UNSW Mathematics Society · 2020-05-01 · Introduction to Vectors Vector Geometry Complex Numbers...

UNSW Mathematics Society

(Higher) Mathematics 1AAlgebra

Presented by: Donald Tang, Shayekh Rouf

Term 1, 2020

Introduction to Vectors Vector Geometry Complex Numbers Linear Equations Matrices

Table of contents

1 Introduction to Vectors

2 Vector Geometry

3 Complex Numbers

4 Linear Equations

5 Matrices

Donald Tang, Shayekh Rouf MATH1131/1141 2 of 153


Introduction to Vectors



OH YEAH



Introduction to Vectors - Addition and Subtraction

Addition of VectorsIf we want to add vectors, say x and y, in any dimension Rn,thenit’s as easy as simply adding together each component of eachvector. The same concept applies with subtraction. The result isalways a new vector. We can do this as follows:

Example 1x1x2...

xn

+

y1y2...

yn

=

x1 + y1x2 + y2

...xn + yn

or

x1x2...

xn

−

y1y2...

yn

=

x1 − y1x2 − y2

...xn − yn

Note:Vector addition is commutative and associative.



Geometric Interpretation - Addition

Visualising Vectors in SpaceVectors can be interpreted as the net movement of a certain

object. For example, consider the vector u =(

23

)in R2.

Example 2We can interpret this as 2 units to the right in the x -axis and 3units up in the y -axis. This would look something like below:

x

yu

This thought process applies to R3 and higher dimensions, thoughthis may be harder to mentally visualise.



Visualising Negative Vectors in SpaceNegative values mean that we need to consider movement in the

opposite direction. For example, consider the vector v =(−1−2

).

Example 3We can interpret this as 1 unit to the left in the x -axis and 2 unitsdown in the y -axis. This would look something like below:

x

y

v



Visualising Vector Addition in Space

We add vectors geometrically from tip to tail, so v + u =(

11

)can

be interpreted as moving v so its tail meets the tip of u (or viceversa) and drawing a vector from the origin to the resultant point.

Example 4

x

y

v

u

u + vor

x

y

v

uu + v

Note:Vector subtraction is the same as vector addition, but we flip thedirection of the vector that we are subtracting, then add it.



Introduction to Vectors - Multiplication

Scalar multiplication of VectorsIf we want to multiply our vector x in any Rn by some real numberλ, then we just multiply each component of the vector by λ. Thislooks like:

Example 5

λ

x1x2...

xn

=

λx1λx2

...λxn

or λ

x1x2...

xn

+

y1y2...

yn

=

λx1 + λy1λx2 + λy2

...λxn + λyn

Note:Scalar multiplication is commutative and associative.



Geometric Interpretation - Multiplication

Visualising Vectors in SpaceMultiplication with some λ can be interpreted as multiplying thelength of the vector by λ. This can preserve the vector’s direction(λ > 0), change its direction to the opposite (λ < 0) or produce

the zero vector (λ = 0). Consider the vector −2v =(

24

).

Example 6We can interpret this as doubling the length of v and changing itsdirection to the opposite. This would look something like below:

x

y −2v



Vector Length

LengthThe length of a vector can be determined through many iterationsof Pythagoras’ theorem through applications in each dimension.Or, we can just use the following formula (which is derived frommany iterations of Pythagoras’ theorem in each dimension):

|x| =√

x21 + x2

2 + ...+ x2n

Also note that:

|λx| = |λ||x| = |λ|√

x21 + x2

2 + ...+ x2n



Vector Components in R2

Angles and MagnitudesGiven a vector m of magnitude |m| with an angle of θ to thex -axis, we can draw it up as a sum of its x - and y -components:

xi

yjm

θ

Using some epic trigonometry, it is clear that:

x = |m| cos θ and y = |m| sin θ

Which means that m as a sum of its components is just:

m = |m|cosθi + |m|sinθj



Lines! Cartesian and Parametric

Parametric LinesA straight line L is just a really long vector, so we can representany line we want in terms of some real number parameter (e.g. λ).We just need a point a on the line and a vector v which is parallelto the line.

L = {x ∈ Rn : a + λv}

Alternatively, we can represent it in some weird Cartesian form:

L ={

x ∈ Rn : x1 − a1b1

= x2 − a1b2

= ... = xn − anbn

}Where (a1, ..., an)T is a point on the line and (b1, ..., bn)T is thedirection of the line.



Lines! Cartesian and Parametric

Note:Parametric vectors can have multiple values depending on whatyour choice of a, v or λ is, so you can have several valid answers.



Lines! Converting Between Cartesian and Parametric

Cartesian to Parametric LinesFrom the Cartesian equation, we can see that each term is equal,so we can actually set them all as some parameter.

Let µ = xk−akbk

for some integer 1 ≤ k ≤ n.

Making xk the subject, xk = ak + µbk .

Thus, a parametric form of our line can be given by:a1a2...

an

+ µ

b1b2...

bn



Lines! Converting Between Cartesian and Parametric

Parametric to Cartesian LinesWe can also go the other way, from parametric to Cartesian.

Example 7 Let’s consider a specific example this time:

v =

243

+ λ

535

. We can equate the components x1, x2 and x3

of v such that: x1 = 2 + 5λ, x2 = 4 + 3λ, x3 = 3 + 5λ.If we rearrange each component to be in terms of λ, then equatethem all, we can find a Cartesian equation for our line:

x1 − 25 = x2 − 4

3 = x3 − 35 .



Lines! An Interval Between Two Points

IntervalsWe can use our parametric representation to represent an interval Ibetween two points, let’s say a and b. This can be done as follows:

I = λa + (1− λ)b

For 0 ≤ λ ≤ 1 - A fact we can use for many geometric applications.

Note:When λ = 0, all the ’weight’ of the vector is at b, and whenλ = 1, all the ’weight’ of the vector is at a.



Planes

Properties of PlanesWhen you really think about it, a plane is just the span of allpoints you get from adding two non-parallel, non-zero vectors.

In other words, it’s a possible linear combination of the vectors.

This means we can represent planes in a bunch of ways involvingvectors. Planes also have some cool properties.

