Tutorial on Number Systems
Q(1)
(a) Convert the following octal and hexadecimal numbers to binary numbers
And check your answers.
(i) 1238 , and (ii) AD.1C16 .
Confirm your answer in the decimal system.
Must remember the 3-bit code and the 4-bit code.
(i) 1238 
1
2
38
001
010
0112
1238  10100112
Check 1238  10100112  1 2 6  1 2 4  1 21  1 2 0 64  16  2  1  8310
while 1238  1  8 2  2  81  3  8 0  64  16  3  8310
(ii) AD.1C16 
=
A
1010
D
1101
.
.
1
0001
C16
1100
 AD.1C16  10101101.0001112 .
Check  AD.1C16  10101101.0001112
 1 2 7  1 25  2 3  2 2  2 0  2 4  2 5  2 6
 128  32  8  4  1  0.0625  0.03125  0.015625
 173.10937510
Also  AD.1C16  10  161  13  16 0  1  16 1  12  16 2
1 12
 160  13   2  173  0.0625  .046875
16 16
=173+0.109375
 173.10937510 which checks with above.
1
Q(2) Convert the octal number 74138 to hexadecimal and confirm your answer.
Solution
Must remember code
4bit Binary code
Decimal
0000
0
0001
1
0010
2
0011
3
0100
4
0101
5
0110
6
0111
7
1000
8
1001
9
1010
10—A
1011
11---B
1100
12---C
1101
13---D
1110
14---E
1111
15---F
Now 74138  111 100 001 0112 in groups of three bits.
 1111 0000 10112 in groups of four bits.
 F
0
B16 .
Check 74138  7  8 3  4  8 2  1  81  3  8 0
 3584  256  8  3  385110
while F 0 B16  F  16 2  0  161  B  16 0
 15  16 2  0  161  11  1  3840  0  11  385110 which checks.
Also note
111 100 001 0112
 1  211  1  210  1  2 9  1  2 8  0  2 7  0  2 6  0  2 5  0  2 4  1  2 3  0  2 2  1  21  1  2 0
 2048  1024  512  256  8  2  1
 385110
2
Alternative method (longer)
74138  7  8 3  4  8 2  1  81  3  8 0
 7  512  4  64  8  3
= 3584  256  11
 385110
Now
16 |3851
____
16| 240
____
16| 15
_____
16| 0
So 74138  F 0 B16
11 -- B
0
-- 0
15
--F
Thus 74138  385110  F 0 B16 .
3
Q3
(a)
Convert 31 and 46 to binary form.
(b)
Convert the following octal and hexadecimal numbers to
decimal.
(i) 603.258 (ii) 2FA.816
(c)
Convert the following binary number to octal and hexadecimal
forms respectively.
Check your answer by converting the 3 numbers to decimal.
1110100110
Solution (a)
2 31
1
2 15
1
2 7
1
2 3
1
2 1
0
1
So going backwards we get 3110  111112
Check 3110  1  2 4  1  2 3  1  2 2  1  21  1 2 0
 16  8  4  2  1
=31.
2 46
0
2 23
1
2 11
1
2 5
1
2 2
0
21
1
0
So going backwards we get 4610  101110 2
Check 4610  1  2 5  0  2 4  1  2 3  1  2 2  1  21  0  2 0
=32 +0+8+4+2+0=32+14=46
4
(b) 603.258  6  8 2  0  81  3  8 0  2  8 1  5  8 2
2 5
 384  0  3  
 391  0.25  0.078125
8 64
387.32812510 .
2FA.816  2  16 2  F  161  A  16 0  8  16 1
8
 512  15  16  10  1 
16
 512  240  10  0.5
 762.5
(c) 1110100110 2  1 110 100 110 2
1 6
4
6 8 remembering 3-bit code.
1110100110 2  0011 1010 0 110 2 inserting 2 zeros at beginning
3
A
616 remembering 4-bit code.
Confirmation
1110100110 2  1 2 9  1  2 8  1  2 7  0  2 6  1  2 5  0  2 4  0  2 3  1  2 2 
1  21  0  2 0 .
 512  256  128  0  32  0  0  4  2  0
 93410 .
1646 8  1  8 3  6  8 2  4  81  6  8 0  512  384  32  6  93410 .
3 A616  3  16 2  10  161  6  16 0  768  160  6  93410
which verifies are answers.
5
Tutorial on Matrices
Q(1)
 1 3
 2 1
 find AB, BA , A  B, and ( A  B) 2 .
 and B  
(a)If A  
  4 5
  2 4
2
2
Show that ( A  B)  A  B 2  AB  BA . Why is ( A  B) 2  A2  2 AB  B 2 ?
Solution
1  15    10 16 
 1 3  2 1   2  12

