Image Processing Reference

In-Depth Information

·

=

If the product of two matrices equals the identity matrix,
A

B

I
, then we say

they are each other's
inverse
. This is denoted as
A
−
1

B
and
B
−
1

=

=

A
, or in other

A
−
1

A
−
1

words
A

·

=

·

A

=

I
.Fora2

×

2 matrix the inverse is calculated as

ac

bd

−
1

d

1

−

c

=

bc
·

(B.24)

−

ba

ad

−

Calculating the inverse for matrices of higher dimensions can be quite compli-

cated. For further information see a textbook on linear algebra.

B.7

Applying Linear Algebra

Say you want to find the equation of a straight line
y
=
αx
+
β
. You know that the

line passes through the point
P
1
(x, y)

β
. Obviously

this is not enough information to find
α
and
β
, or in other words we have one

equation and two unknowns
α
and
β
. So in order to solve the problem we need to

know the coordinates of one more point on the line or in other words we need two

equations to find two unknowns. Say that we then have another point on the line,

P
2
(x, y)
=
(
1
,
1
)
, yielding 1

=

(
2
,
3
)
,sowehave3

=

2
α

+

=
α
+
β
, we can solve the problem in the following

manner. From the last equation we can see that
α
=

1

−
β
. If we insert this into the

first equation we get 3

=

2
(
1

−

β)

+

β

⇔

β

=−

1 and from this follows that
α

=

2.

So the equation for the line is
y

1. This principle can be used to solve simple

problems where we have a few equations and a few unknowns. But imagine we have

10 equations with 10 unknowns; that would require quite an effort (and most likely

we would make mistakes along the way). Instead we can use linear algebra and get

the computer to help us.

Using linear algebra to solve these kinds of problem is carried out by arranging

the equations into the form:
a

=

2
x

−

·
c
, where
a
and
B
are known and
c
contains

the unknowns. The solution is then found by multiplying by the inverse of
B
:

=

B

a

·
c

=

B

⇔

(B.25)

a

·
c

B
−
1

B
−
1
B

=

⇔

(B.26)

a

·
c

B
−
1

=

I

⇔

(B.27)

a

=
c

B
−
1

(B.28)

For the example with the two lines we have

3

1

21

11

α

β

3

=

2
α

+

β

⇒

=

·

(B.29)

1

=

α

+

β

3

1

,

21

11

,

α

β

,

−

1

1

a

c

B
−
1

=

B

=

=

=

−

12

Using Eq.
B.28
we obtain the solution
c

T
.

=[

2

−

1

]