Planes can intersect with other planes. This intersection can beanother plane if they’re both parallel (it’s actually the exact sameplane) or the intersection can occur in the form of a line.

SpanThe span of two vectors can be formally denoted by:

span(v1,v2) = {x : x = λv1 + µv2 where λ, µ ∈ R}



Planes

Cartesian HyperplanesHyperplanes in Rn can be expressed in Cartesian form:

a1x1 + a2x2 + ...+ anxn = d

Hyperplanes in R3 are regular 2D planes, whereas hyperplanes inR4 are three-dimensional!We’ll most commonly be looking at hyperplanes in threedimensions, so it is likely that you’ll see either:

ax + by + cz = d

or

a1x1 + a2x2 + a3x3 = b

or a similar equation for the 2D planes we are familiar with.Donald Tang, Shayekh Rouf MATH1131/1141 19 of 153


Planes

Parametric PlanesA parametric way to express a 2D plane

∏in Rn is:

∏:

c1c2...

cn

+ λ

x1x2...

xn

+ µ

y1y2...

yn

or∏

: c + λx + µy

For some real numbers λ and µ.

Note:x and y are non-parallel, non-zero vectors that are parallel to theplane, and c is the position vector for some point on the plane. Ifthe plane passes through the origin, it’s easiest to just let c = 0.



Converting from Parametric to Cartesian formExample 8

We are given the parametric form of a plane:

∏=

x1x2x3

=

309

+ λ

14−4

+ µ

−207

So,

x1 = 3 + λ− 2µ (1), x2 = 4λ (2) and x3 = 9− 4λ+ 7µ (3).

We can do 7× (1) + 2× (3) to get: 7x1 + 2x3 = 39− λSubstituting (2) into this, we can see that: 7x1 + 2x3 = 39− 1

4x2So a Cartesian equation of this plane is given by:

28x1 + x2 + 8x3 = 156



Converting from Cartesian to Parametric formExample 9

We are given the Cartesian form of a plane:∏: 4x + 3y − 9z = 3

To convert this to some parametric form, we parametrise twochosen variables. In this case, we will let x = λ1 and y = λ2,which means that:

z = 49λ1 + 1

3λ2 −13

So we can see that:

∏=

xyz

=

λ1λ2

49λ1 + 1

3λ2 − 13

=

00−1

3

+ λ1

1049

+ λ2

0113



Vector Geometry



Dot Product

The Dot ProductThe dot product of two vectors x and y can be written as

x1x2...

xn

·

y1y2...

yn

= x1y1 + x2y2 + ...+ xnyn

This is also known as the scalar product, since the result is scalar.The dot product is also equivalent to the expression:

x · y = |x||y| cos θ

Where θ is the angle between two vectors. This means we can usethe dot product to the find the angle between two vectors, and ifvectors are orthogonal! Cool.



Dot Product

Properties of the Dot ProductThe dot product also has some nice properties which we can use:

Commutative Law - a · b = b · aDistributive Law - a · (b + c) = a · b + a · ca · a = |a|2

(λa) · b = λ(a · b)We can also derive some important results using the dot product,notably:−|a||b| ≤ a · b ≤ |a||b| (Cauchy-Schwarz Inequality)|a + b| ≤ |a|+ |b| (Triangle Inequality)



Orthogonality

How do I know I’m RIGHT? (poor joke my bad)If the dot product is equal to 0, then that must mean cos θ = 0,implying θ = π

2 . Using this fact, we know two vectors areorthogonal when their dot product equals zero.

An important definition is that of the orthonormal set of vectors.These describe sets of vectors in Rn that are of unit length andmutually orthogonal to each other. e.g. in R3 the standard basisvectors are a set of orthonormal vectors:1

00

,0

10

and

001



Dot Product



Projections

ProjectionsProjections are given by the simple equation:

projv(u) = v · uv · v v

Where we are projecting u onto v.

The projection is pretty much what we would obtain if we weresliding the vector u along a line perpendicular to v and touchingthe tip of u and forcing u to face the same direction as v. Thislets us find a whole bunch of features, like shortest distances.



Shortest Distance

Distance between a point and line in R2 and R3

We have a point B and we want to find the shortest distancebetween B and x = a + λv. Here is a diagram of the situation:



Shortest Distance

Distance between a point and line in R2 and R3

Our shortest distance is just the magnitude of ~PB. We can alsonote that ~AP is the projection of ~AB onto v.In other words,

~AP = projv(b− a)

We can find the distance by finding an expression for ~PB in termsof our known quantities.Using some vector magic, we note that:

~PB = b− a− projv(b− a)

So our shortest distance is simply the magnitude of this new vector.



Shortest DistanceExample 10 - MATH1141 June 2015, Q4i

Find the shortest distance between the point B(1,2,0) to the linewith parametric vector equation:1

01

+ λ

011

, λ ∈ R

We’ll start by making our vector b − a:

b− a =

120

−1

01

=

02−1



Shortest DistanceExample 10 - MATH1141 June 2015, Q4i (ct’d)

Next, we’ll find the projection of b− a onto v:

projv(b− a) =

( 02−1

)·( 0

11

)( 0

11

)·( 0

11

)0

11

= 12

011

So our vector ~PB is:

~PB =

02−1

− 12

011

=

032−3

2

And | ~PB| =

√(0)2 + (3

2)2 + (−32

2) = 3√

22 units. This is the

shortest distance between the point and our line!



The Cross Product

DefinitionThe cross product is a useful operation that allows us to find avector that is orthogonal to two vectors to which we are applying itto. It is given by:a1

a2a3

×b1

b2b3

=

a2b3 − b2a3b1a3 − a1b3a1b2 − b1a2

Note:The cross product is only applicable if both vectors are in R3 andhas no definition outside of it. The direction of the resultant vectorcan be ascertained using the right hand rule.



The Cross Product

Properties of the Cross Producta × a = 0a × b = − b × a(aλ) × b = a × (bλ) = λ(a × b)Distributive Law - a × (b + c) = a × b + a × ce1 × e2 = e3, e2 × e3 = e1, e3 × e1 = e2, where ek is one ofthe standard basis vectors of R3

|a× b| = |a||b| sin θ, where θ is the angle between a and b.

