A Fast Introduction
What is a matrix ------
(One Matrix many matrices)
Why do they exist
Matrix Terminology
• Elements
• Rows
• Columns
Square Matrix
Adding/Subtracting
Multiplying/ Dividing (Divisions are Multiplications)
The Inverse Matrix (equivalent to 1.0)
inverse matrix
1
Inverse of a Matrix
Solving simultaneous equations
Today’s Goals
—To be able to find the inverse of a Matrix A ~ A’
—To use this for solving simultaneous equations
Recap
The identity Matrix is the equivalent of the number
1 in matrix form
1 0  1

 
0
1

 3
2

4

1 2 


2
4


1

3
inverse matrix
2

4
1 0 


0
1



1 2 


2
4


3
Definition of Inverse
The inverse is Defined as
1

3
2

4
?

?
?

?

AA-1 = A-1A = I
1 0 


0
0


?

?
?

?
1

3
2

4

1 0 


0
0


We will find out how to calculate the inverse for 2x2 matrix
But first why is it important ?
Because it will allow us to solve equations of the form
 a 11

 a 21
a 12   x 1 
 .

a 22   x 2 
 b1 
 

b
 2
inverse matrix
We will only consider 2x2
Matrix systems
That means simultaneous
equations
4
Why will it help us solve equations?
Because if we can express a system of equations in the
form
Ax  b
Then we can multiply both sides by the inverse matrix
A
1
1
Ax  A b
And we can then know the values of X because
A
1
A I
1
x  A b
inverse matrix
5
Recap Multiplication

 4

 6
 6
 1
 5
 4
3
0

2

1
3   2
2   
7   3
 6
2 
 5
 2 

 =
  1
 6 


=
1 
4 
=
 8+ 3+ 12


-12+0+ 6
 12+ 9 
 2+ 6 
 10+ 21 
 24+10
inverse matrix
4+8 
=
 23 
 
 -6 
=
 21
 8

 31




=
 34
12 
6
Shapes and sizes
• Dimensions and compatibility given by the
domino rule
23

31
21
To multiply the columns of the
first must be equal to the rows
of the second
The dimensions of the
result are given by
the 2 outer numbers
Note matrix multiplication is not commutative.
If A is a 3x1 and B is a 1x3 then AB is 3x3 BA is 1x1
inverse matrix
The multiplicative inverse of a matrix
• This can only be done with SQUARE matrices
• By hand we will only do this for a 2x2 matrix
• Inverses of larger square matrices can be calculated but
can be quite time expensive for large matrices, computers
are generally used
Ex A =
4

1
4

1
8

3
8   0 . 75
 
3    0 . 25
then A-1 =
 0 . 75

  0 . 25
 2

1 
 2 
32
  
1   0 . 75  0 . 75
inverse matrix
as AxA-1 = I
 8  8 1
  
 2  3  0
0

1
8
a

c
Finding the Inverse of a 2x2 matrix
b

d
Step-1 First find what is called the Determinant
This is calculated as ad-bc
Step-2 Then swap the elements in the leading diagonal
Step-3 Then negate the other elements
d

