Handbook of mathematics for engineers and scienteists part 34 pdf

5.5. SYSTEMS OF LINEAR ALGEBRAIC EQUATIONS 199
5.5.2-2. General square system of linear equations.
A square system of linear equations has the form
AX = B,(5.5.2.2)
where A is a square matrix.
1
◦
. If the determinant of system (5.5.2.2) is different from zero, i.e. det A ≠ 0, then the
system has a unique solution,
X = A
–1
B.
2
◦
. Cramer rule. If the determinant of the matrix of system (5.5.2.2) is different from zero,
i.e. Δ =detA ≠ 0, then the system admits a unique solution, which is expressed by
x
1
=
Δ
1
Δ
, x
2
=
Δ
2
Δ
, , x
n
=

Δ
n
Δ
,(5.5.2.3)
where Δ
k
(k = 1, 2, , n) is the determinant of the matrix obtained from A by replacing
its kth column with the column of free terms:
Δ
k
=








a
11
a
12
b
1
a
1n
a
21
a

22
b
2
a
2n
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
a
n1
a
n2
b
n
a

nn








.
Example 1. Using the Cramer rule, let us find the solution of the system of linear equations
2x
1
+ x
2
+ 4x
3
= 16,
3x
1
+ 2x
2
+ x
3
= 10,
x
1
+ 3x
2
+ 3x

3
= 16.
The determinant of its basic matrix is different from zero,
Δ =





214
321
133





= 26 ≠ 0,
and we have
Δ
1
=





1614
1021
1633






= 26, Δ
2
=





2164
3101
1163





= 52, Δ
3
=





2116

3210
1316





= 78.
Therefore, by the Cramer rule (5.5.2.3), the only solution of the system has the form
x
1
=
Δ
1
Δ
=
26
26
= 1, x
2
=
Δ
2
Δ
=
52
26
= 2, x
3
=

Δ
3
Δ
=
78
26
= 3.
3
◦
. Gaussian elimination of unknown quantities.
Two systems are said to be equivalent if their sets of solutions coincide.
The method of Gaussian elimination consists in the reduction of a given system to an
equivalent system with an upper triangular basic matrix. The latter system can be easily
solved. This reduction is carried out in finitely many steps. On every step, one performs an
elementary transformation of the system (or the corresponding augmented matrix) and ob-
tains an equivalent system. The elementary transformations are of the following three types:
1. Interchange of two equations (or the corresponding rows of the augmented matrix).
2. Multiplication of both sides of one equation (or the corresponding row of the augmented
matrix) by a nonzero constant.
3. Adding to both sides of one equation both sides of another equation multiplied by a
nonzero constant (adding to some row of the augmented matrix its another row multiplied
by a nonzero constant).
200 ALGEBRA
Suppose that det A ≠ 0. Then by consecutive elementary transformations, the augmented
matrix of the system A
1
[see (5.5.1.4)] of size n × (n + 1) can be reduced to the form
U
1
≡

⎛
⎜
⎜
⎝
1 u
12
··· u
1n
y
1
01··· u
2n
y
2
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
00··· 1

y
n
⎞
⎟
⎟
⎠
and one obtains an equivalent system with an upper triangular basic matrix,
x
1
+ u
12
x
2
+ u
13
x
3
+ ···+ u
1n
x
n
= y
1
,
x
2
+ u
23
x
3

+ ···+ u
2n
x
n
= y
2
,

x
n
= y
n
.
This system is solved by the so-called “backward substitution”: inserting x
n
= y
n
(obtained
from the last equation) into the preceding (n – 1)st equation, one finds x
n–1
. Then inserting
the values obtained for x
n
, x
n–1
into the (n – 2)nd equation, one finds x
n–2
. Proceeding in
this way, one finally finds x
1