Note:The cross product is NOT associative. This is demonstrated bythe fact that:(e1 × e2)× e2 = −e1, bute1 × (e2 × e2) = 0



The Cross Product

Normal to a PlaneRecalling that the parametric form of a plane in R3 can be givenby:

∏:

c1c2c3

+ λ1

x1x2x3

+ λ2

y1y2y3

or∏

: c + λ1x + µ1y

We can deduce that a normal to the plane itself needs to beperpendicular to both vectors x and y that are parallel to theplane. In this case, this means that the normal n can be given by:

n =

x1x2x3

×y1

y2y3

or n = x× y



Areas

Area of a ParallelogramThe area of a parallelogram is simply the magnitude of the crossproduct of its sides.



AreasExample 11 - MATH1131 June 2012 Q2iv

The points C and D have position vectors:

c =( 1

23

)d =

( 415

)Find the area of the parallelogram with adjacent sides OC and OD,where O is the origin.Clearly, A = |c× d| =

∣∣∣( 123

)×( 4

15

)∣∣∣=∣∣∣( 7

7−7

)∣∣∣=√

72 + 72 + (−7)2 =√

147 = 7√

3



Volumes

The Scalar Triple ProductThis cool-sounding product is defined as:

a · (b× c)

And has the following properties:a · (b× c) = (a× b) · ca · (b× c) = −a · (c× b)a · (a× b) = (a× a) · b = 0

Note:The cross product must ALWAYS be performed first because thedot product is a scalar, and the cross product of a vector andscalar has no meaning.



Volumes

ParallelepipedsA parallelepiped is a 3 dimensional object. All its faces areparallelograms (note: rectangles are parallelograms!). Its volume Vis given by the area r of its base times its perpendicular height h.

The length of the projection of a onto n is the perpendicularheight, and the area r of the base is given by |b× c|.Thus, h = |a·n|

|n| = |a·(b×c)||b×c| = |a·(b×c)|

r .Rearranging,

h × r = V = |a · (b× c)|.



Planes

Point-Normal FormWe’ve learnt about Cartesian and parametric forms of a plane, butwhat’s a point-normal form? A point normal form is just:n1

n2n3

·x1

x2x3

−a1

a2a3

= 0

Where the vector n is a normal to the plane and the vector a is apoint on the plane itself.This is a somewhat intuitive formula, because we recall that thedot product of a vector with its normal is equal to 0.Expanding this form, we find a Cartesian equation for the plane:

n1x1 + n2x2 + n3x3 = n1a1 + n2a2 + n3a3



Playing with Planes

Converting between the three formsPoint-Normal to Cartesian - Simply expand.Cartesian to Point-Normal - With some careful inspectionwe can see that the coefficients of x1, x2 and x3 are thecoordinates for a normal to the plane. To find a point on theplane itself, just let 2 variables equal 0. You know the rest.Parametric to Point-Normal - If a plane is defined by∏

1 = a + λ1v1 + λ2v2, then to find a normal we simply findv1 × v2. By definition, a is a point on the plane, so we cannow put the ingredients into point-normal form.



Playing with Planes

Converting between the three forms (ct’d)Parametric to Cartesian - The method with least algebra isto just find the point-normal form, then expand to find theCartesian form.Point-Normal to Parametric - This involves expanding thepoint-normal form into Cartesian form, and then intoparametric form.



Shortest Distance

Shortest Distance Between a Plane and a LineThe process is similar to finding the shortest distance between apoint and a line. Find the normal to a plane through the crossproduct of its parallel vectors or otherwise, then perform vectormagic as before. Here’s a diagram illustrating the situation:



Shortest DistanceExample 12 - MATH1131 June 2015 Q3iiic

Find the shortest distance d between the point P(4,2,2) and theplane

∏: x − 3y + 2z = 1.

We can easily find a normal here by recalling that the variablecoefficients represent such a normal, so we know that:

n =( 1−32

)A point q on the plane can be found if we let y = z = 0, so:

q =( 1

00

)So a vector v from

∏to P is just:

v =( 4

22

)−( 1

00

)=( 3

22

)Donald Tang, Shayekh Rouf MATH1131/1141 44 of 153


Shortest DistanceExample 12 - MATH1131 June 2015 Q3iiic (ct’d)

We have everything we need, so we can just plug our values intoour formula.

d = |projn(v)| =∣∣∣∣v · nn · nn

∣∣∣∣ = |v · n||n|2 |n| = |v · n|

|n|

=

∣∣∣( 322

)·( 1−32

)∣∣∣∣∣∣( 1−32

)∣∣∣ = 1√14



Shortest Distance

Shortest Distance Between 2 Lines in R3 (MATH1141 ONLY)If the lines are parallel, then just find a point on one line, thendo what we did for the distance between a line and point inR2.If the lines are not parallel, find their cross product (to find avector perpendicular to both), then find the magnitude of theprojection of any vector with one end on one line, and theother end on the other line.



Complex Numbers



Complex Arithmetic

Addition and SubtractionIronically, adding and subtracting complex numbers is anything butcomplex. Like adding together vector components, we add togetherthe real components and imaginary components separately.Consider two complex numbers, z and ω, such that:

z = a + bi and ω = c + di , where a, b, c, d ∈ R

z + ω = a + c + (b + d)iz − ω = a − c + (b − d)i



Complex Arithmetic

Multiplication and DivisionMultiplying and dividing complex numbers is similar to multiplyingreal numbers, but keeping in mind that i2 = −1.

zω = (a+bi)(c+di) = ac+adi+bci−bd = ac−bd+(ad+bc)izω = a+bi

c+di = (a+bi)(c−di)(c+di)(c−di) = (ac+bd)+(bc−ad)i

c2+d2

Note:We multiply the denominator by its complex conjugate so we canrealise the denominator, which makes it easier to distinguish certainproperties such as finding Re

( zω

)or Im

( zω

)if we wish to do so.



