

Ex. 5.3 (a) e.g. b1=(−1, 0, 1, 0), b2=(−7, 3, 0, 1), (b) e.g. b1=(−1, 1, 0, 0), (c) e.g.

b1=(−1, 1, 0, 0), b2=(1, 0, 1, 0), b3=(1, 0, 0, 1).

Ex. 5.4 (a) e.g. c1·(−0.5,−0.25, 1)+(0.5, 2.75, 0), (b) no sol., (c) e.g. c1·(−2, 1, 0)+(4, 0, 2).Ex. 5.5 (a) p = ±3, (b) p ∈ R, (c) p = −4, (d) p = 5.

App. 5.1 c1=−10, c2=233, c3=− 5

3; Gauss 3 min., Cramer 5 min., App. 5.2 q= 20

106, k= 67

106,

App. 5.3 x1=30, x2=96, x3=91, x4=298, App. 5.4 n1=100, n2=200, n3=100.

TOPIC 6

Ex. 6.1 (i)[0 0 01 0 1

], (ii)

[1 1 00 1 11 0 1

], (iii)

[1 0 1 1 11 1 1 1 11 1 0 1 1

], (iv)

1 1 10 1 11 1 01 1 11 1 1

, (v)

[1 1 11 1 11 1 1

].

Ex. 6.2 (i) 32768, (ii) 32768, (iii) 455, (iv) 243, (v) 125, (vi) 0, (vii) 60, (viii) 3, (ix) 3.

Ex. 6.3 (a) (i) −2, (ii) 3, (iii) 4, (iv) −1, (v) −2, (vi) −3. (b) (i) 0, (ii) 10, (iii) 5,(iv) −1, (v) 2, (vi) −1. Ex. 6.4 ) (i)

.= −2.58, (ii) out of the domain, (iii)

.= 7.39,

(iv).= 1.79, (v)

.= 1.72, (vi)

.= 1.37. Ex. 6.5 (i) R− {−1, 4}, (ii) R, (iii) (−∞, 2),

(iv) (−∞, 1) ∪ (2,+∞), (v) (−∞, 0), (vi) ⟨−2, 2⟩.Ex. 6.6

(i) -

6

(ii) -

6rbb (iii) -

6rb rb (iv) -

6

(v)-

6rbbrbb (vi) -

6rb rb br rb (vii) -

6

(viii) -

6

-

6

(ix) -

6

Ex. 6.7 (a) (i) y = 2x− 2, (ii) y = 1

2ln(4− x), (iii) y = 2− ex−1, (iv) y = (x+ 3)2, (v)

y = 1−2xx−1

, (i) y = ln(3x− 2

). (b) (i) = x2 + 4, (ii) = x2 + 2x+ 6, (iii) = e2x

2+4, (iv)

= e2x + 2, (v) = e4x + 2, (vi) = x+ 4.

App. 6.1 (iii) . . . = 59·[(10 + n

4

)− 32

]= 5n−440

36. App. 6.2 (a) (i)

.= 1048 EUR, (ii)

.= 27.47 years, (iii) x = 25 ln y

800; time x in years required for the growth of the investment to

the level of y EUR. (b) x = −5 ln y−9004400

; the number of years if we know the resale value of the

machine. App. 6.3 (a) x = ey−4025 − 1; from the blood pressure of the patient y determine

his/her age x. (b) x = −56ln(

8076y

− 476

); from the number of ill y in thousand determine the

number of weeks x from the outbreak of the epidemic. App. 6.4 (a) x = −203ln 50−y

41; 9.41

thousand EUR, (b) x = −1011ln p

10; demand x in thousand of items depending on the price per

item p, (c) x = e15−p

4 ; demand x thousand of T-shirts depending on the price per item p.

TOPIC 7 Ex. 7.1 (a) (i) 1,−1,−1, 1, (ii) 3716, 29

8, 25

4, 23

2, (iii) 5, 6, 14, 23, (iii) 1, 2, 6, 24.

(b) (i) 11, 30, (ii) 37, 97. Ex. 7.2 (i) 90 300, (ii) 100 500, (iii) 8190, (iv) 35 800

2187, Ex.

7.3 (a) (i) 5, (ii) 1, (iii) e. (b) (i) −∞, (ii)+∞, (iii) +∞. (c) (i) 3, (ii) +∞, (iii)

6, (iv) 0. (d) (i) +∞, (ii) 0, (iii) −√

4q, (iv) −∞. Ex. 7.4 (i) 100

11, (ii) 10

3, (iii) +∞.

Ex. 7.5 (a) (i) div., (ii) we can not decide, (iii) conv., (iv) conv., (b) (i) conv., (ii)conv., (iii) we can not decide, (iv) div.

App. 7.1 (a) 1094.86 EUR, (b) 18 years. App. 7.2 (a) 138, (b) 406, (c) 435. App. 7.3p2 = 12, p3 = 10, p4 = 11;D2 = S2 = 4200, D3 = S3 = 4300, D4 = S4 = 4250, p

.= 10.67

EUR, D = S = 4267. App. 7.4 p1988 = 8725.5, q1988 = 8157, r1988 = 23117.5 . App. 7.5.= 91 073 EUR.

29