FOREIGN TRADE UNIVERSITY
LINEAR ALGEBRA
DANG LE QUANG, Ph.D
September - 2022
Introduction
1. Introduction
Dang Le Quang, Ph.D
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Introduction
Set
Some basic definitions (1)
Definition 1.1 (Set)
Set is a fundamental concept of mathematics having no definition. we can
only consider some objects having same property like a set.
Example 1.1
The set of all FTU students passing the exam.
The set Q of all Rational numbers.
The set of all the animals in the forest.
Object in the set is called to be an element of set. To denote x is an element of
set A, write x ∈ A. If x isn’t belong to set A, write x ∈
/ A.
The empty set, denote ∅, is the set having nothing.
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Introduction
Set
Some basic definitions (2)
Definition 1.2 (Denoting set)
There are two way to denote a set
1. List the elements of the set.
2. Describe the special properties of the set.
Example 1.2
Way 1: A = {2, 7, 3}, B = {1, 2, 3, . . .}.
Way 2: C = {x ∈ N | 2 | x}, D = x ∈ R | x 2 − 2x − 3 = 0 .
We have 3 ∈
/ C , 4 ∈ C , 3 ∈ D, 4 ∈
/ D.
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Introduction
Set
Some basic definitions (3)
Definition 1.3 (Subset)
Let two sets A and B, we call B to be a subset of A, denote B ⊂ A, if all
elements belonging to B belong to A.
Convention: The empty set ∅ is a subset of all sets.
Example 1.3
√
Let A = 1, 2, 5, 5, 7, 9 , B = {2, 5, 7, 9},
Then, we have B ⊂ A, C ⊂
̸ A, C ̸⊂ B.
C=
√
5, 7, 9, 10 .
Definition 1.4 (Equal set)
Two sets A and B are called to be equal, denote A = B, if A is a subset of B
and B is a subset of A.
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Introduction
Set
Basic operations - Intersections
Let A and B be two sets. The intersection of A and B, denote A ∩ B, is the set
of all elements belonging to both A and B, i.e.,
x ∈ A ∩ B ⇔ x ∈ A ∧ x ∈ B.
We can describle that set by the following Venn Diagram.
A∩B
Example 1.4
Let A = {1, 2, 3, 4}, let B = {0, 1}. Then A ∩ B = {1}.
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Introduction
Set
Basic operations - Unions
Let A and B be two sets. The Unions of A and B, denote A ∪ B, is the set of all
elements belonging to either A or B, i.e.,
x ∈ A ∪ B ⇔ x ∈ A ∨ x ∈ B.
We can describle that set by the following Venn Diagram.
A∪B
Example 1.5
Let A = {1, 2, 3, 4}, let B = {0, 1}. Then A ∪ B = {0, 1, 2, 3, 4}.
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Introduction
Set
Basic operations - Complements
Let A and B be two sets. The relative complement of B in A, denote A\B, is the
set of all elements that belong to A but don’t belong to B, i.e.,
x ∈ A\B ⇔ x ∈ A ∧ x ∈
/ B.
We can describle that set by the following Venn Diagram.
A∪B
Example 1.6
Let A = {1, 2, 3, 4}, let B = {0, 1}. Then A\B = {2, 3, 4}.
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Introduction
Set
Basic operations - Cartesian product
Let A and B be two sets. The Cartesian product of A and B, denote A × B, is
the set of all ordered pairs (a, b) such that a belongs to A and b belong to B, i.e.,
(a, b) ∈ A × B ⇔ a ∈ A ∧ b ∈ B.
Example 1.7
Let A = {1, 2, 3}, let B = {0, 1}. Then
A × B = {(0, 1), (1, 1), (2, 0), (2, 1), (3, 0), (3, 1)}
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Introduction
Set
Basic properties
Commutative law
A∩B
A∪B
= B ∩ A,
= B ∪ A.
Associative Law
(A ∩ B) ∩ C
(A ∪ B) ∪ C
= A ∩ (B ∩ C ),
= A ∪ (B ∪ C ).
Distribution law
(A ∩ B) ∪ C
(A ∪ B) ∩ C
=
=
Dermorgan law
DANG LE QUANG, Ph.D
(A ∪ C ) ∩ (B ∪ C ),
(A ∩ C ) ∪ (B ∩ C ).
A\(B ∪ C ) = (A\B) ∩ (A\C ),
A\(B ∩ C ) = (A\B) ∪ (A\C ).
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Introduction
Map
Definition of map
Definition 1.5 (Map)
Let X , Y be non-empty sets. The map f from X to Y is the rule that for
every x ∈ X , there is only a unique y ∈ Y , denoted
f : X →Y
x 7→ y = f (x),
Where
X is called to be a domain of f .
y is called to be a image of x through f .
x is called to be a reverse image of y through f .
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Introduction
Map
Example of map (1)
Example 1.8
Let the following rule
f : R→R
x 7→ y = f (x) = x 2 .
That rule is the map because for every x ∈ R, there is only a value x 2 .
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Introduction
Map
Example of map (2)
Example 1.9
Let X be a set. The map
IX : X → X
x 7→ x
is called to be a uniform map.
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Introduction
Map
Example of map (3)
Example 1.10
Let the following rule A = {1, 2, 3}, B = {a, b}, where a, b ∈ R; a ̸= b.
f :A→B
1 7→ a
1 7→ b
2 7→ a
3 7→ b
That rule isn’t a map because there are two images of 1.
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Introduction
Map
Image and Reverse image
Definition 1.6
Let f : X → Y ; C ⊂ X ; D ⊂ Y .
• Image set C of f , denoted f (C ), is the set of all images of x ∈ C , i.e.,
f (C ) = {y ∈ Y | ∃x ∈ C : f (x) = y } ⊂ Y .
• Reverse image set D of f , denoted f −1 (D), is the set of all reverse images of
y ∈ D, i.e.,
f −1 (D) = {x ∈ X | f (x) ∈ D} ⊂ X .
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Introduction
Map
Example of Image and Reverse image
Example 1.11
Let the following map.
a)
b)
f : R → R, f (x) = x 2 . Then
f ([−1, 2)) =
y ∈ R : y = x 2 , x ∈ [−1, 2) = [0, 4).
f −1 ([−1, 1)) =
x ∈ R : x 2 ∈ [−1, 1) = (−1, 1).
f : R → R, f (x) = |x + 1|. Then
f ([−2, 1])
= {y ∈ R : y = |x + 1|, x ∈ [−2, 1]} = [0, 2].
f −1 ([0, 1])
= {x ∈ R : |x + 1| ∈ [0, 1]} = [−2, 0].
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Introduction
Map
Properties of map
Let f : X → Y ; A1 , A2 , A ⊂ X ; B1 , B2 , B ⊂ Y . Then
•
f −1 (f (A)) ⊃ A,
f f −1 (B) ⊂ B.
•
f (A1 ∩ A2 ) ⊂ f (A1 ) ∩ f (A2 ),
f (A1 ∪ A2 ) = f (A1 ) ∪ f (A2 ).
•
f −1 (B1 ∩ B2 ) = f −1 (B1 ) ∩ f −1 (B2 ),
f −1 (B1 ∪ B2 ) = f −1 (B1 ) ∪ f −1 (B2 )
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Introduction
Map
Injection, Surjection, Bijection
Definition 1.7
Let f : X → Y .
f is a injective map if for each y ∈ Y , there isn’t more than a x ∈ X , i.e.
f is a injective map if and only if
∀x1 , x2 ∈ X , f (x1 ) = f (x2 ) ⇒ x1 = x2
⇔ x1 ̸= x2 ⇒ f (x1 ) ̸= f (x2 )
⇔ ∀y ∈ Y , f −1 (y ) has at most one element.
f is a surjective map if for all y ∈ Y , there is at least a reverse image, i.e.
f is a surjective map if and only if
⇔
⇔
∀y ∈ Y , ∃ x ∈ X : y = f (x)
f (X ) = Y
∀y ∈ Y , f −1 (y ) has at least one element.
f is a bijective map if f is both injective and surjective map.
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Introduction
Map
Example
Example 1.12
Let f : R → R, f (x) = x 2 . By the example 1.8, f is neither injective nor
surjective map.
Example 1.13
The map given by the example 1.9 is a bijective map.
Example 1.14
Let X , Y ⊂ R, let f : X → Y , f (x) = sin(x).
π
If X = −π
2 , 2 , Y = [−1, 1], then for all y ∈ Y , there is a unique reverse
image. Thus, f is a bijective map.
If X = R, Y = [−1, 1], then f is surjective map but not bijective map
since for all y ∈ Y , there is more than reverse images.
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Introduction
Real number field
Calculations and Properties
In Q, R there are arithmetic operations: addition, subtraction, multiplication, and
division and have the following basic properties: for all a, b, c ∈ R then
Commutation: a + b = b + a ; ab = ba.
Association: (a + b) + c = a + (b + c) ; (ab)c = a(bc).
Distribution: a(b + c) = ab + ac.
Q is dense in R : ∀a, b ∈ R if a < b, then exists q ∈ Q satisfied a < q < b.
(
x, khi x ≥ 0,
Absolute value: |x| =
−x, khi x < 0.
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Introduction
Real number field
The premise of the supremum
Definition 1.8 (Upper bound, Lower bound, sup, inf)
Subset A ⊂ R is called to be upper bounded if there exists M satisfied
a ≤ M for all a ∈ A.
Futhermore, M is called to be a upper bound of A. The most minimum in
all upper bound of A is called to be supremum of A, denoted sup A.
If sup A ∈ A, then sup A is the maximum of A, denoted max A.
Subset A ⊂ Ris called to be lower bounded if there exists m satisfied
m ≤ a for all a ∈ A.
Futhermore, M is called to be a lower bound of A. The most maximum
in all lower bound of A is called to be infimum of A, denoted inf A.
If inf A ∈ A, then inf A is the minimum of A, denoted min A.
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Introduction
Real number field
The premise of the supremise of the supremum
Subset A ⊂ R is called to be bounded if it is both upper bounded and lower
bounded, i.e. A is bounded if there exists M, m ∈ R (m < M) such that
A ⊂ [m, M]. In other words, A is bounded if there exists α ≥ 0 such that
|a| ≤ α ∀a ∈ A.
The premise of the supremise of the supremum: For all A ⊂ R, A ̸= ∅ be
upper bounded have upper bound belonging to R. Similarly, For all
A ⊂ R, A ̸= ∅ be lower bounded have lower bound belonging to R
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Introduction
Complex number field
Basic definition
Definition 1.9
A complex number z is given that z = a + bi, where
a ∈ R is a real part of z, denoted Re(z) = a.
b ∈ R is a imaginary part of z, denoted Im(z) = b.
i is a complex unit, we have i 2 = −1.
The set of all complex numbers is denoted C.
Definition 1.10
Two complex numbers are equal if and only if their both real part and
imaginary part are equal.
Definition 1.11
Let z = a + bi. Then z = a − bi is a complex conjugate number of z.
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Introduction
Complex number field
Calculation on complex number field
Let z1 = a1 + ib1 , z2 = a2 + ib2 (a1 , b1 , a2 , b2 ∈ R).
• Addition:
z1 + z2 = (a1 + a2 ) + i(b1 + b2 ),
z1 − z2 = (a1 − a2 ) + i(b1 − b2 ).
• Multiplication:
i 2 b1 b2
z1 z2 = (a1 + ib1 )(a2 + ib2 ) = a1 a2 + ia1 b2 + ib1 a2 + |{z}
{z
}
|
=−1
= (a1 a2 − b1 b2 ) + i(a1 b2 + b1 a2 ).
• Division:
a1 + ib1
(a1 + ib1 )(a2 − ib2 )
(a1 a2 + b1 b2 ) + i(−a1 b2 + b1 a2 )
z1
=
=
=
z2
a2 + ib2
(a2 + ib2 )(a2 − ib2 )
a22 + b22
a1 a2 + b1 b2
b1 a2 − a1 b2
=
+i
2
2
a2 + b2
a22 + b22
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Introduction
Complex number field
Trigonometric form of complex numbers
Let z = a + ib (a, b ∈ R).
Modun of z, denoted |z|, is given that |z| = r =
√
a2 + b 2
Argument of z, denoted (z), is the set of all angles φ satisfied
cos φ =
a
,
r
sin φ =
b
.
r
(1)
If φ is a solution of (1), then Arg(z) = φ + k2π (k ∈ Z).
Main Argument of z, denoted arg(z), is a Argument of z satisfied
We have
0 ≤ arg(z) < 2π.
(2)
z = a + ib = r (cos φ + i sin φ)
(3)
Where r = |z|, φ = arg(z).
(3) is a trigonometric form of complex number of z.
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Introduction
Complex number field
Example
Example 1.15
Find the trigonometric form of z = 1 + i.
√
√
1
Proof. We have |z| = r = 12 + 12 = 2 and cos φ = sin φ = √ .
2
One solution of that equation is φ = π4 . Thus
Arg(z) =
π
4
+ k2π,
k ∈ Z.
To imply arg(z), We select k ∈ Z to satisfy (2).
Therefore, by k = 0, we have arg(z) = π4 .
By (3), the trigonometric form of z is given that
π
i
√ h π
z = 1 + i = 2 cos
+ k2π + i sin
+ k2π ,
4
4
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k ∈ Z.
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Introduction
Complex number field
Powers of complex numbers
Multiply z1 times z2 :
Let z1 = r1 (cos φ1 + i sin φ1 ), z2 = r2 (cos φ2 + i sin φ2 ). Then
z1 z2 = r1 r2 [cos(φ1 + φ2 ) + i sin(φ1 + φ2 )]
Moivre law:
Let z = r (cos φ + i sin φ). Then
z n = r n [cos(nφ) + i sin(nφ)], n ∈ N.
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Introduction
Complex number field
Example
Example 1.16
√
Find (1 − i 3)2010 .
q
√
√
proof. Take z = 1 − i 3. We have |z| = 12 +√
(− 3)2 = 2 and
− 3
1
.
cos φ = , sin φ =
2
2
(4)
−π
k ∈ Z.
One solution of (4) is φ = −π
3 . Thus Arg(z) = 3 + k2π,
To imply arg(z), select k = 2 to satisfy (2). We have arg(z) = 5π
3 .
By (3), the trigonometric form of z is given that
−π
z = 2 cos −π
, k ∈ Z.
3 + k2π + i sin
3 + k2π
By Moivre law, we have
−2010π
−2010π
2010
z
= 2 cos
+ k.2.2010π + i sin
+ k.2.2010π ,
3
3
k ∈Z
= 2 [cos(−670π) + i sin(−670π)] = 2(1 + i.0) = 2 + 0i = 2
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Introduction
Complex number field
n-th root of complex numbers
• Opening n − th root of complex number: Let α, β be complex numbers. β
is called to be a n-th root of α if β n = α. Opening n − th root of α is to find all
n − th root of α (i.e. to find all β ∈ C to β n = α).
• Way of opening: Let α = r [cos(φ + k2π) + i sin(φ + k2π)] , k ∈ Z.
Assume that β = s [cos ψ + i sin ψ] is the n − th of α. Then
β n = α ⇒ s n [cos(nψ) + i sin(nψ)] = r [cos(φ + k2π) + i sin(φ + k2π)]

(
√
s = n r
sn = r
⇒
⇒
ψ = φ + k2π
nψ = φ + k2π
n
Thus, the set of the n-th root of is given that
√
n



√
α= βk = n r cos


φ + k2π
φ + k2π
,
+i sin
n
n


k=0,1,...,n−1 .