Complex Notation

The Polar FormPolar forms consider a complex number as a vector, with amagnitude and a direction. We can represent a complex number zin the form:

z = r(cos θ + i sin θ)

Or, with less writing, in the form:

z = reiθ

Where r is the length (modulus) of our complex vector, and θ isthe angle (argument) it makes with the first quadrant of the x -axis.

x = r cos θ and y = r sin θ. Thus, r =√

x2 + y2.



Complex Notation

Note:Since the value of θ is not unique, we can choose a value of

θ such that −π < θ ≤ π. This is known as our principalargument.Also keep in mind, arg(z) is asking for any argument of z ,whereas Arg(z) is asking specifically for the principalargument.cisθ notation from the good ol’ 4U days is no longer used, sobe mindful of this!



Complex Conjugates

The ConjugateIf z = reiθ, then its reflection along the x -axis is simply

z = re−iθ or z = r(cos(−θ) + i sin(−θ)) = r(cos θ − i sin θ)

z is also known as the complex conjugate of z. On an Arganddiagram, it looks like this:

y

x

z

r

zr

θ−θ



Complex Properties

De Moivre’s TheoremThis simple yet highly important theorem is perhaps one of ’De’most versatile and useful concepts related to complex numbers.It states that:

(r(cos θ + i sin θ))n = rn(cos(nθ) + i sin(nθ)) for n ∈ R.

This can be proved by induction, give it a go if you want!

Note:This result also applies to negative and fractional values of n too.



Complex Properties

Simple Complex PropertiesComplex numbers have some pretty cool properties:|z ||ω| = |zω||z||ω| =

∣∣ zω

∣∣|zn| = |z |n

zz = |z |2

z = 1z (ONLY if |z | = 1)

arg(zω) = arg(z) + arg(ω)arg( z

ω ) = arg(z)− arg(ω)arg(zn) = n × arg(z) (a consequence of De Moivre’s theorem)



Complex Roots and Roots of Polynomials

Applications of De Moivre’s TheoremUsing De Moivre’s theorem and the properties of complexnumbers, we can now attempt to find the nth roots of numbers aswell as solving various polynomial roots.



Polynomials

Polynomial theoremsSome important polynomial theorems:

Remainder Theorem: The remainder when the polynomial isdivided by z − α is given by simply evaluating p(α).Factor Theorem: A number α is a root of p(z) only if z − αis a factor of p(z).The Fundamental Theorem of Algebra: A polynomial ofdegree n has precisely n roots over the complex field.Complex roots always occur in conjugate pairs in a polynomialwith real number coefficients. If complex roots exist, there isalways an even quantity (including 0) of them in a realpolynomial.



Polynomials

More polynomial theoremsSome more important polynomial theorems:

Every polynomial can be factorised in the form:

p(z) = a(z − α1)(z − α2)...(z − αn)

Where a is the coefficient of zn and αi are roots of p(z).

Note:As a side note, we can convert linear factors with complexcoefficients into a quadratic with real coefficients (provided thatthe initial polynomial has real coefficients).

(z − α)(z − α) = z2 − 2Re(α)z + |α|2



Complex RootsExample 13

Finding all ninth roots of 1:We begin by stating our required equation:

z9 = 1

i.e. z9 = e0+2kiπ for k = 0, 1, 2, .., 8

z = e(2kiπ)/9

z = 1, e2iπ/9, e4iπ/9, e6iπ/9, e8iπ/9, e10iπ/9, e12iπ/9, e14iπ/9, e16iπ/9

z = 1, e2iπ/9, e4iπ/9, e2iπ/3, e8iπ/9, e−8iπ/9, e−2iπ/3, e−4iπ/9, e−2iπ/9



Roots of Unity and Polynomials ShenanigansExample 14 - MATH1141 June 2014 Q4iv

iv) You may assume that z9 − 1 = (z3 − 1)(z6 + z3 + 1)a) Explain why the roots of

z6 + z3 + 1 = 0 are e±2iπ/9, e±4iπ/9, e±8iπ/9.b) Divide z6 + z3 + 1 by z3 and let x = z + 1

z . Find a cubicequation satisfied by x .

c) Deduce that cos 2π9 + cos 4π

9 + cos 8π9 = 0



Roots of Unity and Polynomials ShenanigansExample 14 - MATH1141 June 2014 Q4iva

For this part, we can actually cheat a little since we already knowthe roots of z9 − 1 = 0 from example 13.

If we factorise z9 − 1 = 0 to (z3 − 1)(z6 + z3 + 1) = 0, then wecan deduce that either the cubic or the sextic factor must equal 0to satisfy the equation.




We also know that the polynomial can be expressed as acombination of 9 linear factors, some with complex constants.

After appropriately expanding these linear factors with complexconstants, we end up with our real cubic and sextic polynomials,which means that when one real polynomial equals 0, the othercannot equal 0 since they have distinct factors.




Finally, we note the roots of z3 − 1 = 0 are z = 1 and e±2iπ/3, sothe roots of z6 + z3 + 1 = 0 must be the remaining roots ofz9 − 1, i.e. z = e±2iπ/9, e±4iπ/9, e±8iπ/9.



Roots of Unity and Polynomials ShenanigansExample 14 - MATH1141 June 2014 Q4ivb

Upon dividing, we get:

z6 + z3 + 1z3 = z3 + 1

z3 + 1

=(

z + 1z

)3− 3

(z + 1

z

)+ 1

= x3 − 3x + 1

As required.



Roots of Unity and Polynomials ShenanigansExample 14 - MATH1141 June 2014 Q4ivc

The new roots of this equation are z + 1z = z + z , which means

that the roots are simply e 2iπ9 + e

−2iπ9 , e 4iπ

9 + e−4iπ

9 , e 8iπ9 + e

−8iπ9 .

Which equal:

2 cos 2π9 , 2 cos 4π

9 , 2 cos 8π9 , respectively.

Taking the sum of roots of this polynomial, we find:

2 cos 2π9 + 2 cos 4π

9 + 2 cos 8π9 = −0

1Which gives us our desired result of:

cos 2π9 + cos 4π

9 + cos 8π9 = 0



PolynomialsExample 15 - MATH1131 June 2015 Q3ii

ii) Consider the complex polynomial

p(z) = z4 − 3z3 + 6z2 − 12z + 8.

a) Given that p(2i) = 0, factorise p(z) into linear andquadratic factors with real coefficients.

b) Find all roots of p.



