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Title: ELEMENTS OF LINEAR ALGEBRA
Description: Comprehensive notes and practicals (Problems and Solutions) covering: -Determinants -Matrix operations. Rank of matrix -Inverse matrix. Matrix equation -Matrix method for solving systems of linear algebraic equations. Cramer's rule.
Description: Comprehensive notes and practicals (Problems and Solutions) covering: -Determinants -Matrix operations. Rank of matrix -Inverse matrix. Matrix equation -Matrix method for solving systems of linear algebraic equations. Cramer's rule.
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ELEMENTS OF LINEAR ALGEBRA
Practical class № 1
...
If the rows of the matrix A aij size m n are made into columns with the
same numbers, we get the matrix AT a ji size n m , which is called transpose to the matrix A
...
Each square matrix of n-th order have a number which is called the determinant
and calculated according following rule:
n 1, det A a11 (the determinant of the first order matrix);
n 2, detA
a11
a21
a12
a11a22 a12 a21 (the determinant of the second ora22
der matrix);
a11 a12 a13
n 3, det A a21 a22 a23 a11a22 a33 a12 a23a31 a13a32 a21
a31 a32 a33
a13a22 a31 a12 a21a33 a11a23a32 (the determinant of the third order matrix)
...
In General, a rule called a determinant decomposition on any row (column) is applied, for example, on the i-th row:
det A
а11
а21
a12
a22
a1n
a2 n
аn1
an 2
an n
ai1 Ai1 ai 2 Ai 2
...
M ij is the determinant obtained from matrix A with i row and column j erased
...
а31 а33
а31 а33
а21 а23
The diagonal of the determinant formed by the elements a11, a22 , a33 ,
...
If elements of matrix above (below) main diagonal are equal to zero, than matrix is called triangular
...
det AT det A
...
The determinant will not change if you add the corresponding elements
of another row (column) multiplied by any number
...
The determinant will change the sign when two rows (columns) are exchanged
...
det A 0 , if it contains a) a zero row (column); b) two equal rows or
two equal columns; C) a row (column) that is a linear combination of other rows
(columns)
...
The determinant of a triangular matrix is equal to the product of diagonal elements
...
1 2
...
cos 2 sin 2
sin 2
cos 2
...
...
4 5 6
...
0
4
2
2
1
3
...
3 4 a
...
15 10 5
...
1 6 5
...
0 1 3
...
0 3
2
7
...
2cos x 2sin x
...
The determinant 4
0
1 is given
...
Calculate the determinants in two ways: using the decomposition of the determinants on the row and transforming them to a triangular form:
3 0
3 2 4
2
13
...
14
...
5 1 3
x2
15
...
2
x 1 1
...
Solve the inequality: 0
Solve following equations and make a check:
2 x3
0
...
4
1
x2
18
...
1 1
2
0 1
19
...
log 2 x 1 3
Solve the inequality:
5
x
20
...
1
1
1
1 0
1
1 3 0
...
log 1 x 1 5
2
1
2
22
...
The solution of typical problems
3 2 5
Problem 1
...
3
12 17
5 17
5 12
2
5
26
...
Calculate the determinant, transforming it to a triangular
1 2 5
form: 1 1 7
...
Then, using the 2 determinants
property, we obtain a triangular form of the determinant whose value is equal to
the product of the main diagonal elements:
1 2 5 S2 S1 1 2 5
1 2 5
1 1 7 0 1
2
0 1
2 12
...
10
...
1
...
7a
...
0
...
25
...
10a 3
...
50
...
45
...
144
...
6
...
4
...
M13 8, A13 8 , M 32 23, A32 23
...
73
...
5
...
0;2
...
1;1
...
3,5
...
2;3
...
0,125
...
4;0
...
0,5;
...
160
...
Matrix operations
...
To multiply matrix A by any number R we use following formulae:
cij aij , i 1, m, j 1, n
...
Re-
sult is m n matrix C cij , where element cij is the sum of products of elements of i th row of matrix A and j th column of matrix B :
k
cij ai1 b1 j ai 2 b2 j
...
s 1
Problems
2 3
3 1
B
and
0 5 are given, E is identity matrix of the
1 2
1
...
