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Файл:Mathematics (Математика). Учебное пособие
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11
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Answer:
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21100
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Reading Questions
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1. Form the matrix vector product of
?
2. Multiply together the two matrices below (in the order given).
121
æ
ç
=
A
ç
è
ö
÷
B
÷
013
ø
31
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æ
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=
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21
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Topic 3. Determinant of a Matrix
Introduction. This lecture is devoted to the discussion of the following
concepts: sub matrix, Determinant of a Matrix, the rules to calculate the
determinant of Matrices of Size Two and Determinant of a 3×3 matrix.
For a full understanding of the topic you need to have an idea about matrices. We'll now turn to the definition of a determinant and do some sample
computations. The definition of the determinant function is recursive, that is,
the determinant of a large matrix is defined in terms of the determinant of
smaller matrices. To this end, we will make a few definitions.
Definition SM: Sub Matrix. Suppose that A is an m×n matrix. Then
the sub matrix M
ing row i and column j.
is the (m−1)×(n−1) matrix obtained from A by remov-
ij
12
12345
33271
54562
A
æö
ç÷
ç÷
ç÷
èø
12.45
.....
54.62
æö
ç÷
ç÷
ç÷
èø
.....
332.1
545.2
æö
ç÷
ç÷
ç÷
èø
11
kk
MA
212223111112121313
aaaaAaAaA
=×-×+×
M
=
=
23
M
=
,
14
.
Example. For the matrix
we have the submatrices
Definition DM: Determinant of a Matrix. Suppose A is a square
matrix. Then its determinant, det(A)=∣A∣, is an element of C defined
recursively by:
1. If A is a 1×1 matrix, then det(A)=[A]11.
2. If A is a matrix of size n with n≥2, then
det(A)=[A]11det(M11)−[A]12det(M12)+[A]13det(M13)−[A]14det(M14)+⋯+
n+1
+(−1)
[A]1n det(M1n)
aaа
æ
ç
ç
А=
ç
ç
ç
è
1
k
+
(1)
-=
...
...
...
21
n
11211
aaa
22221
............
aaa
nnnn
ö
÷
÷
n
det A =
÷
÷
÷
ø
n
å
=
k
+
k
1
-
)1(
1
Ma
,
kk
11
aaa
111213
aaa
313233
So to compute the determinant of a 5×5 matrix we must build 5 sub
matrices, each of size 4. To compute the determinants of each the 4×4 matrices we need to create 4 sub matrices each, these now of size 3 and so on.
To compute the determinant of a 10×10 matrix would require computing
the determinant of 10!=10×9×8×7×6×5×4×3×2=3,628,800 1×1 matrices.
13
Fortunately there are better ways. However this does suggest an ex-
ab
cd
æö
ç÷
èø
1252
1212
-
==×-×+-×-+
4561597623847539421860
=××+××+××-××-××-××=
cellent computer programming exercise to write a recursive procedure to
compute a determinant.
In practice it is a bit silly to decompose a 2×2 matrix down into a
couple of
1×1 matrices and then compute the exceedingly easy determinant of
these puny matrices. So here is a simple theorem.
Theorem DMST: Determinant of Matrices of Size Two. Suppose
that
A
=
. Then det(A)=ad−bc.
=+
For system
aa
aa
ì
í
î
1211
2221
byaxa
11211
a determinant is
=+
byaxa
22221
aaaa
-==D
21122211
Determinant of a 3×3 matrix
311
11
det51231(1)(1)
A
( )
+
112
51
13
1(1)120
+
×-=¹
11
an over way
aaa
131211
aaa
232221
aaa
333231
123
12
+
---++=
233211122133132231213213122331332211
aaaaaaaaaaaaaaaaaa
789
14
121
ö
÷
Then its determinant, det(A)=∣A∣,
-
320
÷
÷
113
ø
-
20
×+×-
1
13
Example. Let А =
121
-
32
×=- )2310()3310(2)3112(
1
320
113
2
11
æ
ç
ç
ç
è
30
13
= -5 + 18 + 6 = 19.
Theorem DER: Determinant Expansion about Rows. Suppose that
A is a square matrix of size n. Then for 1≤i≤n
aaа
...
ö
n
11211
÷
aaa
...
÷
detA =
n
22221
÷
............
÷
÷
aaa
...
nnnn
ø
å
k
n
=
+
ik
-
)1(
1
, i = 1,2,…,n.
Ma
ikik
А=
æ
ç
ç
ç
ç
ç
è
21
which is known as expansion about row i.
