Algebra

Dr. Caroline Danneels

1 Real numbers ........................................................................................................................3

1.1 The power of a real number with an integer exponent............................................................... 3

1.2 The nth rooth of a real number .................................................................................................... 4

1.3 The power of a real number with a rational exponent ............................................................... 5

1.4 Properties of extraction of roots ................................................................................................. 6

1.5 Rationalizing the denominator ................................................................................................... 7

1.6 The logarithm of a real number.................................................................................................. 7

1.7 Exercices .................................................................................................................................... 9

2 Polynomials with real coefficients in 1 variable x ...........................................................11

2.1 Definition ................................................................................................................................. 11

2.2 Zeroes of a polynomial function .............................................................................................. 11 2.3 Special products ....................................................................................................................... 11

2.4 Polynomial division ................................................................................................................. 12

2.5 Exercices .................................................................................................................................. 16

3 Equations and inequalities.................................................................................................17

3.1 Equations in IR ........................................................................................................................ 17

3.2 Inequalities in IR ...................................................................................................................... 20

3.3 Exercises .................................................................................................................................. 26

4 Absolute value of a real number .......................................................................................28

4.1 Definition ................................................................................................................................. 28

4.2 Property .................................................................................................................................... 29

4.3 Operations ................................................................................................................................ 29

4.4 Application: graph of a real absolute value function ............................................................... 29

4.5 Exercises .................................................................................................................................. 31

5 Matrices and determinants ................................................................................................32

5.1 Matrices ................................................................................................................................... 32

5.2 Determinants ............................................................................................................................ 37

5.3 Exercises .................................................................................................................................. 41

6 Systems ................................................................................................................................43

6.1 Systems of n equations in n unknowns .................................................................................... 43

6.2 Systems of linear inequalities in 1 unknown ........................................................................... 45

6.3 Exercises .................................................................................................................................. 46

_____________________________________________________________________________ Algebra 3

1 Real numbers

1.1 The power of a real number with an integer exponent

n0

00

a and n : a a.a....a (n factors)

a : a 1

∀ ∈ ∀ ∈ =

∀ ∈ =

ℝ ℕ

ℝ

( ) ( )n -1-n -1 n0 n

1 a and n : a = a = a =

a∀ ∈ ∀ ∈ℝ ℕ

Properties: a,b and m,n∈ ∈ℝ ℝ

m n m+na a a⋅ =

mm-n

n

aa

a=

( )n n nab a b= ⋅

n n

n

a a

b b

=

( )nm m na a ⋅=

Example

( ) ( )-3 -2-2 4 3 3

5-5 -2

2

a a ab aab

bb

a

⋅ ⋅ =

Rule of signs:

if n is even, then ( )n n-a a=

if n is odd, then ( )n n-a -a=

_____________________________________________________________________________ Algebra 4

1.2 The nth rooth of a real number

The nth ( )0n ∈ ℕ root of a real number a is each real number x of which the nth power is a

or n0 x, a , n : x is the nth root of a x a∀ ∈ ∀ ∈ ⇔ =ℝ ℕ

if n is odd and a∈ℝ then a has in ℝ one nth root, written as:

Examples

3 8 2= since 32 8=

3 8 2− = − since 3( 2) 8− = −

If n is even and :

a has 2 nth roots in ℝ which are each others opposite and are

denoted and .

We agree upon to represent a positive real number; so = 2 > 0.

Example: 4 has 2 square roots and

a = 0 then a has in ℝ one nth root namely 0

then a has in ℝ no nth root.

Convention:

n na a=

n na a− = −

We will only treat with a +∈ℝ , since the other forms can be rewritten to this form since

if < 0 and is even then does not exist

if < 0 and is odd then we write as - || with | |

n

n n

a n a

a n a a a +

∈ ℝ

an

a∈ IR0+

an − a

n

an

164

4 = 2 − 4 = −2

a∈ IR0−

an

_____________________________________________________________________________ Algebra 5

1.3 The power of a real number with a rational exponent

Definition: m

n mn0 0 a , m , n : a a+∀ ∈ ∀ ∈ ∀ ∈ =ℝ ℤ ℕ

Use:

1. 1

nna a=

2. m

-n

n m

1a

a=

3. mp m

np nmp mnp na a a a= = =

Calculation rules: +0 a, b en q, q' :∀ ∈ ∀ ∈ℝ ℚ

q q' q+q'a a a⋅ =

qq-q'

q'

aa

a=

( )q q qa b a b⋅ = ⋅

q q

q

a a

b b

=

( )q'q q q'a a ⋅=

_____________________________________________________________________________ Algebra 6

1.4 Properties of extraction of roots

1.4.1 Multiplication and division of radicals

+ n n n0 a, b , n : a b a b∀ ∈ ∀ ∈ ⋅ =ℝ ℕ

n+ + n

0 0 n

a a a , b ; n :

b b∀ ∈ ∀ ∈ ∀ ∈ =ℝ ℝ ℕ

Examples

44 48 2 16 2= =

4 123 9

12 53 12 4

a aa

a a= =

1.4.2 Exponentiation and extraction of roots of radicals

( )m n+ mn0 a , n , m : a a∀ ∈ ∀ ∈ ∀ ∈ =ℝ ℕ ℕ

+ m nn m n m0 a ; m,n : a a a⋅∀ ∈ ∀ ∈ = =ℝ ℕ

Examples

( )23 2 238 8 2 4= = =

3 38 8 2= =

1.4.3 Adding and subtraction of radicals

( )+ n n n n0 a , n , p, q, r : p a q a r a p q r a∀ ∈ ∀ ∈ ∀ ∈ + − = + −ℝ ℝ ℝ

Example:

243 + 5 33 − 813 = 2 33 + 5 33 −3 33 = 4 33

_____________________________________________________________________________ Algebra 7

1.5 Rationalizing the denominator

a a b

bb=

1 a b

a ba ± b=

−∓

3 32 23

3 3

1 a a b b

a ba b

⋅ +=±±

∓

1.6 The logarithm of a real number

Let a be the base of a logarithmic system 0(a > 0, a 1), x , y ≠ ∈ ∈ℝ ℝ

Then log yay x x a= ⇔ =

We call y the logarithm of x to base a.

