Lectures: on Wed at 12am (Chem LTA); on Thu
11am ( Chem LTB).
Lecturer: Dr Vladimir V Kisil, room 8.18L (Math).
Classes: run on even weeks (i.e., 2,
4 ...) Thu at 3pm
(Textile LT2). They are dedicated to homework and exams
questions.
Classes on odd weeks are shared by MATH1410 and MATH1610.
Attendance: will be collected during the lectures and
classes. A fact: absent from > 35% lectures
have 64% failure rate, in opposite to missing
< 15% lecture with only 9%!
Homework: 6 assignments during the
semester, contribute 15% toward the final mark.
Tutorials: (mainly) on Tuesdays 2pm, run by tutors by groups.
The subject of Linear Algebra is based on the study of systems of
simultaneous linear equations. As far as we are concerned there is
one basic technique - that of reducing a matrix to echelon form. I
will not assume any prior knowledge of such reduction nor even of
matrices. The subject has ramifications in many areas of science,
engineering etc. We shall look at only a minuscule part of it. We
shall have no time for "real" applications.
Why the name Algebra, by
the way?
Very often, in dealing with "real life" problems we find it
easier-or even necessary! -to simplify the problem so that it
becomes mathematically tractable. This
often means "linearising" it so that
it reduces to a system of (simultaneous) linear equations.
We shall deal with specific systems of linear equations in a
moment.
There are essentially three different possibilities which can arise when we
solve systems of linear equations. These may be illustrated by the following
example each part of which involves two equations in two unknowns. Since there
are only two unknowns it is more convenient to label them x and y − instead of
x1 and x2.
Example 1
The first case illustrated by the system:
2x+5y
=
3
(α)
3x−2y
=
14
(β)
To solve this system eliminate x from equation (β)
by replacing (β) by 2· (β)−3·(α), that is
(6x−4y)−(6x+15y)=28−9.
2x+5y
=
3
(α)
0x−19y
=
19
(γ)
So the given pair of equations are changed to
(α)-(γ). Equation (γ) shows
that y=−1 and then (α) shows that x=4. This
system is consistent and has the unique solution.
The second case:
6.8x+10.2y
=
2.72
(α)
7.8x+11.7y
=
3.11
(β)
Replace here (β) by
6.8·(β)−7.8·(α) gives the
following system:
6.8x+10.2y
=
2.72
(α)
0x+0y
=
0.068
(γ)
Then (γ) shows that there can be no such x and
y, i.e. no solution. This means that system is inconsistent.
The third case:
6.8x+10.2y
=
2.72
(α)
7.8x+11.7y
=
3.12
(β)
Replace here (β) again by
6.8·(β)−7.8·(α) gives the
following system:
6.8x+10.2y
=
2.72
(α)
0x+0y
=
0
(γ)
Here (γ) imposes no restriction on possible values of
x and y, so the given system of equations reduces to the
single equation (α).
Taking y to be any real
number you wish, say y=c, then (α) determines a
corresponding value of x, namely
x=
2.72−10.2·c
6.8
.
Thus the given system is consistent and has infinitely many solutions.
This three cases could be seen geometrically if we do three drawings:
Figure 1: Three cases of linear
systems considered in Example 1.1.
(i) Lines are in generic position and intersect in
one point-the unique solution.
(ii) Lines are parallel and
distinct-there is no solution.
(iii) Lines coinside-all points are solutions.
When there are more than two equations we try to eliminate even more
unknowns. A typical example would proceed as follows:
Example 2
(i) We get, successfully, the solution of the following system:
3x+y−z
=
2
(α)
x+y+z
=
2
(β)
x+2y+3z
=
5
(γ)
3x+y−z
=
2
(α)
0x−2y−4z
=
−4
(δ)=(α)−3(β)
0x+y+2z
=
3
(ε)=(γ)−(β)
3x+y−z
=
2
(α)
0x+y+2z
=
2
(ζ)=−
(δ)
2
0x+y+2z
=
3
(ε)
3x+y−z
=
2
(α)
0x+y+2z
=
2
(ζ)
0x+0y+0z
=
1
(η)=(ε)−(ζ)
The last equation η shows that the given system of equations
has no solutions. Could you imagine it graphically in a space?
(ii) Let consider the same system but change only the very last
number:
3x+y−z
=
2
(α)
x+y+z
=
2
(β)
x+2y+3z
=
4
(γ)
3x+y−z
=
2
(α)
0x−2y−4z
=
−4
(δ)=(α)−3(β)
0x+y+2z
=
2
(ε) = (γ)−(β)
3x+y−z
=
2
(α)
0x+y+2z
=
2
(ζ)=−
(δ)
2
0x+y+2z
=
2
(ε)
3x+y−z
=
2
(α)
0x+y+2z
=
2
(ζ)
0x+0y+0z
=
0
(η)=(ε)−(ζ)
This time
equation (η) places no
restrictions whatsoever on x,y and z and so can be ignored.
Equation (ζ) tells us that if we take z to have the
"arbitrary" value c, say, then y must take the
value 2 − 2z = 2 − 2c and then (β) tells us that x
must take the value 2 − y − z = 2 − (2 − 2c) − c = c.
That is, the general solution of the given
system of equations is: x = c, y = 2−2c, z = c, with c
being any real number. So solutions include, for example, (x,y,z) = (1,0,1), (3,−4,3),
(−7π,2−14π,−7π),….
The method of determining y from z and then x from y
and z is the method of back substitution.
The general system of m linear
equations in n "unknowns" takes the
form:
a11x1
+
a12x2
+
…
+
a1j xj
+
…
+
a1nxn
=
b1
a21x1
+
a22x2
+
…
+
a2j xj
+
…
+
a2nxn
=
b2
:
:
:
:
:
:
:
ai1x1
+
ai2x2
+
…
+
aij xj
+
…
+
ainxn
=
bi
:
:
:
:
:
:
:
am1x1
+
am2x2
+
…
+
amj xj
+
…
+
amnxn
=
bm
(L)
Remark 3
The aij and bi are given real
numbers and the
PROBLEM is to find all n-tuples
(c1,c2,…,cj,…,cn ) of real numbers
such that when the c1, c2, ..., cnare
substituted for the x1, x2, ..., xn, each of the
equalities in (L) is satisfied. Each such
n-tuple is called a solution of (the system) (L).
If b1 = b2 = … = bn = 0 we say
that the system
(L) is homogeneous.
Notice the useful double suffix notation in which the
symbol aij denotes the coefficient of xjin thei-th equation.
In this module the aij and the bj will always be
real numbers.
All the equations are linear. That is, in each term
aij xj, each xj occurs to the power exactly
1. (E.g.: no √{xj} nor products such as
xj2 xk are allowed.)
It is not assumed that the number of equations is equal to
the number of "unknowns".
(cf. Example 1.1.1). We easily obtain the
answer u = 4, v = −1 (cf. x = 4, y = −1 in
Example 1.1.1).
This shows that it is not important which letters are
used for the unknowns.
The important facts are what values the m×n coefficients aij and the m coefficients bj have. Thus we can abbreviate the
equations of Example 1.1.1
to the arrays
correspondingly.
Any such (rectangular) array (usually enclosed in brackets
instead of a box) is called a
matrix. More formally we give the following definition:
Definition 4
An array A of m×n numbers arranged in m rows and n columns is
called an m by n matrix (written
"m×n matrix").
Example 5
(
π
−[99/23]
8.1
0
e2
−700
) is a 2×3 matrix with a1,2=
−[99/23] and a2,1=0. What are a2,3 and
a3,2?