M215 Linear Algebra
April 2010 Presentation
Stop Press 2
22 June 2010
1. M215 Final Examination Date
The final exam will be held on 19 August 2010 and the Registry will send you a set of
exam documents. If you have not received any document regarding your

Solutions to the exercises
Solutions to the exercises
1.1 (a)
1.3 We use Strategy 1.1.
(a) First t(0) = 0, so t may be a linear
transformation.
Next we check whether t satises LT1:
t(v1 + v2 ) = t(v1 ) + t(v2 ), for all v1 , v2 R2 .
This is a (2, 3)-stret

Solutions to the exercises
Solutions to the exercises
1.1 (a) This is a linear equation.
(b) This is not a linear equation. The third term
involves the product of x3 and x4 .
(c) This is a linear equation.
(d) This is not a linear equation. For example

Unit LA1 Vectors and conics
Solutions to the exercises
1.1 Using the formula for the equation of a line
when given its gradient and one point on it, we nd
that the equation of this line is
y (1) = 3(x 2).
We can rearrange this in the form
y = 3x + 5.
1.2

Unit AA1 Numbers
Solutions to the exercises
1.1 Since 45 20 = 900 and 17 53 = 901, we have
1.7 (a) Putting all the fractions over the common
45/53 < 17/20. Thus
17
45
45
17
1 <
<
<0<
<
< 1.
20
53
53
20
denominator of 180, we obtain
3
7
1
11
7
<
< <
<
.
2

a %“i’ﬁki M215 Specimen
OF HONG KONG
M215 Linear Algebra
Time Allowed: 3 hours
THIS PAPER )ll'ST BE RETl'RNEI)
Admissible / Inadmissible materials in this examination: Material/Stationery Material to be
provided to returned to the
candidates invigilator

M215 TMA 01 solution
Cut-o date: 24 May 2010
Question 1
(a) Set up the equations
2y + x + 1 = 6
(x y) = 1
and obtain
x = 1; y = 2:
(b)
(i) Every point X on CQ has a position vector of the form
r = q + (1
) c;
2 R and 0
1:
where q is the position vector of

M215
Solutions to Specimen Examination Paper
This is a guide to the type of written solutions required. Some questions may
have alternative answers.
Part I
Question 1
(a) The position vectors c and d are given by
c = (3; 3) (2; 1) = (1; 2) ;
d = 9 (1; 2)

M215 Linear Algebra
April 2010 Presentation
Stop Press 1 (Letter to Students)
12 February 2010
Important note:
Please note that you will NOT be sent printed copies of any supplementary mailings. All
supplementary mailings such as stop presses, tutorial sc

Unit LA5 Eigenvectors
Solutions to the exercises
1.1 We have
t(2, 2) = (2 8, 2 + 4) = (6, 6)
= 3(2, 2)
and
t(7, 7) = (7 + 28, 7 14) = (21, 21)
= 3(7, 7).
In each case the original vector is scaled by the
factor 3.
Let k and l be real numbers which are no