ENG1091 Lectures Notes

8/8/2019 ENG1091 Lectures Notes

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MONASH UNIVERSITY SCHOOL OF MATHEMATICAL SCIENCES

ENG1091 Vectors

Lecture 1 vector arithmetic revision dot product cross productText Reference: 4.1 - 4.2

Vectors and Lines, a quick review.

Many quantities in nature are completely specied by one number (called the magnitude of the

quantity) and are usually referred to as scalar quantities. Some examples are temperature, time,

length, and mass.

However, certain quantities require both a magnitude and a direction to specify them. To

say that a boat sailed 10 kilometers (km) does not specify where it went. It is necessary to

give the direction too; perhaps it sailed 10 km northwest. We then describe the position of the

boat by giving its displacement relative to some point, a quantity that involves distance as

well as direction. Quantities that require both a magnitude and a direction to describe them are

called vectors. Other examples include velocity and force. Vector quantities will be denoted by

boldface type: u; v; w, and so on. In handwritten work vectors are denoted by ve or by !v : Thevector that joins the two points A and B is denoted

!AB or by AB:

A vector v can be represented geometrically as a directed line segment or arrow. The magnitude

of a vector v will be denoted by kvk and is sometimes referred to as the length of v because itis represented by the length of the arrow.

Two vectors v and w are equal (written v = w)

if they have the same length and the same di-rection. Thus, for example, the two vectors in

the diagram are equal even though the initial

and terminal points are dierent!

v

w

There is however one vector that has no direction-the zero vector 0:

Given a vector v the vector that has the same

length as v but opposite in direction is the neg-

ative of vector v, denoted v:v

v

When we multiply a vector by a scalar we

simply multiply the length of the vector by the

relevant amount, without changing its direction

(unless the scalar is negative and then the di-

rection is opposite). Two vectors are parallel if

one is simply a scalar multiple of the other.

For example if a = b then a is parallel to b:

vv

2v

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If u and v are two vectors we dene their sum u + v by adding the vectors head to tail which

is to say we attach the tail of the second vector ; v; to the head of the rst u; the sum u + v is

then the vector drawn from the tail of rst vector to the head of the last.

This method allows also us to add several vec-

tors at once.a + b + c

c

b

a

Should it happen that vectors add together forming a loop, so that the end point is the same as

the initial point, then the vector sum is 0. Thus for example if A;B;C are any three points in

space !AB + !BC + !CA = 0:We can also add two vectors u and v geometrically by drawing them from the same point and

completing a parallelogram with the two vectors as adjacent sides. The diagonal vector drawn

from the common tail to the common head point is then the vector u + v:

From the parallelogram method of vector addition we see that u + v = v + u:

The opposing diagonal, drawn towards v; is the vector v u:The unit coordinate vectors.

Vectors of length one unit are called unit vec-

tors. The unit vectors parallel to the positive

x and y axis in the plane are labelled i and j:

Do not confuse i (or more probably j ) with the

complex number i:

In three dimensional space we add a further

unit vector, k , parallel to the z axis:

Any vector in space can be written as a combi-

nation of multiples ofi;j and k: The coecients

of i;j and k are called its rectangular com-

ponents.

i

k

j

x

z

y

The magnitudes of vectors given in component form.

Using Pythagoras theorem it is an easy matter to nd the lengths of vectors:

In three dimensions where we have v = ai + bj + ck; then kvk = kai + bj + ckk = pa2 + b2 + c2:Example:

ki

j + kk

= q(1)2 + (1)2 + (1)2 =p

3

In two dimensions the length of v = ai + bj is given by jvj = kai + bjk = pa2 + b2:



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The Scalar or Dot Product

In the previous section we saw how vectors can be added/subtracted together, and we saw how

to multiply them by scalars. The question naturally arises: is it possible to multiply two vectors

together?

In fact there are two types of vector multiplication that are useful-the scalar or dot productand the vector or cross product. Now for a word of warning. Many of the rules we take for

granted in ordinary arithmetic dont hold when it comes to vector multiplication. When we look

at the vector cross product later this lecture we will see that a b 6= b a: We will also see thatthere is no such thing as vector division-vectors dont have reciprocals! Of course we dont just

multiply vectors for fun-we do it because it has useful applications.

First, consider the scalar product. One modern use of the scalar product is the projection of

a 3D image on a 2D screen and to do it in such a way as to convince the viewer that he/she is

looking at a 3D image.

Given two vectors a and b then we dene their scalar or dot product as

a b = (kak kbk cos )where is the angle between the two vectors.

Note that a b is a scalar quantity-it is not a vector.Historically the reason that the scalar product was studied is that in physics the work done by

a force F in moving an object a displacement d is the dot product of force with displacement,

i.e. W = F d:

From the denition we immediately get the following:

(i) a a = kak2 (because the angle between a vector a and itself is 0:)(ii) If a ? b then a b = 0

The dot products of the unit coordinate vectors i;j and k:

Given the denition above we see thati j = j k = k i = 0

and i i = j j = k k = 1

Properties of the Dot Product

(i) a b = b a the dot product is commutative(ii) a b = a b = (a b) ; for any scalar (iii) a (b + c) = a b + a c the dot product is distributive

Notice that the expression a (b c) has absolutely no meaning because it is attempting toform a dot product of vector a with the scalar b c.

The expression a (b c) has a meaning though it is better written as (b c) a: The expression(b c) a simply means to multiply vector a by the scalar b c; resulting in a vector having thesame or opposite direction as a and of length: = jb cj kak :



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Notice how we can use the distributive law to simplify the dot product of two vectors given in

component form: Let a = a1i + a2j + a3k, and b = b1i + b2j + b3k then

a b = (a1i + a2j + a3k) (b1i + b2j + b3k)= a1i (b1i + b2j + b3k) + a2j (b1i + b2j + b3k) + a3k (b1i + b2j + b3k)

= a1i b1i + a2j b2j + a3k b3k (since i j = j k = i k = 0)= a1b1 + a2b2 + a3b3 (since i i = j j = k k = 1)

This gives a computational formula for evaluating a b = a1b1 + a2b2 + a3b3Example: This next example should convince you that there is no such thing as being able to

cancel a dot product.

Let a = 2i j + 4k; b = i + 2k; and c = 3i: Show a b = a c:Comment.a

b = (2)(

1) + (

1) (0) + (4) (2) = 6 and a

c = (2) (3) + (

1) (0) + (4) (0) = 6:

However b 6= c so we conclude it is not possible to cancel out vectors (even non-zero vectors)from a dot product like we can in ordinary arithmetic.

As a geometrical application we use the dot product to nd the angle between two vectors:

cos =a b

kak kbk :

Example: Find the angle between the main diagonal of a cube and the diagonal of a face which

it meets:

This angle will be the same regardless of the

size of the cube so lets assume the cube has

side length = 1:

Then face diagonal: a = i + k and main diago-

nal b = i +j + k:

Now a b = (1) (1) + (0) (1) + (1) (1) = 2 andkak =

q(1)2 + (1)2 =

p2 and

kbk =

q(1)2 + (1)2 + (1)2 =

p3 giving

cos = 2

p2p3from which = 35:26

The dot product provides a very easy way of telling when two vectors are perpendicular. If

a b = 0 then = 90o and we write a ? b:Example: Show that the points P (2; 1; 3) ; Q (4; 2; 5) and R (3; 3; 1) are the vertices of aright angled triangle:

!P Q =

!OQ !OP = (4i + 2j 5k) (2i +j 3k) = 2i +j 2k

!P R =

!OR !OP = (3i + 3j k) (2i +j 3k) = i + 2j + 2k

!QR =

!OR !OQ = (3i + 3j k) (4i + 2j 5k) = i +j + 4k;

and from these it is clear that !P Q !P R = 0 so we conclude the triangle is right angled at P:



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The Vector or Cross Product

This is a way of multiplying two vectors to-

gether which results in a vector. Given two

vectors a = a1i + a2j + a3k, and

b = b1i + b2j + b3k then we dene their vectoror cross product as

a b = (kak kbk sin ) n

where is the angle between the two vectors,

and n is the unit vector perpendicular to both

a and b, in a right-hand rule direction:

Note: (i) a b = b a(ii) If = 0o then a b = 0(iii) If = 90o then ka bk = kak kbk

The cross products of the unit coordinate vectors i;j and k:

Given the denition above we see that

i j = kj k = ik i = j

and i i = j j = k k = 0

Properties of the Cross Product

(i) a b = (b a) cross product is anti-commutative(ii) a b = a b = (a b) ; for any scalar (iii) a (b + c) = a b + a c cross product is distributive(iv) a (b c) 6= (a b) c (in general) non-associativity of the cross product

So if a = a1i + a2j + a3k, and b = b1i + b2j + b3k then



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a b = (a1i + a2j + a3k) (b1i + b2j + b3k)= (a1b1) i i+ (a1b2) i j+ (a1b3) i k

+ (a2b1)j i+ (a2b2)j j+ (a2b3)j k+ (a3b1) k i+ (a3b2) k j+ (a3b3) k k

= (a2b3 a3b2) i (a1b3 a3b1)j+ (a1b2 a2b1) k=

a2 a3b2 b3 i

a1 a3b1 b3j+

a1 a2b1 b2k

note the in the j term

=

i j k

a1 a2 a3

b1 b2 b3

A geometrical application of the cross-product:

Two vectors a and b; if drawn from the same point, dene a parallelogram:

Now we can determine the area of the parallelogram by breaking it up into two identical triangles.

Area = 2 12base perpendicular height

A = kak kbk sin = ka bk

Examples

(a) Let P;Q;R be the points P (2; 1; 3) ; Q (3; 4; 7) and R (1; 2; 3). Find the area of theparallelogram which has P Q and P R as adjacent sides.

