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
]
