Yang X.-S. Introductory Mathematics for Earth Scientists
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140 Chapter 10. Matrix Algebra
=
cos ψ cos θ − sin ψ sin θ −[cos ψ sin θ + sin ψ cos θ]
sin ψ cos θ + cos ψ sin θ cos ψ cos θ − sin ψ sin θ
, (10.27)
which is another way of deriving the sine and cosine of the addition of
two angles.
In a special case when first rotating by θ, followed by rotating back
by −θ, a point P (x, y) should reach its or iginal point. That is
x
y
=
1 0
0 1
x
y
= R
−θ
R
θ
x
y
, (10.28)
which means that
R
−θ
R
θ
=
1 0
0 1
= I. (10.29)
In other words, R
−θ
is the inverse of R
θ
. That is to say
R
−θ
=
cos(−θ) −sin(−θ)
sin(−θ) cos(−θ)
=
cos θ sin θ
−sin θ cos θ
, (10.30)
is the inverse of
R
θ
=
cos θ −sin θ
sin θ cos θ
. (10.31)
In genera l, the inverse A
−1
of a square matrix A, if it exis ts, is
defined by
A
−1
A = AA
−1
= I, (10.32 )
where I is a unit matrix which is the same size as A.
Example 10.1: For example, a 2 × 2 matrix A and its inverse A
−1
A =
a b
c d
, A
−1
=
α β
γ κ
,
can be related by
AA
−1
=
a b
c d
α β
γ κ
=
aα + bγ aβ + bκ
cα + dγ cβ + dκ
=
1 0
0 1
= I.
This means that
aα + bγ = 1, aβ + bκ = 0, cα + dγ = 0, cγ + dκ = 1.
These four e q u a tions will solve the four u n knowns α, β, γ and κ. After
some simple rearrangement and calculations, we have
α =
d
∆
, β =
−b
∆
, γ = −c∆, κ =
a
∆
,
10.2 Transformation and Inverse 141
where ∆ = ad−bc is the determinant o f A. Therefore, the inverse becomes
A
−1
=
1
ad − bc
d −b
−c a
.
It is straightforward to verify that A
−1
A =
1 0
0 1
.
In th e case of
R
θ
=
cos θ −sin θ
sin θ cos θ
,
we have
R
−1
=
1
cos θ co s θ − (−sin θ) sin θ)
cos θ sin θ
−sin θ cos θ
=
cos θ sin θ
−sin θ cos θ
,
where we have used cos
2
θ + sin
2
θ = 1. This is the same as (10.30).
We have see n that some special combinations o f the elements such
as the determinant ∆ = ad−bc is very important. We now try to define
it more generally.
The determinant of an n × n square matrix A = [a
i
j] is a numb e r
which can be obtained by cofactor expansio n either by row or by column
det(A) ≡ |A| =
n
X
j=1
(−1)
i+j
a
ij
|M
ij
|, (10.33)
where |M
ij
| is the cofactor or the determinant of a minor matrix M
of A, obtained by deleting row i and column j. This is a recur sive
relationship. For example, M
12
of a 3×3 matrix is obtained by deleting
the first row and the second column
a
11
− − a
12
|− −− a
13
− −
a
21
a
22
| a
23
a
31
a
32
| a
33
=⇒ |M|
12
=
a
21
a
23
a
31
a
33
. (10.34)
Obviously, the determinant of a 1 ×1 matrix |a
11
| = a
11
is the number
itself. The determinant of a 2 × 2 matr ix
det(A) =
a
11
a
12
a
21
a
22
= a
11
a
22
− a
12
a
21
. (10.35)
The determinant of a 3 × 3 matrix is given by det(A) or
a
11
a
12
a
23
a
21
a
22
a
23
a
31
a
32
a
33
= (−1)
1+1
a
11
a
22
a
23
a
32
a
33
+(−1)
1+2
a
12
a
21
a
23
a
31
a
33
+(−1)
1+3
a
13
a
21
a
22
a
31
a
32
= a
11
(a
22
a
33
− a
32
a
23
)
142 Chapter 10. Matrix Algebra
−a
12
(a
21
a
33
− a
31
a
23
) + a
13
(a
21
a
32
− a
31
a
22
). (10.36)
Here we used the expansion along the first row i = 1. We can also
expand it along any o ther rows or columns, and the results are the
same. As the determinant of a matrix is a scalar or a simple number,
it is not difficult to understand the following properties
det(AB) = det(A) det(B), det(A
T
) = det(A). (10.37)
There are many applications of the determinant. For e xample,
det(A) = 0, the square matrix is called singular, and the inverse of
such a matrix do e s not exist. The inverse of a matrix exists only if
det(A) 6= 0. Here we will use it to calculate the inverse A
−1
using
A
−1
=
adj(A)
det(A)
=
1
det(A)
B
T
, B =
(−1)
i+j
|M
ij
|
, (10.38)
where the matrix B
T
is called the adjoint of matrix A with the same
size as A, and i, j = 1, ..., n. Each of the element B is expressed in
terms of a cofactor so that b
ij
= (−1)
i+j
|M
ij
|. B itself is called the
cofactor matrix, while adj(A)=B
T
is s ometimes used to denote the
adjoint matrix. This seems too complicated, and let us compute the
the inverse of a 3 × 3 matrix as an example.