c
b

a
d  b 


 c a 
Step-4 Then multiply the Matrix by 1/determinant
d  b 



c
a
ad  cb 

1
inverse matrix
9
Example Find Inverse of A
Step 1 – Calc Determinant
4
A  
1
8

3
Determinant (ad-cb) = 4x3-8x1 = 4
Step 2 – Swap Elements on leading diagonal
step2
3

1

8

4 
Step 3 – negate the other elements
step3
 3

1

Step 4 – multiply by 1/determinant
step4
1 3

4   1
8

4 
8

4 
check
A
1
 0 . 75
 
  0 . 25
 2

1
AA
1
4 8
 

1 3 

inverse matrix
 0 . 75

  0 . 25
3  2

 0 . 75  0 . 75
 2

1
 8  8

23

1 0 


 0 1
10
Find the inverses and check them
2
A  
1
5
B  
 1
2
C  
0
6

5
20 

2 
2 

 1
A
1 5
 
4 1
1
B
C
1
1
 6   1 . 25
  
2    0 . 25
1 2


10  1
1 1


2 0
inverse matrix
 20   0 . 2
  
 5   0 .1
 2   0 .5
  
2   0
 1 .5 

0 .5 
2 

 0 .5 
1 

 1
11
More inverses to find and check
2
A  
1
4

3
A
 5 10 

B  
1 2 
3
C  
2
 8
D  
1
 2
E  
1
1
F  
 0
C
2

0
 8

4 
0 

 1
1
B
2

1
1
D
E
1
1
1
 4   1 .5
  
2    0 .5
1 3
 
2 1

 10   0 . 1
  
5   0 . 05
1 2

20  1
1  1


1 2
1 0
 
2 1
 2 1
  
3   2
 2  0
  
8   0 .5
 2

1 
 0 .5 

0 . 25 
2 

 3
 1

4 
cannot be found as det  2  4  (  1   8 )  8  8  0
F
inverse matrix
1
11
 
1 0
0  1
  
 1  0
0 
 self inverting
 1
12
Applications of matrices
• Because matrices are clever storage systems for
numbers there are a large and diverse number of
ways we can apply them.
• Matrices are used in to solve equations on
computers
- solving equations
• They are used in computer games and multi-media
devices to move and change objects in space
- transformation geometry
• We only consider solving equations on Maths1 with
using 2x2 matrices
inverse matrix
13
Solving simultaneous equations
•
•
•
We can use our 2x2 matrices to express 2
simultaneous equations (2 equations about the same
2 variables)
First we must put them in the correct format
for the variables x & y the format should be
ax + by = m
cx + dy = n {where a,b,c,d,m & n are constants}
Example
Peter and Jane spend £240 altogether and Peter spends 3 times as
much as Jane.
let p: what Peter spends and j: what Jane spends
then p + j = 240
p=3j
rewrite p-3j = 0
(right format a and b = 1 m = 240)
(wrong format)
(right format c = 1 d = -3 n = 0)
inverse matrix
14
Solving simultaneous equations
We can use our 2x2 matrices to express these
simultaneous equations
x y
 240 

0

x  3y
1

1

1  x1

 3   x 2
constants
from the
left hand
side
Becomes in matrix form
  240 


  0 

 
UNKNOWNS
X ~ x1
Y ~ x2
inverse matrix
constants
from the
right hand
side
15
1

1

1  x1

 3   x 2
  240 


  0 

 
Format is Ax=B
To solve this using the matrix we must get rid of it by using its
inverse!
First find the inverse
1

1
1 

 3
1
1 3


 4  1
 1   0 . 75
  
1   0 . 25
0 . 25 

 0 . 25 
now use it on both sides of the equation
 0 . 75

 0 . 25

0 . 25   1

 0 . 25   1
1  x1

 3   x 2
1

0

  0 . 75

  0 . 25
 
0  x1

1   x 2
  0 . 75

  0 . 25
 
 x1

x
 2
  180 


  60 

 
0 . 25   240

 0 . 25   0




0 . 25   240

 0 . 25   0




So Answer is p = 180 j = 60
inverse matrix
16
Summary of method
1.
Format the simultaneous equations for variable x & y
2.
Rewrite them in matrix form
3.
Find the inverse of the 2x2 matrix
4.
Solve for the variables x,y by multiplying the right hand
side of the equation by the inverse
ax + by = m
cx + dy = n
a

c
b  x   m 
     
d  y   n 
a

c
b

d
1
 d


ad  bc   c
1
x
 d
1
  

 y  ad  bc   c
inverse matrix
 b

a 
 b  m 
  
a  n 
17
Your Turn solve the following
3x +4y = 5
5x = 7-6y
x+7y = 1.24
3y -x = 0.76
8x = 3y -1
x+y =-7
3

5
4  x   5 
     
6  y   7 
x
1  6
  

y  2 5
 1

1
x
1 3
  

 y  10  1
8

1
Answer x = -1 y = 2
 4  5 
1  30  28 
1  2    1
   

 

  

3   7   2   25  21   2   4   2 
7   x   1 . 24 
    

3   y   0 . 76 
Answer x = -0.16 y = 0.2
 7   1 . 24 
1  3 . 72  5 . 32 
1   1 . 6    0 . 16 
 
 

 

  

1   0 . 76  10  1 . 24  0 . 76  10  2   0 . 2 
 3  x    1 
    

1  y    7 
x
1  1
  

 y  11   1
Answer x = -2 y = -5
3   1 
1   1  21 
1   22    2 
 
 

 

  

8    7  11  1  56  11   55    5 
inverse matrix
18