3211
-
4111
-
)1(1
-×=
3132
5013
411
-
2111
++
313
)1()1(
-×-+
501
411
-
312
+
503
=×+×+×-×-×-×-=
411
)1(2
332
-×+-×+
++
)1(3
513
111
-
4131
132
)1504305(1
++---+=
013
)109320(3)103369815(2)10012905(1
=--++-----++++---++
54273461393)17(2613
-=---=×--×+-=
Theorem DT: Determinant of the Transpose. Suppose that A is a
square matrix.
Then det(At)=det(A).
15
When a matrix has all zeros above (or below) the diagonal, exploiting
21
45
æö
ç÷
èø
100
211
021
æö
ç÷
ç÷
ç÷
èø
ij
e
ij
e
the zeros by expanding about the proper row or column makes computing a determinant insanely easy.
Reading Questions
1. Compute the determinant of the matrix
2. Compute the determinant of the matrix using too ways
.
.
Topic 4. Matrix Inverses
and Systems of Linear Equations
Introduction. This lecture is devoted to the discussion of the following concepts: nonsingular matrix, a formula for the inverse, a minor. For
a full understanding of the topic you need to have an idea about matrices
and determinants of the matrix.
Definition IM: Identity Matrix. The m×m identity matrix, Im, is de-
fined by system
= 0, i ¹ j,
= 1, i = j . or
0...01
æ
ç
ç
ç
ç
ç
è
Definition SQM: Square Matrix. A matrix with m rows and n columns is square if m=n. In this case, we say the matrix has size n. To emphasize the situation when a matrix is not square, we will call it rectangular.
We can now present one of the central definitions of linear algebra.
ö
÷
0...10
÷
÷
............
÷
÷
1...00
ø
16
Definition NM: Nonsingular Matrix. Suppose A is a square matrix.
n
I
n
I
n
I
n
I
1
-
A
det
det
ji
A
A
ijij
MA
-=
4,3,2,1
==-=-=
Suppose further that the solution set to the homogeneous linear system of
equations LS(A,0) is {0}, in other words, the system has only the trivial
solution. Then we say that A is a nonsingular matrix. Otherwise we say
A is a singular matrix.
Theorem CINM: Computing the Inverse of a Nonsingular Matrix. Suppose A is a nonsingular square matrix of size n. Create the n×2n
matrix M by placing the n×n identity matrix
to the right of the matrix
A. Let N be a matrix that is row-equivalent to M and in reduced rowechelon form. Finally, let J be the matrix formed from the final n col-
umns of N. Then AJ=
.
Theorem OSIS: One-Sided Inverse is Sufficient. Suppose A and B
are square matrices of size n such that AB=
. Then BA=
.
So Theorem OSIS tells us that if A is nonsingular, then the matrix B
guaranteed by Theorem CINM (yourself) will be both a “right-inverse”
and a “left-inverse” for A, so A is invertible and A
=B.
So if you have a nonsingular matrix, A, you can use the procedure described in Theorem CINM to find an inverse for A.
Usually we use the formula
A Formula For The Inverse
i j
+
-
( )
x
=
ij
M
1
ji
=
,
(1)
ij
+
Some books would call these Mij minor
1) To calculate the determinant of a matrix A,
2) each matrix element to replace its algebraic addition,, divided by
the determinant of A
3) to transpose the result
1 2
æ
Example Let А =
ö
, calculate А-1.
ç
÷
3 4
è
ø
det A = 4 – 6 = -2.
M11=4; M12= 3; M21= 2; M22=1
AAAA
11122122
17
3
2
2
211
2
æö
ç÷
ç÷
ç÷
-
ç÷
èø
-
30
30
30
43
-
æö
1
*
A
=-=
ç÷
-
2
èø
-
2 1
So , А-1=
æ
ç
è
32 12/ /
-
1
ö
, x11= -2; x12= 1; x21= 3/2; x22= -1/2
÷
-
ø
– intermediate result
Properties of Inverse Matrices
1) (A-1)-1 = A;
2) (AB)-1 = B-1A-1
3) (AT)-1 = (A-1)T.
Reading Questions
1) to find the inverse of the 3×3 matrix below
--
115
(D = det A =
5(4-9) + 1(2 – 12) – 1(3 – 8) = -25 – 10 +5 = -30.