!P Q =

!OQ !OP = (3i + 4j + 7k) (2i +j 3k) = i + 3j + 10k

!P R =

!OR !OP = (i 2j + 3k) (2i +j 3k) = i 3j + 6k

So!P Q !P R =

i j k

1 3 10

1 3 6

= i 3 103 6

j 1 101 6

+ k 1 31 3

= ((3)(6) (3) (10)) i ((1) (6) (1) (10))j + ((1) (3) (1) (3)) k= 48i 16jHence Area =

!P Q !P R = q(48)2 + (16)2 = 16p32 + 1 = 16p10:(b) Find area 4QP R = 12

!P Q !P R

= 8p

10:



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ENG1091 Vectors

Lecture 2 lines in 3DText Reference: 4.3.1

1. Revision of straight lines in two dimensional space

We are all quite familiar with the two-dimensional representation of a line as y = mx + b; (called

its Cartesian equation) where m is the slope and b is the y-intercept. Students should also be

familiar with the point-slope equation of a straight line:

(y y0) = m(x x0) (1)

Given any two points (x1; y1) and (x2; y2) in the x-y plane, we can readily get the equation of

the line passing through these two points by nding the slope m = (y1y2)(x1x2) ; and using this value

in the equation (1) above. The basic equation of a straight line is unique up to a scalar factor,

regardless of which point is chosen as (x0; y0) :

It would be natural to try to extend the equation of line in 2D space to 3D space. Perhaps one

might consider z = m1x + m2y + b. Unfortunately, this does not work, indeed, we will see in a

future lecture that this is actually the Cartesian equation of a plane in three-dimensional space.

2. Equations of straight lines in three dimensional space

In three-dimensional space, the concept of a slope is not so easily dened. Instead of slope, a

straight line will have an orientation associated with it that can be represented as a vector. The

line is then fully dened by a point on the line, say A, and an orientation vector, say v: Note

that the magnitude of the orientation vector doesnt actually matter, as long as we travel in the

right direction, we should stay on the line. Working in Cartesian coordinates, we can dene the

point A = (a;b;c), by its position vector and v as a vector with components (p;q;r): Then the

position vector!OP of any point P is given by

!OP =

!OA +

!AP where

!AP = tv for some scalar

t We can dene the equation of a line r(t) as:

r(t) =!OP =

!OA + tv:

This is the vector equation of a line. The variable t; which can take on any real value, is known

as the parametric variable.Breaking this up into the three components we get any point on the

line (x ;y ;z) being dened as:

x(t) = a + pt;

y(t) = b + qt;

z(t) = c + rt:

(Note: students may have actually been informally introduced to parametric variables when

learning trigonometry. We know that the circle of radius a centred at the origin can be representedby the parametric equations x = a cos(t) and y = a sin(t), where t can represent the angle from

the x-axis.)



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If we are given two-points, say A (x1; y1; z1) and B (x2; y2; z2), then the line between these two

points can be readily found by dening the orientation (direction) vector as the vector from A

to B.

Example1: Dene the (vector and parametric) equation of the line between the points A (2; 3; 4)

and B (1; 1; 1) :

!AB = (i +j + k) (2i + 3j + 4k) = i 2j 3k = vEquation of line:

!OP =

!OA + tv = (2i + 3j + 4k) + t (i 2j 3k)

= (2 t) i + (3 2t)j + (4 3t) kParametrically:

x(t) = 2 t;y(t) = 3 2t;

z(t) = 4 3t:

In the previous example, notice how the parametric variable works. Ift = 0; we are at one point,

A (2; 3; 4) ; and if t = 1 we are at the other point, B (1; 1; 1) :

Example 2: From the previous example, nd the value of t that denes the point (0; 1; 2) :Again, any value of t denes some point on the line. The value t = 1=2 denes the mid-point of

AB:

Equate x values: solve 2 t = 0 from which t = 2: If this value of t gives matching y and z valueswe know the point (0; 1; 2) is on the line. Otherwise the point lies o the line.With t = 2; y(2) = 3 4 = 1; and z(2) = 4 6 = 2: Therefore we conclude the point(0; 1; 2) is on the line.With t = 1=2; x(12) = 2 12 = 32 ; y( 12) = 3 1 = 2; and z( 12) = 4 32 = 52 ; so

32 ; 2;

52

is the

midpoint of AB:

Also note, however, that the equation of a line is not unique. The line between the points (2; 3; 4)

and (0; 1; 2) is equivalent to the equation found in the rst example, but the equation looksdierent: v = ((

j

2k)

(2i + 3j + 4k)) =

2i

4j

6k

!OP = !OA+tv = (2i + 3j + 4k)+t ((j 2k) (2i + 3j + 4k)) = (2 2t) i+(3 4t)j+(4 6t) kSo

x(t) = 2 2t;y(t) = 3 4t;z(t) = 4 6t:

The equation looks dierent but is it really?



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Finally note that the equation of a line can be manipulated to eliminate the parametric variable,

t: In this form the equation of the line is:

x ap

=y b

q=

z cr

This is sometimes called the algebraic equation of a line. Students may be apprehensive about

a relationship with two equal signs, instead of just one. Its interpretation is straightforward. Also

note that given this form of the equation of a line, we can immediately read o a point on the

line and its orientation vector.

Example 3: Given the relationx + 2

1=

y

2 =z 3

2

nd any two points on the line.

Solution: By examining the general from in the previous equation we see x = 2; y = 0; z = 3

is one such point (equate each numerator to zero).

Now of course the choice of zero is completely arbitrary; we can of course equate each fraction

to 1 (or any real number)

we do this:

x + 2

1= 1 giving x = 1

y

2 = 1 giving y = 2z 3

2

= 1 giving z = 5

Thus the point (1; 2; 5) is also on the line.Importantly, a direction vector for the line can also be read o namely: v = i 2j + 2k: Thischoice of v is unique up to scalar multiplication, (i.e. the only other direction vectors for this

line are non-zero scalar multiples of i 2j + 2k). We have a problem if the orientation vector is parallel to any of the axes. In such a case p; q

or r would be equal to zero. For that reason it is best to initially work with the parametric

representation and then nd the algebraic form.

After understanding the basic principles of lines, more sophisticated problems can be attempted.

Example 4: Find the minimum distance between the point B (1; 2; 3) and the line dened by

x + 2

1=

y

2 =z 3

2

Which point on the line is closest to the point B (1; 2; 3)?

Solution: A point on the line is A (2; 0; 3) and a direction vector for the line is v = i 2j + 2k:The point B (1; 2; 3) is not on the line. (Check this.)

The shortest distance between the point B and the line is d = !AB sin = !ABvkvk ; (drawa diagram)



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Now!AB = (i + 2j + 3k) (2i + 3k) = 3i + 2j and

!AB v =

i j k

3 2 0

1 2 2

= 4i 6j 8k:The shortest distance is thus d = k4i6j8kkki2j+2kk =

p16+36+64

3 =23

p29.

The closest point to BSolution 1: (Sound but pedestrian)

The closest point P on the line to B has position vector!OP =

!OA + tv where jtj is the distance

along the line from A to P:

Now t =!AB cos = !ABvkvk (draw a diagram):

= 343

= 13 (the negative sign is important and depends on our choice of v)

and v =1kvkv =

13 (i 2j + 2k) =

13 i

23j +

23k

Now!OP =

!OA + tv

= (2i + 3k) 1313 i 23j + 23k

= 199 i + +29j + 259 k

Hence the closest point is P =199 ; 29 ; 259 :

Solution 2: (Slicker)

Converting the equation of the line into parametric form we have:

x = 2 + t y = 2t z = 3 + 2tSo a general point on the line is P (2 + t; 2t; 3 + 2t)Hence

!BP =

!OP !OB = (3 + t) i + (2 2t)j + 2tk

The closest point on the line must satisfy!BP v = 0

Which is (3 + t) (1) + (2 2t) (2) + 2t (2) = 1 + 9t = 0So t = 19 ; hence the closest point is P (2 + t; 2t; 3 + 2t) with t = 19The closest point is 219 ; 29 ; 3 29 = 199 ; 29 ; 259



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ENG1091 Vectors

Lecture 3 planes in 3DText Reference: 4.3.2

1. Planes in three-dimensional space

When dening a straight line in three-dimensional space, we needed a point on the line and an

orientation vector.

To dene a plane in three-dimensional space, we need a point in the plane and a normal vector,

to the plane. Here n is the normal to the plane. For our immediate purpose, the magnitude of

n is not important, only its direction.

So lets assume that we have some point on the plane which we label A (a;b;c) and we have a

normal vector n = pi + qj + rk: We take a general point on the plane P (x ;y ;z) : Now the vector!AP lies in the plane and hence is normal to n:

Thus!AP n = 0:

This equation is the Cartesian equation of the plane

Explicitly this becomes (x a)p+(y b) q+ (z c) r = 0; which can be simplied to the generalform:

Ax + By + Cz = D:

Example 1: Find the equation of the plane that contains the point (2; 2; 3) and is normal to

the vectorh

1; 1; 2i

:

Solution:!AP = hx ;y ;zi h2; 2; 3i = hx 2; y 2; z 3i

!AP n = hx 2; y 2; z 3i h1; 1; 2i = x 6 + y + 2zHence the equation of the plane is x 6 + y + 2z = 0 or x + y + 2z = 6 Example 2: Find the equation of the plane going through the points (1; 0; 4) ; (2; 5; 0) ;(2; 2; 1) :Solution: Label the points A (1; 0; 4) ; B (2; 5; 0) ;and C(2; 2; 1) : A normal vector n is givenby n =

!AB

!AC:

Now!AB = h2; 5; 0i h1; 0; 4i = h3; 5; 4i and !AC = h2; 2; 1i h1; 0; 4i = h3; 2; 5i :

Thus n =

i j k

3 5 43 2 5

= i 5 42 5

j 3 43 5

+ k 3 53 2

= 17i + 3j 9k:Now

!AP = hx ;y ;zi h1; 0; 4i = hx + 1; y ; z 4i and

!AP n = hx + 1; y ; z 4i h17; 3; 9i = 0that is

17x

17 + 3y

9z + 36 = 0

giving the equation of the plane as 17x 3y + 9z = 19:



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We should check that all three points satisfy the planes equation:

A (1; 0; 4) : 17x 3y + 9z = 17 + 36 = 19 XB (2; 5; 0) : 17x 3y + 9z = 34 15 = 19 XC(2; 2; 1) : 17x 3y + 9z = 34 6 9 = 19: X

There are two observations that can be made. Firstly, the equation of a plane in three-dimensional

space is unique (up to multiplication by a scalar constant). Secondly, parallel planes have the

same normal vector and hence will only dier by the constant D:

Example 3: Find the minimum distance between the parallel planes 2x + 3y z = 6 and2x + 3y z = 0:Let P (x1; y1; z1) be any point in the plane 2x + 3y z = 6 andQ (x2; y2; z2) be any point in the plane 2x + 3y z = 0: [Notice that the equations of the planesare arranged so that they have identical coecients. Rearrange the equations if necessary-this

is important for what comes next.]