. This back substitution process is described by the formulas
x
k
= y
k
–
n

s=k+1
u
ks
x
s
(k = n – 1, n – 2, , 1).
Suppose that det A = 0 and rank (A)=r, 0 < r < n. In this case, the system is
either inconsistent (i.e., has no solutions) or has infinitely many solutions. By elementary
transformations and, possibly, reindexing the unknown quantities (i.e., introducing new
unknown quantities y
1
= x
σ(1)
, , y
n
= x
σ(n)
,whereσ(1), , σ(n) is a permutation of
the indices 1, 2, , n), one obtains a system of the form (for the sake of brevity, we retain
the notation x
j
for the reindexed unknown quantities)

c
11
x
1
+ ···+ c
1r
x
r
+ c
1,r+1
x
r+1
+ ···+ c
1n
x
n
= d
1
,

c
rr
x
r
+ c
r,r+1
x
r+1
+ ···+ c
rn

x
n
= d
r
,
0 = d
r+1
,

0 = d
n
,
where the matrix [c
ij
](i, j = 1, 2, , r)ofsizer × r is nondegenerate. If at least one
of the right-hand sides d
r+1
, , d
n
is different from zero, then the system is inconsistent.
If d
r+1
= = d
n
= 0, then the last n – r equations can be dropped and it remains to find all
solutions of the first r equations. Transposing all terms containing the variables x
r+1
, , x
n
to the right-hand sides and regarding these variables as arbitrary free parameters, we obtain

a linear system for the unknown quantities x
1
, , x
r
with the nondegenerate basic matrix
[c
ij
](j, j = 1, 2, , r).
Example 2. Let us find a solution of the system from Example 1 by the Gaussian elimination method.
By elementary transformations of the augmented matrix, we obtain
⎛
⎝
214
16
321
10
133
16
⎞
⎠
→
⎛
⎝
11/22
8
01/2 –5
–14
05/21
8
⎞

⎠
→
⎛
⎝
11/22
8
01–10
–28
00 26
78
⎞
⎠
→
⎛
⎝
11/22
8
01–10
–28
00 1
3
⎞
⎠
.
5.5. SYSTEMS OF LINEAR ALGEBRAIC EQUATIONS 201
The transformed system has the form
x
1
+
1

2
x
2
+ 2x
3
= 8,
x
2
– 10x
3
=–28,
x
3
= 3.
Hence, we find that
x
3
= 3, x
2
=–28 + 10x
3
= 2, x
1
= 8 –
1
2
x
2
– 2x
3

= 1.
4
◦
. Gauss-Jordan elimination of unknown quantities.
This method consists of applying elementary transformations for reducing a system with
a nondegenerate basic matrix to an equivalent system with the identity matrix. On the kth
step (k = 1, 2, , n) the rows of the augmented matrix A

1
obtained on the preceding step
can be transformed as follows:
a

kj
=
a

kj
a

kk
, b

k
=
b

k
a


kk
(j = k, k + 1, , n),
a

ij
= a

ij
– a

ik
a

kj
a

kk
, b

i
= b

i
– a
ik
b

k
a


kk
(i = 1, 2, , n, i ≠ k, j = k, k + 1, , n),
provided that the diagonal element obtained on each step is not equal to zero. After n steps,
the basic matrix is transformed to the identity matrix and the right-hand side turns into the
desired solution.
Example 3. For the linear system from Examples 1 and 2 we have
⎛
⎝
214
16
321
10
133
16
⎞
⎠
→
⎛
⎝
11/22
8
01/2 –5
–14
05/21
8
⎞
⎠
→
⎛
⎝

10 7
22
01–10
–28
00 26
78
⎞
⎠
→
⎛
⎝
100
1
010
2
001
3
⎞
⎠
,
and therefore x
1
= 1, x
2
= 2, x
3
= 3.
The diagonal element obtained on some step of the above elimination procedure may
happen to be equal to zero. In this case, the formulas become more complicated and
reindexing of the unknown quantities may be required.

