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Файл:Mathematics (Математика). Учебное пособие
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21
In general a linear system is of the form
=+++
ì
ï
ï
...
...
í
ï
ï
î
...
2211
bxaxaxa
nn
11212111
=+++
bxaxaxa
nn
22222121
...............................................
=+++
bxaxaxa
mnmnmm
where the xi are variables and the aij and bi are constants. This system
can be represented by the augmented matrix
æ
ç
ç
...
...
ç
ç
ç
è
...
21
baaa
n
ö
111211
÷
baaa
÷
n
222221
.
÷
...............
÷
÷
baaa
mmnmm
ø
Changes to the system of equations as a result of an elementary operations translate into changes of the augmented matrix resulting from a
row operation.
Definition RO The row operations consist of the following
1. Switch two rows.
2. Multiply a row by a nonzero number.
3. Replace a row by a multiple of another row added to it.
Gauss elimination is a systematic procedure to simplify an augmented matrix to a reduced form. In the following definition, the term “lead-
ing entry” refers to the first nonzero entry of a row when scanning the
row from left to right.
The following is the algorithm for obtaining a matrix which is in row
reduced echelon form.
Algorithm.
This algorithm tells how to start with a matrix and do row operations
on it in such a way as to end up with a matrix in row reduced echelon
form.
1. Find the first nonzero column from the left. This is the first pivot
column. The position at the top of the first pivot column is the first pivot
position. Switch rows if necessary to place a nonzero number in the first
pivot position.
2. Use row operations to zero out the entries below the first pivot position.
22
3. Ignore the row containing the most recent pivot position identified
and the rows above it.
Repeat steps 1 and 2 to the remaining sub-matrix, the rectangular array of numbers obtained from the original matrix by deleting the rows
you just ignored. Repeat the process until there are no more rows to
modify. The matrix will then be in echelon form.
4. Moving from right to left, use the nonzero elements in the pivot
positions to zero out the elements in the pivot columns which are above
the pivots.
5. Divide each nonzero row by the value of the leading entry. The result will be a matrix in row reduced echelon form.
Example. Give the complete solution to the system of equations
=-+
52
xxx
321
-=+-
332
xxx
321
=-+
107
xxx
321
-
5112
æ
ö
ç
÷
--
~
3321
ç
÷
ç
÷
-
10117
è
ø
-=+-
332
xxx
321
=-
1175
xx
32
-=-
2
,Þ x3 = 2; x2 = 5; x1 = 1.
--
3321
æ
ö
ç
÷
-
-
~
5112
ç
÷
ç
÷
10117
è
ø
--
3321
æ
ö
ç
-
-
÷
~
11750
ç
÷
ç
÷
3122150
è
ø
--
3321
ö
÷
-
11750
÷
÷
--
2100
ø
А*=
ì
ï
í
ï
î
æ
ç
ç
ç
è
ì
ï
í
ï
x
3
î
In summary,
A system of linear equations is a list of equations
=+++
ì
ï
ï
...
...
í
ï
ï
î
...
2211
bxaxaxa
nn
11212111
=+++
bxaxaxa
nn
22222121
...............................................
=+++
bxaxaxa
mnmnmm
Can be solve by using a algorithm. This procedure is called the
Gauss-Jordan procedure.
23
Reading Questions.
361
3251
236
++=-
1
-
1
-
1. Give the complete solution to the system of equations doing row
=-+
1263
operations
ì
ï
í
ï
î
zyx
-=++
10523
zyx
=-+
6352
zyx
2. Give the complete solution to the system of equations doing row
xyz
+-=
operations
ì
ï
xyz
í
ï
xyz
+-=
î
Topic 6. Systems of Equations
Introduction. This lecture is devoted to the discussion of the following concepts: forms of systems of equations, using the inverse of a matrix to find the solution of systems of equations.
=+++
ì
ï
ï
...
...
í
ï
ï
î
æ
ç
ç
A =
ç
ç
ç
è
More simply, this is of the form A×X = B.
...
