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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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MONASH UNIVERSITY SCHOOL OF MATHEMATICAL SCIENCES
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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MONASH UNIVERSITY SCHOOL OF MATHEMATICAL SCIENCES
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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MONASH UNIVERSITY SCHOOL OF MATHEMATICAL SCIENCES
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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MONASH UNIVERSITY SCHOOL OF MATHEMATICAL SCIENCES
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
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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
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MONASH UNIVERSITY SCHOOL OF MATHEMATICAL SCIENCES
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
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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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MONASH UNIVERSITY SCHOOL OF MATHEMATICAL SCIENCES
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 :
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(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
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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
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MONASH UNIVERSITY SCHOOL OF MATHEMATICAL SCIENCES
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
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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
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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:
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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
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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
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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
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MONASH UNIVERSITY SCHOOL OF MATHEMATICAL SCIENCES
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.
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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
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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
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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:
Thus the eigenvectors for = 0 are
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
3775 which is in echelon form.
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:
Thus the eigenvectors for = 1 are
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
3775 which is in echelon form.
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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MONASH UNIVERSITY SCHOOL OF MATHEMATICAL SCIENCES
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