(5)
The n-th root of α are n differently complex numbers calculated by (5).
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Introduction
Complex number field
Example
Example 1.17
Find all n − th of 1.
Proof. We have
1 = 1 + 0i = 1 [cos(k2π) + i sin(k2π)] ,
k ∈ Z.
Thus, n − th of 1 is the set given that
√
k2π
k2π
n
+ i sin
, k = 0, 1, ..., n − 1 .
1 = cos
n
n
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Introduction
Complex number field
Solving equation
A quadratic equation ax 2 + bx + c = 0 always has two solutions. A n-degree
equation always has n solution in C.
Example 1.18
Solve x 2 + 4x + 7 = 0.
Proof.
∆ = −12 = 12i 2
√
√
−4 + 2 3i
= −2 + 3i
x1 =
2 √
√
−4 − 2 3i
x2 =
= −2 − 3i.
2
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Matrix and Determinants
Section 2: Matrix and Determinants
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Matrix and Determinants
Basic definitions of Matrix
Definition of Matrix
Definition 2.1
Let m, n ∈ N. A m × n matrix A in R consists of m.n numbers in R arranged
in the following m rows and n collums
a11
 a21
A=
 ...
am1

a12
a22
...
am2
...
...

a1n
a2n 

... 
... amn
Where aij ∈ R (i = 1, m, j = 1, n).
Number aij belonging to both row i and colum j of matrix A, denoted (A)ij , is
a elements of matrix A. Matrix A is denoted A = (aij )m×n .
If m = 1, A is row matix. Simply, If n = 1, A is colum matrix.
Two matrix are equal if they have same size and all their corresponding
elements are equal.
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Matrix and Determinants
Basic definitions of Matrix
Some concept of matrices
Definition 2.2
• Transpose of matrix: Let m × n matrix A = (aij )m×n . Transpose of A,
denoted At or AT , is the n × m matrix that for each i-th row, j-th column
element of AT is the j-th row, i-th column element of A, i.e. At = (aji )n×m .
t
Remark: (At ) = A.
• Zero matrix: If aij = 0 ∀i = 1, m, j = 1, n, then A is a zero matrix, denoted
Om×n or O.
• Square matrix: If m = n, then A is n-square matrix, denoted A = (aij )n .
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Matrix and Determinants
Basic definitions of Matrix
Example
Example 2.1
1 2 −5
0 0 0 0
A=
:size 2 × 3. E =
−7 3 8
0 0 0 0


1 2
B = −6 4 :size 3 × 2.
5 9


1 2 0
C = 4 7 −5 :size 3 × 3, square matrix.
0 0 0
0 0
:size 2 × 2, square matrix, O2×2 , O2 .
D=
0 0
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0
0
: O2×5 .
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Matrix and Determinants
Basic definitions of Matrix
Some concept of matrices (2)
Definition 2.3
Let A = (aij )i,j=1,n is a n square matrix. Then
• All a11 , a22 , ..., ann are all elements above main diagonal.
• All a1n , a2(n−1) , ..., an1 are all elements above secondary diagonal.
• If all elements above main diagonal are zero, then A is a upper triangular
matrix.
• If all elements above secondary diagonal are zero, then A is a lower
triangular matrix.
• If all the elements outside the main diagonal are zero, then A is a diagonal
matrix.
• If all elements belonging to main diagonal are equal to 1, all other elements
are equal to 0, then A is a identity matrix of size n, denoted In .
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Matrix and Determinants
Basic definitions of Matrix
Example
Example 2.2


1 2 0
A = 0 2 0 : upper triangular matrix.
0 0 0



1
0 0 1
0
B = 1 0 0 : square matrix, D = 
0
0 0 0
0
−1 0
C=
: lower triangular matrix, E
1 0
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0
1
0
0
0
0
1
0
=
1
3

0
0
 : indentity matrix I4 .
0
1
−5 −10
: matrix
7
4
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Matrix and Determinants
Basic caculation of matrix
Addition, subtraction
Let A = (aij )m×n , B = (bij )m×n . The sum (or subtraction) of A and B, denoted
A + B (or A − B) is the matrix m × n given that
(A + B)ij = (A)ij + (B)ij = aij + bij
(A − B)ij = (A)ij − (B)ij = aij − bij
Where i = 1, m, j = 1, n.
Remark: Two matrices are added/ subtraced if their sizes are equal.
Example 2.3
1 2
Let A =
4 5
Then
3
7
, B=
6
10
A+B =
1 2
4 5
8
11
9
1
,C=
12
3
3
7
+
6
10
8
11
0
2
, D=
−7
1
9
8
=
12
14
−8
0
0
.
0
10 12
14 18
C can not be added with D since they have different size.
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Matrix and Determinants
Basic caculation of matrix
Scalar multiplication
Let A = (aij )m×n and λ ∈ R. The product of λ and A, denoted λA, is the matrix
m × n given that
(λA)ij = λ(A)ij = λaij ,
Example 2.4
0 1 −2
(−3).0
−3
=
−5 0 0
(−3).(−5)
(−3).1
(−3).0
i = 1, m, j = 1, n.
(−3).(−2)
0 −3
=
(−3).0
15 0
6
0
Remark:
If λ = −1, then λA = (−1)A = −A. − A is a opposite matrix of A.
We can also define the subtraction that A − B = A + (−B).
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Matrix and Determinants
Basic caculation of matrix
Matrix multiplication
Let A be a m × n matrix, let B be a n × p matrix. Then the product of A and B,
denoted AB, is the m × p matrix given that
(AB)ij =
n
X
(A)ik (B)kj ,
i = 1, m, j = 1, p.
k=1
Example 2.5
1 2
Let A =
4 5
1 2
AB =
4 5


7
0
3
, B =  1 −2. Then
6
−1 3

7
3 
1
6
−1
DANG LE QUANG, Ph.D

0
1.7 + 2.1 + 3.(−1) 1.0 + 2.(−2) + 3.3
−2 =
4.7 + 5.1 + 6.(−1) 4.0 + 5.(−2) + 6.3
3
6 5
=
27 8
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Matrix and Determinants
Basic caculation of matrix
Matrix multiplication (2)
Remark:
• To AB is defined, both the number of colums of A and the number of rows
of B are equal.
• Each Element of (AB)ij is equal to the sum of the product each
corresponding elements belonging to row i of A and each corresponding
elements belonging to colum j of B
Definition 2.4
For each n-square matrix A and each p ∈ N, we define
A0 = In
Ap = Ap−1 .A,
p ≥ 1.
We also call Ap (p ∈ N is the power p of A.
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Matrix and Determinants
Basic caculation of matrix
Matrix multiplication (3)
Example 2.6


1 2
Let A = 1 0 . Find A.AT ; A2 .
0 −1
Proof.



2 5 1 −2
1
1
0
= 1 1 0 
0
2 0 −1
−1
−2 0 1



1 2
1 2
2



A = A.A = 1 0
can’t calculate.
1 0
0 −1 0 −1

1
T

A.A = 1
0
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Matrix and Determinants
Basic caculation of matrix
Property of calculation
Assume that the following calculations are defined. Then
1)
A + B = B + A.
2)
A + (B + C ) = (A + B) + C ; A(BC ) = (AB)C .
3)
A + O = A.
4)
A + (−A) = O.
5)
λ(A + B) = λA + λB,
λ ∈ R.
6)
(λ + µ)A = λA + µA,
λ, µ ∈ R.
7)
(λµ)A = λ(µA),
8)
1.A = A ; AI = IA = A.
DANG LE QUANG, Ph.D
λ, µ ∈ R.
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Matrix and Determinants
Determinants
Determinant of square matrices (1)
• 1-square matrix: Let A = [a11 ]. Determinant of A, denoted |A| or detA, is
given that detA = a11 .
Example 2.7
Let A = [−5] , B = [2]. Then, detA = 5, detB = 2.
a
• 2-square matrix: Let A = 11
a21
detA =
DANG LE QUANG, Ph.D
a11
a21
a12
. Determinant of A is given that
a22
a12
= a11 a22 − a12 a21 .
a22
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Matrix and Determinants
Determinants
Determinant of square matrices (2)

a11
• 3-square matrix:Let A = a21
a31
a11
detA = a21
a31
a12
a22
a32
a12
a22
a32

a13
a23 . Determinant of A is given that
a33
a13
a23
a33
= (a11 a22 a33 + a12 a23 a31 + a13 a21 a32 ) − (a11 a23 a32 + a12 a21 a33 + a13 a22 a31 ).
To remember that law, we use Sarrus law.
Example 2.8
2
Find 1
5
4 8
6
−1 3 ; −1
4 7
5
DANG LE QUANG, Ph.D
−2
2
4
8
3
5
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Matrix and Determinants
Determinants
Determinant of square matrices (3)
• n-square matrix:
Definition 2.5 (Definition of algebraic complement)
Let n-square matrix A.
a11
a21

 ...
an1

a12
a22
...
an2

... a1n
... a2n 

... ... 
... ann
Denote Mij is a matrix given by A after delecting row i and colum j.
Number (−1)i+j detMij is called to be algebraic complement of aij , denoted Aij .
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Matrix and Determinants
Determinants
Determinant of square matrices (4)
Determinant of A is given that
detA = ai1 Ai1 + ai2 Ai2 + ... + ain Ain =
n
X
aij Aij
(6)
aij Aij
(7)
j=1
or
detA = a1j A1j + a2j A2j + ... + anj Anj =
n
X
i=1
(6) is called extended by row i.
(7) is called extended by colum j.
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Matrix and Determinants
Determinants
Determinant of square matrices (5)
Example 2.9


1 1 1
Find determinant of matrix M = 1 2 2.
1 2 3
Proof. By (6), we expand determinant by row 3
1
1
1
1 1
1
2 2 = 1(−1)1+3
2
2 3
1
1
+ 2(−1)3+2
2
1
1
1
+ 3(−1)3+3
2
1
1
2
= 1.2 − 1.2 − 2(1.2 − 1.1) + 3(1.2 − 1.1)
= 1.
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Matrix and Determinants
Determinants
Property of determinant (1)
det A = detAT .
Determinant is equal to 0 if it satisfies one of the following conditions:
- There is a row (or column) be equal to 0.
- There are two rows (or columns) be equal orproportional.
if we swap the two rows (or columns) of determinant, then the determinant
is swaped sign.
if we multiply a row (or column) of determinant with λ ∈ R, then the
determinant is also multiplied with λ.
In other word, a factor of a row (or a column) can be put outside of
determinant. det(λA) = λn .detA.
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Matrix and Determinants
Determinants
Property of determinant (2)
If determinant has a row (or column) expanded to the sum of two rows (or
columns), then determinant is expanded to two corresponding determinant.
If we multiply a row (or column) of the determinant by any number of λ and
then add it to another row (another column) then the determinant does not
change.
If A, B are two n- square matrices, then det(AB) = detA.detB.
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Matrix and Determinants
Determinants
Way to calculate (1)
• Bringing the determinant to the triangular form: Use property of
determinant to bring it to the triangular form. Then the determinant is equal to
the product of all numbers on main diagonal
Example 2.10
2
Find M = 1
5
4 8
−1 3 ; N =
4 7
DANG LE QUANG, Ph.D
1
2
3
−2
0
−1
1
1
2 0
3 1
0 2
0 3
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Matrix and Determinants
Determinants
Way to calculate (2)
• Expanding determinant by Row or column .
Remark: Expand on row or colum having many zero numbers.
Example 2.11
Proof. By (7), we expand by column 2
5 0 0 10
5
10 0 2 −4
= 1(−1)3+2 10
2 1 4 9
−9
−9 0 0 −7
0 10
2 −4
0 −7
Continue expanding determinant in right. By (7), expand by column 2
5
1(−1)3+2 10
−9
0 10
5
2 −4 = (−1)2(−1)2+2
−9
0 −7
10
−7
= −2 (5(−7) − (−9).10) = −110.
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Matrix and Determinants
Determinants
Determinant of the profuct of matrices
If A, B are n-square matrices, then det(AB) = detA.detB
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Matrix and Determinants
Rank of matrix
Definition
• Sub-determinant: Let A be a m × n matrix. Select the elements that lie on
the intersection of k rows and k columns of A, we get a k-square matrix. The
determinant of this k-square matrix is called k-sub-determinant of A.
• Rank of matrix: Let A be a none zero m × n matrix. Rank of A, denoted
rank(A) or r (A), is the highest level of the non-zero subdeterminators of the
matrix A. Thus, rank of A, rank(A) = r satisfied
- There exists at least one none zero sub-determinator r of A.
- All sub-determinators of A level greater than r (if exists) must be 0.
Convention: If A = O then r(A) = 0.
Example 2.12
Find rank of the following matrices.



1 0 3 −2
2 0 1
A = 0 1 2 −1 ; B = 0 1 2
2 0 6 −4
5 0 6
DANG LE QUANG, Ph.D
LINEAR ALGEBRA

−2
3
4
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Matrix and Determinants
Rank of matrix
Property of rank of matrix
1)
Let A be a m × n matrix. Then 0 ≤ rank(A) ≤ min{m, n} ;
rank(A) = 0 ⇔ A = O ; rank(A) > 0 ⇔ A ̸= O.
2)
If A is a m × n matrix having (at least) a none zero r -sub-determinant
r (0 < r ≤ min{m, n}) then rank(A) ≥ r .
In particular, if A has a none zero r -sub-determinant with r = min{m, n},
then rank(A) = min{m, n}. Then, we calll that A has maximum rank.
In separate case, If n-square matrix A has determinant equal to 0, then
rank(A) = n, i.e. A A has maximum rank. If detA = 0, then rank(A) < n.
3)
rank(AT ) = rank(A).
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Matrix and Determinants
Rank of matrix
Way to calculate rank of matrix (1)
• Elementary transformations on matrices: There are three the following way.
1. Multiply a row by any non-zero real number λ.



a11
a11 a12 ... a1n
a
 ...
 21
... ... ... 




 di →λ.di  ...
 ai1 ai2 ... ain  −−−−−→ 


λai1

 ...
... ... ... 
 ...
am1 am2 ... amn
am1
DANG LE QUANG, Ph.D
LINEAR ALGEBRA
a12
a22
...
λai2
...
am2

... a1n
... a2n 


... ... 