Find:
а) A B ;
b) 2 A 3B ;
c) AB ;
d) A2 3E ;
f) det( AB) det A det B
...
Find: а) A AT ; b) det AT A
...
The matrix A
3
...
Find the rank of the following matrices:
1 2 1 3
...
1 2 3
5
...
3 6 9
2
3
6
...
2
0
7
1 2 1 1 3
7
...
1 1 1 4 13
0
1
8
...
1
0
1
1
0
1
1 2 3 4
2 4 6 8
...
3 6 9 13
5 10 15 21
Material for individual work
4 2
0 3
B
and
1 2 are given
...
The matrices A
а) 3 A 2B ;
b) A ( A B) ;
T T
c) det( A B 2 E ) (E is identity matrix
of the second order)
...
The matrices A 1 3 4 , B 1
, C 0 3 are
0
3
1 1
2 0 1
2 1 1
given
...
Find the product of matrices:
1 2 4 2
12
...
3 3 1 0
1 1
2 3
13
...
2 2
6 2
...
Find f ( A) , if f ( x) x 9 x 20 , A
Find the rank of the following matrices:
1 2 3
15
...
4 8 9
2 3 1 1
3 1
4
2
...
1 2 3 1
1 4 7 5
2 3
3 1 4
17
...
Find matrix X from following equa
2
0
3
0 1
tion AT 2 X 2 B
...
Find matrix X from following equation:
0 4 1
2 8 1
2 2 2 2 X 6 0 2
...
3
0
2
20
...
Find
3 1
0 2
B
,
f ( A) g ( B) , if A
1 1
...
Find f ( A) , if f ( x) 2 x 5 x 9, A
1 2 1
3 1 2
21
...
0
1 1 2
22
...
is given
...
1 10 6 1
The solution of typical problems
1 2
and
3
4
Problem 1
...
7
8
T
1 2
5 6 1 3 10
A 2B
2
7 8 2 4 14
3 4
1 10 3 12 11 15
2 14 4 16 16 20
...
Find the product BA of matrices A
1
T
12
16
5
and
0
4 7
B 1 8
...
4 7
4 2 7 1 4 5 7 0 15 20
2
5
BA 1 8
1 2 8 1 1 5 8 0 10 5
...
Problem 3
...
1 1 1 1 0 0
1 1 0 0 ,
1
1
1 1 1 0 2 2 3 0 1 2
2 A 3E 2
3
,
1 1 0 1 2 2 0 3 2 5
0 0 1 2 1 2
f ( A)
2 5 2 5
...
4
...
0 4 2 4
Any minor of the 3rd order of this matrix is zero, since the second and third
rows are proportional
...
2
Therefore, rangA 2
...
Problem 5
...
The matrices A and B are equivalent : A B
...
Answers
5 2
13 3
6 13
1 2
;
;
;
1
...
1 3
14 10
2
...
3
...
4
...
5
...
22
2
4
6
23
12 12
6
...
7
...
8
...
9
...
10
...
11
...
12
...
13
...
14
...
15
...
16
...
6 10
1 2 0
1
4
4
0 6
0 4
...
2 1 2
...
17
...
20
...
9
3
5
3
3,5
2,5
3,5
1 2 3
2
2
21
...
22
...
3
3
Practical class № 3
...
Matrix equation
Inverse matrix for a square matrix A is the same order matrix A1
such that A A1 A1 A E , where E is identity matrix
...
The method of the cofactor matrix
...
The method of elementary transformations
...
The aim of transformations is to obtain
matrix
(see solution of typical problems, examples 1, 2)
...
1 2
...
...
...
...
0 0 1
...
8
3 6
...
2 1 2
...
1
1
1 1 1
3
11
...
1 1 1
2
4 3
1 1
X
4 4
...
X
8
...
3
2
2 3 5
X
3 1 0
...
...
...
1 3 2
...
4 6
...
3 2
2 X 2 2
...
Find det( A 2 E ) 1 , if A
2
3
...
2
0
3 1
19
...
2
1
1 0
1 2 3 2
X
20
...
1
2
2
3
5 3 1 2
21
...
10 6 3 4
22
...