The distance between two parallel planes with normal n is then (diagram)

d =!P Q cos = !P Q nknk

=

!OQ !OP

n

knk

=

!OQ n!OP n

kn

khowever

!OQ n = 2x2 + 3y2 z2 = 0 and similarly !OP n = 2x1 + 3y1 z1 = 6:

Thus (and taking absolute value since we seek a distance):

d =

0 6q22 + 32 + (1)2 = 6p14

2. Lines and Planes

Combining the knowledge of lines, planes and basic vector operations allows for a wide range of

problems to be addressed in three-dimensional space. For example, we can nd:

the minimum distance from a point to a plane, the minimum distance from a point to line, the angle between two intersecting planes, the minimum distance between two non-intersecting lines.Example 4: Find the line dened by the intersection of the planes x + y + z = 2 and x + 2y = 4and the angle of intersection.

Solution: A direction vector of the line of intersection is easily found: it is normal to both

i +j + k and i + 2j and hence could be obtained using the cross product. To nd the equation



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of the line of intersection is best done using Gauss elimination (next lecture).

A direction vector is

i j k

1 1 11 2 0

= 2i + j 3k: (Of course any non-zero scalar multiple of

this is also a direction vector.)

The angle between two planes is dened as being the angle between its normals (diagram).

(i +j + k) (i + 2j) = 1 + 2 = 1k(i +j + k)k =

q(1)2 + 12 + 12 = p3 and k(i + 2j)k = p5

The angle between the planes is then given by cos = 1p3p5

; hence = 75:04 :

3. Parametric representation of a plane

Recall that straight lines have parametric equations giving x ;y ;z as function of one parametric

variable (usually t). Planes have parametric equations where x ;y ;z are given as functions of two

parametric variables (usually u and v):

Suppose we know a point P0 (a;b;c) in the plane and two non-parallel direction vectors

w1 = pi + qj + rk; and w2 = li + mj + nk also in the plane: (diagram):

w1

r(x,y,z)vw2

w2

O

uw1P0

Let r = xi + yj + zk denote the position vector of an general point P (x ;y ;z) in the plane, so

that r = !OP0 + uw1 + vw2 where u; v are any scalars (parameters).This gives r (x ;y ;z) = ha;b;ci + u hp;q;ri + v hl ;m;ni and hence

x (u; v) = a + pu + lv;

y (u; v) = b + qu + mv;

z (u; v) = c + ru + nv:

Theses 3 equations are the parametric equations of a plane. The fact that two parameters (u

and v) are needed to describe it indicates that a plane is a 2 dimensional surface.

In more advanced mathematics (i.e. 2nd level maths), it will be imperative to represent surfaces

parametrically.



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Example 5: Find a parametric representation of the plane going through the points (1; 0; 4) ; (2; 5; 0)and (2; 2; 1) :Solution:label the points P (1; 0; 4) ; Q (2; 5; 0) and R (2; 2; 1) :Now a choice for w1 is

!P Q = h2; 5; 0i h1; 0; 4i = h3; 5; 4i

and a choice for w2 is !P R = h2; 2; 1i h1; 0; 4i = h3; 2; 5i :Check that these are non-parallel X: (Otherwise the three points are collinear and the ques-

tion cannot be answered properly-there will be an innite number of planes.)

In vector form the parametric equations are

r = (1; 0; 4) + u (3; 5; 4) + v (3; 2; 5)= (1 + 3u + 3v; 5u + 2v; 4 4u 5v)

Hence

x (u; v) = 1 + 3u + 3v;y (u; v) = 5u + 2v;

z (u; v) = 4 4u 5v:



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ENG1091 Systems of Linear Equations

Lecture 4 echelon form Gauss eliminationText Reference: 5.5

Our object in this lecture is to solve a system of equations like

2x + y + z + w = 4

4x + y + 3z + 2w = 7

2x + z w = 9:

Such a system is called linear because each of the equations on the left hand side is a linear

function of the unknown variables x ;y ;z and w. Simple linear systems of 2 or 3 variables

are commonly encountered in secondary school and is instructive to view an example before

discussing a more general procedure.

Suppose we wish to solve a system like

x + 2y = 3 (1)

2x 3y = 8 (2)

One way to proceed is to multiply equation 1 by 2 and subtract this from equation 2:

x + 2y = 3 (1)

7y = 14 (2(a))

The reason why this is eective is that one of the variables is eliminated. Equation (2a) is now

easily solved giving y = 2; and substituting this into equation 1 we nd x = 1: Geometrically,the equations x + 2y = 3 and 2x 3y = 8 represent two straight lines in the x y plane whichintersect at the point (1; 2).

The important point is that both of the systemsx + 2y = 3

2x 3y = 8and

x + 2y = 3

7y = 14have identical

solutions. Think about the operations we could perform on the two original equations. We could

interchange the two equations multiply either equation by any number we choose except zero, and add a multiple of one equation to the other.

Now performing any of these operations without thinking is not guaranteed to be eective but

at least we are assured that the resulting system of equations has an identical set of solutions.

Notice that the names of the variables is irrelevant: solvingx + 2y = 3

2x 3y = 8is exactly the same

as solving the systemu + 2v = 3

2u 3v = 8; only the coecients are important.

1. The rst step in solving a linear system is to write the system in augmented matrix form.

This is simply a way of writing the system using only the coecients.



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For example we write the system

2x + y + z + w 4 = 04x + y + 3z + 2w = 7

2x + z w = 9

as2664 2 1 1 14 1 3 2

2 0 1 1

47

9

3775 :Notice each equation is written as a single row and that coecients belonging to the same variable

are written directly underneath each other. (Equation 3, which appears to have no y; has in fact

a ycoecient of zero.) Each constant term must be placed on the right hand side of the equalssign (the 4 becomes +4 on the right hand side of equation 1) and the vertical partition isused to separate the left hand side from the right hand side. (Think of it as replacing all of the

equals signs.)

Example: Write the system

r + s + 2t = 0

2r 3t = 16s 5t = 0

in augmented matrix form.

Solution:

26641 1 2

2 0 30 6 5

0

1

0

3775 :2. Gaussian elimination

Gaussian elimination is a systematic method of solving linear equations by rst reducing thecorresponding system into an equivalent system, called row echelon form, where the unknowns

can be calculated by back substitution.

Example: Given the system

r + s + 2t = 0

s 3t = 1 5t = 5

nd solutions to each of the variables

using back substitution.

Solution: t = 55 = 1 s = 1 + 3t

= 2

r = 2t s

= 2 + 2= 4

The system of equations in the last example has the augmented matrix

26641 1 2

0 1 30 0 5

0

1

5

3775 andwhich is one that is already in row echelon form. We saw how easy it is to nd solutions of

systems in this form.

Denition: A matrix is in row echelon form when

the leading (non-zero) coecient of each row (called the pivot entry) has zeros belowit, and

the pivot entries of following rows are located in columns further to the right. any rows which have no pivot (and therefore consist entirely of zeros) must come last.



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Example: Given the following partitioned matrices, choose those which are in row echelon form:

A.

26641 1 2

1 1 13

0 0 1

0

1

5

3775 B.2664

1 0 2 0

0 1 3 00 0 0 1

2

1

10

3775 C.2664

1 0 0

0 1 0

0 0 1

0

1

5

3775no yes yes

D.

26642 1 2

0 3 30 0 2

0

6

5

3775 E.2664

1 1 2

0 3 13

0 0 0

0

1

5

3775 F.2664

1 0 0 0

0 1 1 0

1 0 0 1

0

0

0

3775yes yes no

G.

2

6641 2 0 1 3 10 0 0 1 2 3

0 0 0 0 0 1

0

1

5

3

775yes

To obtain the equivalent row echelon form of a system we apply a sequence of the three elementary

row operations on the augmented matrix. As discussed above these row operations do not change

the solution set of the corresponding system of linear equations.

The three elementary row operations are:

Interchanging two rows Multiplying a row by a non-zero scalar Adding to one row a multiple of another

2. Row echelon forms

To reduce a matrix row echelon form systematically we follow these steps:

1. Locate the left-most column that doesnt consist entirely of zeros.

2. Ensure that the top entry of this column is a non-zero entry. If necessary, interchange top

row with another row to achieve this.

3. Multiply this top row by the appropriate constant so that the rst non-zero entry of this

row is 1. This entry is the pivot for that column. (It is not absolutely necessary that the

value of each pivot be 1 but this is certainly the most convenient value to have. As an

alternative to multiplying each row by a constant we can add/subtract multiples of other

rows to obtain a 1.)

4. Add a suitable multiple of this rst row to each row below, so that all entries below this

pivot are 0.

5. Consider the submatrix obtained by removing the top row, and apply to this matrix steps

1 to 4.



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Repeat steps 1-5 until the next submatrix under consideration has no rows left.