2211
aaa
...
n
11211
aaa
...
n
22221
............
aaa
...
21
nnnn
nn
ö
÷
÷
; B =
÷
÷
÷
ø
bxaxaxa
11212111
=+++
bxaxaxa
nn
22222121
...............................................
=+++
bxaxaxa
nnnnnn
b
ö
æ
1
÷
ç
b
÷
ç
2
; X =
÷
ç
...
÷
ç
÷
ç
b
ø
è
n
x
ö
æ
1
÷
ç
x
÷
ç
2
.
÷
ç
...
÷
ç
÷
ç
x
ø
è
n
Suppose you find the inverse of the matrix – A
multiply both sides of this equation by A
Because А-1×А = I, so I×Х = А-1×В
This gives the solution as: Х = А-1×В.
Note that once you have found the inverse, you can easily get the solution for different right hand sides without any effort.
to obtain A
. Then you could
-1
×A×X = A-1×B,
24
Example. In this example, it is shown how to use the inverse of a ma-
1011
1113
1112
жцжцжц
зчзчзч
зчзчзч
зчзчзч
ишишиш
00,50,5
110
10,50,5
æö
ç÷
ç÷
ç÷
èø
00,50,51
1103
10,50,52
æöæö
ç÷ç÷
ç÷ç÷
ç÷ç÷
èøèø
2,521,5
æö
ç÷
ç÷
ç÷
èø
trix to find the solution to a system of equations. Consider the following
system of equations. Use the inverse of a suitable matrix to give the solutions to this system.
x + z = 1
x − y + z = 3
x + y − z = 2
The system of equations can be written in terms of matrices as
x
-=
y
z
the inverse of the matrix is
-×
=
--
-
--
.
-
-
Example. To use the inverse of a matrix to find the solution
=--
05
æ
ç
ç
ç
è
11 --
23
15=-
24
zyx
zyx
=++
1432
=++
16234
zyx
--
115
ö
÷
321
÷
÷
234
ø
= 1; M31 =
M32 =
;14
11 --
= -1;
32
15
-
;16
=
31
to a system of equations.
Х =
det A =
M11 =
x
ö
æ
÷
ç
, B =
y
÷
ç
÷
ç
z
ø
è
32
0
ö
æ
÷
ç
, A =
14
÷
ç
÷
ç
16
ø
è
--
115
5(4-9) + 1(2 – 12) – 1(3 – 8) = -25 – 10 +5 = -30.
=
321
234
= -5; M21 =
23
M12 =
31
24
-=
M22 =
;10
ì
ï
í
ï
î
25
M13 =
30
30
30
1
XBA
-
=
1
11
1
21
1
31
30
21
34
5
;
10
;
30
5
;
-=
12
32
M23 =
;5
1
30
1
22
19
1
1
;
14
30
;
15
-
34
---
aaa
13
;
---
aaa
33
M33 =
;19
=
1
1
;
===
30
16
=-=-=
11
30
;
;
A-1 =
1
---
aaa
23
1
-===
15
-
21
æ
ç
ç
ç
ç
ç
ç
è
;11
=
1
1
1
30
6
1
6
7
1
--
15
3
19
30
ö
÷
30
÷
8
÷
÷
15
÷
11
-
÷
30
ø
Make a check
1
1
A×A
5
æ
-1
æ
=
ç
ç
ç
è
ç
--
115
ö
ç
÷
ç
321
÷
ç
÷
234
ç
ø
ç
è
30
30
10
30
5
30
14
--
30
19
30
ö
÷
30
16
30
11
-
30
æ
÷
ç
1
÷
=
ç
÷
30
ç
÷
è
÷
ø
+--+-+
-++-+-
-++-+-
1
1
Х =
x
ö
æ
÷
ç
y
÷
ç
÷
ç
z
ø
è
= А-1В =
1
æ
ç
ç
ç
ç
ç
ç
è
30
6
7
1
--
15
3
19
1
30
6
ö
÷
0
æ
30
÷
ç
×
8
÷
14
ç
÷
15
ç
16
è
÷
11
-
÷
30
ø
1
æ
0
ç
ö
6
ç
÷
=
1
ç
÷
÷
ø
0
ç
3
ç
1
0
ç
6
è
14
30
98
15
266
30
16
ö
++
÷
30
÷
128
÷
=
+--
÷
15
÷
176
-+
÷
30
ø
In summary, : x =1; y = 2; z = 3.