PolynomialsExample 15 - MATH1131 June 2015 Q3ii

Thanks to the factor theorem, we know if z − 2i is a factor, thenso is z + 2i since all coefficients are real. Thus, we know that:

(z − 2i)(z + 2i) = z2 + 4

Is a factor of p(z).Applying some nifty polynomial division using this quadratic factor,we discover that:

p(z) = (z2 + 4)(z2 − 3z + 2) = (z2 + 4)(z − 2)(z − 1)

Which means the solutions to p(z) are z = ±2i , 1, 2



De Moivre’s Theorem and Powers of Trig FunctionsExample 16 - MATH1141 June 2012 Q2iv

iv) Use De Moivre’s theorem to prove that

cos 4θ = 8 cos4 θ − 8 cos2 θ + 1

First, we note that (cos θ + i sin θ)4

= cos 4θ + i sin 4θ.But also,= cos4 θ + 4i cos3 θ sin θ − 6 cos2 θ sin2 θ − 4i cos θ sin3 θ + sin4 θ.Equating the real parts of both equations,

cos 4θ = cos4 θ − 6 cos2 θ sin2 θ + sin4 θ

cos 4θ = cos4 θ − 6 cos2 θ(1− cos2 θ) + (1− cos2 θ)2

cos 4θ = 8 cos4 θ − 8 cos2 θ + 1



Complex Geometry

Common Loci|z − ω| = a - A circle with centre ω and radius a.|z − a| = |z − b| - The perpendicular bisector of pointsa and b.arg(z −ω) = θ - A ray from the point w at an angle of θ fromthe positive x -axis. The point z = w is excluded (becausearg(0) is undefined).Re(z) = a - The line x = a.Im(z) = a - The line y = a.



Complex Geometry

Common Regions|z − ω| ≥ / ≤ a - Area outside/inside circle.|z − a| ≥ / ≤ |z − b| - Region on the side of b/a.α ≤ arg(z − ω) ≤ β - The region in betweenarg(z − ω) = α and arg(z − ω) = β.Re(z) ≥ / ≤ a - The region to the right/left of x = a.Im(z) ≥ / ≤ a - The region above/below y = a.

Note:Replacing ≥ / ≤ with > / < means that the line used as theborder of each region must be a dotted line.



Complex GeometryExample 17 - MATH1131 June 2015 - Q2iii

Sketch the following region on the Argand diagram:

S = {z ∈ C : |z − i − 1| ≤ 1 or |Im(z)| ≥ 1} .



Linear Equations



What is a Linear Equation?

DefinitionA linear equation is defined as

a1x1 + a2x2 + · · ·+ anxn + b = 0

where x1 . . . xn are variables, whereas a1 . . . an and b are thecoefficients.



Linear Equations

Systems of linear equations are very frequently used in the world ofmathematics, engineering, physics, computer science, etc. Areason for this is because systems that are non-linear are oftenapproximated well with linear equations (which makes computationfar easier).



Solving systems of linear equations

Hopefully by now, we are all able to solve systems of equationssuch as

2x + y = 03x − y = 7.

We can recognise that this will have a unique intersection as it isthe intersection of two lines.



Cute cat



A less subtle example

How about this pair of simultaneous equations?

x + 2y + z = 1 (1)

2x + 3y − 2z = −2. (2)

From a quick glance, we can see that since there are 3 unknowns(x , y , z) and only 2 equations, there will be infinitely manysolutions. Now how do we solve this system of equations?




There are many different approaches to this question, but a quickmethod is to set any of the variables equal to a parameter. In thiscase, let z = λ, λ ∈ R. This gives us

x + 2y + λ = 1 (1)

2x + 3y − 2λ = −2. (2)

Now doing 2(1)− (2) gives us

y + 4λ = 4y = 4− 4λ

Subbing this into (1) gives us x = −7 + 7λ.




Hence, our solution is x = −7 + 7λ, y = 4− 4λ, z = λ, which wecan see is a line. A prettier form may be:

` :

−740

+ λ

7−41

.Note: this is the case because the geometrical interpretation forthis system is finding the intersection of two planes (which gives usthe line of intersection).



...

What if we had a lot more equations and variables? For example...

2a + 3b − c + d = 6x − 4b + 5c = 2

1− 6c = 5a + 2b − d = 7.

This is where matrices come in to make the process a lot moreeasier on the eyes.



The augmented matrix

DefinitionThe augmented matrix is defined as

a11 a12 a13 . . . a1n b1a21 a22 a23 . . . a2n b2

......

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

am1 am2 am3 . . . amn bm

Where m, n are the columns and rows. Each row is just a simpleway of expressing a1x1 + a2x2 + · · ·+ anxn = b.



Elementary Row Operations

The three elementary row operations1 Interchange of equations (swapping rows)2 Adding a multiple of one equation to another3 Multiplication by a scalar

Any combination of the three can be applied in one step as well.Each of these operations will give us a new matrix that isequivalent to the old one : their solutions will remain the same.



Row-echelon Form

A combination of all these techniques is called GaussianElimination. We use Gaussian elimination to try and obtain arow-echelon form of our matrix.DefinitionA matrix is considered in row-echelon form if

1 All zero rows are at the bottom of the matrix.2 In every leading row, the leading entry is further to the right

than the leading entry in any row higher up in the matrix (itforms a diagonal downwards from left to right as best as itcan).

Row-echelon form can help us determine if a matrix has nosolutions, a unique solution or infinitely many solutions.



An example

An example of a matrix in row-echelon form is a x x d0 b x e0 0 c f



Reduced Row-Echelon form

Reduced row-echelonA matrix is said to be in reduced row echelon form if it is inrow-echelon form and

1 Every leading entry is 12 Every leading entry is the only non-zero entry in its column.



An example

An example of a matrix in reduced row-echelon form is 1 0 0 d0 1 0 e0 0 1 f



A question

Now let’s try applying these techniques.