Example: Reduce the following matrix to row echelon form:

2664

0 0 2 0 123 6

15 9 42

2 4 5 6 1

3775

Solution:

26640 0 2 0 123 6 15 9 422 4 5 6 1

377513

R2 ! R2

R1 $ R2

26641 2 5 3 140 0 2 0 122 4 5 6 1

3775

R3 2R1 ! R3

26641 2 5 3 140 0 2 0 120 0 5 0 29

3775 12R2 ! R22664

1 2 5 3 140 0 1 0 60 0 5 0 29

3775

R3 5R2 ! R32664 1 2 5 3 140 0 1 0 6

0 0 0 0 1

3775 row echelon formExercise: Find a row echelon form of the matrix

2666664

1 0 1 02 1 0 8

0 1 2 0

1 1 2 6

3777775

Solution:

26666641 0 1 02 1 0 8

0 1 2 01 1 2 6

3777775R2 2R1 ! R2R4 R1 ! R4

26666641 0 1 00 1 2 8

0 1 2 00 1 1 6

3777775

R3 R2 ! R3R4+ R3

!R4

2666664

1 0 1 00 1 2 8

0 0

4

8

0 0 1 2

3777775

14R3 !R3

2666664

1 0 1 00 1 2 8

0 0 1 2

0 0 1 2

3777775

R4R3 !R4

26666641 0 1 00 1 2 8

0 0 1 2

0 0 0 0

3777775 row echelon form



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3. Solving a system using Gaussian elimination: To solve the system

x + 3y + 2z = 1

2x + 7y + 3z = 2

3x 10y 6z = 5

1. we write the augmented matrix:

26641 3 2

2 7 3

3 10 6

1

2

5

37752. by performing appropriate row operations we nd an equivalent row echelon form:

R2 2R1 ! R2R3 + 3R1 ! R3

26641 3 2

0 1 10 1 0

1

0

2

3775 R3+R2 !R32664

1 3 2

0 1 10 0 1

1

0

2

3775

(1)R3 !R32664 1 3 20 1 1

0 0 1

10

2

37753. Use back substitution to nd the values of the unknowns, in this case:

z = 2; y = z = 2 and x = 1 2z 3y = 1 4 6 = 9So the three planes intersect in a single point: x = 9 y = 2 z = 2Note: The pivot in a column does not need to be equal to 1 any non-zero number would do.Exercise:

Solve the systems:

(a) 2a 2b + 3c = 12a 2b + c = 1

a + b c = 3

ANS: Solution is: a = 52 ; b = 112 ; c = 5(b) r + s + 2t = 0

2r + 4s

3t = 1

3r + 6s 5t = 0

ANS: Solution is: fr = 17; s = 11; t = 3gExample: Find a vector equation for the line which forms the solution set of x + y z = 3

2x + y + 2z = 1

(You will recall an example similar to this at the end of lecture 3.)

Writing the augmented matrix of this system and taking the system to row echelon form:

" 1 1 12 1 2 31 # R2 2R1 !R2 " 1 1 10 1 4 35 # (1)R2 !R2 " 1 1 10 1 4 35 #Here is a system of equations with an innite solution set.



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Notice that the pivot entries correspond to variables x and y:"1 1 10 1 4

35#

The non-pivot variable, z; is said to be free and is set equal to a parameter t:

Let z = t

y 4z = 5 hence y = 5 + 4z = 5 + 4tx + y z = 3 hence x = 3 + z y = 2 3t

9>>=>>; ::::parametric formThe solution can be written in vector form as

(x ;y ;z) = (2 3t; 5 + 4t; t) = (2; 5; 0) + t (3; 4; 1)or in algebraic form:

x + 2

3 =y 5

4=

z 01

:

This shows that the solution is a straight line passing through the point (2; 5; 0) and withdirection vector3i + 4j + k:

Example: (from last lecture) Find the line dened by the intersection of the planes x +y + z = 2 and x + 2y = 4:

Augmented matrix:

"1 1 1 21 2 0 4

#

R2 + R1 ! R2"

1 1 1 20 3 1 6

#(now in echelon form)

z is free, y = 2 13z; x = 2 + z + y = 23zset z = 3t; y = 2

t; x = 2t and hence (x ;y ;z) = (0; 2; 0) + t (2;

1; 3) : (Compare with the

direction vector found in that example.)



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ENG1091 Consistency of Linear Equations

Lecture 5 no solution case innite solution caseText Reference: 5.6

The equation systems given in the last lecture were rather special in the sense that they all had

solutions.

An example of this is the equation systemx + 2y = 3

2x 3y = 8; which consists of two straight lines

intersecting in the point (1; 2).

But of course straight lines do not always intersect. The equation systemx + 2y = 3

2x + 4y = 1represents

two parallel straight lines and has no solution.

The question is how can we do this systematically:

How do we use Gauss elimination to recognise when a system of

equations has no solution.

Notice what happens when we employ Gauss elimination to solve the system of equations like

x + 2y = 3

2x + 4y = 1

Augmented matrix:

"1 2

2 4

3

1

#

Converting to row echelon form: (one step only)," 1 2

0 0

35#

.

Notice that in the last row all entries left of the partition are zero, and that there is a non zero

number to the right of the partition. Since it is impossible for 0x + 0y = 5 we know that thesystem has no solution.

Denition: A linear system of equations without solution is called inconsistent.

Now of course the previous example didnt need Gauss elimination to demonstrate its inconsis-

tency. However, a system of 3 equations in 3 unknowns (represented by three planes in space) is

rather more complex. A 3 3 system of equations will be inconsistent if either

the three planes are parallel

two planes are parallel and are intersected by the third,

neither of the planes is parallel but each pair of planes intersects in a line parallel to theothers.

Geometrically the situation for higher dimensions (>3 unknowns) is even more complex still but

algebraically very easy to sort out provided we apply Gauss elimination.



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The great advantage of Gauss elimination is that it takes the guess work out of equation manip-

ulation by being systematic. We can tell whether equations are inconsistent or not by using the

following very simple test:

When the augmented matrix corresponding to a system of inconsistent equations

is converted into a row echelon form, there will be at least one row whereall entries left of the partition are zero and there is a non-zero entry to the right

of the partition.

To put it another way, the row echelon form of an inconsistent linear system will have a row of

typeh

0 0 0 0 i where is some non-zero number.

Moreover, the test is completely diagnostic: if no such row exists then the equation system must

have solutions.

Example: The following partitioned matrices are row echelon forms corresponding to varioussystems of linear equations. Which linear systems are inconsistent?

A.

26641 1 2

0 1 13

0 0 0

0

1

5

3775 B.2664

1 0 2 0

0 1 3 00 0 0 0

2

1

0

3775 C.2666664

1 1 30 2 1

0 0 0

0 0 0

0

4

1

0

3777775 D.2664

2 1 2

0 0 0

0 0 0

1

0

0

3775

E.

2664

1 0 2

0 2 0

0 0 0

0

1

0

3775 F.

2664

1 0 0 0

0 1 1 0

0 0 0 1

0

0

0

3775 G.

2664

1 2 0 1 3 10 0 0 1 2

3

0 0 0 0 0 0

0

1

5

3775

Example: Show that the following system of equations is inconsistent by forming its augmented

matrix and then using row operations convert it to a matrix in row echelon form:

x + 2z = 1

y z = 0x + y + z = 2

Solution

Augmented matrix:

26641 0 2

0 1 11 1 1

1

0

2

3775 (not in echelon from)

R3 R1 ! R3

26641 0 2

0 1 10 1 1

1

0

1

3775

R3 R2 ! R3 26641 0 2

0 1 10 0 0

1

013775

The shaded row indicates inconsistency.



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What is the geometric interpretation of this inconsistent system?

Answer: Since none of the three planes are parallel (why?) we conclude that each pair of planes

intersects in a line parallel to the others.

[Examine the normal vectors (1; 0; 2) ; (0; 1; 1) ; (1; 1; 1) : Since no two of these is parallel neitheris there a pair of parallel planes.]

A system of linear equations that does not have solutions is said to be inconsistent, so obviously

a consistent system is one that does have solutions.

Now we encounter a remarkable fact: either a consistent linear system has a unique solution

(exactly one solution for each of the unknowns) or else it possesses innitely many! To put it

another way, if a linear system of equations is known to have two dierent solutions (say) then

that system must have innitely many solutions.

2. Systems with innitely many solutions:

The augmented matrix of the system

x 3y + z = 12x 6y + 3z = 4x + 3y = 1

reduces to the following equivalent row-echelon form:

Working: Augmented matrix:

2664

1 3 12 6 3

1 3 0

1

4

1

3775

R2 2R1 ! R2R3 + R1 ! R3

26641 3 10 0 1

0 0 1

1

2

2

3775 R3 R2 ! R32664

1 3 10 0 1

0 0 0

1

2

0

3775

row echelon form:

26641 3 10 0 1

0 0 0

1

2

0

3775The echelon form matrix gives us all the information concerning the original system. First of all

we notice there is no row of the typeh

0 0 0 0 i where is non-zero, so we knowthat the system has solutions.

The third row is entirely zero and in eect is totally redundant. We simply ignore rows that

consist entirely of zeros.

From 2nd row we have z = 2:

Solving the rst row for x we have x = 1 z + 3y = 1 + 3y (since z = 2):So z = 2 and x = 1 + 3y where the choice for y is completely arbitrary. There are innitelymany solutions, one for each value of y:

It is customary to assign a parameter to the free variable y: We can then write the solution set



24/157

as y = t; x = 1 + 3t; z = 2; where t is arbitrary.What is the graphical interpretation of this consistent system?

Answer: The three planes x 3y + z = 1; 2x 6y + 3z = 4; and x + 3y = 1 intersectin a straight line in 3D space. This line has a vector equation (x ;y ;z) = (1 + 3t;t; 2) =(

1; 0; 2)+ t (3; 1; 0) ; and therefore passes through the point (

1; 0; 2) and points in the direction

of the vector 3i +j + 0k:

Example: Solve the 3 4 system of linear equations:

2x + y + z + w = 4

4x + y + 3z + 2w = 7

2x + 2y + z w = 9

Solution:

We write the system in augmented matrix form and use elementary row operations to convert

the system to an equivalent one in echelon form. (Gauss elimination.)