... λain 

... ... 
... amn
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Matrix and Determinants
Rank of matrix
Way to calculate rank of matrix (2)
2. Multiply a row by any real number λ and add it to another



a11 a12 ... a1n
a11
a12
 ...


... ... ... 
...
...



 di →λ.di +dj 
 ai1 ai2 ... ain  −−−−−−−→ λai1 + aj1 λai2 + aj2



i̸=j
 ...

... ... ... 
...
...
am1 am2 ... amn
am1
am2
DANG LE QUANG, Ph.D
LINEAR ALGEBRA
row.

...
a1n

...
...


... λain + ajn 


...
...
...
amn
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Matrix and Determinants
Rank of matrix
Way to calculate rank of matrix (3)
3. Swap two consecutive rows.



a11 a12 ... a1n
a11
 ...

 ...
...
...
...






di ↔dj
 ai1 ai2 ... ain 
 aj1

 −−−−−−−−→ 
 aj1 aj2 ... ajn  i,j consecutive  ai1



 ...
 ...
... ... ... 
am1 am2 ... amn
am1
DANG LE QUANG, Ph.D
LINEAR ALGEBRA
a12
...
aj2
ai2
...
am2

... a1n
... ... 


... ajn 

... ain 

... ... 
... amn
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Matrix and Determinants
Rank of matrix
Way to calculate rank of matrix (4)
Example 2.13

1 1
1 2
1 2

1
d2 ↔d3
−−
−−→  0
−1
1
0
−2
DANG LE QUANG, Ph.D


1
1
d2 →(−1)d2
2 −−−−−−−→ −1
3
1


1
1
d →d1 −d2
−−→  0
1  −−1−−−
d3 →d3 +2d2
−2
−1
1
−2
2
1
0
−2
LINEAR ALGEBRA


1
1
d3 →d3 +d2
−−−−→ −1
−2 −−
3
0


0
0
d1 →d1 +d3
−−−−−→  0
1 −−−
0
d3 →d3 +2d1new
1
−2
0
−1
September - 2022

1
−2
1

−1 0
0 1
−4 0
59 / 178
Matrix and Determinants
Rank of matrix
Way to calculate rank of matrix (5)
• ladder matrix (row): Let A be a none zero m × n matrix. A is row ladder
matrix if it satisfies all the following conditions.
1)
Non-zero lines above zero lines (if any).
2)
The first non-zero element from left to right in the lower row is always to the
right of the column containing the element first non-zero of the above line.
These first non-zero elements are called marked elements of A.
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Matrix and Determinants
Rank of matrix
Way to calculate rank of matrix (6)

1

A= 0
0

1

D= 0
0
0
1
0
0
0
2


2
1


0 ; B= 0
0
0
1
1
0
1
1
0


1
0
0
0 ; E = 
0
0
0
DANG LE QUANG, Ph.D


1 2 0
1
0 0 2
0 ; C = 
0 0 0
1
0 0 0


0 0
1 2
0 0
 ; F = 0 1
0 1
0 −3
0 0
LINEAR ALGEBRA

0
0

0
is ladder matrix.
0

3 4
2 0 , isn’t ladder matrix.
0 0
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Matrix and Determinants
Rank of matrix
Way to calculate rank of matrix (6)
way to find the rank of matrix by elementary transformations
Theorem 2.1
The elementary transformations on the line do not change the rank of the matrix.
The rank of a row ladder matrix is equal to its number of non-zero rows.
Therefore, to find the rank A, we use elementary transformations to return the
ladder matrix A′ . Then rank of A is equal to rank of A′ equal to the number of
non-zero lines of A′ .
Example 2.14

1
A = 4
7


2 3
1
d2 →d2 −4d1
−−−−−→ 0
5 6 −−
d3 →d3 −7d1
8 9
0
2
−3
−6


3
1
d3 →d3 −2d2
−−−−−→ 0
−6  −−
−12
0
2
−3
0

3
−6 = A′ .
0
Thus, We have rank(A) = rank(A′ ) = 2.
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Matrix and Determinants
Inverse matrix
Definition
Let A be n-square matrix. A n-square matrix is called inverse matrix of A if
AB = BA = In .
Example 2.15
1 2
3 −2
Let A =
; B=
.
1 3
−1 1
Then, we have AB = BA = I2 , which imply A−1 = B.
Property: If both A and B have inverse matrix, then
A−1
−1
T −1
A
(AB)
DANG LE QUANG, Ph.D
−1
= A,
= A−1
=B
−1
LINEAR ALGEBRA
T
,
−1
A
.
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Matrix and Determinants
Inverse matrix
Conditions of existence and uniqueness
Theorem 2.2
A square matrix A has a inverse matrix if and only if detA ̸= 0. The inverse
matrix of A if exists is unique.
Definition 2.6 (Invertible Matrix, non-degenerate matrix)
Matrix A having inverse matrix is called to be a Invertible Matrix.
Matrix A having detA ̸= 0 is called to be a non-degenerate matrix.
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Matrix and Determinants
Inverse matrix
Way to find inverse matrix (1)
• Way to find inverse matrix by auxiliary matrix
Definition 2.7 (auxiliary matrix)
(Review algebraic complement by Definition 2.5). Let cij = (−1)i+j detMij be a
algebraic complement of aij .
Let C = (cij ) be square matrix. Matrix C T is called to be a auxiliary matrix
of A, denoted PA = C T .
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Matrix and Determinants
Inverse matrix
Way to find inverse matrix (2)
Algorithm to find A−1 (if any) by the auxiliary matrix:
Let A be square matrix in R.
Step 1: Find detA.
Step 2: Argue:
- If detA = 0, then A isn’t invertible, i.e. doesn’t exist A−1 .
- If detA ̸= 0 then A is invertible. Find A−1 by step 3.
Step 3: Find PA .
- Find Cij = (−1)i+j detMij and find
c11
 c21
C =
 ...
cm1

- Then, PA = C T and A−1 =
DANG LE QUANG, Ph.D
c12
c22
...
cm2

... c1n
... c2n 

... ... 
... cmn
1
detA PA .
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Matrix and Determinants
Inverse matrix
Way to find inverse matrix (3)
Example 2.16
Find inverse matrix (if any) of the following matrix.

1
A = 1
1
1
2
2

1
2
3
Proof. By Sarrus law, we have detA = 1 ̸= 0, it implies A is invertible. We
have
c11 = (−1)1+1
2
2
1
2
= 2 ; c12 = (−1)1+2
1
3
c21 = (−1)2+1
1
2
1
1
= −1 ; c22 = (−1)2+2
3
1
c31 = (−1)3+1
1
2
1
1
= 0 ; c32 = (−1)3+2
2
1
DANG LE QUANG, Ph.D
2
1
= −1 ; c13 = (−1)1+3
3
1
1
1
= 2 ; c23 = (−1)2+3
1
3
1
1
= −1 ; c33 = (−1)3+3
2
1
LINEAR ALGEBRA
2
=0;
2
1
= −1 ;
2
1
=1;
2
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Matrix and Determinants
Inverse matrix
Way to find inverse matrix (4)
Which imply

2
C = −1
0

−1 0
2 −1
−1 1

⇒ PA = C T
2
= −1
0
−1
2
−1

0
−1
1
Thus

A−1
DANG LE QUANG, Ph.D
2
1
=
PA = −1
detA
0
LINEAR ALGEBRA
−1
2
−1

0
−1
1
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Matrix and Determinants
Inverse matrix
Way to find inverse matrix (5)
• Find the inverse matrix by using elementary transformations
Algorithm to find A−1 (if any) by using elementary transformations
Let A be a n-square matrix in R.
Step 1: Make a matrix [A|In ] by appending to the right A the indenity matrix In .
Step 2: Use elementary transformations on row to get [A|In] to the form [In|B].
Then A is invertible and A−1 = B.
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Matrix and Determinants
Inverse matrix
Way to find inverse matrix (6)
Example 2.17
Find inverse matrix (if any) of the following matrix.
1
A=
2
1
3
Proof. We have
1
[A|I2 ] =
2
1 1
3 0
0 d2 →d2 −2d1 1
−−−−−−−→
1
0
Thus, A−1 = B =
DANG LE QUANG, Ph.D
3
−2
1 1
1 −2
0 d1 →d1 −d2 1
−−−−−−→
1
0
0 3
1 −2
−1
= [I2 |B ]
1
−1
.
1
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Matrix and Determinants
Inverse matrix
Way to find inverse matrix (7)
Example 2.18
Find inverse matrix (if any) of the following matrix.

1
A = 4
7
2
5
8

3
6
9
Proof. We have

1 2
[A|I3 ] = 4 5
7 8
3 1
6 0
9 0
0
1
0


0
1
d3 →d3 −7d1
−−−−−→ 0
0 −−
1
d2 →d2 −4d1
0

...
d3 →d3 −2d2
−−
−−−−−→ ...
0
2
−3
−6
...
...
0

0 0
1 0
−1 1

... ...
... ... ̸= [I3 |B] .
... ...
3
1
−6 −4
−12 −7
... ...
... ...
0 ...
Thus, A isn’t invertible.
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Linear system
Section 3: Linear system
Dang Le Quang, Ph.D
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Linear system
Basic definitions
General Linear system (1)
Linear system (n variables, m equations) is system given that


a11 x1 + a12 x2 + ... + a1n xn = b1



a x + a x + ... + a x = b
21 2
22 2
2n n
2

.........




am1 x1 + am2 x2 + ... + amn xn = bm
(8)
Where
aij , bi (i = 1, m, j = 1, n are the coefficients, bi are the free coefficients.
x1 , x2 , ..., xn are variables.
a11
 a21
Matrix A = 
 ...
am1
a12
a22
...
am2

Matrix B = b1
b2
DANG LE QUANG, Ph.D
...
...
...
...
...

a1n
a2n 
 is coefficient matrix of system (8).
... 
amn
bm
T
is a free coefficient matrix (8).
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Linear system
Basic definitions
General Linear system (2)

... a1n b1

... a2n b2 
..  is additional coefficient matrix
... ... . 
... amn bm
or expansion coefficient matrix of system (8).
T
Matrix X = x1 x2 . . . xn is variable matrix or variable column.
a11
 a21
Matrix A = [A|B] = 
 ...
am1

a12
a22
...
am2
System (8) can be written that AX = B.
System (8) is Cramer system if it has number of equations equal to number of
variables (n = m) and detA ̸= 0.
System (8) is homogenous system if free column bi = 0 for all i = 1, m.
All n variables (x1 , x2 , ..., xn ) are solutions of system (8) if when we substitute
them into the system (8) we get the correct equality.
Solving a system of linear equations is to find the solution of the system.
Two systems of linear equations with the same variables are equivalent if their
solution sets are equal.
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Linear system
Basic definitions
Condition for existence of solutions (1)
Theorem 3.1 (Kronecker-Capelli)
Lineat system (8) exists solution if and only if rank(A) = rank(A).
Futhermore, assume that rank(A) = rank(A) = r (0 ≤ r ≤ min{m, n}). Then
- If r = n (n is number of variables), then (8) exists unquie solution.
- if r < n, then (8) exists there are infinitely many solutions depending on
n − r parameters.
Example 3.1
Do the following linear system exist solution?
a)



x1 + 2x2 + 3x3 = 1
4x1 + 5x2 + 6x3 = 4


7x + 8x + 9x = 8
1
2
3
DANG LE QUANG, Ph.D
b)



x1 + 2x2 + 3x3 = 1
4x1 + 5x2 + 6x3 = 4


7x + 8x + 9x = 7
1
2
3
LINEAR ALGEBRA
c)
(
x + 2y = 1
2x + 3y = −1
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Linear system
Basic definitions
Condition for existence of solutions (2)
Proof. a)

1
4
7
2
5
8


3 1
1
d2 →d2 −4d1
6 4 −−−−−−−→ 0
d3 →d3 −7d1
9 8
0


3 1
1
d3 →d3 −2d2
−6 0 −−−−−−−→ 0
−12 1
0
2
−3
−6

3 1
−6 0
0 1
2
−3
0
Since rank(A) = 2 ̸= rank(A) = 3 then system doesn’t exist solution.
b)

1
4
7


3 1
1
d2 →d2 −4d1
6 4 −−−−−−−→ 0
d3 →d3 −7d1
9 7
0
2
5
8
2
−3
−6


3 1
1
d3 →d3 −2d2
−6 0 −−−−−−−→ 0
d2
d2 → −3
−12 0
0
2
1
0

3 1
2 0
0 0
Since rank(A) = 2 = rank(A) < number of variables = 3 then system has
infintely solutions depending on 3 − 2 = 1 paramaters.
c)
1
2
2 1 d2 →d2 −2d1 1
−−−−−−−→
3 −1
0
2 1 d1 →d1 +2d2 1
−−−−−−−→
−1 −3
0
0 −5
−1 −3
Since rank(A) = 2 = rank(A) = number of variables then system has unique
solution.
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Linear system
Method to solve Cramer system
Method to solve Cramer system (1)
Definition 3.1 (Linear system Cramer )
By (8), if m = n, i.e. system has number of equations equal to number of
variables, is Linear system Cramer.
Let linear system Cramer be fomal matrix AX = B (A is square matrix, detA ̸= 0)
• Inverse matrix method: System has uniquie solution X = A−1 B.
Example 3.2
Solve



x1 + x2 + x3 = 1
x1 + 2x2 + 2x3 = −1


x + 2x + 3x = 2
1
2
3
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Linear system
Method to solve Cramer system
Method to solve Cramer system (2)
Proof. This system is written that
AX = B
Where

1
A = 1
1
1
2
2

 
 
1
x1
1
2 ; X = x2  ; B = −1
3
x3
2
We have

A−1
2
= −1
0
−1
2
−1
DANG LE QUANG, Ph.D


0
2
−1 ⇒ X = −1
1
0
−1
2
−1

   


0
1
3
x1 = 3
−1 −1 = −5 ⇒ x2 = −5


x = 3
1
2
3
3
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Linear system
Method to solve Cramer system
Method to solve Cramer system (3)
Example 3.3
Solve


2x1 + x2 − x3 = 1

x2 + 3x3 = 3


2x + x + x = −1
1
2
3
• Determinant method: Recall that system Cramer is system having number of
equations equal number of variables (m = n), i.e. system (8) now becomes


a11 x1 + a12 x2 + ... + a1n xn = b1


a x + a x + ... + a x = b
21 2
22 2
2n n
2

.........




an1 x1 + an2 x2 + ... + ann xn = bn
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(9)
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Linear system
Method to solve Cramer system
Method to solve Cramer system (4)
 

b1
a11 a12 ... a1n

b2 
a21 a22 ... a2n 

 ; B=
Step 1: Let A = 

..  ; and Ai be the matrix
 ... ... ... ... 
.
an1 an2 ... ann
bn
obtained from A by replacing column i with column B.