The fact that loga x is the inverse of ax can be expressed with the following two identities:

Logarithmic identy 1: x

alog a x=

Logarithmic identy 2: alog xa x=

Names:

• If a = e (number of Euler), the logarithm is called a natural logarithm, notation ln x.

ln 1 = 0 ln e = 1

• If a = 10, the logarithm is called a Briggs logarithm, notation log x.

log 1 = 0 log 10 = 1

Properties:

1. If the base a is larger than 1 (a > 1), the function loga is increasing everywhere. If the base a is between 0 and 1 (0 < a < 1), then the function loga is decreasing everywhere.

2. Because a1 = a, logarithmic identity 1 above implies loga a = 1

3. Because a0 = 1, logarithmic identity 1 above implies

loga1= 0

_____________________________________________________________________________ Algebra 8

Operations:

0 x, y : log x y = log x + log ya a a+∀ ∈ ⋅ℝ

0

x x, y : log = log x log y

ya a a+∀ ∈ −ℝ

10 x : log (x ) = log x a a+ −∀ ∈ −ℝ

0 x ; z : log (x ) = z log x za a

+∀ ∈ ∀ ∈ ⋅ℝ ℤ

n0 0

1 x ; n : log x = log x

na a+∀ ∈ ∀ ∈ℝ ℕ

_____________________________________________________________________________ Algebra 9

1.7 Exercices

1.7.1 Simplify ( +0a,b,c ∈ ℝ )

1. ( )

( ) ( )

27 2 3 4

3 210 3 19

a a b c

abc b c

−

− 9 16

6

b c

a

2. ( ) ( )

3 2

2 2

2 3

2 2

− −

−+

− + −

1 18

3. ( ) ( )

( ) ( ) ( )

3 23 4

64 4 2

2a b 4a

a a 2ab

− −

− − − 3

2 a

4. n n n 1 n 22 a b+ + n 2 2ab ab

5. 4 5

46 8

32a b

81c d 4

2 2

2ab 2b 3cd c

6. 2 4( 25a) b a, b IR− − ∈

1.7.2 Calculate ( +0a,b ∈ ℝ )

1. 3 4 3 128a b 4 b 2a

2. a a a 8 7 a

3. 4 4

3

4a

2a 6 4 2a

4. 3

3 2

5 a 30

5a 60a× 6

3

2a

5. 1,54 8

6.

32 43a

−

12 a

−

_____________________________________________________________________________ Algebra 10

1.7.3 Rationalize the denominator ( +a ∈ ℝ )

1.

2.

3.

1.7.4 Calculate, even without knowledge of a

1. 3 log 4

log 2a

a

⋅ 6

2. 35 25 5 1

log log 3log log36 21 6 2a a a a+ − + 0

1

a − b

3+ 5

5 −1

1

3+ 7

_____________________________________________________________________________ Algebra 11

2 Polynomials with real coefficients in 1 variable x

2.1 Definition

n n-1n n-1 1 0a x + a x ... a x a+ + + with n n-1 1 0

n

a ,a ,....,a ,a

n ,a 0

∈∞ ∈ ≠ ℕ

is a polynomial of degree n in x,

notation V(x).

2.2 Zeroes of a polynomial function

a ∈ ℝ is a zero of the associated polynomial function f : : x V(x)→ →ℝ ℝ if the polynomial expression evaluated to a, equals 0.

In symbols: a is a zero of V(x) V(a) = 0.

Example

is a zero of V(x) = 3 26x 7x 7x 1− − −

since 3 2

1 1 1 1V 6 7 7 1 0

2 2 2 2 − = ⋅ − − ⋅ − − ⋅ − − =

2.3 Special products

If A, B and C are polynomials:

⇔

−1

2

A + B( ) A −B( ) = A2 − B2

A + B( )2 = A 2 + 2AB + B2 A − B( )2 = A 2 − 2AB + B2

A + B+ C( )2 = A 2 + B2 + C2 + 2AB+ 2BC+ 2AC

A + B( )3 = A3 + 3A2B+ 3AB2 + B3

A − B( )3 = A3 −3A2B+ 3AB2 − B3

A 3 − B3 = A − B( ) A2 + AB + B2( )

A 3 + B3 = A + B( ) A2 − AB + B2( )

_____________________________________________________________________________ Algebra 12

2.4 Polynomial division

For any polynomials A(x) and B(x) ( )B(x) 0, degree A(x) degree B(x)≠ ≥ , there exist unique

polynomials Q(x) and R(x) such that

A(x) = B(x)Q(x) + R(x) and degree R(x) < degree B(x).

2.4.1 Polynomial long division

A(x) (dividend) B(x) (divider)

Q(x)(quotient)

R(x)(remainder)

If R(x) = 0 we say that B(x) evenly divides A(x).

Example

x4 − 2 x3 + 9 x2 − 2 x − 2 x2 + 2

− 6 x4 − 12 x2 6 x2 − 2 x − 3

− 2 x3 − 3 x2 − 2 x − 2

2 x3 + 4 x

− 3 x2 + 2 x − 2

3 x2 + 6

2 x + 4

⇒⇒⇒⇒ 4 3 2

22 2

6 2 9 2 2 2 46 2 3

2 2

x x x x xx x

x x

− + − − += − − ++ +

_____________________________________________________________________________ Algebra 13

2.4.2 Division of a polynomial by x-d with d ∈∈∈∈ IR: IR: IR: IR: synthetic division (Horner algorithm)

Example:

Q(x) = 4x2 − 8x + 11

R(x) = − 16

⇒3

24 5 6 164 8 11

2 2

x xx x

x x

− + = − + −+ +

2.4.3 Divisibility by x-d with d ∈∈∈∈ IR

2.4.3.1 Remainder theorem

If a polynomial f(x) is divided by x – d, then the remainder is equal to V(d).