5
◦
. Method of LU-decomposition.
This method is based on the representation of the basic matrix A as the product of a
lower triangular matrix L and an upper triangular matrix U, i.e., in the form A = LU.This
factorization is called a triangular representation or the LU-representation of a matrix (see
also Paragraph 5.2.3-1).
GivensuchanLU-representation of the matrix A, the system AX = B can be represented
in the form LUX =B, and its solution can be obtained by solving the following two systems:
LY = B, UX = Y .
Due to the triangular structure of the matrices L ≡ [l
ij
]andU ≡ [u
ij
], these systems can be
solved with the help of the formulas
y
i
=
1
l
ii

b
i
–
i–n

j=i
l

ij
y
j

(i = 1, 2, , n),
x
k
= y
k
–
n

s=k+1
u
ks
x
s
(k = n, n – 1, , 1),
provided that l
ii
≠ 0.
202 ALGEBRA
There exist various methods for the construction of LU-decompositions. In particular
if the following conditions hold:
a
11
≠ 0,




a
11
a
12
a
21
a
22



≠ 0, ,detA ≠ 0,
then the elements of the desired matrices L and U can be calculated by the formulas
l
ij
=
⎧
⎪
⎨
⎪
⎩
a
ij
–
j–1

s=1
l
is
u

sj
for i ≥ j,
0 for i < j,
u
ij
=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
1
l
ii

a
ij
–
i–1

s=1
l
is
u
sj

for i < j
1 for i = j,

0 for i > j.
5.5.2-3. Solutions of a square system with different right-hand sides.
1
◦
. One often has to solve a system of linear equations with a given basic matrix A
and different right-hand sides. For instance, consider the systems AX
(1)
= B
(1)
, ,
AX
(m)
= B
(m)
.Thesem systems can be regarded as a single matrix equation AX = B,
where X and B are matrices of size n × m whose columns coincide with X
(j)
and B
(j)
(j = 1, 2, , m).
Example 5. Suppose that we have to solve the equation AX = B with the given basic matrix A and
different right-hand sides:
A =

12–3
3 –21
–21 3

, B
(1)

=

7
1
5

, B
(2)
=

10
6
–5

.
Using the Gauss-Jordan procedure, we obtain
⎛
⎝
12–3
710
3 –21
16
–21 3
5 –5
⎞
⎠
→
⎛
⎝
12–3

710
0 –810
–20 –24
05–3
19 15
⎞
⎠
→
⎛
⎝
10–1/2
24
01–5/4
5/23
0013/4
1/32 0
⎞
⎠
→
⎛
⎝
100
34
010
53
001
20
⎞
⎠
.

Therefore,
X
(1)
=

3
5
2

, X
(2)
=

4
3
0

.
2
◦
.IfB = I,whereI is the identity matrix of size n × n, then the solution of the matrix
equation AX = I coincides with the matrix X = A
–1
.
Example 6. Find the inverse of the matrix
A =

210
–307
–541


.
Let us transform the augmented matrix of the system, using the Gauss-Jordan method. We get
⎛
⎝
210
100
–30 7
010
–54–1
001
⎞
⎠
→
⎛
⎝
1
1/2
0
1/2
00
0
3/2
7
3/2
10
0
13/2
–1
5/2

01
⎞
⎠
→
→
⎛
⎝
10
–7/3
0
–1/3
0
01
14/3
1
2/3
0
00
–94/3
–4
–13/3
1
⎞
⎠
→
⎛
⎝
100
14/47 –1/94 –7/94
010

19/47 1/47 7/47
001
6/47 13/94 –3/94
⎞
⎠
.
5.5. SYSTEMS OF LINEAR ALGEBRAIC EQUATIONS 203
5.5.2-4. General system of m linear equations with n unknown quantities.
Suppose that system (5.5.1.1) is consistent and its basic matrix A has rank r.First,in
the matrix A, one finds a submatrix of size r × r with nonzero rth-order determinant
and drops the m – r equations whose coefficients do not belong to this submatrix (the
dropped equations follow from the remaining ones and can, therefore, be neglected). In
the remaining equations, the n – r unknown quantities (free unknown quantities) that are
not involved in the said submatrix should be transferred to the right-hand sides. Thus, one
obtains a system of r equations with r unknown quantities, which can be solved by any of
the methods described in Paragraph 5.5.2-2.
Remark. If the rank r of the basic matrix and the rank of the augmented matrix of system (5.5.1.1) are
equal to the number of the unknown quantities n, then the system has a unique solution.
5.5.2-5. Solutions of homogeneous and corresponding nonhomogeneous systems.
1
◦
. Suppose that the basic matrix A of the homogeneous system (5.5.1.3) has rank r and
its submatrix in the left top corner, B =[a
ij
](i, j = 1, , r), is nondegenerate. Let
M =detB ≠ 0 be the determinant of that submatrix. Any solution x
1
, , x
n
has n – r