Step 2: Find ∆ = detA ; ∆1 = detA1 ; ... ; ∆n = detAn .
Step 3: Argument
- If ∆ ̸= 0 then system (9) has unique solution
∆2
∆n
∆1
, x2 =
, . . . , xn =
x1 =
∆
∆
∆
- If ∆ = 0 and ∆i ̸= 0 for all i = 1, n, then system(9) hasn’t solution.
- If ∆ = ∆1 = ∆2 = ... = ∆n = 0, there is no general conclusion (either no
solution or infinitely many solutions, now we use the method Gauss which
will be mentioned in ??).
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Linear system
Method to solve Cramer system
Method to solve Cramer system (5)
Example 3.4
Solve the folowing systems by using the Cramer system method




(
x
+
2y
+
3z
=
1



x + 2y + 3z = 0
x + 2y = 1
a)
; b) 4x + 5y + 6z = 4
; c) 4x + 5y + 6z = 0


3x − 4y = 2


7x + 8y + 9z = 8
7x + 8y + 9z = 0
1
1 2
1 2
Proof. a) Ta có A =
; B=
∆ = detA =
= −10 ̸= 0.
2
3 −4
3 −4
1 2
1 1
A1 =
⇒ ∆1 = detA1 = −8., A2 =
⇒ ∆2 = detA2 = −1.
2 −4
3 2

∆1
−8
4


=
=
x1 =
∆
−10
5
Thus system has unique solution

∆
−1
1

y = 2 =
=
.
∆
−10
10
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Linear system
Method to solve Cramer system
Method to solve Cramer system (6)

1

b) We have A = 4
7
2
5
8

 
3
1


6 ; B = 4
9
8
1 2 3
∆ = detA = 4 5 6 = 0 (using Sarrus law to find det).
7 8 9


1 2 3
A1 = 4 5 6 ⇒ ∆1 = detA1 = −3 ̸= 0.
8 8 9
Thus, system hasn’t solution.
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Linear system
Method to solve Cramer system
Method to solve Cramer system (7)

1
c) We have A = 4
7
2
5
8

 
3
0
6 ; B = 0
9
0
∆ = detA = 0
0 2 3
A1 = 0 5 6 ⇒ ∆1 = detA1 = 0.
0 8 9


1 0 3
A2 = 4 0 6 ⇒ ∆2 = detA2 = 0.
7 0 9


1 2 0
A3 = 4 5 0 ⇒ ∆3 = detA3 = 0.
7 8 0
Thus, we don’t know specifically whether the system has no solution or infinite
solutions, i.e, we cannot use the determinant method to solve this system.
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Linear system
Method to solve general linear system
Method of returning to the caramel system
Find rank of A and A
- If rank(A) ̸= rank(A) then system has no solution.
- If rank(A) = rank(A) = r , then there exists sub-determinant of order r
of the nonzero matrix A.
Dr
We remove all equations that do not involve Dr (m − r equations).
Variables corresponding to columns sticked with Dr are kept to the left as hidden.
Variables for columns that don’t stick to Dr are switched to the right as a
parameter. Then we have the system Cramer.
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Linear system
Method to solve general linear system
Gauss’s method
Denote matrix A. Using elementary transformations on the rows to returns A to
the ladder form.
- If during the transformation, a row on the left is zero, the right side is non-zero.
The system has no solution.
- If return A to the ladder form, then the variables corresponding to the columns
containing the marker element are kept as variables, the variables corresponding
to the columns that do not contain the marked element are moved to the right as
a parameter, then solve the equation reversed process from bottom row to row 1.
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Linear system
Method to solve general linear system
Example (1)
Example 3.5
Solve the following system.
(
x1 + x2 + x3 + x4 = 2
a)
x1 + x2 + 2x3 + 2x4 = 4
(
b)
x1 + x2 − x3 + x4 − x5 = 1
x5 = 2
Proof.
a) Use Gauss’s method
1 1
1 1
1
2
1 2 d2 →d2 −d1 1
−−−−−−→
2 4
0
1
0
1
1
1 2
1 2
By the theorem 3.1, r = rank(A) = rank(A) = 2, which implies that system
exists solution. Futhermore, since (r = 2 < n = 4), system has infinite solutions
depending on 2 parameters.
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Linear system
Method to solve general linear system
Example (2)
Thus


x1 = 2 − x2 − x3 − x4 = −a



x = a (a ∈ R)
2

x
 3 = 2 − x4 = 2 − b



x4 = b (b ∈ R)
Thus, set of system’s solution is {(−a, a, 2 − b, b) : a, b ∈ R}.
b) Use Gauss’s method
1
0
1
0
−1
0


x1 = 3 − a + b − c






x2 = a (a ∈ R)
1 −1 1
⇒ x3 = b (b ∈ R)

0 1 2




x4 = c (c ∈ R)

x = 2.
5
Thus, set of system’s solution is {(3 − a + b − c, a, b, c, 2) : a, b, c ∈ R}.
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Linear system
Homogenous linear system
Homogenous linear system (1)
Let the following homogenous linear system


a11 x1 + a12 x2 + ... + a1n xn = 0



a x + a x + ... + a x = 0
21 2
22 2
2n n

.........




am1 x1 + am2 x2 + ... + amn xn = 0
(10)
with the matrix form AX = O. They have solution since r (A) = r ([A|O]) .
Sets (0, 0, . . . , 0) is always solutions of (10), called to be a trivial solution.
Non-zero solutions, if any, are called non-trivial solutions of the system (10).
By the theorem(3.1) Kronecker-Caperrli, we have
- if r (A) = n then (10) has unique solution being trivial solution.
- if r (A) = r < n then (10) has infinte solutions depending n − r parameters.
A homogeneous system is a special case of a general system, so a general solution
method can be applied to it with the note that in the transformation process,
instead of the expanded matrix, we only need to transform the coefficient matrix.
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Linear system
Homogenous linear system
Homogenous linear system (2)
Example 3.6
By Example 3.4, system c) is both a Cramer system and a homogeneous
system. Furthermore, as shown in Example 3.4, this system is not solvable by
the determinant method.



x + 2y + 3z = 0
4x + 5 y + 6 z = 0


7x + 8y + 9z = 0
Review that the Cramer system, which is homogeneous, is a special case of
the general system, so a general solution can be applied to it, but instead of
transforming the extended matrix (as Example 3.5) then we only need to
transform the coefficient matrix. Using the Gauss method, we get

1 2
A = 4 5
7 8
DANG LE QUANG, Ph.D


3
0
d2 →d2 −4d1
−−−−−→ 0
6 −−
9
d3 →d3 −7d1
0
2
−3
−6


1
3
d3 →d3 −2d2
−−−−−→ 0
−6  −−
−d
d2 → 3 2
−12
0
LINEAR ALGEBRA
2
1
0
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
3
2
0
89 / 178
Linear system
Homogenous linear system
Homogenous linear system (3)
Since r (A) = 2 < n = 3, it has no solution.


x = −2y − 3z = a

y = −2a


z = a (a ∈ R)
Remark:
The homogeneous system (10) has a non-trivial solution if and only if the rank of
the coefficient matrix is less than the variable number (rank(A) < n).
If the homogeneous linear system has the same number of equations as the
variables (m = n), the coefficient matrix is square matrix. Then
- The system has a trivial unique solution if and only if detA ̸= 0.
- The system has a non-trivial solution if and only if detA = 0.
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Vector space
Section 4: Vector space
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Vector space
Basic Definition
Basic definition (1)
Let Rn = x = (x1 , x2 , ..., xn ) : xi ∈ R, i = 1, n .
Number xi is called to be a i-th elements of x.
The two elements x and y of Rn are called to be equal, denoted x = y , if all
corresponding elements are equal.
In Rn , We give the following two operations:
• Addition of two elements of Rn :
For all x = (x1 , x2 , ..., xn ), y = (y1 , y2 , ..., yn ) ∈ Rn , we have
x + y = (x1 + y1 , x2 + y2 , ..., xn + yn ).
• Multiplication of a real number by an element of Rn :
For λ ∈ R and x = (x1 , x2 , ..., xn ) ∈ Rn , we have
λx = (λx1 , λx2 , ..., λxn ).
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Vector space
Basic Definition
Basic definition (2)
The set Rn together with those two operations has the following 8 properties:
1)
∀x, y ∈ Rn : x + y = y + x
2)
∀x, y , z ∈ Rn : (x + y ) + z = x + (y + z)
3)
∃O = (0, 0, ..., 0) ∈ Rn such that
x + O = O + x = x∀x ∈ Rn .
4)
∀x = (x1 , x2 , ..., xn ) ∈ Rn , ∃(−x) = (−x1 , −x2 , ..., −xn ) ∈ Rn , we have
x + (−x) = (−x) + x = O.
5)
∀x, y ∈ Rn , ∀λ ∈ R, λ(x + y ) = λx + λy .
6)
∀λ, µ ∈ R, ∀x ∈ Rn , (λ + µ)x = λx + µx.
7)
∀λ, µ ∈ R, ∀x ∈ Rn , (λµ)x = λ(µx).
8)
x ∈ Rn , 1.x = x.
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Vector space
Basic Definition
Basic definition (3)
Rn with these two operations satisfying the above 8 characteristic properties is
called a vector space in R. The Elements of this space are called vectors.
The element O = (0, 0, ..., 0) is called zero vector.
The vector −x = (−x1 , −x2 , ..., −xn ) is called inverse vector of x = (x1 , x2 , ..., xn ).
• Subtraction of two vectors:
Let x, y ∈ Rn . We define x − y = x + (−y ).
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Vector space
Property of vector space
Property of vector space
Vector space has the following properties
1)
Vector space is unique. Vector 0 of Rn , denoted 0Rn or 0.
0Rn = 0 ; 0R2 = (0, 0) ; 0R3 = (0, 0, 0)
2)
Inverse vector of x is also unique.
Convention: Inverse vector of x is denoted −x
3)
Addition has a reduction rule. It mean x, y , z ∈ Rn , x + y = x + z ⇒ y = z.
4)
Multiplication has a reduction rule for a non-zero number. It mean
x, y ∈ Rn ; λ ∈ R, λ ̸= 0 ; λx = λy ⇒ x = y .
5)
6)
Let x, y ∈ Rn . Then λ(x − y ) = λx − λy , λ ∈ R.
λ=0
n
Cho x ∈ R , λ ∈ R. Then λx = 0 ⇔
x =0
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Vector space
Linear relationship between vectors
linear combination (1)
Let {α1 , α2 , ..., αm } ⊂ Rn . Then α ∈ Rn is linear combination of α1 , α2 , ..., αm if
exist a1 , a2 , ..., am ∈ R such that α = a1 α1 + a2 α2 + ... + am αm .
m
P
Linear combination
ai αi of (αi )i=1,m is trival if a1 = a2 = ... = am = 0 ∈ R.
i=1
Else if there is at least one coefficient aj ̸= 0(1 ≤ j ≤ n), then the linear
m
P
combination
ai αi is called no trivial.
i=1
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Vector space
Linear relationship between vectors
linear combination (2)
Example 4.1
In R3 , let the following vector
α1 = (1, 3, −2) ; α2 = (0, 1, −1) ; α3 = (2, 0, 3) ; α = (−2, 1, −1) ;
Is α a linear combination of α1 , α2 , α3 ?
Proof. The vector α is a linear combination of the vectors α1 , α2 , α3 because
α = −6α1 + 19α2 + 2α3 .
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Vector space
Linear relationship between vectors
linear combination (3)
How to check if α is a linear combination of α1 , α2 , ..., αm ?
We have α is a linear combination of α1 , α2 , ..., αm when the following equation
exists solution.
α = a1 α1 + a2 α2 + ... + am αm
(11)
In particular, in the case of the space Rn . Suppose
α = (b1 , b2 , ..., bn )
α1 = (a11 , a21 , ...., an1 )
α2 = (a12 , a22 , ...., an2 )
......
αm = (a1m , a2m , ...., anm )
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Vector space
Linear relationship between vectors
linear combination (4)
Then


a11 α1 + a12 α2 + ... + a1m αm = b1



a α + a α + ... + a α = b
21 1
22 2
2m m
2
(11) ⇔

.......




an1 α1 + an2 α2 + ... + anm αm = bn
a11
a21
Matrixize the (12), we have 
 ...
an1

a12
a22
...
an2
(12)

... a1m b1
... a2m b2 
.
... ... . 
... anm bn
That mean
T
α1
DANG LE QUANG, Ph.D
α2T
...
T
αm
| αT .
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Vector space
Linear relationship between vectors
linear combination (5)
Thus, to check that α is a linear combination of α1 , α2 , ...αm in Rn , we apply the
following steps:
Step 1: Make an extended matrix
T
α1
α2T
...
T
αm
| αT
(13)
Step 2: Solve linear system (13)
- If (13) has no solution, then α is not a linear combination of α1 , α2 , ..., αm .
- if (13) has solution a1 , a2 , ..., am then α is a linear combination of
α1 , α2 , ..., αm and has the following form
α = a1 α1 + a2 α2 + .... + am αm .
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Vector space
Linear relationship between vectors
linear combination (6)
Example 4.2
In R3 , let α1 = (1, 2, 5) ; α2 = (1, 3, 7) ; α3 = (−2, 3, 4) ; α = (4, 3, 5).
Is α a linear combination of α1 , α2 , α3 ?
Proof. We have
h
T
T
T
α1 α2 α3 | α
T
i