MATH1131 June 2011 q3 (ii)A pet shop has x hamsters, y rabbits and z guinea pigs.Each hamster eats 50g of dry food and 40g of fresh vegetables,and needs 1m2 of space.Each rabbit eats 300g dry food and 320g of fresh vegetables, andneeds 5m2 of space.Each guinea pig eats 100g of dry food and 200g of freshvegetables, and needs 3m2 of space.Altogether they eat 2900g of dry food and 3920g of freshvegetables, and need 63m2 of space.(a) Explain why 5x + 30y + 10z = 290.



Solution

(a)

We examine the equation regarding how much dry fruit has beeneaten. In this case, we have

50x + 300y + 100z = 2900.

Dividing by 10 will give us the solution to part (i), or5x + 30y + 10z = 290.



A question

MATH1131 June 2011 q3 (ii)A pet shop has x hamsters, y rabbits and z guinea pigs.Each hamster eats 50g of dry food and 40g of fresh vegetables,and needs 1m2 of space.Each rabbit eats 300g dry food and 320g of fresh vegetables, andneeds 5m2 of space.Each guinea pig eats 100g of dry food and 200g of freshvegetables, and needs 3m2 of space.Altogether they eat 2900g of dry food and 3920g of freshvegetables, and need 63m2 of space.(b) Write down a system of linear equations that determine x , yand z .



Solution

(b)

We do the same thing we did for part (a) with fresh vegetables andspace. This will give us the system of equations:

5x + 30 + 10 = 2904x + 32y + 20z = 392

x + 5y + 3z = 63.



A question

MATH1131 June 2011 q3 (ii)A pet shop has x hamsters, y rabbits and z guinea pigs.Each hamster eats 50g of dry food and 40g of fresh vegetables,and needs 1m2 of space.Each rabbit eats 300g dry food and 320g of fresh vegetables, andneeds 5m2 of space.Each guinea pig eats 100g of dry food and 200g of freshvegetables, and needs 3m2 of space.Altogether they eat 2900g of dry food and 3920g of freshvegetables, and need 63m2 of space.(c) Reduce your system to echelon form and solve to find thenumber of hamsters, rabbits and guinea pigs.



Solution

(c)

We can form an augmented matrix using part (b) like so:

5 30 10 2904 32 20 3921 5 3 63

R1↔R3=⇒

1 5 3 634 32 20 3925 30 10 290

R3=5R1−R3=⇒

1 5 3 634 32 20 3920 −5 5 25

R2=4R1−R2=⇒

1 5 3 630 −12 −8 −1400 −5 5 25



Solution

(c) continued

1 5 3 630 −12 −8 −1400 −5 5 25

R2=− 14 R2=⇒

R3= 15 R3

1 5 3 630 3 2 350 −1 1 5

R3=R2+3R3=⇒

1 5 3 630 3 2 350 0 5 50

and now the matrix is in row-echelon form. But! We haven’tfinished yet. We still need to solve for x , y , z .



Back Substitution

Back SubstitutionThis is the final step in solving a matrix in row-echelon form. First,if there are any non-leading variables, assign an arbitrary parameterto them.Then, from the bottom row, we can deduce the solutions for eachvariable and substitute them for the equations above.



Solution

From our previous question, our final matrix is like so: 1 5 3 630 3 2 350 0 5 50

From the matrix above, we can clearly see that 5z = 50 and thusz = 10. In the second equation, we have 3y + 2z = 35.Substituting z in will give us 3y + 20 = 35 and thus y = 5. Finally,we do the same for the last (or first, depending how you look at it)equation, x + 5y + 3z = 63. This gives us x + 25 + 30 = 63 andthus x = 8.



Deducing the number of solutions from row-echelon form

Deducing solubility from row-echelon formBy examining the leading-terms of a matrix in row-echelon form,we can determine if it has a unique, infinite or no solution.

1 If the right-most column is a leading column, there are nosolutions (there is a row of zeros equalling a number).

2 There is a unique solution if every variable is a leadingvariable (every column has a leading variable).

3 Infinite solutions if and only if there is at least one non-leadingvariable on the left hand side.



A question

1131 June 2012 q3 (iii)A system of three equations in three unknowns x , y and z has been

reduced to the following echelon form:

1 2 3 50 2 4 80 0 α2 − 9 α− 3

(a) For which value of α will the system have no solution?



Solution

(a)

A matrix has no solutions if, in row-echelon form, the right-mostcolumn is a leading column. Therefore, if we let α = −3, theentire bottom row will be 0s and the right-most column will be -6.This will give us a matrix with no solutions as any linearcombination of 0s will never give us −6.



A question

MATH1131 June 2012 q3 (iii)A system of three equations in three unknowns x , y and z has been


1 2 3 50 2 4 80 0 α2 − 9 α− 3

(b) For which value of α will the system have infinitely manysolutions?



Solution

(b)

A matrix has infinitely many solutions if, in row-echelon form,there is at least one non-leading variable on the left-hand side.Therefore, if we let α = 3, this will result in the third column beingnon-leading (as well as the fourth).



A question

1131 June 2012 q3 (iii)A system of three equations in three unknowns x , y and z has been


1 2 3 50 2 4 80 0 α2 − 9 α− 3

(c) For the value of α determined in part (b), find the generalsolution.



Solution

(c)

From part (b), we let α = 3. Therefore, this gives us the matrix: 1 2 3 50 2 4 80 0 0 0

Since the z variable is non leading, we assign it an arbitraryparameter, such as λ. The second row will thus give us theequation 2y + 4λ = 8. Rearranging, we get y = 4− 2λ.Back-substituting this into the first row will give usx + 2(4− 2λ) + 3λ = 5. Therefore, x = −3 + λ.



Solution

xyz

=

−3 + λ4− 2λλ

=

−340

+ λ

1−21

,where λ ∈ R.



Geometric Applications - Problems involving lines andplanes

QuestionIs the line x−1

2 = y3 = z+1

−1 parallel to the plane

x = λ1

110

+ λ2

01−1

, where λ1, λ2 ∈ R?