Consequence: V(x) is divisible by x−d

Examples

V(x) = 3x2 – 5x + 7 is not divisible by x – 2 since V(2) = 9 ≠ 0

V(x) = 2x3 + 3x2 – 5x + 12 is divisible by x + 3 since V(–3) = 0

2.4.3.2 Factoring polynomials

Factoring polynomials is a very important issue in algebra. To factor a polynomial put your polynomial in standard form, from highest to lowest power:

V(x) = anxn + … + a0

After you have factored out the greatest common factor and special products, you can apply the Rational Root Test.

The Rational Root (or Rational Zeroes) Test is a handy way of obtaining a list of useful first guesses when you are trying to find the zeroes (roots) of a polynomial. Given a polynomial with integer coefficients, the possible (or potential) zeroes are found by listing the factors of the constant (last) term over the factors of the leading coefficient, thus forming a list of fractions. This listing gives you a list of potential rational (fractional) roots to test - hence the name of the Test.

4x3 − 5x + 6( ): x + 2( )

4 0 − 5 6

− 2 − 8 16 − 22

4 − 8 11 − 16

⇔ R = V d( ) = 0

_____________________________________________________________________________ Algebra 14

Caution: Don’t make the Rational Root Test out to be more than it is. It doesn’t say those rational numbers are roots, just that no other rational numbers can be roots. And it doesn’t tell you anything about whether some irrational or even complex roots exist. The Rational Roots Test is only a starting point.

Example

Factor the polynomial: 3 2 2x 3x 11x 6− − +

1. The factors of the leading coefficient (2) are {1,2} 2. The factors of the constant term (6) are {1,2,3,6} 3. Therefore the possible rational zeroes are ±1,2,3 or 6 divided by 1 or 2: ±1/1, 2/1, 3/1,

6/1, 1/2, 2/2, 3/2, 6/2 reduced: ±1, 2, 3, 6, 1/2, 3/2

4. Check for which values V(x) equals 0: as soon as you find one value which satisfies this criterium, apply synthetic division to find the reduced polynomial that you’ll use to find the remaining roots: V(1) ≠ 0 V(−1) ≠ 0 V(−2) ≠ 0 V(2) ≠ 0 V(3) = 0

This means that x – 3 is a factor of the given polynomial. The reduced polynomial can be produced by applying the synthetic division:

2 −3 −11 6 d = 3 6 9 −6

2 3 −2 0

We find: 2x3 − 3x2

− 11x + 6 = (x – 3)(2x2 + 3x – 2)

5. Again, search for the remaining candidate roots in the list ±1, 1/2, 2: As we have already checked ±1 and ±2 (see above), we check V(1/2) = 0 Applying synthetic division to 2x2 + 3x – 2 gives: 2x2 + 3x – 2 = (x – 1/2)(2x + 4)

6. All together: 2x3

−3x2 − 11x + 6 = (x – 3)(x – 1/2)(2x + 4)

= (x – 3)(2x – 1)(x + 2)

_____________________________________________________________________________ Algebra 15

2.4.3.3 Coefficientrules

1. A polynomial of degree n is divisible by (x – 1) if the sum of the coefficients (included the constant term) equals zero.

Example

x5 – 2x3 + 2x2 – 1 is divisible by (x – 1)

2. If the sum of the coefficients of the terms with odd power of x equals the sum of the coefficients of the terms with even power of x (included the constant term), then the polynomial is divisible by (x + 1).

Example

x5 + x4 – x3 + x2 – x – 3 is divisible by (x + 1)

_____________________________________________________________________________ Algebra 16

2.5 Exercices

Simplify (be efficient):

1. ( )( ) ( )2x 2 x 2 x 4+ − + 4 x 16−

2. ( ) ( )23x 2 9x 6x 4+ − + 3 27x 8+

3. ( ) ( )4 3 29x x 1 : x x 1+ + − + + 2 Q 9x 10x 19,R 29x 20= − − − = +

4. ( ) ( )2 320x 7x 3x 2 : x 3 − + − − 2 Q 7x x,R 2= − − = −

(use synthetic division)

5. ( ) ( )3 28x 20x +2x 2 : 2x 1 − − − 2 Q 4x 8x 3,R 5= − − = −

(use synthetic division)

6. For which value of n ∈ ℝ the polynomial 3 2(3x 2nx 5nx 10)+ − + is divisible by (x + 1)? Determine for the obtained value of n the quotient.

2 n 1,Q 3x 5x 10= − = − +

7. Factor: 4 3 28x 20x 18x 7x 1− + − + ( ) ( )3x 1 2x 1− −

_____________________________________________________________________________ Algebra 17

3 Equations and inequalities

3.1 Equations in IR

3.1.1 Definition

An equation is an expression that represents the equality of two expressions involving constants and/or variables.

3.1.2 Solving an equation

To solve an equation means to find the value or values of the unknown quantity (or quantities) that satisfy the equation. The solution of an equation is the set of all values which, when substituted for unknowns, make an equation true. Let’s call this set S.

Example

2 2S(7x 2 5x) 1,

7 − = = −

Two equations are equivalent if their solution set is the same.

Example

( ) 94x 3 0 3x

4 + = ⇔ − =

since 9 3

S(4x 3 0) S 3x4 4

+ = = − = = −

Properties:

Following properties make it possible to simplify and solve equations:

1. ( ) ( )A B A C B C= ⇔ + = + . We will apply this property to bring over terms.

2. ( ) ( ) 0A B mA mB with m = ⇔ = ∈ℝ

3. ( ) ( )A B C 0 A 0 B 0 C 0= ⇔ = ∨ = ∨ =i i

4. ( ) ( )A C B C A B C 0= ⇔ = ∨ =i i

_____________________________________________________________________________ Algebra 18

Avoid following errors:

a) If you apply A B A C B C= ⇒ =i i , then it’s possible that you import solutions.

Example

2

2

2

x 41 2x

x 2 x 2x (x 2) 4 2x(x 2)

x 5x 6 0

(x 3)(x 2) 0

(x 3 x 2)

imported

+ = +− −

⇒ + − = + −

⇔ − + =⇔ − − =⇔ = ∨ =

↓

So you should solve this equation as follows:

{ }

2x (x 2) 4 2x(x 2) x 2

x 3

S 3

⇔ + − = + − ∧ ≠⇔ =⇒ =

Remember: before removing the denominator which contains the unknown, mention that the donominator is different from zero.

b) If you apply A C B C A B= ⇒ =i i then it is possible that you loose solutions.