1
= 2
5
−2
3
4

1
d →d1 −d2
−−1−−−
−−→ 0
d3 →d3 −2d2
0
1
3
7


4
1
d2 →d2 −2d1
3 −−−−−−−→ 0
d3 →d3 −5d1
5
0

0 −9 9
1
7 −5 .
0
0 −5
1
1
2

−2 4
7 −5 
14 −15
This system has no solution since
0α1 + 0α2 + 0α3 = −5.
Thus α isn’t a linear combination of α1 , α2 , α3 .
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Vector space
Linear relationship between vectors
linear independence and linear dependence (1)
Definition 4.1 (linear independence and linear dependence)
Let α1 , α2 , ..., αm ∈ Rn . Consider the equation
a1 α1 + a2 α2 + ... + am αm = 0
(14)
- If (14) only has a trival solution a1 = a2 = ... = am = 0 then
α1 , α2 , ..., αm (or {α1 , α2 , ..., αm }) are linear independence.
- If (14) has a non-trivial solution, then we say α1 , α2 , ..., αm (or
{α1 , α2 , ..., αm }) is linear dependence.
In other word
- If (14) has a unique solution, then α1 , α2 , ..., αm is linear independence.
- If (14) has infinite solutions, then α1 , α2 , ..., αm is linear dependence.
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Vector space
Linear relationship between vectors
linear independence and linear dependence (2)
Recall: Let a homogeneous linear system AX = 0 having m variables. Then
rank(A) = rank([A|O]) = rank(A), with A is a extended matrix (Review in 89).
Furthermore, applying Theorem 3.1 Kronecker-Capelli, we have
- If rank(A) = rank(A) = m, then system only has a trivial solution.
- If rank(A) = rank(A) < m then system has infitine solutions.
Recall: (review the remark in page 89) Let A be a square matrix of order n.
Then the following statements are equivalent:
i)
rank(A) = n.
ii)
detA ̸= 0.
iii)
System AX = 0 only has a trivial solution.
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Vector space
Linear relationship between vectors
linear independence and linear dependence (3)
How to check for linear independence/dependence of vectors
α1 , α2 , ..., αm in Rn .
Step 1: Make matrix A A by stacking α1 , α2 , ..., αm into columns or into rows.
Step 2: Find rank of A, i.e. find rank(A).
- If rank(A) = m (rank = vector numbers) then α1 , α2 , ..., αm are linear
independence.
- If rank(A) < m (rank ¡ vector numbers) then α1 , α2 , ..., αm are linear
dependence.
In case m = n (ie vector numbers = all element vector numbers), we have A as
square matrix. Then you can replace Step 2 with the following Step 2’
Step 2’: Find detA.
- If detA ̸= 0 then α1 , α2 , ..., αm are linear independence.
- If detA = 0 then α1 , α2 , ..., αm are linear dependence.
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Vector space
Linear relationship between vectors
linear independence and linear dependence (4)
Example 4.3
In R3 , let
α1 = (2m+1, −m, m+1); α2 = (m−2, m−1, m−2); α3 = (2m−1, m−1, 2m−1).
Find the condition of m so that α1 , α2 , α3 are linear independence.


2m + 1 −m
m+1
Proof. Make A =  m − 2 m − 1 m − 2 . We have
2m − 1 m − 1 2m − 1
2m + 1
detA = m − 2
2m − 1
= m(−1)1+1
−m
m−1
m−1
m
m+1
m−2 = 0
0
2m − 1
m−1
m−1
m−2
2m − 1
−m
m−1
m−1
m+1
m−2
2m − 1
(take c1 → c1 − c3 )
(expand the determinant by column 1)
= m(m − 1)(m + 1)
Then, α1 , α2 , α3 are linear independence if and only if
detA ̸= 0 ⇔ m(m − 1)(m + 1) ̸= 0 ⇔ m ̸= 0 và m ̸= ±1.
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Vector space
The rank of the vector system and the number of dimensions of the vector space
The rank of the vector system (1)
• Maximally Linearly Independent Subsystem: Let vectors α1 , α2 , ..., αm in
Rn , or we can also understand that let a system consisting of m vectors
{α1 , α2 , ..., αm }
(α)
Subsystem of vector system (α) is a vector system consisting of some (or all) of
the system’s vectors.
The subsystem {αi1 , αi2 , ..., αik } of the system (α) is called the maximally linearly
independent subsystem if the system {αi1 , αi2 , ..., αik } is linear independence, and
if we add any other vectors of the system (α) to that subsystem, we get a system
be a linear dependence.
Remark: A vector system can have many different maximally linearly
independent subsystems, but the number of vectors of the maximally linearly
independent subsystems are always equal. That number is called rank of the
system (α), denoted rank(α).
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Vector space
The rank of the vector system and the number of dimensions of the vector space
The rank of the vector system (2)
How to find the maximally linearly independent subsystem,
the rank of the vector system in Rn
In Rn , let system consist m vectors
{α1 , α2 , ..., αm }
(α)
To find the maximally linearly independent subsystem of the system (α), we do
the following:
- Make the matrix A, the rows of A are the vectors αi .
- Use elementary transformations on the row to return A to the ladder matrix
A′ .
Then the rank of the system (α) is equal to the rank of the matrix A and the
maximally linearly independent subsystem of (α), consists of corresponding
vectors to the non-zero rows of the ladder matrix A′ .
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Vector space
The rank of the vector system and the number of dimensions of the vector space
The rank of the vector system (3)
Example 4.4
In R5 , let α1 = (1, 2, −3, 5, 1); α2 = (1, 3, −13, 22, −1); α3 = (3, 5, 1, −2, 5).
Find rank of system (α) = {α1 , α2 , α3 } and point out their maximally linearly
independent subsystem.
Proof. Make
  
α1
1
A = α2  = 1
α3
3
2
3
5
−3
−13
1
5
22
−2


1
1
d2 →d2 −d1
−1 −−−−−−−→ 0
5
d3 →d3 −3d1
2
1
−1
0

1 2
d3 →d3 +d2
−−
−−−−→ 0 1
0 0
−3
−10
10
−3
−10
0
5
17
0
5
17
−17

1
−2
2

1
−2 = A′
0
Then, the rank of the system (α) = rank(A) = 2 and the maximally linearly
independent subsystem of the system (α) is (1, 2, −3, 5, 1), (0, 1, −10, 17, −2).
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Vector space
The rank of the vector system and the number of dimensions of the vector space
The rank of the vector system (4)
Example 4.5
In R4 , let α1 = (1, 1, 2, 2); α2 = (2, 3, 6, 6); α3 = (3, 4, 8, 8); α4 =
(8, 11, 22, 22); α5 = (5, 7, 14, 14).
Find rank of system (α) = {α1 , α2 , α3 , α4 , α5} and point out their maximally
linearly independent subsystem.
Example 4.6
Find rank of system and their maximally linearly independent subsystem.
{β1 = (1, −1, 0, 1); β2 = (1, 0, −1, −2); β3 = (0, 1, −1, 2)} in R4 .
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Vector space
The rank of the vector system and the number of dimensions of the vector space
Bases, dimension numbers, coordinates (1)
• Bases, dimension numbers:
Definition 4.2 (bases)
Vector system
{α1 , α2 , ..., αm }
(α)
in Rn is called a base of Rn if system (α) is linear independence and all
vectors in Rn are linear combination of (α).
Example 4.7
In R3 , let
(α) = {α1 = (1, 1, 1) ; α2 = (1, 2, 1) ; α3 = (2, 3, 1)}.
Is (α) a base in R3 ?
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Vector space
The rank of the vector system and the number of dimensions of the vector space
Bases, dimension numbers, coordinates (2)
Proof. We check if (α) = {α1 , α2 , α3 } is linearly independent?
  
1
α1
α2  = 1
2
α3
1
2
3

1
1
1
We have rank(A) = 3 (or detA = −1). It follows that (α) is linearly independent.
Next, put A = (x, y , z) ∈ R3 , we check if A is a linear combination of α1 , α2 , α3 ?
(review method in page 97). Make an extended matrix
T
α1
α2T

1
T
T

α3 |A = 1
1
1
2
1


1
2 x


3 y −→ 0
0
1 z
1
1
0

2
x
1 −x + y 
−1 −x + z
The system has a solution, so A is a linear combination of {α1 , α2 , α3 }.
So (α) is the basis of R3 .
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Vector space
The rank of the vector system and the number of dimensions of the vector space
Bases, dimension numbers, coordinates (3)
Remark 4.1
- In Rn , consider system vector
e1 = (1, 0, ..., 0)
e2 = (0, 1, ..., 0)
..
.
en = (0, 0, ..., 1)
System {e1 , e2 , e3 } is a basic of Rn . This basis is called canonical basis of
Rn , denoted (Cn ).
- In the vector space Rn , the system of n linearly independent vectors is a
basis of that space.
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Vector space
The rank of the vector system and the number of dimensions of the vector space
Bases, dimension numbers, coordinates (4)
Definition 4.3 (dimesion numbers)
A vector space can have many bases, but the number of vectors in each basis
is the same. The number of vectors in the base is called dimensionnumbers of
the space.
The space Rn has dimension numbers equal to n, denoted dimRn = n.
Definition 4.4 (Vector coordinates )
Let (α) = {α1 , α2 , ..., αn } be a base of Rn . Then every vector u ∈ Rn can be
uniquely written in the form
u = a1 α1 + a2 α2 + ... + an αn ,
with a1 , a2 , ..., an ∈ R
We call the multiple (a1 , a2 , ..., an ) be the coordinates of the vector u in the
base (α), denoted by u/(α) = (a1 , a2 , ..., an ).
T
We also denote [u]/(α) = a1 a2 . . . an or [u](α) .
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Vector space
The rank of the vector system and the number of dimensions of the vector space
Bases, dimension numbers, coordinates (5)
Remark 4.2
In Rn with canonical basis
(Cn ) = {e1 = (1, 0, ..., 0) ; e2 = (0.1, ..., 0) ; ... ; en = (0, ..., 0, 1)} every vector
x = {x1 , x2 , ..., xn } ∈ Rn has coordinates in (Cn ) is itself
[x](Cn ) = (x1 , x2 , ..., xn )T
because x = (x1 , x2 , ..., xn ) = x1 e1 + x2 e2 + ... + xn en .
Method to find coordinates of vector u in base (α) in Rn , i.e. find [u](α)
Assume (α) = {α1 , α2 , ..., αn } is a base and u ∈ Rn . Solve α1T α2T . . . αnT |u T .
T
Assume (a1 , a2 , ..., an ) is the solution. Then [u](α) = a1 a2 . . . an .
Convention: Coordinates in Linear Algebra are written in columns.
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Vector space
The rank of the vector system and the number of dimensions of the vector space
Bases, dimension numbers, coordinates (6)
Example 4.8
In R3 , let (α) = {α1 = (1, 2, 1), α2 = (1, 3, 1), α3 = (2, 5, 3)}.
a) Show that (α) is a basis of R3 .
b) Find the coordinates of the vector u = (a, b, c) on the basis (α).
Proof. a) (do it yourself)
b) With u = (a, b, c) ∈ R3 , to find [u](α) we make a system of equations
T
α1
α2T
α3T |u
T

1
= 2
1
1
3
1


2 a
1
5 b  −→ 0
3 c
0
0
1
0

0 4a − b − c
0 −a + b − c 
1
−a + c

4a − b − c
= −a + b − c  .
−a + c

So [u](α)
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Vector space
The rank of the vector system and the number of dimensions of the vector space
Bases, dimension numbers, coordinates (7)
• Base transfer matrix, Change of basis formula
Definition 4.5 (Base transfer matrix)
In Rn , let
(α) = {α1 , α2 , ..., αn },
(β) = {β1 , β2 , ..., βn
be 2 bases of Rn . Set
Tαβ = [β1 ](α)
[β2 ](α)
... [βn ](
alpha)
.
Then P is called base transfer matrix from base (α) to base (β), denoted
[(α) → (β)] = Tαβ .
Remark 4.3
If (α) = {α1 , α2 , ..., αn } is a base of Rn and
(Cn ) = {e1 = (1, 0, ..., 0), e2 = (0, 1, ..., 0), e3 = (0, 0, ..., 1)} is the canonical
basis of Rn then
Tαβ = α1T α2T ... αnT
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Vector space
The rank of the vector system and the number of dimensions of the vector space
Bases, dimension numbers, coordinates (8)
Way to find Tαβ in Rn :
Assume (α) = {α1 , α2 , ..., αn } ; (β) = {β1 , β2 , ..., βn } are two bases of Rn .
Step 1: Make the extended matrix α1T α2 ... αnT | β1T β2T ... βnT .
Step 2: Using elementary transformations on the row to return this matrix to the
form [In |Tαβ ]. Then Tαβ is the base transfer matrix from base (α) to base (β).
• Coordinate conversion formula: In Rn , let (α), (β) be 2 bases of Rn . Then
[u](α) = Tαβ [u](β) ,
∀u ∈ Rn .
Theorem 4.1
In Rn , let (α), (β), (γ) are the bases of Rn . Then
i)
Tαα = In .
ii)
[u](α) = Tαβ [u](β) ,
iii)
−1
Tβα = Tαβ
.
iv)
Tαθ = Tαβ .Tβθ .
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∀u ∈ Rn .
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Vector space
The rank of the vector system and the number of dimensions of the vector space
Bases, dimension numbers, coordinates (9)
Example 4.9
In R3 , let
(α) = {α1 = (1, 1, 3), α2 = (1, −2, 1), α3 = (1, −1, 2)}
(β) = {(β1 = (1, −2, 2), β2 = (1, −2, 1), β3 = (1, −1, 2)}.
a) Prove that (α) and (β) are two bases of R3 .
b) Find the base transfer matrix from (α) to (β).
T
c) Let u ∈ R3 satisfied [u](β) = 2 −3 −2 . Find [u](α) .
Proof. a) (Do it yourself)
b) Make an extended matrix
h
T
T
T
T
T
α1 α2 α3 | β1 β2 β3
DANG LE QUANG, Ph.D
i

1
= 1
3
1
−2
1
1 1
−1 −2
2 2
1
−2
1
LINEAR ALGEBRA


1
1
−1 → 0
2
0
0
1
0
0 −1
0 −1
1 3
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1
0

0
0
1
118 / 178
Vector space
The rank of the vector system and the number of dimensions of the vector space
Bases, dimension numbers, coordinates (10)