Example 10.2: In order to co m p u te the inverse of
A =
1 1 −2
1 0 2
2 1 1
,
we first construct its adjoint matrix B
T
with
B = [b
ij
] =
(−1)
i+j
|M
ij
|
.
The first element b
11
can be obtained by
b
11
= (−1)
1+1
0 2
1 1
= (−1)
2
× (0 × 1 − 2 × 1) = −2.
The element b
12
is
b
12
= (−1)
1+2
1 2
2 1
= −1 × (1 ×1 − 2 × 2) = 3,
while the element b
21
is
b
21
= (−1)
2+1
1 −2
1 1
= (−1)
3
× (1 × 1 − 1 × (−2)) = −3.
10.2 Transformation and Inverse 143
Following a similar procedure, we have B and its transpose B
T
as
B =
−2 3 1
−3 5 1
2 −4 −1
, or B
T
=
−2 −3 2
3 5 −4
1 1 −1
.
Then, the determinant of A is
det(A)=
1 1 −2
1 0 2
2 1 1
= 1 ×
0 2
1 1
−1 ×
1 2
2 1
+ (−2) ×
1 0
2 1
= 1 × (0 × 1 − 2 × 1) − 1 × (1 × 1 −2 × 2) − 2 × (1 ×1 −2 × 0)
= 1 × (−2) − 1 × (−3) − 2 × 1 = −1.
Finally, the inverse becomes
A
−1
=
B
T
det(A)
=
1
−1
−2 −3 2
3 5 −4
1 1 −1
=
2 3 −2
−3 −5 4
−1 −1 1
.
This result will be use d in the next example.
A linear system can be written as a large matrix equation, and
the solution of such a linear system will become straightforward if the
inverse of a square matrix is used. Let us demonstrate this by an ex-
ample. For a linear system consisting of three simultaneous equations,
we have
a
11
x+a
12
y+a
13
z = b
1
, a
21
x+a
22
y+a
23
z = b
2
, a
31
x+a
32
y+a
33
z = b
3
,
which can be written as
a
11
a
12
a
13
a
21
a
22
a
23
a
31
a
32
a
33
x
y
z
=
b
1
b
2
b
3
, (10.39)
or more compactly as
Au = b, (1 0.40)
where u =
x y z
T
. By multiplying A
−1
on both sides, we have
A
−1
Au = A
−1
b. (10.41)
Therefore, its solution can be written as u = A
−1
b.
Example 10.3: In order to solve the following system
x + y − 2z = −6, x + 2z = 8, 2x + y + z = 5,
144 Chapter 10. Matrix Algebra
we first write it as Au = b, or
1 1 −2
1 0 2
2 1 1
x
y
z
=
−6
8
5
.
We know from the earlier example that the inverse of A
−1
is
A
−1
=
2 3 −2
−3 −5 4
−1 −1 1
,
we now have u = A
−1
b or
x
y
z
=
2 3 −2
−3 −5 4
−1 −1 1
−6
8
5
=
2 × (−6) + 3 × 8 + (−2) × 5
−3 × (−6) + (−5) × 8 + 4 × 5
−1 × (−6) + (−1) × 8 + 1 × 5
=
2
−2
3
,
which gives a unique set of solutions x = 2, y = −2 and z = 3.
In general, a linear system of m equations for n unknowns can be
written in the compact form as
a
11
a
12
... a
1n
a
21
a
22
... a
2n
.
.
.
.
.
.
a
n1
a
n2
... a
nn
u
1
u
2
.
.
.
u
n
=
b
1
b
2
.
.
.
b
n
, (10.42)
or simply
Au = b. (1 0.43)
Its solution can be obtained by inverse
u = A
−1
b. (10.44)
You may wonder how you can get the inverse A
−1
of a large r ma-
trix A more efficiently. For large systems, direct inverse is no t a good
option. There are many other more efficient methods to obtain the
solutions, including the powerful Gauss-Jordan elimination, matrix de-
composition, and iteration methods. Interested r e aders can refer to
more advanced literature for details.