Example

( ) ( ) ( )( )x 1 x 2 3x 2 x 2

x 1 3x 2

2x 3 0

3x :1 solution is lost

2

− + = + +⇔ − = +⇔ + =

⇔ = −

So you should solve this equation as follows:

( )( )( ) ( )x 2 2x 3 0

x 2 0 2x 3 0

3x 2 x

23

S , 22

⇔ + + =

⇔ + = ∨ + =

⇔ = − ∨ = −

⇒ = − −

_____________________________________________________________________________ Algebra 19

3.1.3 Discussion of a linear equation that contains parameters

Consider ax b 0; a, b+ = ∈ℝ

1. a ≠ 0 ax + b = 0 ⇔ x = −b/a (= the only solution) 2. a = 0 0x + b = 0 ⇔ 0x = −b

if b 0 then S = (false equation)

if b = 0 then S = (identical equation)

≠ ∅ ℝ

Example

px – m2 = 3x – 9 p and m are real parameters

⇔ (p – 3)x = m2 – 9

1. 2m - 9

p 3 1 solution: x = p - 3

≠ ⇒

2. 2p = 3 0 x = m 9⇒ −

if m = 3 m = 3 then 0x = 0 S = IR

if m 3 m 3 then S =

∨ − ⇒ ≠ ∧ ≠ − ∅

3.1.4 Solving a second degree equation in 1 unknown

Standard form: 2ax bx c 0+ + = met 0a ;b,c∈ ∈ℝ ℝ

Determine the discriminant 2b 4ac∆ = −

If

1 2

1 2

> 0 then ax² + bx + c = 0 has roots

b b x and x

2a 2a = 0 then ax² + bx + c = 0 has roots

- x = x =

∆

− − ∆ − + ∆= =

∆

two different

two equal

b

2a < 0 then ax² + bx + c = 0 has roots

∆ no real

If b is even, then one can use the simplified formulas:

If b = 2b' then 2 24b' 4ac 4(b ' ac) 4 '∆ = − = − = ∆

⇒ 1 2

b ' ' b ' 'x , x

a a

− − ∆ − + ∆= =

_____________________________________________________________________________ Algebra 20

3.2 Inequalities in IR

3.2.1 Definition

An inequality is a statement that a mathematical expression is less than or greater than some other expression A < B (or A ≤ B, A > B, A ≥ B). A and B are expressions of which at least one contains a variable.

3.2.2 Solving inequalities

To solve an inequality means to find the value or values of the unknown that satisfy the inequality. Let’s call this set of solutions S.

Example

] [ ] [S(2x(x 1) 0) ,0 1,− > = −∞ ∪ +∞

Two inequalities are equivalent if their solution set is the same.

Properties:

Following properties make it possible to simplify and solve inequalities:

1. (A B) (A C B C)< ⇔ + < +

2. ( ) 0

0

mA mB if m A B

mA mB if m

+

−

< ∈< ⇔ > ∈

ℝ

ℝ

In an inequality we never remove the variable from the denominator!

_____________________________________________________________________________ Algebra 21

3.2.3 Discussion of the linear inequality that contains parameters

Consider ax b 0;a, b+ > ∈ℝ

ax b 0 ax b+ > ⇔ > −

1. b ba 0 ax b x S x | x

a a > > − ⇔ > − = ∈ > −

ℝ

2. b ba 0 ax b x S x | x

a a < > − ⇔ < − = ∈ < −

ℝ

3. b 0 S

a 0 ax b 0x bb 0 S

> == > − ⇔ > − ≤ = ∅

ℝ

Example

px 2m 3 2x p (p,m )

(p 2)x 2m p 3

− + < − ∈⇔ − < − −

ℝ

Discussion:

1. 2m p 3

p 2 S x xp 2

− −> = ∈ < − ℝ

2. 2m p 3

p 2 S x xp 2

− −< = ∈ > − ℝ

3. if 2m 5 0 S

p 2 0x 2m 5if 2m 5 0 S

− > == < − − ≤ = ∅

ℝ

_____________________________________________________________________________ Algebra 22

3.2.4 Solving quadratic inequalities in 1 unknown

Consider 20ax bx c 0 with a ;b,c+ + ≥ ∈ ∈ℝ ℝ

We first examine the evolution of the sign of the left hand side. Therefore we look at the graph of the function . This function represents a parabola with axis of symmetry parallel to the y-axis.

The intersection points of the parabola with the y-axis are obtained by solving the set of

equations: 2y ax bx c

x 0

= + +

= . ⇒ { }S (0,c)=

The intersection points of the parabola with the x-axis are obtained by solving the set of

equations: 2y ax bx c

y 0

= + +

= :

b b

if > 0 ,0 and ,02a 2a

− − ∆ − + ∆∆ ⇒

are the intersection points with the x-axis

if = 0 ∆ the parabola touches the x-axis in ,02

b

a −

if < 0 ∆ doesn’t intersect with the x-axis

Furthermore we know that if a > 0 the concave side of the parabola is directed upwards and if a < 0 the concave side of the parabola is directed downwards.

y = ax2 + bx+ c

_____________________________________________________________________________ Algebra 23

Summary:

This summary leads us to the following sign chart of y = ax2 + bx + c :

x sign of y

sign of a 0 opposite sign

of a 0 sign of a

1. If a > 0 ⇒ 21 2S(ax bx c 0) ] , x ] [x , [+ + ≥ = − ∞ ∪ +∞

2. If a < 0 ⇒ 21 2S(ax bx c³ 0) [x , x ]+ + ≥ =

Example: 2x 2x 3 0− − ≤

⇒ roots of the left hand side of this inequality are –1 , 3

⇒ sign chart:

x 2x 2x 3− − + 0 − 0 +

⇒ S = [−1,3]

a > 0

a < 0

D > 0 D = 0 D < 0

x x

1 2

x = x 1 2

x = x 1 2

x x 1 2

x1 x2

x1 x2

_____________________________________________________________________________ Algebra 24

3.2.5 Solving rational inequalities

Solving rational inequalities is very similar to solving polynomial inequalities. But because rational expressions have denominators (and therefore may have points where they're not defined), you have to be a little more careful in finding your solutions.