free components x
r+1
, , x
n
and its fi rst components x
1
, , x
r
are expressed via the free
components as follows:
x
1
=–
1
M
[x
r+1
M
1
(a
i(r+1)
)+x
r+2
M
1
(a
i(r+2)
)+···+ x
n
M

1
(a
in
)],
x
2
=–
1
M
[x
r+1
M
2
(a
i(r+1)
)+x
r+2
M
2
(a
i(r+2)
)+···+ x
n
M
2
(a
in
)],

x

r
=–
1
M
[x
r+1
M
r
(a
i(r+1)
)+x
r+2
M
r
(a
i(r+2)
)+···+ x
n
M
r
(a
in
)],
(5.5.2.4)
where M
j
(a
ik
) is the determinant of the matrix obtained from B by replacing its jth column
with the column whose components are a

1k
, a
2k
, , a
rk
:
M
j
(a
ik
)=








a
11
a
12
a
1k
a
1r
a
21
a

22
a
2k
a
2r
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
a
r1
a
r2
a
rk
a

rr








.
2
◦
. Using (5.5.2.4), we obtain the following n – r linearly independent solutions of the
original system (5.5.1.3):
X
1
=

–
M
1
(a
i(r+1)
)
M
–
M
2
(a
i(r+1)

)
M
··· –
M
r
(a
i(r+1)
)
M
10 ··· 0

,
X
2
=

–
M
1
(a
i(r+2)
)
M
–
M
2
(a
i(r+2)
)
M

··· –
M
r
(a
i(r+2)
)
M
01 ··· 0

,
X
n–r
=

–
M
1
(a
in
)
M
–
M
2
(a
in
)
M
··· –
M

r
(a
in
)
M
00 ··· 1

.
Any solution of system (5.5.1.3) can be represented as their linear combination
X = C
1
X
1
+ C
2
X
2
+ ···+ C
n–r
X
n–r
,(5.5.2.5)
where C
1
, C
2
, , C
n–r
are arbitrary constants. This formula gives the general solution of
the homogeneous system.

204 ALGEBRA
3
◦
. Relations between solutions of the nonhomogeneous system (5.5.1.1) and solutions of
the corresponding homogeneous system (5.5.1.3).
1. The sum of any solution of the nonhomogeneous system (5.5.1.1) and any solution of
the corresponding homogeneous system (5.5.1.3) is a solution of system (5.5.1.1).
2. The difference of any two solutions of the nonhomogeneous system (5.5.1.1) is a solution
of the homogeneous system (5.5.1.3).
3. The sum of a particular solution X
0
of the nonhomogeneous system (5.5.1.1) and the
general solution (5.5.2.5) of the corresponding homogeneous system (5.5.1.3) yields the
general solution X of the nonhomogeneous system (5.5.1.1).
5.6. Linear Operators
5.6.1. Notion of a Linear Operator. Its Properties
5.6.1-1. Definition of a linear operator.
An operator A acting from a linear space V of dimension n to a linear space W of dimension
m is a mapping A : V→Wthat establishes correspondence between each element x of the
space V and some element y of the space W. This fact is denoted by y = Ax or y = A(x).
An operator A : V→Wis said to be linear if for any elements x
1
and x
2
of the space
V and any scalar λ, the following relations hold:
A(x
1
+ x
2