=
321
234
M11 =
M12 =
M13 =
1
11
1
21
1
31
30
32
23
21
34
5
;
10
;
30
5
;
= -5; M21 =
31
-=
M22 =
;10
24
M23 =
;5
-=
1
1
12
30
1
22
19
1
32
;
14
30
;
11 --
= 1; M31 =
23
15
;14
=
24
15
=
M33 =
;19
34
1
1
---
aaa
13
;
---
aaa
33
;
===
30
16
1
---
;
=-=-=
aaa
23
30
11
1
;
-===
M32 =
--
115
321
234
= -1;
ö
÷
÷
÷
ø
æ
ç
ç
ç
è
11 --
32
15
-
31
15
-
;16
=
;11
=
21
18
1
æ
ç
6
ç
A-1 =
1
ç
ç
3
ç
1
ç
6
è
2) For A =
30
7
--
15
19
30
æ
ç
ç
ç
è
30
8
15
11
-
30
ö
÷
÷
.
÷
÷
÷
÷
ø
--
115
ö
÷
and A-1 =
321
÷
÷
234
ø
1
æ
ç
ç
ç
ç
ç
ç
è
30
6
7
1
--
15
3
19
1
30
6
ö
÷
30
÷
8
÷
÷
15
÷
11
-
÷
30
ø
to find A×A
-1
1
1
1
1
3) What is the easiest way to recognize if a square matrix is nonsingu-
lar or not?
Topic 5. Systems of Equations, Algebraic
Procedures (Gauss-Jordan procedure)
Introduction. This lecture is devoted to the discussion of the following concepts: an augmented matrix, augmented matrix as a resulting
from a row operation, Gauss elimination, Gauss-Jordan procedure. For a
full understanding of the topic you need to have an idea about methods
of solution to the system of equations.
Definition EO Elementary operations are those operations consisting of the following.
1. Interchange the order in which the equations are listed.
2. Multiply any equation by a nonzero number.
3. Replace any equation with itself added to a multiple of another
equation.
Example. To illustrate the third of these operations on this particular
system, consider the following.
x + y = 7
2x − y = 8
The system has the same solution set as the system
x + y = 7
−3y = −6
19
To obtain the second system, take the second equation of the first system and add -2 times the first equation to obtain −3y = −6.
Of course a linear system may involve many equations and many variables. The solution set is still the collection of solutions to the equations.
In every case, the above operations of Definition EO do not change the
set of solutions to the system of linear equations.
Example. Find the solutions to the system.
x + 3y + 6z = 25
2x + 7y + 14z = 58
2y + 5z = 19
To solve this system replace the second equation by (−2) times the
first equation added to the second. This yields the system
x + 3y + 6z = 25
y + 2z = 8
2y + 5z = 19
Now take (−2) times the second and add to the third. More precisely,
replace the third equation with (−2) times the second added to the third.
This yields the system
x + 3y + 6z = 25
y + 2z = 8
z = 3
At this point, you can tell what the solution is. This system has the
same solution as the original system and in the above, z = 3. Then using
this in the second equation, it follows y + 6 = 8 and so y = 2. Now using
this in the top equation yields x+6+18 = 25 and so x = 1. This process is
called back substitution.
Alternatively, you could have continued as follows. Add (−2) times
the bottom equation to the middle and then add (−6) times the bottom to
the top.
This yields
x + 3y = 7
y = 2
z = 3
Now add (−3) times the second to the top. This yields
x = 1
y = 2
z = 3
a system which has the same solution set as the original system. This
avoided back substitution and led to the same solution set.
20
A less cumbersome way to represent a linear system is to write it as
1 36| 25
2 714| 58
0 25| 19
æö
ç÷
ç÷
ç÷
ç÷
èø
an augmented matrix. For example the linear system, can be written as
It has exactly the same information as the original system but here it
is understood there is an x column
1
2
0
a y column,
3
7
2
and a z column,
6
14
5
The rows correspond to the equations in the system. Thus the top row
in the augmented matrix corresponds to the equation, x + 3y + 6z = 25.
Now when you replace an equation with a multiple of another equation added to itself, you are just taking a row of this augmented matrix
and replacing it with a multiple of another row added to it. Thus the first
step in solving would be to take (−2) times the first row of the augment-
ed matrix above and add it to the second row,
1 3 6 | 25
0 1 2 | 8
0 2 5 | 19
Next take (−2) times the second row and add to the third,
1 3 6 | 25
0 1 2 | 8
0 0 1 | 3
This augmented matrix corresponds to the system
x + 3y + 6z = 25
y + 2z = 8
z = 3
By back substitution you obtain the solution x = 1, y = 6, and z = 3.
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