To solve a rational inequality, you first find the zeroes (from the numerator) and the undefined points (from the denominator). You use these zeroes and undefined points to divide the number line into intervals. Then you find the sign of the rational on each interval.

Example 1: 3

2

x(1 x)(x 3)f (x) 0

(3 x) (2x 3)

− += >− +

⇒ Sign chart:

x 3− 3

2− 0 1 3

x − − − − − 0 + + + + +

1 x− + + + + + + + 0 − − −

3(x 3)+ − 0 + + + + + + + + +

2(3 x)− + + + + + + + + + 0 +

(2x 3)+ − − − 0 + + + + + + +

− 0 + − 0 + 0 − −

⇒ ( ) ] [3S f (x) 0 3, 0,1

2 > = − − ∪

Example 2: 2

2

(x 1)(2x x 1)f (x) 0

x x 6

− + += ≤+ −

First find the zeroes of the nominator and of the denominator:

zeroes nominator: 1; zeroes denominator: -3, 2

⇒ Sign chart:

x -3 1 2

− − − 0 + + +

+ + + + + + +

+ 0 − − − 0 +

− + 0 − +

⇒ ] [ ] [S(f (x) 0) , 3 1,2≤ = −∞ − ∪

f x( )

x −1

2x2 + x + 1

x2 + x − 6

f x( )

_____________________________________________________________________________ Algebra 25

Example 3: 4 3

3 2

x x x 1f (x) 0

x x x

− + − += <+ +

First, factor the nominator and the denominator: 2

2

(x 1)(x 1)( x x 1)f (x)

x(x x 1)

− + − + −=+ +

⇒ Sign chart:

x −1 0 1

− − − − − 0 +

− 0 + + + + +

x − − − 0 + + +

− − − − − − −

+ + + + + + +

+ 0 − + 0 −

⇒ ] [ ] [S(f (x) 0) 1,0 1,< = − ∪ + ∞

Example 4: x 2 x 3

x 1 x 1

+ +≤+ −

x 2 x 3 3x 50 0

x 1 x 1 (x 1)(x 1)

+ + − −⇔ − ≤ ⇔ ≤+ − + −

⇒ Sign chart:

x 1− 1

3x 5− − + 0 − − − − −

x 1+ − − − 0 + + +

x 1− − − − − − 0 +

f (x) + 0 − + −

] [5S(f (x) 0) , 1 1,

3

⇒ ≤ = − − ∪ − +∞

x −1

x +1

− x2 + x −1

x 2 + x + 1

f x( )

−5

3

_____________________________________________________________________________ Algebra 26

3.3 Exercises

3.3.1 Solve

1. ( ) ( ) ( )( )x 5 3x 7 x 5 5x 3− + = − + { }2,5

2. ( )2x 5 9 0− − = { }2,8

3. 2

6 1 2

3x 1 x 3x x− =

− −

∅

4. 1 1 2

x 2 x 2 3+ =

− + { }-1,4

5. ( )

12x 2x 3 120

3x 1 2 5 1 3x

−= −− −

173,

6

6. 2x 7x 3(11 a x) 8a 0+ + − ⋅ + =

For which value of a ∈ ℤ there exists 1 solution? Determine the solution for that value.

a 1, x 5= − = −

7. 3x 3

4x 32 5

− > − 24

x25

>

8. 4x 3 x 1 x 5

5 8 2

− + −− > 71

x7

> −

9. 3x 1 2x

3x 2 2x 3

+ <+ −

2 3 3

, ,3 11 2

− − ∪ +∞

10. 1 2 3

x 1 x 2 x 3+ ≤

+ − − ] [ [ [ ] [, 1 1,2 3,−∞ − ∪ ∪ +∞

11. ( ) ( ) ( )2 2x 2 2x x 3 x 1 2x 0− + − − − > ] [3, 1,2

2 −∞ − ∪

12. x 1 x 1

x 1 x 1

− +<+ −

] [ ] [1,0 1,− ∪ +∞

_____________________________________________________________________________ Algebra 27

3.3.2 Determine the number of solutions for which value(s) of m and solve

1. 2m x 2 4x m− = −

2. x x

2 0p m p m

+ − =+ −

3. ( )2p p x 2x m p 2− = − ⋅ −

4. mx 1 x m− ≤ +

5. 2x 4 m x 8+ > ⋅ +

6. 2m (x 4) 4 x− < −

7. 2p x 5m 2m m x m p⋅ − + < ⋅ − ⋅

_____________________________________________________________________________ Algebra 28

4 Absolute value of a real number

4.1 Definition

We define the absolute value (or modulus) of a real number x as follows:

x x if x 0

x :x x if x 0

= ≥∀ ∈ = − ≤ℝ

Remark: 1) for each real number different from 0 one of the 2 parts of the functiondefinition is applicable. Only for 0 the 2 parts are applicable since they both give the same functionvalue.

2) x +∈ℝ

Graph:

To obtain the graph of y = you can reflect that part of the graph of y = x that is under the x-axis around the x-axis.

x

x

_____________________________________________________________________________ Algebra 29

4.2 Property

x , a : x a

a x a

+∀ ∈ ∀ ∈ ≤

− ≤ ≤

ℝ ℝ

⇕

4.3 Operations

x, y : x y x y∀ ∈ ⋅ = ⋅ℝ

110x : x x

−−∀ ∈ =ℝ

0

xxx , y :

y y∀ ∈ ∈ =ℝ ℝ

x, y : x y x y∀ ∈ + ≤ +ℝ

4.4 Application: graph of a real absolute value function

Given: f (x) x 3= −

Asked: make a graph of f

Solution:

Method 1: starting from the definition of f without absolute value signs:

( )( )x 3 if x 3 0

f (x)x 3 if x 3 0

= − − ≥= = − − − ≤

x 3 if x 3 (1)f (x)

3 x if x 3 (2)

= − ≥⇔ = = − ≤

(1) represents a halfline through (3, 0), (4, 1)

(2) represents a halfline through (3, 0), (2, 1)

f x( ) = x − 3

_____________________________________________________________________________ Algebra 30

Method 2: starting from the graph of y = x − 3

• y = x − 3 is represented by a line through (0, −3), (3, 0)

• to get the graph of y x 3= − , reflect that part of the graph of y = x − 3 which is under the

x-axis around the x-axis.