−1
= −1
3

0 0
Which implies Tαβ
1 0.
0 1

−1
c) We have [u](α) = Tαβ [u](β) = −1
3
   
0 0
2
−2
1 0 −3 = −5 .
0 1 −2
4
Example 4.10
In R3 , let (C3 ) be the canonical basis and let the base
(β) = {(β1 = (1, −1, 1)), β2 = (2, 3, 1), β3 = (1, 2, 1)}
a) Find the base transfer matrix from (C3 ) to (β).
b) Find the base transfer matrix from (β) to (C3 ).
c) Let α = (1, 2, 3) ∈ R3 . Find [α](β) .
d) Find the vector β4 ∈ R3 whose coordinates in (β) are (2, 3, 5).
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Vector space
vector subspace
vector subspace (1)
Let U be a non-empty subset of Rn . The set U is called vector subspace
(subspace) of Rn if
i)
∀α, β ∈ U then α + β ∈ U.
ii)
∀a ∈ R, α ∈ U then aα ∈ U.
Remark 4.4
In Rn , let U be a non-empty subset of Rn
- If U is a subspace of Rn then the vector 0 ∈ U.
- If vector 0 ∈
/ U then U is not a subspace of Rn .
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Vector space
vector subspace
vector subspace (2)
How to check the Subspace:
Let U be a subset of Rn . To check that U is a subspace of Rn , we perform the
following steps:
Step 1: Check vector 0 ∈ U.
- If vector 0 ∈
/ U, then conclude U is not a subspace of Rn −→. Stop.
- If vector 0 ∈ U, then go to Step 2.
Step 2: For every α, β ∈ U and every a ∈ R.
- If α + β ∈ U and aα ∈ U, then conclude U is a subspace of Rn .
- On the contrary, we need to give a contradiction, i.e., we show a concrete
example to show that α, β ∈ U but α + β ∈
/ U; or α ∈ U, a ∈ R but
n
aα ∈
/ U. Then U is not a subspace of R .
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Vector space
vector subspace
vector subspace (3)
Example 4.11
Let U = {(x1 , x2 , x3 ) ∈ R3 |x1 = 2x2 x3 }. Is U a subspace of R3 ?
Proof. With α = (2, 1, 1) and β = (4, 2, 1). We have α, β ∈ U but
α + β = (6, 3, 2) ∈
/ U (because 6 ̸= 2.3.2). So U is not a subspace of R3 .
Example 4.12
Let U = {(x1 , x2 , x3 ) ∈ R3 |x1 + 3x2 + x3 = 1}. Is U a subspace of R3 ?
Proof. We have the vector 0 = (0, 0, 0) ∈
/ U because 0 + 3.0 + 0 = 0 ̸= 1. So U
3
is not a subspace of R .
Example 4.13
Let U = {(x1 , x2 , x3 ) ∈ R3 | 2x1 + x2 − x3 = 0}. Is U a subspace of R3 ?
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Vector space
vector subspace
vector subspace (4)
Proof. We have
• Vector 0 = (0, 0, 0) ∈ U (because 2.0 + 0 − 0 = 0). So U ̸= ∅.
• For all a ∈ R, α = (α1 , α2 , α3 ), β = (β1 , β2 , β3 ) ∈ U, i.e.
2α1 + α2 − α3 = 0,
2β1 + β2 − β3 = 0.
We have α + β = (α1 + β1 , α2 + β2 , α3 + β3 ). Futhermore,
2(α1 + β1 ) + (α2 + β2 ) − (α3 + β3 ) = 0.
Which implies α + β ∈ U. We have aα = (aα1 , aα2 , aα3 ). Futhermore,
2aα1 + aα2 − aα3 = a(2α1 + α2 − α3 ) = a0 = 0.
Which impies aα ∈ U. Thus U is a subspace of R3 .
Example 4.14
Which of the following sets is a subspace of R2 ?
a) U1 = {(x1 , x2 ) ∈ R2 |x2 = 3x1 }.
b) U2 = {(x1 , x2 ) ∈ R2 |x2 = 2 + 3x1 }.
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Vector space
vector subspace
Subspace generated by a vector system (1)
Definition 4.6 ( subspace generated by a vector system)
In Rn , let the vector system {α1 , α2 , ..., αm }. Then the set of all linear
combinations of vectors α1 , α2 , ..., αm , denoted ⟨α1 , α2 , ..., αm ⟩ is a subspace in
Rn . That space is called subspace of Rn generated by the vector system
{α1 , α2 , ..., αm } (also called linear bound of the vector system {α1 , α2 , ..., αm }).
⟨α1 , α2 , ..., αm ⟩ = {a1 α1 + a2 α2 + ... + am αm |ai ∈ R}.
Note: A base of ⟨α1 , α2 , ..., αm ⟩ is the maximally linearly independent subsystem
of the system {α1 , α2 , ..., αm }.
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Vector space
vector subspace
Subspace generated by a vector system (2)
Example 4.15
In R2 , we consider (α) = {α = (1, 2)}. Then
⟨α⟩ = {a(1, 2) |a ∈ R} = {(a, 2a) |a ∈ R}.
Example 4.16
In R3 , we consider
(α) = {α1 = (1, 2, 1), α2 = (−1, 2, 0)}.
Then
⟨α1 , α1 ⟩ = {tα1 + sα2 | t, s ∈ R} = {(t − s, 2t + 2s, t) |t, s ∈ R}.
Example 4.17
In R3 , find a base, the dimension numbers of the subspace generated by the
following vector system
{α1 = (1, 1, 1), α2 = (2, 3, 4), α3 = (4, 5, 6)}.
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Vector space
vector subspace
Solution space of a homogeneousl linear system (1)
Consider the following homogeneousl linear system


a11 x1 + a12 x2 + ... + a1n xn = 0



a x + a x + ... + a x = 0
21 2
22 2
2n n
.........




am1 x1 + am2 x2 + ... + amn xn = 0
with the matrix form
AX = 0
(15)
Solution space: Let the solution set of the system (15) be
L = {x = (x1 , x2 , ..., xn ) ∈ Rn : AX = 0}.
Theorem 4.2
L is a subspace of Rn and if rank(A) = r then dimL = n − r .
We call L the solution space of the homogeneous linear system.
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Vector space
vector subspace
Solution space of a homogeneousl linear system (2)
Method to find the base of the solution space:
Step 1: Solve the system (15), find the general solution.
Step 2: Give the free implicit set the values in turn
(1, 0, ..., 0), ...., (0, 0, ..., 1).
we get the basic solutions x1 , x2 , ..., xn .
Step 3: Then, the base of the solution space is {x1 , x2 , ..., xn }.
Example 4.18
Find the base and dimension numbers of the solution space of the linear system.


x1 + 2x2 − 3x3 + 5x4 = 0



x + 3x − 13x + 22x = 0
1
2
3
4

3x
+
5x
+
x
−
2x
=
0
 1
2
3
4



2x1 + 3x2 + 4x3 − 7x4 = 0
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Vector space
vector subspace
Solution space of a homogeneousl linear system (3)
Proof. Matrixize the system of equations, we have

1 2
1 3
A=
3 5
2 3
DANG LE QUANG, Ph.D


−3
5
1
−13 22 
d2 →d2 −d1 , d3 →d3 −3d1 0
 −−−−−−−−−−−−−−→ 
0
d4 →d4 −2d1
1
−2
4
−7
0

1
d1 →d1 −2d2 , d3 →d3 +d2 0
−−−−−−−−−−−−−−→ 
0
d4 →d4 +d2
0
LINEAR ALGEBRA
2
1
−1
−1
0
1
0
0

−3
5
−10 17 

10 −17
10 −17

17 −29
−10 17 

0
0 
0
0
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Vector space
vector subspace
Solution space of a homogeneousl linear system (4)
The solution of the system is
x = (x1 , x2 , x3 , x4 ) = (−17t + 29s, 10t − 17s, t, s),
with t, s ∈ R.
Given t = 1, s = 0 and t = 0, s = 1, respectively, the basic solutions of the
system are
X1 = (−17, 10, 1, 0) ; X2 = (29, −17, 0, 1).
Therefore, if L is a solution space then B = {X1 , X2 } is the base of L and
dimL = 2.
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Linear Model in economic analysis
Section 5: Linear Model in economic analysis
Dang Le Quang, Ph.D
DANG LE QUANG, Ph.D
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September - 2022
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Linear Model in economic analysis
Interdisciplinary equilibrium model (Input-Output Leontief)
Interdisciplinary equilibrium model (I/0 Leontief) (1)
This model is also known as the I/O model. It refers to determining the level of
aggregate demand for the products of each industry in the economy as a whole.
Within the framework of the model, the concept of industry is considered in a
purely productive sense. The following assumptions are made:
1)
Each industry produces a homogeneous product or produces a combination
of goods a certain rate. In the second case, we consider each of these
fixed-proportion combinations of goods to be one side row.
2)
The inputs of production within an industry are used in a fixed proportion.
Aggregate demand for each industry’s product includes:
- Intermediary demand from manufacturers who use that type of product for
the production process.
- The final demand from users to use the product for consumption or export,
including households the family, the state, the exporters.
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Linear Model in economic analysis
Interdisciplinary equilibrium model (Input-Output Leontief)
Interdisciplinary equilibrium model (I/0 Leontief) (2)
Suppose an industry economy consists of n industries: industry 1, industry 2, . . . ,
industry n and in addition one other part of the economy (called the open
economy), it does not produce goods like the n industry above, but only
consumption of the product of this industry n. To facilitate the cost of factors of
production, we express the quantity demanded of all goods in terms of value, i.e.
measured in money (assuming the market is stable determined). The aggregate
demand for industry goods i is calculated by the formula:
xi = xi1 + xi2 + ... + xin + bi ,
i = 1, 2, ..., n.
(16)
Where
xi is the industry’s aggregate demand for goods i.
xik is the value of the good of industry i that industry k needs to use for
production (intermediate demand).
bi is the value of industry goods i that is consumed and exported (final
demand).
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Linear Model in economic analysis
Interdisciplinary equilibrium model (Input-Output Leontief)
Interdisciplinary equilibrium model (I/0 Leontief) (3)
Transform (16)
xi =
xi1
xi2
xin
x1 +
x2 + ... +
xn + b i ,
x1
x2
xn
i = 1, 2, ..., n.
Put
aik =
xik
,
xk
i, k = 1, 2, ..., n.
(17)
We get the system of equations (the Input-Output Liontief model or the
production equation):


x1 = a11 x1 + a12 x2 + ... + a1n xn + b1


x = a x + a x + ... + a x + b
2
21 1
22 2
2n n
2

...




xn = an1 x1 + an2 x2 + ... + ann xn + bn


(1 − a11 )x1 − a12 x2 − ... − a1n xn = b1



−a x + (1 − a )x − ... − a x = b
21 1
22 2
2n n
2
⇔
...




−an1 x1 − an2 x2 − ... + (1 − ann )xn = bn
(18)
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Linear Model in economic analysis
Interdisciplinary equilibrium model (Input-Output Leontief)
Interdisciplinary equilibrium model (I/0 Leontief) (4)
The matrix form is
X = AX + B
or
(I − A)X = B
(5.3’)
Where
a11
a21
A=
 ...
an1

a12
a22
...
an2

... a1n
... a2n 
;
... ... 
... ann
is the input cost factor matrix or the engineering coefficientmatrix
T
X = x1 x2 ... xn
is the total demand matrix (or production vector);
T
B = b1 b2 ... bn
is the terminal bridge matrix .
From (17), the element aik of A is the fraction of the cost of industry k paid for
the purchase of goods by industry i per unit value of the good of industry.
industry k (factorial cost of production).
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Linear Model in economic analysis
Interdisciplinary equilibrium model (Input-Output Leontief)
Interdisciplinary equilibrium model (I/0 Leontief) (5)
Example aik = 0.2 means that to produce 1USD worth of its goods (on average),
industry k must buy 0.2USD of industry goods i.
Under assumption 2 we have aik constant. We call aik the factor cost factor or
technical coefficient, so 0 ≤ aik < 1.
In the matrix A, the elements of the line i are the coefficients of the value of
goods of industry i sold to all industries. intermediate goods (including industry
i), and column k is the coefficient of commodity value of industry k purchased by
industries for use used for themselves to produce their goods (including industry
k). Sum of all elements of column k is the cost level industry’s k has to pay for
the purchase of factors of production per 1 of its value, thus:
a1k + a2k + ... + ank < 1,
k = 1, 2, ..., n.
The equation (5.3’) allows us to determine the aggregate demand for a good
across all industries. This has important implications for production planning to
ensure smooth running of the economy. Avoid excess or shortage of goods.
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Linear Model in economic analysis
Interdisciplinary equilibrium model (Input-Output Leontief)
Interdisciplinary equilibrium model (I/0 Leontief) (6)
Theorem 5.1
Assume A is the input coefficient matrix of an economy and B is the final
demand. If the elements of A and B is not negative and the sum of the elements
per column of A is less than 1 then (I − A)−1 exists and the aggregate demand
matrix X = (I − A)−1 B. The matrix I − A is called the matrix Liontief or the
technology coefficient matrix.
Example 5.1
Suppose in an economy, there are 3 manufacturing industries: industry 1,
industry 2, industry 3. Show the matrix of technical coefficients

0.2
0.4
0.1
0.3
0.1
0.3

0.2
0.2
0.2
a) Explain the meaning of the number 0.4 in the matrix A.
b) Indicate that the final demand for goods of industries 1, 2, 3 is 10,
respectively; 5; 6 million USD. Let’s Determine the level of aggregate demand
DANG LE
QUANG, Ph.D
LINEAR ALGEBRA
September - 2022
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for each
industry.
Linear Model in economic analysis
Interdisciplinary equilibrium model (Input-Output Leontief)
Interdisciplinary equilibrium model (I/0 Leontief) (7)
Proof. b) We have

0.8
I − A = −0.4
−0.1
−0.3
0.9
−0.3



−0.2
0.66 0.30 0.24
1
0.34 0.62 0.24
−0.2 ⇒ (I − A)−1 =
0, 384
0.8
0.21 0.27 0.60
Aggregate demand matrix is

0.66
1 
X = (I − A)−1 B =
0.34
0, 384
0.21
0.30
0.62
0.27
  

0.24
ten
24.84
0.24  5  = 20.68
0.60
6
18.36
Thus, the aggregate demand for goods of industry 1 is 24.84; for goods of
industry 2 is 20.68; for goods of industry 3 is 18.36 (million USD).
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Linear Model in economic analysis
Market equilibrium model n related goods
Market equilibrium model n related goods (1)
• The model of market equilibrium for a commodity: In economic analysis,
we use the supply function QS and the demand function QD to express the
dependence quantity supplied , quantity demanded in the price of the good P
assuming all other factors remain unchanged. Assume QS and QD have linea form
QS = −a + bP ; QD = c − dP,



QS = −a + bP
Market equilibrium model is
QD = c − dP


Q = Q
S
a, b, c, d > 0.



