FP1 matrices lesson 1

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Further Pure 1
Lesson 1 – Matrices
Working & Transformations
Definitions
Wiltshire
 A matrix is just a rectangle of numbers.
 It’s a bit like a two-way table.
 You met this concept briefly in D1.
 The matrix below shows how many arcs exist
between each node.
Definitions Definitions
Wiltshire
 Here are more examples of matrices.
 3  1

 = 2 × 2
2 5 
 2 
 
 0 
  3
 
 1 
 
=4×1
 2 3 6 2  3

 = 2 × 5
4  4 0 1 5 
 1 0 3 


 2 9 0  = 3 × 3
 5 2  7


 You can see that each matrix is a different size.
 The size (or order) of a matrix is given as rows × columns.
 What is the order of each of the above matrices?
Definitions
Wiltshire
 Here are two special types of matri that you need to
be familiar with.
 The identity matrix, I.
 1 0 0


 1 0
 I3   0 1 0 
I2  
 0 1
 0 0 1


 1 0 .. 0 


0
1
:


In  
:
:


 0 .. .. 1 


 The zero matrix, O.
0 0

O2  
0
0


0 0 0


O3   0 0 0 
0 0 0


 0 0 .. 0 


0
0
:


On  
:
:


 0 .. .. 0 


Definitions
Wiltshire
 Matrices with the same number of rows and
columns are known as square matrices.
 2 2


 1 6 
 Identity matrices are
always square as the
1`s on the diagonal
must run from corner
to corner.
 1 0 3 


 2 9 0 
 5 2  7


1

0
0

0

0
1
0
0
0
0
1
0
0

0
0

1 
Definitions
Wiltshire
 Two matrices are equal if:


They have the same order
Each element in one matrix is equal to the
corresponding element in the other matrix.
Using Matrices (+/-)
Wiltshire
 We can add and subtract matrices only if they have
the same order.
 All you do is add or subtract corresponding elements.
 2  2 1   2 0 0   4  2 1

  
  

 3 0  5   7  1 8  10  1 3 
1 
2  2 1  2 0 0  0  2

  
  

 3 0  5   7  1 8    4 1  13 
 Why can you not add matrices with different orders?
Using Matrices (×)
Wiltshire
 You can multiply a matrix by a number as
illustrated in the example below.
3 
2  2 1  6  6

  
3
 3 0  5   9 0  15 
 All that has happened is each element has been
multiplied by the number outside the matrix.
 In general for any 2 × 2 matrix.
 p q   ap aq 

  
a
 r s   ar as 
 Remember that this will work for any matrix of any
order.
Problems
Wiltshire
 Explain why matrix addition is


Commutative, i.e. A + B = B + A
Associative, i.e. A + (B + C) = (A + B) + C
 Addition in elements is both commutative and
associative.
 Do Ex 1A pg 3
Using Matrices (×)
Wiltshire
 Sometimes you can multiply two matrices together.
 However not all matrices can be multiplied together.
 Lets imagine 2 matrices called A and B.
 If we want to calculate A × B then A must have the
same number of columns as B has rows.
 The sum you do is multiply each element in the 1st
row of A by each element in the first column of B,
then add together your answers.
 You then do the same for all the row and column
combinations.
 On the next slide is a worked example.
Using Matrices (×)
Wiltshire
 Let A be a 2 × 3 matrix and B be a 3 × 2 matrix.
 a1 a2 a3 

A  
 a 4 a5 a 6 
 So A × B is given by
 a1 a2

 a 4 a5
 b1 b 4 


B   b 2 b5 
b b 
6
 3
 b1 b 4 
  a1b1  a2b2  a3b3
a3 
 b2 b5   
a6 
a 4b1  a5b2  a6b3


 b3 b 6 
a1b 4  a2b5  a3b6 

a 4b 4  a 5b 5  a 6b 6 
 Now take every element in the first row of A and multiply them
by every element in the first column of B, adding your answers.
 Now repeat with the 2nd row of A and 1st column of B.
 Next 1st row of A and 2nd column of B.
 Finally 2nd row of A and 2nd column of B.
Using Matrices (×)
Wiltshire
 Try the numerical example below.
2 3 1 

A  
  1 2  4
 So A × B is given by.
 1  2


B 2
0 
 4 5 


 1  2
 4
0 
 2 3 1 

 2

0   
  1 2  4   4 5  19  18 


 Now take every element in the first row of A and multiply them
by every element in the first column of B, adding your answers.
 Now repeat with the 2nd of A and 1st column of B.
 Next 1st row of A and 2nd column of B.
 Finally 2nd row of A and 2nd column of B.
Using Matrices (×)
Wiltshire
 What would happen if you found B × A.
2 3 1 

A  
  1 2  4
 So B × A is given by.
 1  2


B 2
0 
 4 5 


9 
 1  2
 4 1

 2 3 1  

   4
0 
6
2 
 2
  4 5   1 2  4    9  2  24 




 What do you notice about this answer compared to the
last?
 From these examples we can conclude that AB = BA
 So matrix multiplication is not commutative
Using Matrices (×)
Wiltshire
 If two matrices A and B have orders p × q and q × r
respectively then A × B does exist, and will have
order p × r.
p
q ×
q
r
=
p
 Note: In this case B × A does not exist.
q
p ×
r
q
r
Associative
Wiltshire
 Use the following matrices to show that matrix
multiplication is associative.
i.e. A(BC) = (AB)C
 2 1  3 0

A  
 4 0 4 3
0

3
B
4

 6

5 1

  2 4 0


 9 0
C   2 3 9

0 1
 0  5 1




2 6
Using Matrices (×)
Wiltshire
 Note that any matrix multiplied by the identity
matrix is itself.
 2 1  6  1 0 0   2 1  6 


 

  3 1 8  0 1 0     3 1 8 
 6 9 3  0 0 1   6 9 3 


 

 And any matrix multiplied by the zero matrix
is the zero matrix.
 2 1  6  0 0 0   0 0 0 


 

  3 1 8  0 0 0    0 0 0 
 6 9 3  0 0 0   0 0 0 


 

Questions
Wiltshire
 Try some of the following multiplications.
1   1 4    2 10 
 2


  


3
2
0
2
3

8


 

2  5   8 
 1 3

  

4  1    7 
1 0
 0
0
 2  3    6 

 3 


1  0   20 
6 1 0


    8 
2
7
2

3

4



 
 2 


1 5  1 5  4   10
 6


 

2
0
3
4

1
3


   2
 1
1  2  0 0  2   5

29
 10
4
 31

2 
3 
Example
Wiltshire
 Below is a league table for the group stage of the
world cup 2006.
 The top 2 teams in each group progress through to
the next round.
Team
MP
W
D
L
England
3
2
1
0
Paraguay
3
1
0
2
Sweden
3
1
2
0
3
0
1
2
Trinidad and
Tobago
 Use matrix multiplication to calculate the final points
and hence state who progressed through to the next
round.
Solution
Wiltshire
 We can write the league table as a matrix.
 Next we can add the
w d l
matrix that represents
E 2 1 0
7
the points awarded.

 3   
P  1 0 2    3 
 Its important to make
 1   


sure that the correct
S 1 2 0  
5

 0   
points line up with the

 1
T  0 1 2 
 
appropriate column.
 Now we can multiply the two matrices
together.
 This shows us that England and Sweden
progressed through to the next group.
Summary
Wiltshire
 Matrix addition is Commutative
 Matrix addition is Associative
 Matrix multiplication is not Commutative
 Matrix multiplication is Associative
 AI = A
 AO = O
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