10.3 Eigenvalues and Eigenvectors
A specia l case of a linear system Au = b is when b = λu, and this
becomes an eigenvalue pro blem. An eige nvalue λ and corresponding
eigenvector u of a square matrix A satisfy
Au = λu, or , (A − λI)u = 0. (10.45)
10.3 Eigenvalues and Eigenvectors 145
Any nontrivial solution requires that
det |A − λI| = 0 , (10.46)
or
a
11
− λ a
12
... a
1n
a
21
a
22
− λ ... a
2n
.
.
.
.
.
.
a
n1
a
n2
... a
nn
− λ
= 0, (10.47)
which is equivalent to
λ
n
+ α
n−1
λ
n−1
+ ... + α
0
= (λ − λ
1
)(λ − λ
2
)...(λ − λ
n
) = 0. (10.48)
In general, the characteristic equation has n solutions. Eigenvalues
have interesting connections with the matrix. The trace of any square
matrix is defined as the sum of its diagonal elements, i.e.,
tr(A) =
n
X
i=1
a
ii
= a
11
+ a
22
+ ... + a
nn
. (10.49)
The sum of all the eigenvalues o f a square matrix A is equivalent to
the trace of A. That is
tr(A) = a
11
+ a
22
+ ... + a
nn
=
n
X
i=1
λ
i
= λ
1
+ λ
2
+ ... + λ
n
. (10.50)
In addition, the eigenvalues are als o related to the determinant by
det(A) =
n
Y
i=1
λ
i
. (10.51)
Example 10.4: For a simple 2 × 2 matrix
A =
1 5
2 4
,
its eigenvalues can be determine d by
1 − λ 5
2 4 − λ
= 0,
or
(1 − λ)(4 − λ) −2 ×5 = 0,
which is equivalent to
(λ + 1)(λ − 6) = 0.
146 Chapter 10. Matrix Algebra
Thus, the eigenvalues are λ
1
= −1 and λ
2
= 6. The trace of A is
tr(A) = a
11
+ a
22
= 1 + 4 = 5 = λ
1
+ λ
2
. In order to obtain the
eigenvector for each eigenvalue, we assume
v =
v
1
v
2
.
For the eige n valu e λ
1
= −1, we plug this into
|A − λI|v = 0,
and we have
1 − (−1) 5
2 4 − (−1)
v
1
v
2
= 0,
2 5
2 5
v
1
v
2
= 0,
which is equivalent to
2v
1
+ 5v
2
= 0, o r v
1
= −
5
2
v
2
.
This equation has infinite solutions; each corresponds to the vector parallel
to the unit eigenvector. As the eigenvector should be normalised so that
its mo d u lus i s unity, this additional condition requires
v
2
1
+ v
2
2
= 1,
which means
(
−5v
2
2
)
2
+ v
2
2
= 1.
We have v
1
= −5/
√
29, v
2
= 2/
√
29. Thus, we have the first set of
eigenvalue and eigenvector
λ
1
= −1, v
1
=
−
5
√
29
2
√
29
!
. (10.52)
Similarly, the second eigenvalue λ
2
= 6 gives
1 − 6 5
2 4 − 6
v
1
v
2
= 0.
Using the normalisation condi tion v
2
1
+ v
2
2
= 1, the above equation has the
following solution
λ
2
= 6, v
2
=
√
2
2
√
2
2
!
.
Furthermore, the trace and determinant of A are tr(A) = 1 + 4 =
5, and det(A) = 1 × 4 − 2 × 5 = −6. The sum of th e eige n valu e s i s
P
2
i=1
λ
i
= −1 + 6 = 5 = tr(A), while the prod u c t of the eigenvalues is
Q
2
i=1
λ
i
= −1 ×6 = −6 = det(A). Indeed, the above relationships about
eigenvalues are true.
10.4 Harmonic Motion 147
10.4 Harmonic Motion
Harmonic oscillations or mechanical vibrations occur in many processes
related to earth sciences. For e xample, the Earth itself has free oscilla-
tions. The natural freq uencies of a system can often be calculated using
eigenvalue methods be c ause the natural frequencies are the eigenvalues
if the system equations are formulated properly. Let us demonstrate
this using a simple system with three mass blocks connected by two
springs as shown in Figure 10.2. This system can be thought of as a
tectonic plate pushing two other plates, or a c ar towing two caravans
on a flat road, ignoring friction.