Augmented matrix:

[A j b] =

26642 1 1 1

4 1 3 2

2 2 1 1

4

7

9

3775

26642 1 1 1

4 1 3 2

2 2 1 1

4

7

9

3775 R2 2R1 !R2 26642 1 1 1

0 1 1 02 2 1 1

4

19

3775 R3+ R1 !R3 26642 1 1 1

0 1 1 00 3 2 0

4

113

3775

R3+ 3R2 !R3

26642 1 1 1

0 1 1 00 0 5 0

4

110

3775This time the pivot variables are x; y and z (since the pivot entries occur in columns 1,2, and 3,

corresponding to the variables x ;y ;z):

The free variable is w:

w = free = t (say)

from row 3: 5z = 10 ) z = 2

from row 2: y + z = 1 ) y = z + 1 = 3from row 1: 2x + y + z + w = 4 ) x = 2 12z 12y 12w = 12 12 t

Writing the solutions in vector form:

hx ;y ;z ;wi = 12 12t; 3; 2; t = D 12 ; 3; 2; 0 E + tD 12 ; 0; 0; 1 E :



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Exercise:

The row echelon form of a system with unknowns r;s;t; and u; is

2666664

1 1 0 1

0 0 1 1

0 0 0 0

0 0 0 0

1

1

0

0

3777775

Describe the solutions of the system.

ANS: innite number of solutions with s and u free t = 1 u; r = 1 t s) (r;s;t;u) = (1 t s;s; 1 u; u) where s; u are arbitrary.Exercises: Solve the following systems of linear equations:

(a)

x

y

2z = 3

x + 2y z = 02x y + z = 5

x y z = 3

ANS: unique solution x = 2; y = 1; z = 0(b)

x + y + z = 2

x y + z = 12x + 2z = 4

ANS: no solution

(c)

a + b + c + 2d + e = 0a c + d e = 1

2b + c d 2e = 1

ANS: innite solution set; d and e are free

a = 2 + 6d + 3e; b = 1 3d; c = 3 + 7d + 2ei.e. (a;b;c;d;e) = (2 + 6d + 3e; 1 3d; 3 + 7d + 2e;d;e) = (2; 1; 3; 0; 0)+ d (6; 3; 7; 1; 0)+e (3; 0; 2; 0; 1) showing that the solution set is a plane in 5D space



26/157


ENG1091 Matrices

Lecture 6 matrices matrix arithmeticText Reference: 5.1-5.2

A matrix is a rectangular arrangement of numbers or variables, which can be either real or

complex, enclosed in square brackets. It is usual to denote matrices using capital letters. For

example:

A =

26642 34 5 07

p2 1

0:18 7 20 78

3775; B ="

1 2 3

4 5

#is not a matrix

A matrix has rows, running left to right, and columns running form top to bottom.

The matrix A has three rows and four columns and consists of 12 entries.

A matrix with m rows and n columns is called a m n matrix; the matrix A in theexample is a 3 4 matrix. The position of each entry is determined by the column and row numbers. We usesubindices to indicate this: for example,

a24 is the entry in row 2 , column 4. In this example, a24 = 1:a13 is the entry in row 1 , column 3. In this example, a13 = 5:

Sometimes we use the notation A = [aij] which simply indicates that A is a matrix (hence the

square brackets) whose entries are generically indicated as aij: The notation A = [aij]mn means

that A is an m n matrix.Special matrices

1. A 1 n matrix is a row matrix or row vector, e.g.h

1 2 4 3i

is a 1 4 row vector.

2. An m 1 matrix is a column matrix or column vector; e.g.2

6641

2

3

3

775is a 3 1 column vector.

3. A matrix with the same number of rows and columns is called a square matrix; e.g."1 3

2 4

#is a 2 2 matrix

Operations with matrices

1. Addition and subtraction

Addition and subtraction are possible only between matrices of the same order. These



27/157

are performed by adding or subtracting the corresponding entries respectively.

Example: 2664

1 13 5

4 8

3775

+

2664

7 12

6 1

3 5

3775

=

2664

8 11

9 6

7 13

3775The addition of matrices is commutative i.e. A + B = B + A

2. Multiplication by scalars

Given a matrix A and a number k; the multiplication of A by the scalar k and is obtained

by multiplying each entry of A by k: For example let k = 3 and A =

26641 13 5

4 8

3775, then

3A = 3

2664

1 13 5

4 8

3775

=

2664

3 39 15

12 24

3775

Note that subtraction can be expressed in terms of a scalar product (k = 1) and anaddition: A B = A + (B)There is a special matrix, called the zero matrix O = [Oij] where Oij = 0 for all i and j:

For any matrix A; A A = O:

3. Multiplication

Two matrices A and B can be multiplied together only when the number of columns in

A equals the number of rows in B: To nd the ij entry in the product AB we multiply

the entries along the ith row of A pairwise with entries on the jth column of B and then

add:

A =

26641 13 5

4 8

3775, B ="

1 1 32 4 2

#, C =

26641

2

3

3775(a)

AB = 26641 1

3 54 8

3775" 1 1 32 4 2 # = 26641 1 + 1 2 1 1 + 1 4 1 3 + 1 2

3 1 + 5 2 3 1 + 5 4 3 3 + 5 24 1 + 8 2 4 1 + 8 4 4 3 + 8 2

3775=

26641 5 513 17 112 36 28

3775(b) AC is not dened



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(c)

BA =

"1 1 32 4 2

#26641 13 5

4 8

3775 ="

1 1 + 1 3 + 3 4 1 1 + 1 5 + 3 82 1 + 4 3 + 2 4 2 1 + 4 5 + 2 8

#

= " 14 1822 2

#This example demonstrates something very important: matrix multiplication is not

usually commutative, i.e. AB 6= BA in general. In fact as we have seen AB and BAneed not be of the same order, or even if the product AB is dened, the other product

BA need not be.

In general if A = [aij]mp and B = [bij ]pn then AB is dened and AB = C = [cij]mn

where cij

= ai1

b1j

+ ai2

b2j

+

+ aip

bpj

=p

Pk=1 aikbkj :To illustrate: 2666664

: : : : : : : : : : : :

: : : : : : : : : : : :

: : : cij : : : : : :

: : : : : : : : : : : :

3777775mn

=

2666664: : : : : : : : : : : :

: : : : : : : : : : : :

ai1 ai2 : : : aip

: : : : : : : : : : : :

3777775mp

2666664: : : b1j : : : : : :

: : : b2j : : : : : :

: : : : : : : : : : : :

: : : bpj : : : : : :

3777775pn

= 2666664: : : : : : : : : : : :

: : : : : : : : : : : :

: : : ai1b1j + ai2b2j + + aipbpj : : : : : :: : : : : : : : : : : :

3777775So cij = ai1b1j + ai2b2j + + aipbpj =

pXk=1

aikbkj

4. Examples:

"2 3

1 5 #2 2

"1 2

2 3 #2 2 = 2 2

"1 2 3

4 5 6 #2664

1 1 30 2

1

3 5 4

3775

2 3 3 3 = 2 3

=

"2 (1) + 3 (2) 2 ( 2) + 3 (3)

1 (1) + 5 (2) 1 (2) + 5 (3)

#=

"1 + 9 1 4 + 15 3 + 2 + 124 + 18 4 + 10 + 30 12 5 + 24

#

=

"8 139 13

#=

"8 12 17

22 36 7

#



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5. Important:

We stress again that to be able to perform A B there is a size restriction:the number of columns in A (the matrix on the left) must equal the number of rows

in B (the second matrix in the product). We then say that AB is dened.

If A is a m p matrix, and B is a p n matrix, then AB is a m n matrix.

6. Properties of matrix multiplication

If A;B; and C are matrices of appropriate sizes, and k is a scalar then:

A(B + C) = AB + AC (B + C)A = BA + CA (AB)C = A(BC)

k(AB) = (kA)B = A (kB) AB 6= BA in general.

Exercises

1. Find the following product of matrices"2 1

3 5

#"

3 12 4

#=

"4 2

19 23

#

2. The product in the reverse order, although possible, leads to a dierent matrix:"3 1

2 4

#"

2 1

3 5

#=

"9 2

16 18

#

3. Given

A = " 1 31 2

#; B = 26640

7

8

3775; C = " 2 4 68 10 12 #; D = 26649 8 7 6

5 4 3 2

1 0 9 8

3775determine which of the following are dened and give their sizes (orders).

(a) AB not dened

(b) AC 2 3

(c) CD 2

4

(d) AD not dened

(e) DC not dened

(f) CB 2

1

(g) BC not dened

(h) (AC)D 2 4(i) A(CD) 2 4



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ENG1091 Matrices

Lecture 7 transpose matrix inversesText Reference: 5.4

The transpose of a matrix

The transpose of a matrix is obtained by interchanging its rows and columns. That is, the entries

of the ith row become the entries of the ith column.

So, if A is a m n matrix, then its transpose, denoted AT , is a n m matrix.Example:

Let A be the 3 2 matrix

26641 3

2 4

5 8

3775 :

Then AT is the 2 3 matrix: AT =2664 1 32 4

5 8

3775T

=" 1 2 5

3 4 8

#:

If A is conformable with respect to B i.e. if AB is dened then (AB)T = BTAT :

First we illustrate this with an example.

Let

A =

"1 23 0

#; B =

"2 1 40 1 3

#

we have

AB =

"1 23 0

#"2 1 40 1 3

#=

"2 3 26 3 12

#so that

(AB)T =

"2 3 26 3 12

#T=

26642 6

3 32 12

3775On the other hand

BTAT =

2

6642 0

1 14 3

3

775"

1 3

2 0

#=

2

6642 6

3 32 12

3

775This is a general rule:

The transpose of a matrix product is equal to the product of individual transposes taken

in the reverse order.

Now we show why (AB)T = BTAT is true in general.

Let A = [aij]mp and B = [bij]pn:

Recall that AB is dened and (AB)ij = ai1b1j + ai2b2j +

+ aipbpj =

p

Pk=1 aikbkj :Now the (i; j) entry of(AB)T is the the (j;i) entry of(AB); this is found by swapping

is and js :



31/157

(AB)Ti;j =pPk=1

ajkbki = aj1b1i + ai2b2i + + ajpbpi

= b1iaj1 + b2iaj2 + + bpiajp =pPk=1

bkiajk

= the sum of products found by multiplying, term by term, the

ith

row of BT

with the jth

column of AT

and this is the (i; j)entry of BTAT

Special matrices

Some types of matrices appear more often and so they have their own name:

A zero or null matrix contains all zero entries.