18
...
Find A1 for matrix A 0 1 2
...
Therefore, A1 exists
...
1 2
0 2
0 1
A11
2, A12
2, A13
1,
1 0
1 0
1 1
A21
A31
0
1 1
1 1
1 1
1, A22
1, A23
0,
1 0
1 0
1 1
1 1
1 1
1 1
3, A32
2, A33
1
...
3 2 1
13
3) Transpose the cofactormatrix: A
2
1
4) A1
A 2
det A
1
2 1
5) Check: A1 A 2
1
1 0
T
1
1
0
T
2 1 3
2 1 2
...
1
3 1 1 1 1 0 0
2 0 1 2 0 1 0 ,
1 1 1 0 0 0 1
1 1 1 2 1 3 1 0 0
AA1 0 1 2 2 1 2 0 1 0
...
Using elementary transformations find inverse matrix for
the matrix A from the previous problem
...
1 0 1
Problem 3
...
1 1
2
1 0
1 1
1 2 ; then det B 10, det A 1
...
Then
1
1
2
This equation is rewritten as B X A C , where С
B 1B X A A1 B 1CA1 or X B 1CA1
...
1 0 1 10 8 2 11 0,8 0,2 1,1
Answers
1 0 0
3 2 3
1
2
2 1
...
0 1 1
...
0 1 2
...
4
...
3,5
...
4
...
3 2
1,5 0,5
0 1 0
4 3 5
1
1
3
2
7
...
...
8
...
10
...
3
15 15
1
2
2,5
11
...
12
...
13
...
15
...
...
3 3
1
0,4 0,2 0,2
7
2 4
0,1 0,3
13 8
16
...
17
...
18
...
19
...
20
...
0,3
0,1
1 6
a
22
...
2
15
Practical class № 4
...
Cramer's rule
...
32 2
33 3
3
31 1
The matrix form of the system (1
...
1)
or AX B
...
1) det A 0 , then this system has a unique solu-
X A1 B
tion
or
(1
...
3)
, x2 2 , x3 3 ,
where i i 1,3 are determinants, obtained from the determinant with
x1
column i replaced by B
...
3) is called the Cramer’s rule
...
2) is called the matrix method
...
Write the system x 2 z 8,
in matrix form
...
numbers 2; 1;3 is solution of this system
...
4 x 5 y 3
...
2 x y 0,
3x y z 0
...
x 2 y 4 z 3,
3x y 5z 2
...
3x 2 y 2 z 2, 6
...
x 3 y 4 z 6
...
3x 5 y 9
...
2 x y z 6,
x 5 y 3
...
3x 3 y z 13,
x 2 y 3z 9
...
At what values of the parameter a the system
inconsistent?
ax 2 y 5,
is indetermi2 x ay 5
11
...
Find the unknown coefficients of the function f x a 3x b x c, satisfying the conditions: f 0 2, f 1 1, f 2 4
...
Find the unknown coefficients of the polynomial f x a x 2 b x c , satisfying the conditions: f 2 8, f 1 4, f 2 4
...
Solve the system 3x1 2 x2 2 x3 1,
x 4 x 7 x 3
2
3
1
Use Cramer’s rule
...
1 4 7
Find
1 1 1
2 1 1
2 1 1
1 1 2 2 3; 2 3 1 2 9; 3 3 2 1 6
...
3) , we obtain
x1
1
1, x2 2 3, x3 3 2
...
Solve the system from the previous example using matrix
method
...
Find A1
...
1 4 7
3
A11
2 2
6,
4 7
1 1
3,
4 7
A21
A31
1 1
0,
2 2
3 2
19,
1 7
A12
A22
2 1
13,
1 7
A32
2 1
1,
3 2
A13
3 2
10,
1 4
A23
A33
2 1
7,
1 4
2 1
1
...
Then according the formula (1
...
Answers
2 3 1
2
1
...
2
...
3
...
4
...
z 1
5
...
6
...
7
...
8
...
9
...
10
...
11
...
12
...
13
...
Practical class № 5
...
a1n xn b1 ,
a x a x
...
amn xn bm
...