Reading Questions.
1. Give the complete solution to the system of equations using the
=-+
1263
zyx
-=++
10523
zyx
=-+
6352
zyx
inverse of a matrix
2.
Is it thru?
ì
ï
í
ï
î
;
111651914551025
ö
=I
÷
333215728115205
÷
÷
2248438424103020
ø
1
ö
æ
÷
ç
.
2
÷
ç
÷
ç
3
ø
è
26
Topic 7. The Cramer’s rule
Introduction. This lecture is devoted to the discussion of the follow-
ing concepts: the Cramer’s rule and Procedure of The Cramer’s rule.
For large systems Cramer’s rule is less than useful if you want to find
an answer. This is because to use it you must evaluate determinants.
However, you have no practical way to evaluate determinants for large
matrices other than row operations and if you are using row operations,
you might just as well use them to solve the system to begin with. It will
be a lot less trouble. Nevertheless, there are situations in which Cramer’s
rule is useful.
Procedure of The Cramer’s rule
=+++
ì
ï
ï
í
ï
ï
î
D
=
i
Example. Let
Then
æ
ç
A =
ç
ç
è
...
...
...
2211
ì
ï
í
ï
î
aaa
ö
131211
÷
; D
=
aaa
aaa
1
÷
232221
÷
333231
ø
bxaxaxa
nn
11212111
=+++
bxaxaxa
nn
22222121
xi =
D
/D,, D = det A, det A¹0
i
...............................................
=+++
bxaxaxa
nnnnnn
aabaa
......
nii
11111111
+-
aabaa
......
niii
2122221
+-
.........
aabaa
......
nnninnin
111
+-
=++
bxaxaxa
1313212111
=++
bxaxaxa
2323222121
=++
bxaxaxa
3333232131
aab
13121
;
aab
23222
aab
33323
27
aba
Þ
D
x
=
2
= D
1
/detA; x
1
aba
aba
= D
2
13111
23221
33331
; D
=
3
/detA; x
2
3
= D
baa
11211
baa
22221
baa
33231
/detA;
3
;
Example. Give the complete solution to the system of equations using the Procedure of The Cramer’s rule
=--
05
ì
ï
í
ï
î
zyx
=++
1432
zyx
=++
16234
zyx
115 --
D =
= 5(4 – 9) + (2 – 12) – (3 – 8) = -25 – 10 + 5 = -30;
321
234
110 --
D
1
=
= (28 – 48) – (42 – 32) = -20 – 10 = -30.
3214
2316
x
= D1/D = 1;
1
105 -
D
2
=
= 5(28 – 48) – (16 – 56) = -100 + 40 = -60.
3141
2164
x
= D2/D = 2;
2
015 -
D
3
=
= 5( 32 – 42) + (16 – 56) = -50 – 40 = -90.
1421
1634
x
= D3/D = 3.
3
If b
= 0, and D¹0
i
x1 = x2 = … = xn = 0.
Reading Questions.
1. Give the complete solution to the system of equations using the
Procedure of The Cramer’s rule
=-+
1263
ì
ï
í
ï
î
zyx
-=++
10523
zyx
=-+
6352
zyx
28
2. Tell whether the statement is true or false:
a) If Ax = 0 for some x = 0, then det (A) = 0.
b) Cramer’s rule is useful for finding solutions to systems of linear
equations in which there is an infinite set of solutions.
3. Use Cramer’s rule to find the solution to system
x + 2y = 1
2x − y = 2
Topic 8. А analytical geometry
Introduction. This lecture is devoted to the discussion of the follow-
ing concepts: different equations of straight lines on the plane and in
space, conditions parallelism and perpendicularity direct in space, the
cross product and its geometrical meaning.