Solution

We first convert the line to parametric form by setting everythingequal to a parameter. So we have x−1

2 = y3 = z+1

−1 = λ. Therefore,xyz

=

1 + 2λ3λ−1− λ

=

10−1

+ λ

23−1

.We then have to take the direction vector of this line and see if itcan be expressed as a linear combination (or alternatively, in thespan) of the direction vectors of the plane.



SolutionThis gives us the equation (LHS is direction vector of line, RHS isplane):

23−1

= λ1

110

+ λ2

01−1

.We can then put everything into an augmented matrix like so:

2 3 11 1 30 −1 −1

We can immediately perform back substitution to give λ2 = 1.Substituting this into the second row gives us λ1 = 2. Substitutingthese 2 values into the first row confirms these values as true,which means they are parallel.



Matrices



Matrices

Important facts about matricesA m× n matrix is one with m rows and n columns (in that order).

1 A matrix is square if the number of rows = number ofcolumns (it looks like a square).

2 Addition can only be performed when the two matrices havethe same number of rows AND columns. We just add thecorresponding elements together into a new matrix (and samefor subtraction).

3 Multiplying a matrix by a scalar means multiplying every entryby that scalar.

4 A zero matrix is one where every entry is 0.



Matrices are non-commutative!

Careful!Most matrices are NOT commutative, that is, AB 6= BA, where Aand B are matrices. Be careful when ordering them.



Matrix Multiplication

More facts about matrices1 We can multiply two matrices together IF the number of

columns in the first matrix equals the number of rows in thesecond matrix.

2 The resulting matrix will have the number of rows from thefirst matrix, and the number of columns from the secondmatrix.

3 To multiply them together, go through the rows of the firstmatrix and multiply them by the columns of the second.



Question

QuestionFind AB, where

A =

−2 13 4−1 5

and

B =(

1 0 21 −2 −3.

)



Solution

Since A has 2 columns and B has 2 rows, we can multiply themtogether. We can also tell that the end result will be a 3× 3matrix.

−2 + 1 0− 2 −4− 33 + 4 0− 8 −2− 15−1 + 5 0− 10 −2− 15

=

−1 −2 −77 −8 −174 −10 −17

.



The identity matrix

DefinitionThe identity matrix is a very important matrix that consists of 1’salong the diagonal and 0’s elsewhere. An example of a 3× 3identity matrix is 1 0 0

0 1 00 0 1

.We can view this as the value of 1 in terms of matrixmultiplication, that is, IA = AI = A.



Transpose of a matrix

DefinitionThe transpose (denoted as AT ) of an m × n matrix is the n ×mmatrix with entries

[AT ]ij = [A]ji .

In other words, we swap the columns and the rows.



Properties of the transpose

Properties of the transpose1 (AT )T = A2 (λA + µB)T = λAT + µBT

3 (AB)T = BT AT

4 If AT = A, the matrix is symmetric5 If AT = −A, the matrix skew symmetric



More on the transpose

Examples of transpose1 2 3 1

1 1 30 −1 −1

T

=

2 1 03 1 −11 3 −1

2 A symmetric matrix: 2 3 4

3 1 54 5 −1



The determinant of a matrix

The determinantThe concept of a matrix determinant is a very important topic inmathematics - and is something you will revisit (many times) againin 1B and later year math courses. An example of something coolabout determinants is that an n× n matrix has a unique solution ifand only if its determinant is not equal to 0.

The first thing to note is: the determinant is only defined forsquare matrices.




DefinitionThe determinant of a matrix is written as det(A) or |A|. For a1× 1 matrix, it is its sole entry. For a 2× 2 matrix(

a bc d

)

the determinant is given as det(A) = ad − bc.




The determinant of a 3× 3 matrix a b cd e fg h i

is given as

det(A) = a(ei − hf )− b(di − gf ) + c(dh − ge).




Sign changesBe very careful when calculating the determinant of a 3× 3matrix! Note the equation:

det(A) = a(ei − hf )− b(di − gf ) + c(dh − ge)

has a negative sign in front of the middle coefficient. Be carefulabout these sign changes! The signs are alternating, starting withpositive.



:(




DefinitionThe exam will (hopefully) not examine you on the determinants ofmatrices greater than 3× 3, but... the determinant of an n × nmatrix is defined as

det(A) =n∑

k=1(−1)1+ka1k |A1k |




ExampleThe calculation of the determinant of a 3× 3 matrix is:∣∣∣∣∣∣∣

5 1 7−2 3 −46 −1 2

∣∣∣∣∣∣∣ = 5∣∣∣∣∣ 3 −4−1 2

∣∣∣∣∣− 1∣∣∣∣∣ −2 −4

6 2

∣∣∣∣∣+ 7∣∣∣∣∣ −2 −4

6 2

∣∣∣∣∣= 5(6− 4)− (−4 + 24) + 7(2− 18)= −122.




Properties of the determinant1 If any two rows of a determinant swap, the sign of the

determinant is reversed.2 If a row of a matrix is multiplied by a scalar, then the entire

determinant will be multiplied by that scalar.3 The determinant is not changed by adding two rows.4 If a matrix contains a zero row/column, then the determinant

is 0.5 If a row/column of a matrix is a scalar multiple of another

row/column in the same matrix, the determinant is 0.




More properties1 det(AT ) = det(A)2 If AB exists, then det(AB) = det(A)det(B)3 det(A + B) 6= det(A) + det(B)4 det(mA) = mndet(A)




An efficient way to calculate the determinantIf A is a square matrix in row-echelon form, then det(A) isequivalent to the diagonals multiplied together!



Question

2016 MATH1141 q2 viiLet

A =

1 1 40 3 80 2 6

(a) Calculate the determinant of A.



Solution

det(A) =

∣∣∣∣∣∣∣1 1 40 3 80 2 6

∣∣∣∣∣∣∣3× 2× det(A) =

∣∣∣∣∣∣∣1 1 40 6 160 6 18

∣∣∣∣∣∣∣6det(A) =

∣∣∣∣∣∣∣1 1 40 6 160 0 2

∣∣∣∣∣∣∣det(A) = 1

6 × 12 = 2



Question

June 2015 MATH1141 q3 iv

Consider the matrix A =

a b cd e fg h i

. Suppose that det(A) = 7.