4)
a1n
a11 a12
a
a
a
21
22
2
n
,
Matrix A is the matrix of the system : A
amn
am1 am 2
a1n b1
a11 a12
a
a22
a2n b2
21
is extended matrix of the system (1
...
A | B
a
a
a
b
mn
m
m1 m 2
Kronecker-Capelli Theorem
...
4) is consistent if and only if the rank of the matrix of the system is equal to the rank of
the extended matrix of the system: rangA rang ( A | B)
...
The essence of the
method consists in the sequential elimination of the unknowns in order to obtain
a step system
...
e
...
It is obvious that the resulting systems will be
equivalent
...
(see the solution of typical problems)
...
2 x1 x2 3 x3 3,
3x x 5 x 0
...
7 x1 4 x2 x3 3x4 5,
5 x 7 x 4 x 6 x 3
...
x1 2 x2 3x3 0,
x x 3x 12
...
8 x1 3x2 6 x3 0,
4 x x 3x 0
...
x
5 y z 10
...
2 x 3 y 7 z 16,
5 x 2 y z 16
...
2 x y 5z 1,
x y z 2
...
x 2 y z 1,
x y 2 z 1
...
10
...
7 x1 5 x2 9 x3 10 x4 8
...
In
case the system is consistent, find its solution:
x1 x2 2 x3 5,
2 x1 x2 8,
12
...
x1 2 x2 4 x3 10
...
Solve system by Gauss method:
11
...
2
3
1
1 1 1 0
We write the extended matrix of the system: 2 1 3 7
...
To do this, we subtract the first row multiplied by 2 from the
second row and the first row multiplied by 3 from the third row:
1
1
0 S2 2S1 1 1 1 0
1
2 1 2 1 1 2 3 (1) 2 7 0 2
0 3 5 7
3 1 3 2 1 3 1 (1) 3 7 0 3 S 3S 0 1 4 7
1
3
In order to get one on the place of the next leading element a22 , swap the
second and third rows, and then multiply the elements of the new second row
by (-1):
20
1 1 1 0 1 S2 1 1 1 0 S3 3S2
S3
0 1 4 7
0 1 4 7
0 3 5 7
0 3 5 7
1
1 1 1 0 7 S3 1 1 1 0
0 1 4 7
0 1 4 7
0 0 1 2
0 0 7 14
S2
The matrix of the system is given to a triangular form
...
But we will get the solution of the system using the modified Gauss-Jordan method, i
...
we will get zeros not only lower but also higher than the diagonal elements
...
1 0 0 1
0 1 0 1
...
Solve system by Gauss method:
x1 x2 x3 4,
x1 2 x2 3x3 0,
2 x
2 x3 16
...
2 0 2 16 S 2 S 0 2 4 8
0 0 0 0
1
3
Since rangA rang ( A | B) 2 3 n , that system is consistent and indefinite
(i
...
has infinitely many solutions)
...
These variables are
basic, the variable x3 is called free variable
...
x
2
x
4
...
Let x3=t, then
21
x1 x2 t 4,
x2 2t 4
...
Substituting the expression
for x2 in the first equation, we obtain that x1 t 8, t
...
The particular solution of the system is obtained, for example, when t 0 : 8;4;0
...
Solve system by Gauss method:
x1 x2 x3 4,
x1 2 x2 3x3 0,
2 x
2 x3 3
...
2 0 2 3 S 2 S 0 2 4 5 S 2 S 0 0 0 13
1
2
3
3
Since rangA 2 3 rang A | B , that system is inconsistent (it has no solutions)
...
Answers
1
...
2
...
3
...
4
...
5
...
6
...
7
...
8
...
9
...
10
...
11
...
12
...
22
; at
Title: ELEMENTS OF LINEAR ALGEBRA
Description: Comprehensive notes and practicals (Problems and Solutions) covering: -Determinants -Matrix operations. Rank of matrix -Inverse matrix. Matrix equation -Matrix method for solving systems of linear algebraic equations. Cramer's rule.
Description: Comprehensive notes and practicals (Problems and Solutions) covering: -Determinants -Matrix operations. Rank of matrix -Inverse matrix. Matrix equation -Matrix method for solving systems of linear algebraic equations. Cramer's rule.