_____________________________________________________________________________ Algebra 31

4.5 Exercises

Draw the function graph of following functions in an orthogonal coordinate system:

1. y 2x 1= +

2. y 2 x 3 4= + +

3. 2y x 4= −

4. 2y 3x 4x 1= − + +

5. 2y x 9 9= − +

6. y x 2 x 2= − + +

7. 2y x 1 x 4= − − −

_____________________________________________________________________________ Algebra 32

5 Matrices and determinants

5.1 Matrices

5.1.1 Definition

An m×n-matrix A is a rectangular array of m×n real (or complex) numbers, containing m rows and n columns.

In general:

( )

11 12 13 1n

ij 21 22 23 2n

mxn

m1 m2 m3 mn

a a a ... a

a a a a ... a=

... ... ... ... ...Aa a a ... a

The horizontal and vertical lines in a matrix are called rows and columns, respectively. The numbers in the matrix are called its entries or its elements. Entries are often denoted by a variable with two subscripts, as shown in the general notation above. The first index denotes the row, while the second one denotes the column.

To specify a matrix's size, a matrix with m rows and n columns is called an m-by-n matrix or m × n matrix, while m and n are called its dimensions.

The set of all m×n matrices is denoted as m n×ℝ or as m n×ℂ depending on the elements aij are real or complex numbers.

5.1.2 Equal matrices

Two matrices A and B are called equal if and only if:

• They have the same dimensions,

• Elements on the same position are equal.

( ) ( )ij ij ij ij

i 1,2,3,...mA a and B b then A B a b with

j 1,2,3,...m

== = = ⇔ = =

There are a number of operations that can be applied to modify matrices called matrix addition, scalar multiplication and transposition. These form the basic techniques to deal with matrices.

_____________________________________________________________________________ Algebra 33

5.1.3 Addition of matrices with the same dimension

The sum A+B of two m-by-n matrices A and B is calculated entrywise:

( ) ( ) ( )ij ij ij ija b a b where 1 i m and 1 j n+ = + ≤ ≤ ≤ ≤

Example

2 4 6 1 4 0 2 1 4 4 6 0 3 8 6

3 5 7 0 2 3 3 0 5 2 7 3 3 7 4

+ + + + = = − + + −

Properties:

1.Closure: mxn mxnA,B : A B∈ + ∈ℝ ℝ

2. Commutative property: A + B = B + A

3. Associative property: (A + B) + C = A + (B + C)

4. The zero Matrix (all aij = 0) is the identity element. A + 0 = A

5. Inverse element: for each matrix A=(aij) there exists –A=(−aij) such that A + (−A) = 0

⟹ m n, is a commutative group× +ℝ

5.1.4 Scalar multiplication of a matrix and a real number

ij ijr : r(a ) (r a )∀ ∈ =ℝ i

r is called a scalar. This operation is called scalar multiplication.

Example

1 0 3 0

3 2 3 6 9

0 1 0 3

= − −

_____________________________________________________________________________ Algebra 34

Properties:

1. Closure: m n m nr , A : r A× ×∀ ∈ ∀ ∈ ⋅ ∈ℝ ℝ ℝ

2. 1 is identity element: m nA :1 A A×∀ ∈ ⋅ =ℝ

3. Associative property: m nr,s , A : (r s)A r(s A)×∀ ∈ ∀ ∈ ⋅ = ⋅ℝ ℝ

4. Scalar multiplication is distributive over the addition in m n : r(A B) r A r B× + = ⋅ + ⋅ℝ

5. Scalar multiplication is distributive over the addition in : (r s)A r A s A+ = ⋅ + ⋅ℝ

Since m n,× +ℝ is also a commutative group we call m n, ,+×ℝ ℝ is a real vector space.

5.1.5 Transposed matrix

The transpose of an m-by-n matrix A is the n-by-m matrix AT formed by turning rows into columns and vice versa:

(AT)ij = Aji

11 12 1n 11 21 m1

21 22 2n 12 22 m2T

m1 m2 mn 1n 2n mn

a a ... a a a ... a

a a ... a a a ... aA= A =

... ... ... ... ... ... ... ...

a a ... a a a ... a

5.1.6 Matrix multiplication

Multiplication of two matrices is defined only if the number of columns of the left matrix is the same as the number of rows of the right matrix. If A=(aij) is an m-by-n matrix and B=(bij) is an n-by-p matrix, then their matrix product C=A∙B is the m-by-p matrix of which entries are given by dot-product of the corresponding row of A and the corresponding column of B:

( )ijC c= where p

ij i1 1j i2 2 j ip pj ik kjk 1

c a b a b ... a b a b , i 1...m, j 1...n=

= + + + = = =∑ .

_____________________________________________________________________________ Algebra 35

Example

11 12 1311 12

11 12 13 21 22 2321 22

21 22 23 31 32 3331 32

41 42 4341 42

c c ca a

b b b c c ca a=

b b b c c ca a

c c ca a

4 x 2 2 x 3 4 x 3

with

Properties:

1. Associative property: (A∙B)∙C = A∙(B∙C)

2. Right distributivity: A∙(B + C) = A∙B + A∙C

3. Left distributivity: (A + B)∙C = A∙C + B∙C

4. Not-commutative property: A∙B ≠ B∙A in general.

Example

[ ] [ ] [ ]1 1 1 2 3

1 2 3 1 9 while 1 1 2 3 1 2 3

2 2 2 4 6

= =

5.1.7 Square matrices

A square matrix is a matrix which has the same number of rows and columns. An n-by-n matrix is known as a square matrix of order n. The elements aii with i = 1, 2, 3, ... n form the main diagonal of the matrix.