A symmetric matrix is the same as its transpose:

Diagonal matrices are matrices where any non-zero entries occur on the main diagonal:

Identity matrices for example I2 ="

1 0

0 1

#, I3 =

26641 0 0

0 1 0

0 0 1

3775 :If A is a square matrix, of size n n say, then AI = IA = A where I is the n n identitymatrix. Identity matrices play a role analogous to the number 1 for ordinary numbers.

The inverse of a matrix

Denition: The inverse of a square n n matrix A is an n n matrix B;(if one exists), such that AB = BA = I where I is the nn identity matrix.If such a B exists it is unique and we write it as A1:

Warning: A1 does not mean 1A :

If A has an inverse, then we say that matrix A is invertible or nonsingular.

We can calculate A1 by forming the augmented matrix consisting of A and I; theidentity matrix, and applying the following elementary row operations until A be-

comes I: Correspondingly, I will have become A1:

Schematically: [A jI] row operations!

I j A1Recall that elementary row operations are:

1. Interchanging two rows

2. Multiplying a row by a non-zero scalar

3. Adding to one row a multiple of another

STAGE 1: Forward elimination process



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1. Let C = [A j I] :

2.If there is a stage where C has a column consisting entirely of zeros, we stop immediately:

A has no inverse.

3. Ensure that the top left entry ofC is a non-zero entry, which we will label as a: (If necessary,

interchange the top row with another row to achieve this.)

4. Multiply this row by 1a so that the rst non-zero entry of this row is 1. This entry is the

pivot for that column. (Alternatively this can sometimes be aected by row interchange.)

5. Add a suitable multiple of this rst row to the rows below row so that all entries in the

column below the pivot become 0.

If there is a stage where there the sub-matrix ofC left of the partition has a row consisting

entirely of zeros, we stop immediately: the matrix A has no inverse.

6. Consider the submatrix of C found by removing its 1st row and 1st column, regard this

as a new matrix C: Repeat steps 2-6 until the next submatrix under consideration has no

rows left.

7. Provided the algorithm has not been exited at steps 2 or 5 the full matrix is now in echelon

form. The pivots are all 1 and located on the main diagonal of the matrix left of the

partition.

STAGE 2: Backward elimination process

1. Notice that all pivots are 1 and are located on the main diagonal of the matrix left of the

partition. Locate the row containing the right-most pivot, (which must be in the bottom

row).

2. Add suitable multiples of this row to the rows above so that all entries in the column above

become 0.

3. Locate the next pivot by moving up the diagonal and repeat steps 2 and 3.

4. This procedure is repeated until the top left pivot is reached, at which point the full matrix

is

I j A1 :



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Examples:

Find inverses of the following (if they exist).

1. A =

2

6640 1 1109

0p

2 40 3

3

775, A has a column of zeros and hence no inverse.

2. A =

26641 1 11 1 0

1 1 1

3775

We form [A j I] = C =

26641 1 11 1 0

1 1 1

1 0 0

0 1 0

0 0 1

3775 Step 1We note that C has a pivot in the top left entry and that this pivot is 1. Steps 3-4

Subtract row 1 from row 2:2664 1 1 10 0 1

1 1 1

1 0 0

1 1 00 0 1

3775

Subtract row 1 from row 3:

26641 1 10 0 1

0 0 2

1 0 0

1 1 01 0 1

3775 : Step 5 is now complete.Step 7. We apply the algorithm again to the submatrix of C found by deleting its 1st row and

column (shaded)

26641 1 10 0 1

0 0 2

1 0 0

1 1 01 0 1

3775

but since this new matrix has a column of zeros, we conclude the matrix

26641 1 11 1 0

1 1 1

3775 has noinverse. (Exiting the algorithm at step 2.)



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3. Find the inverse of A =

26642 7 1

1 4 11 3 0

3775

Solution: [A

jI] =

2664

2 7 1

1 4

1

1 3 0

1 0 0

0 1 0

0 0 1

3775

R1 $ R2

26641 4 12 7 1

1 3 0

0 1 0

1 0 0

0 0 1

3775 R2 2R1 ! R22664

1 4 10 1 31 3 0

0 1 0

1 2 00 0 1

3775

R3 R1 ! R3

2664

1 4 10 1 30 1 1

0 1 0

1 2 00 1 1

3775

(1)R2 ! R2

2664

1 4 10 1 30 1 1

0 1 0

1 2 00 1 1

3775

R2 + R3 ! R3

26641 4 10 1 30 0 2

0 1 0

1 2 01 1 1

3775 12R3 ! R32664

1 4 10 1 30 0 1

0 1 0

1 2 012 12 12

3775

R2 + 3R3 ! R2

26641 4 10 1 0

0 0 1

0 1 012

12 32

12 12 12

3775 R1 + R3 ! R12664

1 4 0

0 1 0

0 0 1

12

12 12

12

12 32

12 12 12

3775

R1 4R2 ! R1 26641 0 0

0 1 0

0 0 1

32 32 11212 12 3212 12 12

3775 = I j A1 :

Hence A1 =

266432 32 11212

12 32

12 12 12

3775 = 122664

3 3 111 1 31 1 1

3775

Check: 12

2664

3 3 111 1 31

1

1

3775

2664

2 7 1

1 4 11 3 0

3775

= 12

2664

6 3 + 11 21 12 + 33 3 + 3 + 02 + 1 3 7 + 4 9 3 3 + 02

1

1 7

4

3 1 + 1 + 0

3775

= 12

26642 0 0

0 2 0

0 0 2

3775 =2664

1 0 0

0 1 0

0 0 1

3775 :

Strictly speaking we should also check that

26642 7 1

1 4 11 3 0

37752664

32 32 11212

12 32

12 12 12

3775 =2664

1 0 0

0 1 0

0 0 1

3775,however it is a known fact for matrices that a left inverse is also a right inverse and vice versa,

so a one sided check is sucient.



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Inverses of 2 2 matrices

Example: nd the inverse of the matrix

"2 4

1 3

#:

Solution: [A j I] ="

2 4

1 3

1 0

0 1

#12R1 ! R1

"1 2

1 3

12 0

0 1

#R2 R1 ! R2

"1 2

0 1

12 0

1

2

1

#

R1 2R2 ! R1"

1 0

0 1

32 212 1#

: Hence

"2 4

1 3

#1=

" 32 2

12 1

#:

However there is also a simple formula for 2 2 matrices.

The inverse of a 2 2 matrix: A ="

a b

c d

#, A1 =

1

ad bc

"d b

c a

#:

Notes: The matrix A = " a bc d #is invertible provided ad bc 6= 0: This number is called thedeterminant of A and is denoted by

a bc d or det(A) :

The determinant of any square matrix A is also dened (see next lecture) and this number

determines whether or not A is invertible:

A square matrix A is invertible if and only if det(A) 6= 0:

Using matrix methods to solve linear systems of equations

Consider the 3 3 linear system:

2x1 + 7x2 + x3

x1 + 4x2 x3x1 + 3x2

=

=

=

1

4

5

which can also be written in matrix form

2

6642 7 1

1 4 1

1 3 0

3

775

2

664x1

x2

x3

3

775=

2

6641

4

5

3

775:

Any n n linear system can be written in the form Ax = b; where x and b are column vectors(matrices).

If A is invertible we can multiply on the left by A1 and so obtain the unknown

matrix x :

Ax = b

A1Ax = A1b

Ix = A1b

giving x = A1b



36/157

This method is somewhat more restrictive than Gauss elimination. It only works

for n n systems and either produces a unique solution (when det(A) 6= 0) but isincapable of distinguishing between the no solution or innite solution cases which

occur when det(A) = 0:

The main advantage to using matrix inverse method occurs when working with mul-

tiple equations with the same set of coecients.

Example:

Solve: (a)

2x1 + 7x2 + x3 = 1

x1 + 4x2 x3 = 4x1 + 3x2 = 5

and (b)

2x1 + 7x2 + x3 = 2x1 + 4x2 x3 = 4x1 + 3x2 = 6

In (a) we have 2664x1

x2

x3

3775 = 26642 7 1

1 4 11 3 0

37751

26641

4

5

3775, and in (b) 2664x1

x2

x3

3775 = 26642 7 1

1 4 11 3 0

37751

266424

6

3775 :

Now

26642 7 1

1 4 11 3 0

37751

= 12

26643 3 111 1 31 1 1

3775 (shown above),

giving the solution to (a): 2664x1

x2x3

3775 = 12 26643 3 11

1 1 31 1 1

377526641

453775 = 2664

20

54

3775

and to (b):

2664x1

x2

x3

3775 = 122664

3 3 111 1 31 1 1

37752664

24

6

3775 =2664

30

86

3775 :Exercise: Solve the following system of equations using matrix inversion followed by matrix

multiplication:2x + 3y = 7

4x + y = 3

In matrix form:

"2 3

4 1

#"x

y

#=

"7

3

#

If exists

"2 3

4 1

#1we may write

"2 3

4 1

#1 "2 3

4 1

#"x

y

#=

"2 3

4 1

#1 "7

3

#

Now

"2 3

4 1

#1=

[not zero soh

2 34 1

i1

exists] %

1212

"1 3

4 2

#=

" 110 310

25 15

#

So " 1 00 1

#" xy# = " 110 3102

5 15#" 7

3# = " 1511

5

# ; giving x = 1=5 and y = 11=5



37/157


ENG1091 Determinants

Lecture 8 Determinants Cramers RuleText Reference: 5.3

Determinants

The determinant of a 2 2 matrix A = " a bc d

#is dened by det A =

a bc d = ad bc:As we noted in the previous lecture determinants are used to determine whether a square matrix

is invertible or not.