  
 = 
 .
Q(1)(a) AB  
  2 4   4 5    4  16  2  20    20 18 
6  4   0 10 
 2 1  1 3   2  2

  
  
 .
BA  

4
5

2
4

4

10

12

20

14
8


 
 

 1 3  2
  
A  B  
  2 4   4
3 4  3

 A  B 2  
  6 9   6
1  3 4

.
5    6 9 
4   9  24
12  36    15 48 
  
 

9    18  54  24  81   72 57 
3  12    5 15 
 1 3  1 3   1  6

  
  
 .
A 2  
  2 4   2 4    2  8  6  16    10 10 
25   0
7
 2 1  2 1   4  4

  
  
 .
B 2  
  4 5   4 5    8  20  4  25    28 21
7
  5 15    10 16   0 10   0
  
  
  

A 2  AB  BA  B 2  
  10 10    20 18    14 8    28 21
15  16  10  7    15 48 
  5  10  0  0
  

 
  10  20  14  28 10  18  8  21   72 57 
2
 A  B
Hence  A  B   A 2  AB  BA  B 2 and  A  B   A 2  2 AB  B 2 since AB  BA .
N.B. Matrix multiplication is non-commutative.
2
2
6
Q(2) (a)
 1 2 1
 3  2  1




Show that if A   2 3 1 then B    4 3
1  and hence
  2 0 1
 6  4  1




find (i) AB and (ii) BA
Q(2) Solution
 1 2 1 3  2  1



1
(i) AA   2 3 1  4 3
  2 0 1 6  4  1



1(2)  2(3)  1(4)
1(1)  2(1)  1(1) 
 1(3)  2(4)  1(6)


  2(3)  3(4)  1(6)
2(2)  3(3)  1(4)
2(1)  3(1)  1(1) 
  2(3)  0(4)  1(6)  2(2)  0(3)  1(4)  2(1)  0(1)  1(1) 


1
 3  8  6  2  6  4  1  2  1  1 0 0 

 

  6  12  6  4  9  4  2  3  1   0 1 0 
 6  0  6  4  0  4
2  1   0 0 1 

 3  2  1 1 2


BA    4 3
1  2 3
 6  4  1  2 0


 (3)1  (2)( 2)  (1)( 2) 3(2)  (2)(3)  (1)0

   4(1)  3(2)  1(2)
(4)2  3(3)  1(0)
 6(1)  (4)2  (1)( 2) 6(2)  (4)(3)  (1)0

1

1
1
66
3  2  1   1 0 0
 3 4 2

 

   4  6  2  8  9  4  3  1   0 1 0 
 6  8  2  12  12 6  4  1   0 0 1 

 

Here matrix multiplication is commutative.
7
3(1)  (2)1  (1)1 

(4)1  3(1)  1(1) 
6(1)  (4)(1)  (1)1
Q(3)
 4  5
1 0 




(a)If A   2
0  and B   0
7  evaluate
 6 3 
 2  5




(i) A  B , (ii) B  A , (iii) AB and (iv) 5B .
(b)Find AB where
1 3 3 
 7  3  3




A  1 4 3  and B    1 1
0 .
1 3 4 
 1 0
1 



 5 6
14 23 18 7 1 
 and P  
 find C  QP .
(c)Given the matrix Q  
 4 5
 5 16 15 18 13
Q(3)Solution
(a)
 4  5


A 2
0 
 6 3 


1 0 


B 0
7 
 2  5


 4  1  5  0  3  5

 

07   2
7 
(i) A  B   2  0
  6  2 3  5    4  2

 

  1  4 0  (5)    5 5 

 

7  0    2 7 
(ii) B  A   0  2
 2  (6)  5  3   8  8 

 

(iii) AB is not possible since the order of A is 3 2 while the order of B is
3 2 that is the number of columns of A  number of rows of B .
0 
 1 0    5

 

7  0
35 
(iv) 5B  5 0
 2  5   10  25 

 

8
1 3 3  7  3  3   7  3  3  3  3  0  3  0  3 


 

(b) AB  1 4 3   1 1
0  =  7  4  3  3  4  0  3  0  3
1 3 4   1 0
1   7  3  4  3  3  0  3  0  4 


 1 0 0


=  0 1 0 .
 0 0 1


.
(c)
matrix C  QP is given by
 5 6 14 23 18 7 1 


C  QP  
 4 5  5 16 15 18 13
 5(14)  6(5) 5(23)  6(16) 5(18)  6(15) 5(7)  6(18) 5(1)  6(13) 

= 
 4(14)  5(5) 4(23)  5(16) 4(18)  5(15) 4(7)  5(18) 4(1)  5(13) 
100 211 180 143 83 
 .
C  
 81 172 147 118 69 
9
Q(4)
 1 2 3
7 8 9
7 8 
 , B  
 and C  