Any two square matrices of the same order can be added and multiplied.

c11 = a11 b11 + a12 b21

c23 = a21 b13 + a22 b23

_____________________________________________________________________________ Algebra 36

Special square matrices:

Identitymatrix of order n:

The identity matrix or unit matrix of order n is the n-by-n square matrix with ones on the main diagonal and zeros elsewhere. It is denoted by In:

ii

ij

a 1 i, j 1,2,...,n :

a 0 if i j

=∀ = = ≠

The important property of matrix multiplication of identity matrix is that for m-by-n A:

m nI A A I A⋅ = ⋅ =

Zero matrix of order n:

The zero matrix of order n is the n-by-n square matrix with all its elements 0. It is denoted by On. The zero matrix acts as an absorbing element for matrix multiplication:

for m-by-n A:

m n m,nO A A O O⋅ = ⋅ =

Om,n is the zero m-by-n matrix with all its entries 0.

Zero Divisors:

There exist matrices A and B for which A∙B = 0 and A ≠ 0 en B ≠ 0. Such matrices are called zero divisors.

Example: 1 1 1 2 0 0

1 1 1 2 0 0

− = −

Invertible matrix:

A square matrix A is called invertible or non-singular if there exists a matrix B such that A∙B = In. This is equivalent to B∙A = In. Moreover, if B exists, it is unique and is called the inverse matrix of A, denoted A−1.

_____________________________________________________________________________ Algebra 37

Diagonal matrix:

If all entries outside the main diagonal are zero, A is called a diagonal matrix.

5.2 Determinants

5.2.1 Definitions

The determinant is a special number associated with any square matrix. The determinant of a matrix A is denoted det(A), or without parentheses det A or |A| or |a|. An alternative notation, used in the case where the matrix entries are written out in full, is to denote the determinant of a matrix by surrounding the matrix entries by vertical bars instead of the usual brackets or parentheses.

The determinant of a square 2-by-2 matrix 11 12

21 22

a a

a a

is given by 11 1211 22 12 21

21 22

a aa a a a

a a= −

To calculate the determinant of a square 3-by-3 matrix A

A = 11 12 13

21 22 23

31 32 33

a a a

a a a

a a a

continue as follows:

Minor of an element:

To find the minor of aij

1. Cross out the entries that lie in the corresponding row i and column j. 2. Rewrite the matrix without the marked entries. 3. Obtain the determinant �ij of this new 2-by-2 matrix.

M ij is called the minor for element aij.

Cofactor of an element:

�ij = (−1)i+j Mij

The cofactor �ij is independent of the elements of the ith row and the elements of the jth column.

_____________________________________________________________________________ Algebra 38

Example

12 13 12 132 121

32 33 32 33

a a a a( 1)

a a a aα += − = −

Definition:

The determinant of a 3-by-3 matrix is the real number obtained by taking the elements of a chosen row or column and multiplying each term by the corresponding cofactor and then adding these products.

So, for a chosen fixed row value i, the determinant can be calculated emanating from the ith row:

|A| = ai1 . αi1 + ai2 . αi2 + ai3 . αi3 + ... ain . αin

Example

3 2 1

A = 2 1 2

3 2 4

−

1 111

1 2( 1) 4 4 8

2 4α += − = + =

−

1 212

2 2( 1) (8 6) 2

3 4α += − = − − = −

1 313

2 1( 1) 4 3 7

3 2α += − = − − = −

−

⟹ det A = 3∙8 + 2∙(−2) + 1∙(−7) = 13 (emanating from the first row).

_____________________________________________________________________________ Algebra 39

5.2.2 Properties of determinants

1. A square matrix A and its transposed AT have the same determinant.

det A = det AT

2. If B results from the matrix A by interchanging two rows (or columns) of A, then det B = − det A.

Consequences:

• a determinant with 2 equal rows (columns) is zero.

• the sum of the products of the elements of a row (column) and the corresponding cofactors of another row (column) is zero. Example: check the previous example:

21 11 22 12 23 13a a a 2 8 1 ( 2) 2 ( 7) 0α α α+ + = ⋅ + ⋅ − + ⋅ − =

3. If the elements of a row (column) can be written as a sum of 2 elements, the determinant can be considerd in the same way as the sum of 2 determinants.

' '1 1 1 1 1 1 1 1 1 1

' '2 2 2 2 2 2 2 2 2 2

' '3 3 3 3 3 3 3 3 3 3

a a b c a b c a b c

a a b c a b c a b c

a a b c a b c a b c

++ = ++

Example

2 1 2 1 2 2 1 1 2 1

0 2 1 2 0 1 2 2 1 2 13

1 4 2 4 1 2 4 4 2 4

++ = + =

− + − − − −

4. If we multiply one row (column) with a constant, the determinant of the new matrix is the determinant of the old one multiplied by the constant.

1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

a kb c a b c

a kb c k a b c

a kb c a b c

=

Consequences:

• a common factor of all the elements of a row (column) of A can be ‘taken outside the determinant’.

• a determinant in which one row (column) is a multiple of another row (column) is zero.

_____________________________________________________________________________ Algebra 40

5. To any row of A we can add any multiple of any other row without changing det(A).

1 1 1 1 1 1 1

2 2 2 2 2 2 2

3 3 3 3 3 3 3

a b c a kb kb c

a b c a kb kb c

a b c a kb kb c

+= +

+

6. det (A.B) = det A . det B

7. The previous properties allow us to calculate a determinant by reducing the order: by applying the previous properties try to get a row or column which contains only one value different from zero. Example

6 9 2

2 3 1

3 5 2

=r1 − r3

3 4 0

2 3 1

3 5 2

=r3 − 2r2

3 4 0

2 3 1

−1 −1 0

= −13 4

−1 −1= −1 −3+ 4( ) = −1

_____________________________________________________________________________ Algebra 41

5.3 Exercises

1. Calculate x, y, z if 2

2

4 x x y+5z =

y 0 2x+5 x+y+z

Solution: x = 2, y = −3, z = 1

2. Fill in the missing numbers:

2 . 5 1 . 6

. 3 2 3 . 11 5

4 3 . 3 4 .