Determinants can also be used to solve n n systems of linear equations using a rule known asCramers rule.

For example the system :2x + 3y = 5

7x + 11y = 13has the solution:

x =

5 313 11 2 37 11

; y =

2 57 13 2 37 11

The denominator is always the determinant of coecients and the determinant on the top line

replacing the column containing the coecients of the variable in question with the numbers on

the right hand side.

Evaluating the determinants gives

x =55 3922 21 ; y =

26 3522 21

i.e.

x = 16; y = 9:

Cramers rule works provided determinant of coecients (the denominator) is non-zero, when it

is zero Cramers rule fails and the system of equations has either no solutions or innitely many.

Determinants of larger matrices

The determinant is a number we assign to any square matrix. It plays an important role in

nding the inverse of a matrix, solving systems of equations, multiplication of vectors, nding

areas of triangles, etc.

To nd the determinant of larger matrices we need to know about cofactors. A cofactor of a

particular entry in a matrix is the (smaller) determinant consisting of those elements which

remain if we removed the row and column belonging to that entry, together with a sign, + or ;depending on where the entry is located.

Example In the matrix A =2664 1 3 74 2 2

5 6 9

3775 the cofactor of the (2; 3) entry, namely 2; is theENG1091 Mathematics for Engineering page 37


38/157

signed determinant 1 35 6

:

The minus sign comes from the sign matrix:

2666664

+ + : : : + : : :+

+ : : :

......

...

3777775

and the minor determinant

1 35 6 is obtained by removing row 2 and column 3

We refer to the cofactor of the (i; j) entry as Cij:

In the example above, C23 = 1 35 6

= (6 15) = 21:

Example Find, but do not evaluate, C41 in the matrix 26666641 4 3 72 3 9 11 8 6 1

1 2 1 10

3777775 :

This is clearly

4 3 7

3 9 18 6 1

:Note that the comes from the position not the sign of the entry.

How to evaluate determinants.

1. Choose any row or column.

2. For each position in the selected row or column, calculate the corresponding cofactor.

3. Form the product of each cofactor with the corresponding entry. The determinant is the

sum of these products.

Example Find det A = 1 3 74 2 25 6 9

:We choose to expand along the second row.

det A = 4 3 76 9

+ 2 1 75 9

2 1 35 6

= 4 (27 42) + 2 (9 35) + 2 (6 15) = 70:

If we chose instead to expand along the 1st column the answer is the same.

det A = 1 2 26 9

+ 4 3 76 9

+ 5 3 72 2

= 30 + 4 15 + 5 20 = 70ENG1091 Mathematics for Engineering page 38


39/157

Which row or column? It is a remarkable fact that the answer is independent of the row and

column we choose.

As a practical consideration we would do well to choose that row/column that has the greatest

number of zeros.

Example: Find the 4 4 determinant

1 4 3

7

0 3 9 10 0 6 10 0 0 10

:

An obvious choice is to expand along the 1st column:

1 4 3 70 3 9 10 0 6 10 0 0 10

= 1

3 9 10 6 10 0 10

+ other terms all zero.

And again: 1

3 9 10 6 10 0 10

= 1 3 6 10 10

+ other terms all zero= 1 3 (6 10 0) = 1 3 6 10 = 180

We should state this as a general rule:the determinant of a triangular matrix (either upper or lower)

is the product of the entries along the main diagonal.

Exercise: Find det B where B =

26642 1 11 3 3

10 5 2

3775 :

det B =

2 1 11 3 3

10 5 2

(notice the dierent bracketing which distinguishes

a matrix from its determinant)

=

(1)

1 1

5 2 + (3) 2 1

10 2 (3) 2 1

10 5 (expanding along row 2)= (3) + 3 14 3 20= 3 42 + 60 == 21:



40/157

Properties of Determinants

To illustrate the following properties of determinants we will work with an arbitrary 33 matrix

A =

2664

a1 a2 a3

b1 b2 b3

c1 c2 c3

3775

: We stress that the all the following properties are true regardless of size.

1. Transpose property: det(A) = det ATa1 a2 a3

b1 b2 b3

c1 c2 c3

=

a1 b1 c1

a2 b2 c2

a3 b3 c3

2. Scaling property: to multiply a determinant by a number we multiply any row or column

by that number.

k

a1 a2 a3b1 b2 b3

c1 c2 c3

=

a1 a2 ka3b1 b2 kb3

c1 c2 kc3

=

ka1 ka2 ka3b1 b2 b3

c1 c2 c3

(we may pick any row or column)

Hence det(kA) = det

0BB@k2664

a1 a2 a3

b1 b2 b3

c1 c2 c3

37751CCA =

ka1 ka2 ka3

kb1 kb2 kb3

kc1 kc2 kc3

= k3

a1 a2 a3

b1 b2 b3

c1 c2 c3

If A is n n then det(kA) = kn det(A)

3. Interchange property: Swapping any two rows, or two columns, changes the sign of thedeterminant

e:g:

a1 a2 a3

b1 b2 b3

c1 c2 c3

=

a1 a2 a3

c1 c2 c3

b1 b2 b3

Hence a matrix with two identical rows or columns has determinant = zero

a1 a2 a3

a1 a2 a3

c1 c2 c3

= 0:

4. Elimination property: Adding a multiple of a row to another row does not alter the

value of a determinant. Similarly for columns.

e.g.

a1 a2 a3

b1 + kc1 b2 + kc2 b3 + kc3

c1 c2 c3

=

a1 a2 a3

b1 b2 b3

c1 c2 c3

:

5. Matrix multiplication property: Let A and B be square matrices of the same size (both

n n); thendet(AB) = det A det B



41/157

Special case: if A is invertible then AA1 = I so that det

AA1

= det A det A1 = det I = 1:

In particular det A 6= 0; anddet

A1

=

1

det A:

It can be shown that the condition det A 6= 0 is also sucient for invertibility, i.e.

A is invertible if and only if det A 6= 0:

Application: Simplify the 3 3 Vandemonde determinant:1 a a2

1 b b2

1 c c2

=

1 a a2

0 b a b2

a2

1 c c2

R2 R1 ! R2

=

1 a a2

0 b a b2 a20 c a c2 a2

R3 R1 ! R3

= (b a) (c a)

1 a a2

0 1 b + a

0 1 c + a

taking out common factor of (b a) from row 2and (c a) from row 3

= (b a) (c a)

1 a a2

0 1 b + a

0 0 c b

R3 R1 ! R2= (b a) (c a) (c b) multiplying down the main diagonal to evaluate the determinant



42/157

Cramers Rule

Example: Solve the system of equations below using Cramers rule:

x + 2y + z = 1

2x 3y + 7z = 4x + y 3z = 1

Solution:

x =

1 2 1

4 3 71 1 3

1 2 1

2

3 7

1 1 3

; y =

1 1 1

2 4 71 1 3

1 2 1

2

3 7

1 1 3

; z =

1 2 1

2 3 41 1 1

1 2 1

2

3 7

1 1 3

:

1 2 1

4 3 71 1 3

=

1 2 1

0 5 11

0 3 2

= 5 113 2

= 43

1 1 1

2 4 71 1 3

=

1 1 1

0 6 50 0 2

= 12

1 2 1

2 3 4

1 1

1

=

1 2 1

0 7 60 3 0

= 18

1 2 1

2 3 7

1 1

3

=

1 2 1

0 7 50 3

2

= 14 15 = 1

giving x =431 = 43 y =

12

1 = 12 z =18

1 = 18



43/157


ENG1091 Eigenvalues & Eigenvectors

Lecture 9&10 EigenvaluesText Reference: 5.7

1. Eigenvalues and eigenvectors

Generally speaking, when the determinant of an n n system of equations is zero, we can onlydeduce that the system has no solutions or innitely many.

A homogeneous system of equations, introduced earlier (i.e. Ax = 0), always has the trivial

solution x = 0: If the determinant of coecients of a homogeneous system is zero the system

must have innitely many solutions.

Example 1: Let

A = 2664a 1 3

2 2 12 a 1

3775 ;nd the values of a; such that Ax = 0 has non-trivial solutions.

Denitions: Let A be an nn matrix and x be an n1 vector. Any scalar satisfying Ax = xfor some non-zero x is called an eigenvalue of A: The corresponding non-zero vectors x for

which Ax = x are called the eigenvectors of A corresponding to .

To quote from the textbook, such problems arise naturally in many branches of engineering.

For example, in vibrations the eigenvalues and eigenvectors describe the frequency and mode ofvibration respectively, while in mechanics they represent principal stresses and the principal axes

of stress in bodies subject to external forces.

Example 2: Show

x =

"1

1

#is an eigenvector of

A =

"2 1

1 2 #corresponding to the eigenvalue = 3: This is straightforward matrix arithmetic :"

2 1

1 2

#"1

1

#=

"3

3

#= 3

"1

1

#

Note also that if we multiply of the sides of this equation by the scalar t we get"2 1

1 2

#"t

t

#=

"3t

3t

#= 3

"t

t

#

This demonstrates that any non-zero multiple of an eigenvector corresponding to is also an

eigenvector.



44/157

2. Finding eigenvalues

It is tempting to rewrite the equation Ax = x as (A)x = 0, but this cannot possibly becorrect why?

We write instead: (AI)x = 0; this is a homogeneous system of equations. Now we know thetrivial solution x = 0 is always available, but we are interested only in the non-zero solutions

(called eigenvectors). This is the requirement that a homogeneous system has innite number of

solutions which happens precisely when

det(A I) = (1)n ( 1) ( 2) ( n) = 0:

Here the eigenvalues are simply 1; 2; : : : ; n. The common convention is to label from the

largest in magnitude to the smallest in magnitude. Note that it is possible for the roots of the

polynomial to be repeated or complex.