Find the value of the determinant of the following matrices:

B =

d e fa b c

d − 3a e − 3b f − 3c

C =

g − 3a 2a a − dh − 3b 2b b − ei − 3c 2c c − f



SolutionIn matrix B, we can see that the third row is just a linearcombination of the first two rows. Therefore, the determinant willbe 0.

In matrix C , we have to recall that det(AT ) = det(A). Thetranspose will thus be: g − 3a h − 3b i − 3c

2a 2b 2ca − d b − e c − f

.We can then perform two swaps to obtain the matrix: 2a 2b 2c

−(d − a) −(e − b) −(f − c−)g − 3a h − 3b i − 3c

.Donald Tang, Shayekh Rouf MATH1131/1141 130 of 153


Solution continued

2a 2b 2c−(d − a) −(e − b) −(f − c−)g − 3a h − 3b i − 3c

.We can observe that we have matrix A, but the 1st row is scaledby 2, the second row was multiplied by −1, the second and thirdbeing just addition of rows (which doesn’t affect the det). Thus,2×−1× (−1)2 × det(A) = −14 = det(C).



The inverse of a matrix

DefinitionThe inverse of a matrix A, denoted A−1, is a matrix which followsthe rules:

1 AA−1 = I or A−1A = I2 All invertible matrices are square3 (A−1)−1 = A4 det(A−1) = (det(A))−1 or 1

det(A)Your exam will always have a couple of inverse matrix questions, soit’s good to be prepared for them!




ConditionA matrix has an inverse if and only if

det(A) 6= 0.




DefinitionThe inverse of a 2× 2 matrix is defined as

A−1 =(

a bc d

)−1

= 1det(A)

(d −b−c a

).




DefinitionTo calculate the general inverse of any matrix, we can use rowreduction.

1 We write the matrix in the form (A|I), where I is the identitymatrix.

2 Row-reduce the matrix to obtain I on the left hand side.3 We will thus have the form (I|A−1), the RHS being the

inverse matrix.




QuestionFind the inverse of the matrix

A =(

2 33 4

)



Solution

In order to apply the formula on A =(

2 33 4

), we first evaluate

its determinant: det(A) = 2× 4− 3× 3 = −1. Hence, itsdeterminant is

1det(A)

(4 −3−3 2

)=(−4 33 −2

)




MATH1131 2017 q4 iiCalculate the inverse of the matrix

A =

1 1 10 1 10 0 1



Solutions

1 1 1 1 0 00 1 1 0 1 00 0 1 0 0 1

=

1 0 0 1 −1 00 1 1 0 1 00 0 1 0 0 1

=

1 0 0 1 −1 00 1 0 0 1 −10 0 1 0 0 1

Therefore the inverse matrix is the RHS, or

1 −1 00 1 −10 0 1



Questions

MATH1131 June 2013 q3 ii

Let M =

1 2 02 5 10 2 α

.(a) Evaluate the determinant of M.



Solutions

(a)

det(M) = 1∣∣∣∣∣ 5 1

2 α

∣∣∣∣∣− 2∣∣∣∣∣ 2 1

0 α

∣∣∣∣∣+ 0∣∣∣∣∣ 2 5

0 2

∣∣∣∣∣= 1(5α− 2)− 2(2α− 0) + 0= α− 2.



Questions


Let M =

1 2 02 5 10 2 α

.(b) Determine the value(s) of α for which M does not have aninverse.



Solutions

(b)

Since the inverse of a matrix does not exist if and only if thedeterminant is 0, then we have to solve the equationdet(M) = α− 2 = 0.Clearly, α = 2 for which M does not have an inverse.



Questions


Let M =

1 2 02 5 10 2 α

.(c) Find the inverse of M when α = 1.



Solutions(c)

1 2 0 1 0 02 5 1 0 1 00 2 1 0 0 1

=

1 2 0 1 0 00 1 1 −2 1 00 2 1 0 0 1

=

1 2 0 1 0 00 1 1 −2 1 00 0 1 −4 2 −1

=

1 2 0 1 0 00 1 0 2 −1 10 0 1 −4 2 −1

=

1 0 0 −3 2 −20 1 0 2 −1 10 0 1 −4 2 −1

Therefore, the inverse matrix is the RHS, or

−3 2 −2−2 −1 1−4 2 −1

.Donald Tang, Shayekh Rouf MATH1131/1141 145 of 153


Very important learning



Questions

MATH1141 2019 2ivLet A be an n × n real matrix with the property AT = −A. Such amatrix is called skew-symmetric.

(a) Show that A is not invertible if n is odd.



Solutions

(a)

A matrix is not invertible if the determinant is 0. Since the onlything we really have to go by is AT = −A, we apply the equate thedeterminants on both sides.

det(AT ) = det(−A)det(A) = (−1)ndet(A)

If n is an odd number, then (−1)n will be -1. Therefore,det(A) = −det(A) and thus det(A) can only be 0. This meansthat for odd n’s, there will be no inverse.



Questions


(b) Let x be a solution to the linear system

Ax = x.

Show that xT x = 0.



Solutions

(b)

x = AxxT = (Ax)T

xT = xT AT

xT = −xT AxT x = −xT AxxT x = −xT x.

Therefore, xT x = 0 as xT x = −xT x (similar to the previousquestion).



Questions


(c) Show that the matrix In − A is invertible.



Solutions

(b) Begin by recalling that a non-zero matrix M is invertible if andonly if Mx = 0 has only the zero solution. This is because we canmultiply both sides of the equation Mx = 0 by M−1. So weresume with the aim to prove that (In − A) x = 0 has only the zerosolution.(Proof by contradiction) Suppose that there is some x 6= 0 suchthat (In − A) x = 0. Then

(I − A)x = 0Ax = x.

However from part (b), this means that xT x = |x|2 = 0 whichimplies that x = 0. This is a contradiction and therefore In − A isinvertible.



Farewell

Sadly, this is the end of our seminar! We hope you all learn a lotand good luck for your exams :)


UNSW Mathematics Society · 2020-05-01 · Introduction to Vectors Vector Geometry Complex Numbers...

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