+ =

3. If A is a p-by-q matrix and B is a r-by-s matrix, under what condition both matrixproducts exist?

Solution: q = r and s = p

4. Following matrices are given:

0 11 2 1 0 1

A = 1 0 B = C = 3 4 2 1 3

2 2

Calculate A∙B, (A∙B)∙C, B∙C, A∙(B∙C) and check the associative property.

5. Proof that (A∙B)T = BT∙AT if

g ha b c

A = , B = i jd e f

k l

6. Determine all matrices B for which A∙B = B∙A if 1 2

A = 3 4

Solution: 2

x z3

z x z

+

7. Which relationship exists between matrices B (of order 3) and A∙B if

1.

1 0 0

A = 0 k 0

0 0 1

2.

0 1 0

A = 1 0 0

0 0 1

_____________________________________________________________________________ Algebra 42

8. Calculate the following determinants (by reducing the order):

1.

3 2 0

4 2 1

1 3 4

−−

2.

2 1 1 3

0 0 3 1

2 3 4 1

1 1 0 1

3.

1 0 0 0 a

0 0 0 a 1

0 0 a 1 0

0 a 1 0 0

a 1 0 0 0

4.

9 5 1

2 3 1

1 6 2

−−

−

5.

8 1 9 1

6 3 4 5

5 4 7 2

2 3 4 1

6.

a b c

c a b

b c a

Solutions:

49 −5 1 + a5 10 20 3 3 3a b c 3a b c+ + − ⋅ ⋅

9. Proof that:

a b c k a b c

p q r k p q r

x y z x k y k z

⋅⋅ =

⋅ ⋅

_____________________________________________________________________________ Algebra 43

6 Systems

6.1 Systems of n equations in n unknowns

6.1.1 Matrixnotation

11 1 12 2 1n n 1

21 1 22 2 2n n 2

n1 1 n2 2 nn n n

a x + a x +...+ a x = b

a x + a x +...+ a x = b

...

a x +a x +...+ a x = b

ij ia ,b∈ ∈ℝ ℝ

Set

11 12 1n

21 22 2n n n

n1 n2 nn

a a ... a

a a ... aA =

... ... ... ...

a a ... a

×

∈

ℝ

1 1

2 2n 1 n 1

n n

x b

x bX = , B =

... ...

x b

× ×

∈ ∈

ℝ ℝ

⟹ Matrixnotation: A∙X = B

Example

-2x + 5y - z = 1 -2 5 -1 x 1

x + z = 0 A = 1 0 1 , X = y , B = 0

-2y + 3z = -2 0 -2 3 z -2

⇒

_____________________________________________________________________________ Algebra 44

6.1.2 Solving a system of equations by elimination

The goal is to reduce the system until you are left with an equivalent system in which the first equation contains n unknown, the second equation contains n - 1 unknown, the third n – 2, the last but one contains 2 unknown, the last contains 1 unknown. This is done by combining equations to cancel out unknown. When you have reduced the system, you solve the last equation, and back-substitute this value into either of the other equations and then solve the remaining system for the other unknown.

Example

6.1.3 Solving a system of equations by the method of Cramer

The system A.X = B has one unique solution ⇔ | A | ≠ 0 .

This solution is given by

ii

Ax

A=

With Ai is the n-by-n matrix defined as:

Ai =

11 12 1 i-1 1 1 i+1 1n

21 22 2 i-1 2 2 i+1 2n

n1 n2 n i-1 n n i+1 nn

a a ... a b a ... a

a a ... a b a ... a

... ... ... ... ... ... ... ...

a a ... a b a ... a

x + y − z = 9 1 −3

x − 2y − 3z = 1 −1

3x + 6y + z = 37 1

⇒

x + y − z = 9

3y + 2z = 8 −1

3y + 4z = 10 1

⇒

x + y − z = 9

3y + 2z = 8

2z = 2

⇒

x + y = 9+1

3y = 8 − 2

z =1

⇒

x = 10− 2

y = 2

z = 1

⇒

x = 8

y = 2

z = 1

_____________________________________________________________________________ Algebra 45

Example

Take the example of 6.1.2.

A =

1 1 1

1 2 3

3 6 1

− − −

⇒ |A| = − 6

A1 =

9 1 1

1 2 3

3 6 1

− − −

⇒ |A1| = − 48

A2 =

1 9 1

1 1 3

3 37 1

− −

⇒ |A2| = − 12

A3 =

1 1 9

1 2 1

3 6 37

−

⇒ |A3| = − 6

1 32A AA

x 8, y 2,zA A A

⇒ = = = = =

6.2 Systems of linear inequalities in 1 unknown

Solve each inequality separately and the solution of the system is formed by those values which satisfy all inequalities.

Example

x 3 5234 2x

x2 3 92x x 2

x 515 3

− − < − − < ⇔ − <− <

⟹ S = 23

x x 59

∈ − < <

ℝ

- 23/9 5

_____________________________________________________________________________ Algebra 46

6.3 Exercises

6.3.1 Solve the following systems of equations

1.

x 2y 3z 2

2x y 3z 1

4x 7y 3z 7

− + − = + + = + + =

9 3

k,1 k,k5 5

− +

2.

y 3z 1

2x 2y z 2

4x 5y z 5

− + = + + = + − =

∅

3.

3x y z 16

x 3y 4z 72

x y z 8

− + = + + = − + = −

{ }(12,20,0)

4.

x 2y 1

3y 5z 11

2z 3x 13

+ = − = − − = −

{ }(5,-2,1)

5.

1 1 3

x y 2

1 1 2

y z

1 1 1

z x

+ = + =

+ =

4 4

4, ,5 3

6.3.2 Solve the following systems of inequalities

1. 2x 4 0

(x 3)(x 5) 0

− ≥

− + < ] [ [ [-5,-2 2,3∪

2.

3 1

3x 2 x1

0(4 x 1)(x 3)

< − <

− +

1

0,4

3. 3x 17

1 25

−− < < ] [4,9