Equation of a Line (point – slope form).
Definition: A straight line on the coordinate plane can be described
by the equation
y = m(x-Pх) + Pу
where m is the slope of the line and Pх, Py are the coordinates of a given
point on the line.
Recall that the slope (m) is the "steepness" of the line. In the figure
above, adjust m with the slider and drag the point P to see the effect of
changing the two givens.
Equation of a Line (slope and intercept form)
Definition: A straight line on the coordinate plane can be described
by the equation y = mx+b where m is the slope of the line and b is the
intercept.
Recall that the slope (m) is the "steepness" of the line and b is the intercept – the point where the line crosses the y-axis. In the figure above,
adjust both m and b with the sliders to see the effect of these variables.
Both in the plane and in space, any line can be defined as the set of
points whose coordinates to some selected in the space coordinates satisfy the equation
F(x, y, z) = 0.
29
Let F(x, y, z) = 0 and
(,,)0
Ô xyz
=
-=-=-
tersecting line L. Then a couple of equations
is the equation of the surfaces in-
0),,(
ì
í
î
zyxF
let's call the
==0),,(
zyxФ
equation of a line in space. Some more equations of line:
zz
000
p
m
xx
yy
n
This is the canonical equation of the straight line in space.
xx
-
1
=
xx
-
12
yy
-
1
yy
-
12
zz
-
1
=
This is the equation of a straight line
zz
-
12
passing through the two points in space.
Conditions parallelism and perpendicularity direct in space.
So that the two lines are parallel it is necessary and sufficient that
guides the vectors of these were direct collinear, i.e. their corresponding
coordinates were proportional.
m
m
1
n
2
1
1
2
==
p
2
p
n
So that the two lines are perpendicular it is necessary and sufficient
that guides the vectors of these were direct perpendicular i.e. the cosine
of the angle between them is zero.
0
=++ ppnnmm
212121
Example. To lead to the canonical form of the equation of a straight
=--+
071632
line, specified in the form:
ì
í
î
zyx
=-+
zyx
0173
To find an arbitrary point straight, which is the line of intersection of
the above planes will take z = 0. Then:
zyx
071632
ì
í
î
=--+
zyx
0173
=-+
;
xy
3;
-=
2x – 9x – 7 = 0;
x = -1; y = 3;
A(-1; 3; 0).
30
The direction vector:
rrr
1
235
7
14335
-=+
-
+
r
r
rrr
rrr
r
r
r
r
nnS
21
kji
1632
-=´=
1713
-
r
rr
.
kji
71435
---=
1 zyxzyx
=
-
-
;
=
-
1
;
=
From now on, the vectors, i, j, k will always form a right handed system. To repeat, if you extend the fingers of your right hand along i and
close them in the direction j, the thumb points in the direction of k.
Definition Let a and b be two vectors in R3. Then a × b is defined by
the following two rules.
1. |a × b| = |a| |b| sin θ where θ is the included angle.
2. a × b · a = 0, a × b · b = 0, and a, b, a × b forms a right hand sys-
tem.
Note that |a × b| is the area of the parallelogram determined by a and b.
kji
´
=
a
b
zyx
aaa
zyx
bbb
Example.
r
r
ba
=´ 717
kji
=
152
-
321
rr
15
-
-
32
r
12
-
31
52
21
r
rr
kjikji
-+-=+
You notice the area of the base of the parallelepiped, the parallelogram determined by the vectors, a and b has area equal to |a × b| while
the altitude of the parallelepiped is |c| cos θ where θ is the angle shown in
the picture between c and a × b. Therefore, the volume of this parallelepiped is the area of the base times the altitude which is just
zyx
r
r
r
),,(
cba =
111
zyx
222
zyx
333
,
),,(
zyxa =
111
),,(),,,(
zyxczyxb ==
333222
Example. To prove that point And(5; 7; 2), B(3; 1; -1), C(9; 4; -4),
D(1; 5; 0) lie in the same plane.
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