(a)If A  
 4 5 6
10 11 12 
 9 10 
compute (i) A  B , (ii) A  C and (iii) 3B .
 1 7
4 0 
 2 
 , E  
 , F    and G  3  9
(b) If D  
  5 11
 2  1
  8
compute (i) DE , (ii) FD and (iii) GF .
 3 13 9 5 
6 7
 and the matrix H  

P  
15 16 12 27 
5 6
find C  HP .
(c) If
10
 1 2 3  7 8 9    6  6  6
  
  

 4 5 6  10 11 12    6  6  6 
Q(4)Solution (a) (i) A  B  
(ii) A  C is not possible as A is a 2 3 matrix while C is a 2  2 matrix.
 21 24 27 
 .
(iii) 3B  
 30 33 36 
1(0)  7(1)  18  7 
 1 7  4 0   1(4)  7(2)

  
  
 .
(b) (i) DE  
  5 11 2  1  4(5)  11(2)  5(0)  11(1)   2  11
(ii) FD is not possible since F is a 2  1 matrix and D is a 2  2 matrix.
 2
(iii) GF  3  9   3(2)  (9)( 8)   6  72  78 .
  8
(c)
 3 13 9 5 
 .
P  
15 16 12 27 
we get
 6 7  3 13 9 5  123 190 138 219 

  
 .
C  HP  
 5 6 15 16 12 27  105 161 117 187 
11
Q(5)Consider the following matrices
2 1
3 2
 a 2b 
 , B  
 , C  
 ,
A  
 3 2
1 4
 3c d 
a
 x
D    , E  x y , F    ,
 y
b
2 4 1


 2x y 
 2 3 4
 and J  
 .
G   3 2 4  , H  
y
z
1
4
2




 1 3 2


(a) If possible perform the following calculations
(i) A  C , (ii) J  C , (iii) C  D , (iv) A  H ,
1
(v) 2C  B , (vi) G .
2
(b) If possible perform the following calculations
(i) AC , (ii) JC , (iii) CJ , (iv) EB and (v) BE .
12
Solution
Q(5)
Solution
 2 1   a 2b   2  a 1  2b 
  
  
 .
(a) (i) A  C  
3
2
3
c
d
3

3
c
2

d

 
 

(ii) J  C is not possible since order of matrix J is 2 3 while C is of order
2 2.
(iii) C  D is not possible since order of matrix C is of order
2  2 while D is 2  1 matrix.
 2 1   2x y   2  2x 1  y 
  
  

(iv) A  H  
 3 2  y z   3  y 2  z 
 a 2b   3 2   2a  3 4b  2 
  
  
 .
(v) 2C  B  2
 3c d   1 4   6c  1 2d  4 

1
 2 4 1 

1
1
3
(vi) G   3 2 4   
2
2
2

 1 3 2  1

2
1

2
2.


1

2
1
3
2
 2 1  a 2b   2a  3c 4b  d 

  
 .
(b)(i) AC  
 3 2  3c d   3a  6c 6b  2d 
(ii) JC is not possible since J is a 2 3 while C is a 2  2 matrix.
 a 2b  2 3 4   2a  2b 3a  8b 4a  4b 

  
 .
(iii) CJ  
 3c d  1 4 2   6c  d 9c  4d 12c  2d 
 3 2
  3x  y 2 x  4 y  .
y 
1 4
(v) BE is not possible since B is a 2  2 matrix while E is a 1 2 matrix.
(iv) EB  x
13
Tutorial
on Matrices or Arrays-Storing and manipulating data
Example 1 A distributor records the weekly sales of personal computers(PC’s) in three
retail outlets in different parts of the country and given in Table below.
Shop A
Shop B
Shop C
Number of computers sold in each shop
Pentium(basic)
Pentium(special)
150
320
170
420
201
63
Pentium(latest)
180
190
58
The cost price of each model is Pentium(basic) €480, Pentium(special) €600 and
Pentium(latest) €1020.
The retail price of each model in each of the three shops is given in Table below.
Shop A
Shop B
Shop C
Selling Price of computers in each shop
Pentium(basic)
Pentium(special)
560
750
520
690
590
720
Pentium(latest)
1580
1390
1780
Use matrix multiplication to calculate
(a) Total weekly cost of computers to each shop.
(b) Total weekly revenue of each model for each shop
(c) Total weekly profit for each shop.
Whish shop makes the greatest overall profit?
Solution
(a) The number sold may be written in a matrix Q
150 320 180 


Q  170 420 190 
 201 63 58 


and the cost of each type of computer ca be written as a column matrix C
 480 


C   650 
1020 


14
150 320 180   480 



Hence Total cost  QC  170 420 190   650 
 201 63 58  1020 



150  480  320  600  180  1020   447600 

 