Example 3: Find the eigenvalues of

A =" 0 1

1 0

#

Solution: The characteristic polynomial: det(A I) = 11

= 2 + 1 = 0 for = i:Example 4: Find the eigenvalues of

A =

2

6641 1 2

1 2 10 1

1

3

775Solution: A I =

26641 1 2

1 2 10 1 1

3775 2664

1 0 0

0 1 0

0 0 1

3775 =2664

1 1 21 2 10 1 1

3775

The characteristic polynomial: det(A I) =

1 1 21 2 10 1 1

=

1 2 11 1 2

0 1 1 R1 $R2

=

1 2 10 (1 ) (2 ) + 1 1 0 1 1

R2 + (1 )R1 !R2= (1)

(1 ) (2 ) + 1 1 1 1

expanding along col1

= (1 ) (1 ) (2 ) + 1 11 1 factoring col2= (1 )[(1 ) (2 ) + 1 1] = (1 ) (1 ) (2 ) : Hence eigenvalues: = 1; 2; 1



45/157

3. Finding the eigenvectors of unique eigenvalues

Having solved the nth order polynomial (characteristic equation) for the n roots (eigenvalues),

it still remains to nd the corresponding eigenvectors. For the moment lets assume that the

eigenvalues are distinct (non-repeated.)

Let ej

be an eigenvector that corresponds to the eigenvalue j

so that Aej

= j

ej

Example 3 (again): Find the eigenvectors of

A =

"0 11 0

#

Solution: = i :

"0 11 0

#"x1

x2

#= i

"x1

x2

#is equivalent to

"i 1

1 i

# "x1

x2

#=

"0

0

#The 2 2 case always leads to two identical equations, in this example x1 = ix2 and ix1 = x2:Thus the eigenvectors are t (i; 1) where t

6= 0:

= i :"

0 11 0

#"x1

x2

#= i

"x1

x2

#is equivalent to

"i 11 i

#"x1

x2

#=

"0

0

#hence

x1 = ix2 giving eigenvectors t (i; 1) where t 6= 0:Example 4 (again): Find the eigenvector corresponding to the dominant eigenvalue of

A =

26641 1 2

1 2 10 1 1

3775Solution: The dominant eigenvalue is = 2:

Solving Ax = 2x :26641 1 2

1 2 10 1 1

37752664

x1

x2

x3

3775 = 22664

x1

x2

x3

3775

this is the homogeneous system:

x1 + x2 2x3 = 2x1x1 + 2x2 + x3 = 2x2

x2

x3 = 2x3

equivalent to

x1 + x2 2x3 = 0x1 + x3 = 0x2

3x3 = 0

The augmented matrix is (we dont need to include the column of zeros):26641 1 21 0 10 1 3

3775 (1)R1 !R12664

1 1 21 0 10 1 3

3775 R1+R2 !R22664

1 1 20 1 30 1 3

3775 R3+R2 !R32664

1 1 20 1 30 0 0

The non-pivot variable x3 is chosen free, so x3 = t

x2 = 3x3 = 3t from row 2, and x1 = x2 2x3 = t from row 1, giving (x1; x2; x3) = t (1; 3; 1)hence an eigenvector corresponding to = 2 is h1; 3; 1i :

Check2664 1 1 21 2 1

0 1 1

37752664 131

3775 = 2664 262

3775 = 2 2664 131

3775 :ENG1091 Mathematics for Engineering page 45


46/157

Note that the eigenvectors are only unique up to multiplication by scalars. Also, while the

0 vector is always a solution to the system (AI)x = 0 the eigenvectors are the non-zerosolutions. The vector 0 is never an eigenvector.

4. Finding the eigenvectors of repeated eigenvalues

If the eigenvalues of the matrix A are distinct, then it can be shown that the corresponding

eigenvectors are linearly independent. If, however, the eigenvalues are repeated, it may not be

possible to nd n linearly independent eigenvectors. By repeated roots of the characteristic

equation, we simply mean that two or more of the eigenvalues are the same.

Example 5: Find the eigenvalues of the matrix

A =

2664

0 0 1

0 1 2

0 0 1

3775

:

Solution: Characteristic polynomial is det(A I) =

0 1

0 1 20 0 1

= (1 )2

which has roots = 0 and = 1 (multiplicity = 2)

In the previous example, it is not clear how many independent eigenvectors exist when = 1: The

eigenvalue has a multiplicity of 2, but that doesnt assure us that there will be two independent

eigenvectors.

Example 6: In the previous example, nd the eigenvector(s) corresponding to each eigenvalue.

Solution: Eigenvectors for = 1 :

A I = A I =

26641 0 10 0 2

0 0 0

3775 which is in echelon form.

Solving

2664

1 0 10 0 2

0 0 0

3775

2664

x1

x2

x3

3775

=

2664

0

0

0

3775

we have x3 = 0; x2 = t and x1 + x3 = 0 so x1 = 0 also.

Thus the eigenvectors for = 1 are

2664x1

x2

x3

3775 =2664

0

t

0

3775 = t2664

0

1

0

3775 ; that is, the non-zero multiplesof the vector x = (0; 1; 0) :



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Eigenvectors for = 0 :

A I = A =

26640 0 1

0 1 2

0 0 1

3775 R1 $ R22664

0 1 2

0 0 1

0 0 1

3775 R3 R2 ! R32664

0 1 2

0 0 1

0 0 0

3775 (now in ech-elon form)

Solving

26640 1 2

0 0 1

0 0 0

37752664

x1

x2

x3

3775 =2664

0

0

0

3775 we have x3 = 0; x2 + 2x3 = 0 (and hence x2 = 0) whilex1 is free and we set x1 = t:


2664x1

x2

x3

3775 =2664

t

0

0

3775 = t2664

1

0

0

3775 ; that is, the non-zero multiplesof the vector x = (1; 0; 0) :

Consider the next example.

Example 7: Find the eigenvalues and corresponding eigenvectors for the matrix

A =

26640 0 0

0 1 0

1 0 1

3775

Solution: Characteristic polynomial is det(A I) =

0 00 1 01 0 1

= (1 )2

which has roots = 0 and = 1 (multiplicity = 2)

The eigenvectors:

= 1 :

A I = A I =

26641 0 00 0 0

1 0 0

3775 R3+ R1 ! R32664

1 0 00 0 0

0 0 0


Solving 2664 1 0 0

0 0 0

0 0 0

37752664x1

x2

x3

3775 = 26640

0

0

3775 we have x1 = 0; x2 and x3 are free so we set x2 = sand x3 = t:


2664x1

x2

x3

3775 =2664

0

s

t

3775 = s2664

0

1

0

3775+ t2664

0

0

1

3775 ; that is, the sums ofnon-zero multiples of (0; 1; 0) and (0; 0; 1) :

This time we do have 2 independent eigenvectors for = 1:



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= 0 :

A I = A =

26640 0 0

0 1 0

1 0 1

3775 R3 $ R12664

1 0 1

0 1 0

0 0 0


Solving 26641 0 1

0 1 0

0 0 0

37752664x1

x2

x3

3775 = 26640

0

0

3775 we have x2 = 0; and x1 + x3 = 0 with x3 as free.

We set x3 = t; giving

2664x1

x2

x3

3775 =2664

t0

t

3775 = t2664

10

1

3775 that is, the non-zero multiples of thevector x = (1; 0; 1) :5. Properties of eigenvalues

Let us assume that we have an n n matrix A with the eigenvalues 1; 2; 3; : : : n: (Notnecessarily distinct.)

Property 1: The sum of the eigenvalues ofA is equal to the sum of the elements of the diagonal

of A:nX1

i = 1 + 2 + ::: + n =nX1

aii

(The right-hand summation is known as the trace of A:)

Property 2: The product of the eigenvalues of A is equal to the determinant of A.

na1

i = 1 2 ::: n = det(A)

Property 3: The eigenvalues of the inverse matrix A1, provided they are non-zero, are:

1

1;

1

2; : : : ;

1

n

Property 4: The eigenvalues of the transposed matrix AT are the same as those of A.

Property 5: If k is a scalar then the eigenvalues of kA are simply k1; k2; k3; : : : ; k n:

Property 6: If k is a scalar and I is an n n identity matrix then eigenvalues of A kI arerespectively

1 k; 2 k; 3 k ; : : : n k:

Property 7: If k is a positive integer than the eigenvalues of Ak are

k1; k2;

k3; : : : ;

kn



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ENG1091 Further Calculus

Lecture 11 Implicit dierentiation Logarithmic dierentiationText Reference: 8.3.14&8.4

Functions such as f(x) = x sin x express f(x) explicitly in terms of x: Expressions of the form

x2 + y2 = 4; or 2y + x = 11; dene an implicit relationship between x and y:

Implicit dierentiation is a method of nding the derivatives of functions dened implicitly,

(indeed the expression may not even dene a function), without solving for the dependent variable

y:

Illustrative example

The implicit relation x2 + y2 = 4 is the equation of a circle, centre (0; 0), radius 2. Solving for

y gives y2 = 4 x2 and hence y = p4 x2 or y = p4 x2: The equation x2 + y2 = 4 thusrepresents two functions of x : y

1=

p4

x2

and

y2 = p

4 x2

.

Both these forms are readily dierentiated using the chain rule:

dy1dx

:

let u = 4 x2 dudx

= 2xdy1

dx=

dy1

du du

dx

=1

2u1=2 2x

= x 4 x21=2= x

y1

dy2dx

:

let u = 4 x2 dudx

= 2xdy2

dx=

dy2

du du

dx

= 12

u1=2 2x

= x

4 x21=2= x

y2

Note that both results can be expressed in terms ofy; indeed as the same expression. (dy2dx

= xy

:)

Examples

1. x2 + y2 = 4:

This equation implicitly denes y as a function of x; so we can dierentiate both sides with

respect to x :

2x + ddx

y2

= 0

2x + ddy

y2 dydx = 0 by the chain rule

2x + 2y dydx = 0

Now solve for dydx :

ANS:dy

dx= x

y: Compare this answer with that obtained in the illustrative example above



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2. The equation 5x2 6xy + 5y2 = 16 denes an ellipse, and while it is possible it is neitherdesirable nor necessary to nd y explicitly in terms of x:

-3 -2 -1 1 2 3

-3

-2

-1

1

Documents

ENG1091 Lectures Notes