QS = −a + bP
D
⇔
QD = c − dP


−a + bP = c − dP
Derive the equilibrium price is
P=
a+c
b+d
Balance amount is
QS = QD =
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cd − ad
b+d
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Linear Model in economic analysis
Market equilibrium model n related goods
Market equilibrium model n related goods (2)
• Market equilibrium model fo n related goods: In a multi-commodity
market, the value of the good can affect the quantity supplied and the demand
supplied for other commodities. The linear demand and supply functions of the
market for n goods have the form:
QSi = ai 0 + ai 1 P1 + ai 2 P2 + ... + ain Pn
QDi = bi 0 + bi 1 P1 + bi 2 P2 + ... + bin Pn ,
i = 1, 2, ..., n.
Where QSi , QDi , Pi are respectively the quantity supplied, the quantity
demanded, the price of the good i. The market equilibrium model for n goods is
QSi = QDi ,
i = 1, 2, ..., n.
Converting cik = aik − bik we get the system

c11 P1 + c12 P2 + ... + c1n Pn = −c10




c21 P1 + c22 P2 + ... + c2n Pn = −c20

...



cn1 P1 + cn2 P2 + ... + cnn Pn = −cn0
(19)
Solving the (19) system we get the equilibrium price of n of the good, from
which the equilibrium supply and demand are found.
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Linear Model in economic analysis
Market equilibrium model n related goods
Market equilibrium model n related goods (3)
Example 5.2
The market consists of three types of goods. The supply and demand
functions at prices are determined by:
QD1 = 8 − 2P1 + P2 + P3
QD2 = 10 + P1 − 2P2 + P3
QD3 = 14 + P1 + 2P2 − 2P3
QS1 = −5 + 4P1 − P2 − P3
QS2 = −2 − P1 + 4P2 − P3
QS3 = −1 − P1 + P2 + 4P3
Determine the market equilibrium.
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September - 2022
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Linear Model in economic analysis
Market equilibrium model n related goods
Market equilibrium model n related goods (4)
Proof. System of equations for determining the equilibrium price




8 − 2P1 + P2 + P3 = −5 + 4P1 − P2 − P3


6 − 2P2 − 2P3 = 13
10 + P1 − 2P2 + P3 = −2 − P1 + 4P2 − P3


14 + P + 2P − 2P = −1 − P + P + 4P
1
2
3
1
2
3
⇔
−2P1 + 6P2 − 2P3 = 12


−2P − P + 6P = 15.
1
2
3
Solving this system we get the equilibrium point
P1 =
DANG LE QUANG, Ph.D
425
52
391
; P2 =
; P3 =
.
72
9
72
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Linear Model in economic analysis
Market equilibrium model n related goods
National income equilibrium model (1)
Consider the model given the form



Y
C


T
= C + I0 + G0
= a + b (Y − T ),
= d + tY ,
(20)
a > 0, 0 < b < 1
d > 0, 0 < t < 1.
Where
Y is the gross national income,
C is residential consumption,
I0 is the planned fixed investment.
G0 is a fixed level of government spending.
T is tax.
Transform (20) we get a system of equations with three variables



Y − C = I0 + G0
(21)
−bY + C + bT = a


−tY + T = d
Solving the (21) system we get the equilibrium level of national income,
consumption and tax rates.
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Linear Model in economic analysis
Market equilibrium model n related goods
National income equilibrium model (2)
Example 5.3
Let gross national income Y , consumption C and tax rate T determine by
Y = C + I0 + G0
C = 15 + 0.4(Y − T )
T = 36 + 0.1Y .
where I0 = 500 is fixed investment, G0 = 20 is fixed expenditure
Find the equilibrium level of national income, consumption, and tax rate
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Linear Model in economic analysis
Market equilibrium model n related goods
National income equilibrium model (3)
Proof. We have



Y = C + 500 + 20
C


T
⇔
= 15 + 0.4(Y − T ) ⇔
= 36 + 0.1Y .



C = Y − 520
T = 0.1Y + 36


0.64Y = 520.6
⇔



C = Y − 520
T = 0.1Y + 36


Y − 520 = 15 + 0.4(Y − 0.1Y − 36)



C = 293.4375
T = 117.34375


Y = 813.4375
So Y = 813.4375 ; C = 293.4375 ; T = 117.34375.
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Linear Model in economic analysis
Equilibrium market for goods and money (IS-LM model)
Equilibrium market for goods and money (IS-LM model) (1)
The IS-LM model is used to analyze the market equilibrium of the economy in
both market: commodity market and money market.
In the presence of the money market, the investment of I depends on the interest
rate of r . Suppose
I = a1 − b1 r ,
a1 , b1 > 0.
Consider the equilibrium income and consumption model of the form

(a)
 Y = c + I + G0
I = a1 − b1 r
(a1 , b1 > 0)
(b)

C = a + bY
(a > 0, 0 < b < 1)
(c)
(22)
Substituting (b) and (c) into (a), we get
Y = a + bY + a1 − b1 r + G0
⇔ b1 r = a + a1 + G0 − (1 − b)Y
(23)
The equation (23), which represents the relationship between interest rates and
income when the goods market is in equilibrium, is known as the IS equation.
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Linear Model in economic analysis
Equilibrium market for goods and money (IS-LM model)
Equilibrium market for goods and money (IS-LM model) (2)
In the money market, the quantity demanded of money L depends on income Y
and the interest rate r . Suppose
L = a2 Y − b2 r ,
a2 , b2 > 0.
Assume that the money supply is fixed at M0 . The money market equilibrium
condition is
M0 = a2 Y − b2 r ⇔ b2 r = a2 Y − M0
(24)
The equation (24) representing the equilibrium condition of the money market is
called the LM equation.
IS-LM model is a model that combines IS and LM into one system
IS
LM
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Linear Model in economic analysis
Equilibrium market for goods and money (IS-LM model)
Equilibrium market for goods and money (IS-LM model) (3)
From the model you determine the level of income Y and the interest rate r that
ensures equilibrium in both markets goods and currency.
For example, solve the system
(
b1 r = a + a1 + G0 − (1 − b)Y
b2 r
= a2 Y − M0
We found
(a + a1 + G0 )b2 + b1 M0
b1 a2 + (1 − b)b2
(a + a1 + G0 )a2 + (1 − b)M0
r=
b1 a2 + (1 − b)b2
Y =
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Linear Model in economic analysis
Equilibrium market for goods and money (IS-LM model)
Equilibrium market for goods and money (IS-LM model) (4)
Example 5.4
Let
G0 = 250 ; M0 = 4500 ; I = 34 − 15r
C = 10 + 0.3Y ; L = 22Y − 200r .
a) Make the IS equation.
b) Make the equation LM.
c) Find the equilibrium level of income and interest rate in the commodity
and money markets.
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Linear Model in economic analysis
Equilibrium market for goods and money (IS-LM model)
Equilibrium market for goods and money (IS-LM model) (5)
Proof. a) We have
Y = C + I + G0 ⇔ Y = (10 + 0.3Y ) + (34 − 15r ) + 250.
So the IS equation is 15r = 294 − 0.7Y .
b) The LM equation has the form
L = M0 ⇔ 22Y − 200r = 4500 ⇔ 200r = 22Y − 4500.
c) The equilibrium level of income Y and the interest rate r are the solution of
the system of equations
(
(
15r
= 294 − 0.7Y
15(0.11Y − 22.5) = 294 − 0.7Y
⇔
200r = 22Y − 4500
r = 0.11Y − 22.5
(
(
2.35Y = 631.5
Y = 268.72
⇔
⇔
r = 7.06.
r = 0.11Y − 22.5
So Y = 268.72 ; r = 7.06.
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Linear mapping and Quadratic form
Section 6: Linear mapping and Quadratic form
Dang Le Quang, Ph.D
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Linear mapping and Quadratic form
Linear map
Definition (1)
A mapping f : Rn → Rm is called linear mapping if f satisfies both of the
following properties:
f (x + y ) = f (x) + f (y ),
f (λx) = λf (x),
∀x, y ∈ Rn
(a)
n
∀x ∈ R , ∀λ ∈ R
(b)
If f is a linear map from Rn to Rn (f : Rn → Rn ) then f is called linear transform
on Rn .
Condition (a) in the definition is addition conservation, and condition (b) is
multiplication conservation. We can combine the above two conditions with the
following condition
f (λ1 x + λ2 y ) = λ1 f (x) + λ2 f (y ),
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LINEAR ALGEBRA
∀x, y ∈ Rn ; ∀λ1 , λ2 ∈ R.
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Linear mapping and Quadratic form
Linear map
Definition (2)
Example 6.1
Let the map f : R2 → R2 be defined by
f (x, y ) = (3x − 2y , x),
∀x, y ∈ R2
Is the f linear map?
Proof. ∀x, y ∈ R2 , i.e., x = (x1 , x2 ), y = (y1 , y2 ); ∀λ1 , λ2 ∈ R. we have
f (λ1 x + λ2 y ) = f (λ1 x1 + λ2 y1 , λ1 x2 + λ2 y2 )
= (3(λ1 x1 + λ2 y1 ) − 2(λ1 x2 + λ2 y2 ), λ1 x1 + λ2 y1 )
= λ1 (3x1 − 2x2 , x1 ) + λ2 (3y1 − 2y2 , y1 )
= λ1 f (x) + λ2 f (y ).
So f is a linear mapp.
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Linear mapping and Quadratic form
Linear map
Definition (3)
Example 6.2
Let A be a square matrix of order n in R, denoted Mn (R). Let Map
ϕ : Mn (R) → Mn (R) defined by ϕ(X ) = XA + AX , with X ∈ Mn (R). Prove
that ϕ is a linear map
Proof. For all X , Y ∈ Mn (R); λ1 , λ2 ∈ R. We have
ϕ(λ1 X + λ2 Y ) = (λ1 X + λ2 Y )A + A(λ1 X + λ2 Y )
= (λ1 X )A + (λ2 Y )A + A(λ1 X ) + A(λ2 Y )
= λ1 (XA) + λ1 (AX ) + λ2 (YA) + λ2 (AY )
= λ1 (XA + AX ) + λ2 (YA + AY )
= λ1 ϕ(X ) + λ2 ϕ(Y )
So ϕ is a linear map.
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Linear mapping and Quadratic form
Linear map
Definition (4)
Example 6.3
Let the map f : R2 → R2 defined by
√ √
f (x, y ) = ( 3 x, 3 y ),
∀x, y ∈ R2
Is the f linear map?
Proof. For all λ ∈ R, we √
consider
√
√ √
f (λ(x, y )) = f (λx, λy ) = ( 3 λx, 3 λy ) ̸= λ( 3 x, 3 y ) = λf (x, y ).
So f is not a linear map.
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Linear mapping and Quadratic form
Linear map
Definition (5)
Example 6.4
Let a map f : M2 (R) → R defined by f
a
c
map?
Proof. For all λ ∈ R, we have
a b
λa λb
a
2
f λ
= det
= λ .det
c d
λc λd
c
b
d
a
=det
c
b
a
̸ λ.det
=
d
c
b
. Is f a linear
d
b
a
= λf
d
c
b
d
the ”=” sign occurs when λ = 1, but by definition, we must consider ∀λ ∈ R.
So f is not a linear map.
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Linear mapping and Quadratic form
Linear map
Matrix of the linear map (1)
Definition 6.1
Let a linear map f : Rn → Rm and let (α) = {α1 , α2 , ..., αn } is the base of Rn ;
(β) = {β1 , β2 , ..., βn } is the base of Rm .
Then, the matrix
[f ](α),(β) = [f (α1 )](β) [f (α2 )](β) ... [f (αn )](β)
is called matrix of linear map f .
Example 6.5
Let f : R3 → R2 ; f (x1 , x2 , x3 ) = (x1 + x2 + x3 , x1 − x2 − x3 ) and bases
(A) = {e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1)}
(B) = {ε1 = (1, 0), ε2 = (0, 1)}
(α) = {α1 = (1, 1, 1), α2 = (1, 2, 2), α3 = (1, 2, 3)}
(β) = {β1 = (1, 1), β2 = (1, 2)}.
Find [f ](A),(B) and [f ](α),(β) .
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Linear mapping and Quadratic form
Linear map
Matrix of the linear map (2)
Proof. We have
[f ](A),(B) =
[f (e1 )](B)
=
[(1, 1)](B)
1
=
1
1
−1
[f (e2 )](B)
[f (e3 )](B)
[(1, −1)](B) [(1, −1)](B)
1
(see Comments 4.2 page 114)
−1
[f ](α),(β) =
[f (α1 )](β)
=
[(3, −1)](β)
[f (α2 )](β)
[f (α3 )](β)
[(5, −3)](β)
[(6, −4)](β)
(∗)
We calculate (∗) as follows: (see Definition ?? page ??). Let X = (x1 , x2 ).
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LINEAR ALGEBRA
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Linear mapping and Quadratic form
Linear map
Matrix of the linear map (3)
a
Assume [X ] = 1 . We infer
a2
X = a1 β1 + a2 β2 = a1 (1, 1) + a2 (1, 2)
⇔(x1 , x2 ) = (a1 , a1 ) + (a2 , 2a2 ) = (a1 + a2 , a1 + 2a2 ) ⇔
⇔
(
a1 + a2 = x1
a1 + 2a2 = x2
(
a1 = 2x1 − x2
a2 = x2 − x1
⇒[X ](β) = [(x1 , x2 )](β)
⇒[f ](α),(β) =
2x1 − x2
=
x2 − x1
2.3 + 1 2.5 + 3 2.6 + 4
7
=
−1 − 3 −3 − 5 −4 − 6
−4
DANG LE QUANG, Ph.D
LINEAR ALGEBRA
13
−8
16
−10
September - 2022
158 / 178
Linear mapping and Quadratic form
Linear map
Matrix of the linear map (4)
Example 6.6
Let f : R3 → R3 ; f (x1 , x2 , x3 ) = (x1 − x3 , x1 + x2 , −x2 − x3 ), and let bases
(α) = {α1 = (1, 1, 1), α2 = (−1, 1, 2), α3 = (1, 2, 3)}
(β) = {β1 = (0, 1, 1), β2 = (1, 0, 1), β3 = (1, 1, 0)}
Find the matrix of f in the base pair (α), (β).
Example 6.7
Let f : Rn → Rm ; f (x1 , x2 , ..., xn ) =
(a11 x1 +a12 x2 +...+a1n xn , a21 x1 +a22 x2 +...+a2n xn , ..., am1 x1 +am2 x2 +...+amn xn ).
Find the matrix of f in the canonical base pair (Cn ), (Cm ).
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LINEAR ALGEBRA
September - 2022
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Linear mapping and Quadratic form
Eigenvalues and eigenvectors
Definitions (1)
• Eigenvalues and eigenvectors of matrix: Let A be a square matrix of order n
a11
a21
A=
 ...
an 1

a12
a22
...
an 2
...
...
...
...