)=Ax
1
+ Ax
2
(additivity of the operator),
A(λx)=λAx (homogeneity of the operator).
A linear operator A : V→Wis said to be bounded if it has a finite norm,whichis
defined as follows:
A =sup
x V
x≠0
Ax
x
=sup
x=1
Ax ≥ 0.
Remark. If A is a linear operator from a Hilbert space V into itself, then
A =sup
x V
x≠0
Ax
x
=sup
x=1
Ax =sup
x,y≠0
|(x, Ay)|
xy
=sup
x=y=1

|(x, Ay)|.
THEOREM.
Any linear operator in a finite-dimensional normed space is bounded.
The set of all linear operators A : V→Wis denoted by L(V, W).
A linear operator O in L(V, W) is called the zero operator if it maps any element x of
V to the zero element of the space W: Ox = 0.
A linear operator A in L(V, V) is also called a linear transformation of the space V.
A linear operator I in L(V, V) is called the identity operator if it maps each element x
of V into itself: Ix = x.
5.6.1-2. Basic operations with linear operators.
The sum of two linear operators A and B in L(V, W) is a linear operator denoted by A + B
and defined by
(A + B)x = Ax + Bx for any x
V.
The product of a scalar λ and a linear operator A in L(V, W) is a linear operator denoted
by λA and defined by
(λA)x = λAx for any x
V.
5.6. LINEAR OPERATORS 205
The opposite operator for an operator A
L(V, W) is an operator denoted by –A and
defined by
–A =(–1)A.
The product of two linear operators A and B in L(V, V) is a linear operator denoted by
AB and defined by
(AB)x = A(Bx)foranyx
V.
Properties of linear operators in L(V, V):
(AB)C = A(BC) (associativity of the product of three operators),
λ(AB)=(λA)B (associativity of multiplication of a scalar and two operators),

(A + B)C = AC + BC (distributivity with respect to the sum of operators),
where λ is a scalar; A, B,andC are linear operators in L(V, V).
Remark. Property 1 allows us to define the product A
1
A
2
A
k
of finitely many operators in L(V, V)
and the kth power of an operator A,
A
k
= AA A

 
k times
.
The following relations hold:
A
p+q
= A
p
A
q
,(A
p
)
q
= A
pq

.(5.6.1.1)
5.6.1-3. Inverse operators.
A linear operator B is called the inverse of an operator A in L(V, V)ifAB = BA = I.The
inverse operator is denoted by B = A
–1
. If the inverse operator exists, the operator A is said
to be invertible or nondegenerate.
Remark. If A is an invertible operator, then A
–k
=(A
–1
)
k
=(A
k
)
–1
and relations (5.6.1.1) still hold.
A linear operator A from V to W is said to be injective if it maps any two different
elements of V into different elements of W, i.e., for x
1
≠ x
2
,wehaveAx
1
≠ Ax
2
.
If A is an injective linear operator from V to V, then each element y V is an image of
some element x V: y = Ax.

T
HEOREM.
A linear operator
A : V→V
is invertible if and only if it is injective.
5.6.1-4. Kernel, range, and rank of a linear operator.
The kernel of a linear operator A : V→Vis the set of all x in V such that Ax = 0.The
kernel of an operator A is denoted by ker A and is a linear subspace of V.
The range of a linear operator A : V→Vis the set of all y in V such that y = Ax.The
range of a linear operator A is denoted by im A and is a subspace of V.
Properties of the kernel, the range, and their dimensions:
1. For a linear operator A : V→Vin n-dimensional space V, the following relation holds:
dim (im A)+dim(kerA)=n.
2. Let V
1
and V
2
be two subspaces of a linear space V and dim V
1
+dimV
2
=dimV.Then
there exists a linear operator A : V→Vsuch that V
1
=imA and V
2
=kerA.
A subspace V
1
of the space V is called an invariant subspace of a linear operator

A : V→Vif for any x in V
1
, the element Ax also belongs to V
1
. A linear operator A : V→V
is said to be reducible if V can be represented as a direct sum V = V
1
⊕ ···⊕ V
N
of two or
more invariant subspaces V
1
, , V
N
of the operator A,whereN is a natural number.
Example 1. ker A and im A are invariant subspaces of any linear operator A : V→V.