Let u
1
, u
2
, u
3
be the displacement of the three mass blocks m
1
, m
2
,
m
3
, respectively. Then, their accelerations will be ¨u
1
, ¨u
2
, ¨u
3
where
¨u = d
2
u/dt
2
. From the balance of forces and Newton’s law, we have
m
1
¨u
1
= k
1
(u
2
− u
1
), (10.53)
m
2
¨u
2
= k
2
(u
3
− u
2
) − k
1
(u
2
− u
1
), (10.54)
m
3
¨u
3
= −k
2
(u
3
− u
2
). (10.55)
These equations can be wr itten in a matrix form as
m
1
0 0
0 m
2
0
0 0 m
3
¨u
1
¨u
2
¨u
3
+
k
1
−k
1
0
−k
1
k
1
+ k
2
−k
2
0 −k
2
k
2
u
1
u
2
u
3
=
0
0
0
,
or
M ¨u + Ku = 0, (10.56)
where u
T
= (u
1
, u
2
, u
3
). The mas s matrix M and stiffness matrix K
are
M =
m
1
0 0
0 m
2
0
0 0 m
3
, K =
k
1
−k
1
0
−k
1
k
1
+ k
2
−k
2
0 −k
2
k
2
. (10.57)
Equation (10.56) is a second- order or dinary differential equation
in terms of matrices . We will learn more about ordinary differential
equations in the next chapter. At the moment, we assume that the
motion of our system is ha rmonic, therefore, we write their solution in
the form u = U cos(ωt) where U = (U
1
, U
2
, U
3
)
T
is a constant vector
related to the amplitudes of the vibrations. Here the unknown ω is the
natural frequency or frequencies of the system.
We know that
¨
u = −U ω
2
cos(ωt). Then, equation (10.56) becomes
(K −ω
2
M)U = 0. (10.58)
This is essentially an eigenvalue problem because any non-trivial solu-
tions for U require
|K − ω
2
M| = 0. (10.59)
148 Chapter 10. Matrix Algebra
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-
u
1
-
u
2
-
u
3
m
1
k
1
m
2
m
3
k
2
Figure 10.2: Harmonic vibrations and their natural frequencies.
Therefore, the eigenvalues of this equation give the natural frequencies.
Example 10.5: For the simple st case when m
1
= m
2
= m
3
= m and
k
1
= k
2
= k, we have
k − ω
2
m −k 0
−k 2k −ω
2
m −k
0 −k k − ω
2
m
= 0,
or
−ω
2
(k − ω
2
m)(3km − ω
2
m
2
) = 0.
This is a cubic equation in terms of ω
2
, and it has three solutions. There-
fore, the three natural frequencies are
ω
2
1
= 0, ω
2
2
=
k
m
, ω
2
3
=
3k
m
.
For ω
2
1
= 0, we h a ve U
T
= (U
1
, U
2
, U
3
) =
1
√
3
(1, 1, 1), which is the rigid
bod y motion. For ω
2
= k/m, the eig e n vector is determined by
0 −k 0
−k k −k
0 −k 0
U
1
U
2
U
3
=
0
0
0
,
which leads to U
2
= 0, and U
1
= U
3
. Written in normalised form, it
be c omes (U
1
, U
2
, U
3
) =
1
√
2
(1, 0, −1). This means that block 1 mo ves in
the opposite direction away from bl ock 3 , and block 2 remai n s stationary.
For ω
2
3
= 3k/m, we have (U
1
, U
2
, U
3
) =
1
√
6
(1, −2, 1). Tha t is to say,
block 2 moves i n the different directio n from block 3 whi c h is at the same
pace with block 1 .
Chapter 11
Ordinary Differential
Equations
Differential equations are very important in science, and many geo-
physical and geological processes can be modelled in terms of differen-
tial equations. In this chapter, we will introduce ordinary differential
equations and their basic solution techniques, and in the next chapter
will discuss the more complicated partial differential equations.
Differential equations have b e e n applied to almost every branch
of earth sciences. As illustrative ex amples for this chapter, we will
apply them to study the variations of air pressure with altitude, climate
changes, flexural deflection of the lithosphere, and post-glacier isosta tic
adjustment.
11.1 Differential Equations
In the introduction to basic equations such as x
3
− x
2
+ x − 1 = 0, we
know that the relationship is a function f (x) = x
3
− x
2
+ x − 1 and
the only unknown is x. The aim is to find values of x which satisfy
f(x) = 0. It is easy to verify that the equation has three solutions
x = 1, ±i.
A differential equation, on the other hand, is a relationship that
contains functions and their derivatives. For example, the following
equation
dy
dx
= x
3
− x, (11.1)
is a differential equation because it provides a relationship betwee n the
derivative dy/dx and the function f(x) = x
3
− x. The unknown is a
function y(x) and the aim is to find a function y(x) (not a s imple value)
which satisfies the above equation. Here x is the independent variable.
149