 170  480  420  600  190  1020    527400 
 201  480  63  600  58  1020   193440 

 

Total cost of computers in Shop A =£447600,
Total cost of computers in Shop B =£527400,
Total cost of computers in Shop C =£193440.
(b) Total revenue=price  quantity
Quantities are given by matrix Q and prices from Table . Matrix multiplication is
carried out by muptiplying rows  columns.Hence to multiply quantity  price for each
computer, the rows of table must be written as columns in a price matrixThat is the price
matrix must be transposed before premultiplying by the quantities matrix Q .
T
 560 520 590 
 560 750 1580 




P   520 690 1390    750 690 720 
 590 720 1780 
1580 1390 1780 




Multipy Q by P
150 320 180   560 520 590 



QP  170 420 190   750 690 720 
 201 63 58  1580 1390 1780 



Shop A
Shop B
Shop C
NA
NA
150  560  320  750  180  1580



NA
170  520  420  690  190  1390
NA



NA
NA
201  590  63  720  58  1780 

The total revenue for each shop is given by the elements on the main diagonal.
15
 84000  240000  284000 
 604400 




Total Revenue   88400  289800  264100    642300 
 118590  45360  103240 
 267190 




 604400   447600  156800 

 
 

(c) Profit=Total Revenue-Total Cost=  642300    527400   114900  .
 267190   193440   73750 

 
 

Shop A has the largest profit.
16
Tutorial on functions
Exercise Let f : (, )  (, ) where f is given by 28x-7y=-70. Is this a function?
Draw its graph and state its domain, co-domain and range.
Solution Rewriting 28 x  7 y  70  7 y  28 x  70  y  4 x  10 .
Its domain is (, ) ,its co-domain is (, ) and its range is (, ) .
Yes it is a function as every real number in its domain (, ) is mapped to an element
of its co-domain (, ) e.g.  x  (, ) is mapped to real number
y  4 x  10 (, ) .
Its domain is (, ) ,co-domain (, ) and range (, ) .
Its graph is given below.
Graph of linear function y=4x+10
20
15
10
4x 10
5
0
5
 10
 10
5
0
5
10
x
17
15
20
Exercise Consider the function f : 1,2,3  a, b, cgiven by f (1)  a, f (2)  b and
f (3)  c . Draw its arrow diagram and state its domain, codomain and range.
Is it (i) onto and (ii) one-to-one ? Explain. Does it have an inverse f 1 and if so
what is it ?
Solution
Arrow diagram of f
1 a
2 b
3c
domain  1,2,3 codomain  a, b, c range  a, b, c.
f is onto since rangea, b, c = codomaina, b, c ,
It is one-to-one as no two distinct elements of the domain have the same image,since
f (1)  a,
f (2)  b
and f (3)  c .
Now since f is (i) onto and (ii)one-to-one then by theorem by the above function
f has an inverse f 1 and it is given by f 1 : a, b, c  1,2,3where
f 1 (a)  1
f 1 (b)  2
f 1 (c)  3 .
You just reverse the process.
18
Exercise Are the following
(i) f : A, B, C  11,12,13where f ( A)  11, f ( B)  13
(ii ) g : A, B, C  11,12,13where g ( A)  11, g ( B)  13, g (C )  13
(iii) h : A, B, C  11,12,13where h( A)  11, h( B)  13, h(C )  12
functions. Give your reasons. For the ones that are functions give their
domain, codomain and range. Are they onto and one-to-one .Do the have an
inverse and if so what is it.
Solution (i) f is not a function since not every element of the domain is
mapped to an element of the codomain, for example C is not mapped to any
element. By definition for a function to be a function every element must be
mapped to an unique element of the codomain{Note two elements of the domain
can be mapped to the same element of the codomain)
(ii) g is a function since every element of the domain A, B, Cis mapped to an
element of the codomain 11,12,13.
The domain is A, B, C, the codomain 11,12,13and the range 11,13.
g is not onto since range 11,13  codomain 11,12,13.
g is not one-to-one since two distinct elements of the domain have the same
image, namely g ( B)  13, g (C )  13 .
Since neither g is not onto nor g is not one-to-one then by theorem
g has no inverse. (Note if a function has not got one of these properties
then it will not have an inverse.)
(iii) h is a function since every element of the domain A, B, Cis mapped to an
element of the codomain 11,12,13.
The domain is A, B, C, the codomain 11,12,13and the range 11,12,13 .
h is onto since range 11,12,13  codomain 11,12,13.
h is one-to-one since no two distinct elements of the domain have the same
image, namely h( A)  11, h( B)  13, h(C )  12 .
Since h is onto nor h is one-to-one then by theorem
h has an inverse given by h 1 : 11,12,13  A, B, C where
h 1 (11)  A
h 1 (13)  B
h 1 (12)  C .
19