a1n
a2n 

... 
ann
The number λ ∈ R is the eigenvalue of A if there exists a vector
x = (x1 , x2 , ..., xn ) ̸= 0 satisfied
T
T
A = x1 x2 . . . xn = λ x1 x2 . . . xn
That vector x ̸= 0 is called eigenvector of A, corresponding to the eigenvalue λ.
a11 − λ
a12
...
a1n
a21
a22 − λ ...
a2n
Polynomial PA (λ) =
is a polynomial of degree n
...
...
...
...
an1
an2
... ann − λ
for λ , called characteristic polynomial of the matrix A.
The equation Pa (λ) = 0 is called characteristic equation of the matrix A.
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Linear mapping and Quadratic form
Eigenvalues and eigenvectors
Definitions (2)
How to find eigenvectors, eigenvalues:
Step 1: Solve the characteristic equation Pa (λ) = 0. All solutions of this equation
are eigenvalues of the matrix A.
Step 2: Assume λ0 is an eigenvalue of A. Then all non-trivial solutions of a
homogeneous linear system
T T
(A − λ0 In ) x1 . . . xn = 0 . . . 0
is the eigenvector of the matrix A corresponding to the eigenvalue λ0 .
Theorem 6.1
If u1 , u2 , ..., uk are eigenvectors corresponding to distinct eigenvalues λ1 , λ2 , ..., λk
of A then {u1 , u2 , ..., uk } are linearly independent.
Example 6.8
3
Find eigenvalues, eigenvectors of a matrix A =
0
DANG LE QUANG, Ph.D
LINEAR ALGEBRA

2 1
5

;B = 0 1
7
0 2
September - 2022

0
−1
4
161 / 178
Linear mapping and Quadratic form
Eigenvalues and eigenvectors
Diagonalize a square matrix (1)
Definition 6.2
Let A be a square matrix of order n. The matrix A is said to be diagonalizable
if there exists a square matrix that is non singular T of order n such that
T −1 AT = D is a diagonal matrix.
Matrix diagonalization A means finding the matrices T and D such that
T −1 AT = D.
How to diagonalize matrix:
To diagonalize the matrix A we find linearly independent eigenvectors of A.
- If the number of linearly independent eigenvectors is less than n: then A is
not diagonalizable.
- If A has enough linearly independent eigenvectors n then A is diagonalizable.
The matrix T is the matrix that the columns of T are linearly independent
eigenvectors.
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Linear mapping and Quadratic form
Eigenvalues and eigenvectors
Diagonalize a square matrix (2)
Example 6.9
Check if the following matrices are
them

2
3 5

A=
; B= 0
0 7
0
DANG LE QUANG, Ph.D
diagonalizable? If possible, diagonalize
1
1
2


0
−3


−1 ; C = −7
4
−6
LINEAR ALGEBRA
1
5
6

−1
−1
−2
September - 2022
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Linear mapping and Quadratic form
Quadratic form
Definitions (1)
A quadratic form in Rn is a map q : Rn → R defined by
n
P
aij xi xj ; x = (x1 , x2 , ..., xn ) ∈ Rn
q(x) =
i,j=1
where aij is a constant, aij = aji ∀i, j = 1, 2, ..., n.
Since aij = aji , the quadratic form is also written as
n
P
P
aii xi2 + 2 aij xi xj .
q(x) =
i<j
i=1
a11
 a21
Square Matrix A = 
 ...

am1
a12
a22
...
am2
... a1n
... a2n 
 is called matrix of quadratic form of q.
... ... 
... amn

We have a matrix A that satisfies aij = aji so A = AT . This square matrices
whose property is called symmetric matrix.
T
If we sign [x] = x1 x2 . . . xn then we can write the quadratic form as a
matrix as follows q(x) = [x]T A[x].
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LINEAR ALGEBRA
September - 2022
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Linear mapping and Quadratic form
Quadratic form
Definitions (2)
Example 6.10
Give the quadratic form in R3
q(x) = 2x12 + 3x22 − x32 − x1 x2 + 4x2 x3 + 6x1 x3 .
Find the matrix A of q(x).
Proof. The coefficients x12 , x22 , x32 are the elements on the main diagonal of A.
The coefficients of the product xi xj (i ̸= j) are twice the value of aij of A. Thus
2a12 = −1 ⇒ a12 =
−1
= a21
2
2a13 = 6 ⇒ a13 = 3 = a31
2a23 = 4 ⇒ a23 = 2 = a32

 2

−1
So A = 

 2
3
−1
2
3
2
DANG LE QUANG, Ph.D

3


2

−1
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Linear mapping and Quadratic form
Quadratic form
Definitions (3)
• Canonical form of the quadratic form: The canonical form of the quadratic
form in Rn is the quadratic form containing only the squares of the variables
q(x) = b1 x12 + b2 x22 + ... + bn xn2
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Linear mapping and Quadratic form
Quadratic form
Return quadratic form to canonical form (1)
n
P
• Lagrange Method: Let the quadratic form q(x) =
aij xi xj .
i,j=1
If a11 ̸= 0, we write
q (x ) = a11 x12 + 2a12 x1 x2 + ... + 2a1n x1 xn + ...
2
a12
a1n
= a11 x1 +
x2 + ... +
xn + g1
a11
a11
Sets x1′ = x1 +
a12
a1n
x2 + ... +
xn , we have
a11
a11
q(x) = a11 x1′2 + g1
where g1 is a quadratic form that does not contain x1 .
If a11 = 0 but a12 ̸= 0, sets
x1 = x1′ + x2′ ,
x2 = x1′ − x2′ .
Then a12 x1 x2 = a12 x1′2 − a12 x2′2 . In the case of a11 ̸= 0 we also have
q(x) = b1 x1′2 + g1 where g1 is the quadratic form that does not contain x1′ .
Continuing this process, we will get q(x) to canonical form.
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Linear mapping and Quadratic form
Quadratic form
Return quadratic form to canonical form (2)
Example 6.11
Return the following quadratic form to the canonical form
q(x) = x12 + 5x22 + 8x32 + 4x1 x2 + 6x1 x3 + +8x2 x3
Proof.
q (x ) = (x12 + 4x1 x2 + 6x1 x3 ) + 5x22 + 8x32 + 8x2 x3
= x12 + 2x1 (2x2 + 3x3 ) + (2x2 + 3x3 )2 + x22 − x32 − 4x2 x3
= (x1 + 2x2 + 3x3 )2 + (x22 − 4x2 x3 ) − x32
= (x1 + 2x2 + 3x3 )2 + (x22 − 4x2 x3 + 4x32 ) − 5x32
= (x1 + 2x2 + 3x3 )2 + (x2 − 2x3 )2 − 5x32
Put



y1 = x1 + 2x2 + 3x3
y2 = x2 − 2x3


y = x
3
3
We get q = y12 + y22 − 5y32 .
DANG LE QUANG, Ph.D
⇔



x1 = y1 − 2y2 − 7y3
x2 = y2 + 2y3


x = y
3
3
LINEAR ALGEBRA
September - 2022
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Linear mapping and Quadratic form
Quadratic form
Return quadratic form to canonical form (3)
Example 6.12
Return the following quadratic form to the canonical form
q(x) = 2x1 x2 + 2x1 x3 − 6x2 x3 .
Proof. Changing the variable, we set
x1 = y1 + y2 ,
x2 = y1 − y2 ,
x3 = y3 .
Then
q = 2(y1 + y2 )(y1 − y2 ) + 2(y1 + y2 )y3 − 6(y1 − y2 )y3
= 2y12 − 2y22 − 4y1 y3 + 8y2 y3
Change
q = (2y12 − 4y1 y3 ) − 2y22 + 8y2 y3 = 2 y12 − 2y1 y3 + y32 − 2y22 + 8y2 y3 − 2y32
= 2(y1 − y3 )2 − 2(y22 − 4y2 y3 + 4y32 ) + 6y32 = 2(y1 − y3 )2 − 2(y2 − y3 )2 + 6y32
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LINEAR ALGEBRA
September - 2022
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Linear mapping and Quadratic form
Quadratic form
Return quadratic form to canonical form (4)
Put



z1 = y1 − y3
z2 = y2 − y3


z = y
3
3
⇔



y1 = z1 + z2 + z3
y2 = z2 + z3


y = z
3
3
We get q = 2z12 − 2z22 + 6z32 .
So, with the variable change



variablex1 = z1 + 2z2 + 2z3
.
x2 = z1


x = z
3
3
The quadratic form q has the canonical form q = 2z12 − 2z22 + 6z32 .
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LINEAR ALGEBRA
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Linear mapping and Quadratic form
Quadratic form
Return quadratic form to canonical form (5)
• Jacobi Method: This method applies to
matrix of order n

a11 a12
a21 a22
A=
 ... ...
an1 an2
all quadratic form having square

... a1n
... a2n 

... ... 
... ann
Which satisfy
D1 = a11 ̸= 0, D2 =
DANG LE QUANG, Ph.D
a11
a21
a11
a12
̸= 0, ..., Dk = ...
a22
ak1
LINEAR ALGEBRA
... a1k
... ... ̸= 0,
... akk
k = 1, 2, ..., n −
September - 2022
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Linear mapping and Quadratic form
Quadratic form
Return quadratic form to canonical form (6)
Then change the variable according to the formula


x1 = y1 + y2 b21 + ... + yn bn1



x = y + y b + ... + y b
2
2
3 32
n n2

...




xn = yn
Where
bji = (−1)i+j
Dj−1,i
,
Dj−1
(j > i)
Dj−1,i is the determinant of a matrix whose elements lie on the intersection
of rows 1, 2, ..., j − 1 and columns 1, 2, . . . , i − 1, i + 1, . . . , j (remove column
i ) of matrix A.
Then
q = D1 y12 +
DANG LE QUANG, Ph.D
D2 2
Dn 2
y2 + ... +
y .
D1
Dn−1 n
LINEAR ALGEBRA
September - 2022
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Linear mapping and Quadratic form
Quadratic form
Return quadratic form to canonical form (7)
Example 6.13
Return the following quadratic form to the canonical form
q(x) = 2x12 + x22 + x32 + 3x1 x2 + 4x1 x3 .
Proof. Matrix of q is

2

3
A=

2
2
3
2
1
0

2


0

1
We have
2
D 1 = 2 ; D2 =
DANG LE QUANG, Ph.D
3
2
2
3
−1
2 =
; D3 = 3
4
2
1
2
LINEAR ALGEBRA
3
2
2
1
0 =
0
1
−17
4
September - 2022
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Linear mapping and Quadratic form
Quadratic form
Return quadratic form to canonical form (8)
We calculate bji
b21 = (−1)2+1
−3
−3
D1,1
= 2 =
D1
2
4
3
2
b31 = (−1)3+1
b32 = (−1)3+2
DANG LE QUANG, Ph.D
2
ten
D2,1
=
−1
D2
4
D2,2
=
D2
−
LINEAR ALGEBRA
2
3
2
−1
4
=8
2
0
= −12
September - 2022
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Linear mapping and Quadratic form
Quadratic form
Return quadratic form to canonical form (9)
Change variable

3

x1 = y1 − y2 + 8y3



4
x2 = y2 − 12y3



x3 = y3
we get
q = 2y12 +
DANG LE QUANG, Ph.D
−1
4
2
y22 +
−17
4
−1
4
1
y32 = 2y12 − y22 + 17y32
8
LINEAR ALGEBRA
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Linear mapping and Quadratic form
Quadratic form
Return quadratic form to canonical form (10)
• Gauss Method: Assume A is a matrix of q in a base, then by matrix language,
to return q to the canonical form, we find the matrix T such that T T AT is a
diagonal matrix.
How to find the matrix T :
Step 1: Write the matrix [A | In ].
Step 2: Performs the elementary transformations on the row, and repeats the
same transformations on the column of [A | In ] to return A to diagonal form.
Then the right matrix is T T .
Example 6.14
Return the following quadratic form to the canonical form
q(x) = x12 + 5x22 + 8x32 + 4x1 x2 + 6x1 x3 + 8x2 x3
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LINEAR ALGEBRA
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Linear mapping and Quadratic form
Quadratic form
Return quadratic form to canonical form (11)
Solve. Find the matrix T T . We have



1 2 3 1 0 0
1 2
3 1 0
d2 →d2 −2d1
[A | I3 ] = 2 5 4 0 1 0 −−
−−−−−→ 0 1 −2 −2 1
d3 →d3 −3d1
3 4 8 0 0 1
0 −2 −1 −3 0



1 0
0 1 0 0
1 0
c →c2 −2c1
d →d3 +2d2
−−2−−−
−−→ 0 1 −2 −2 1 0 −−3−−−
−−→ 0 1
c3 →c3 −3c1
0 −2 −1 −3 0 1
0 0


1 0 0 1 0 0
c3 →c3 +2c2
−−
−−−−→ 0 1 0 −2 1 0
0 0 −5 −7 2 1
DANG LE QUANG, Ph.D
LINEAR ALGEBRA

0
0
1
0 1
−2 −2
−5 −7
September - 2022

0 0
1 0
2 1
177 / 178
Linear mapping and Quadratic form
Quadratic form
Return quadratic form to canonical form (12)
The left matrix block is diagonal, so we get

TT

1
⇒ T = 0
0
Put



x1 = y1 − 2y2 − 7y3
x2 = y2 + 2y3


x = y
3
3
DANG LE QUANG, Ph.D
−2
1
0
1
= −2
−7

0 0
1 0
2 1


−7
1
2  ; T T AT = 0
1
0
0
1
0

0
0 .
−5
, we get q = y12 + y22 − 5y32 .
LINEAR ALGEBRA
September - 2022
178 / 178