LA-Ch-4-SGLee
Chapter 4
Determinant
4.1 Definition and Properties of the Determinants
4.2 Cofactor Expansion and Applications of the Determinants
4.3 Cramer's Rule
*4.4 Application of Determinant
4.5 Eigenvalues and Eigenvectors
(English Textbook) http://matrix.skku.ac.kr/2015-Album/Big-Book-LinearAlgebra-Eng-2015.pdf
(e-book : Korean) http://matrix.skku.ac.kr/2015-Album/BigBook-LinearAlgebra-2015.pdf
http://matrix.skku.ac.kr/LA-Sage/
선형대수학 http://matrix.skku.ac.kr/LinearAlgebra.htm
Credu http://matrix.skku.ac.kr/Credu-CLA/index.htm
OCW http://matrix.skku.ac.kr/OCW-MT/index.htm
행렬 계산기 http://matrix.skku.ac.kr/2014-Album/MC-2.html
그래프 이론 http://matrix.skku.ac.kr/2014-Album/Graph-Project.html
행렬론 http://matrix.skku.ac.kr/MT2010/MT2010.htm
JFC http://matrix.skku.ac.kr/JCF/index.htm
The concept of determinant was introduced 150 years before the use of modern matrix, and we have used the determinant to solve the systems of linear equations for over 100 years. In late 19th century, Sylvester introduced the concept of matrix and the method for solving systems of equations by using an inverse matrix, where the determinant is used to check if an inverse of a matrix exists or not. Also, the determinant can be used to find area, volume, equations of lines or planes, and exterior product. It also helps in geometric interpretation of vectors.
In this chapter, we first define the determinant and review its properties. Then we study how to compute the determinant by cofactor expansion. We also study Cramer's rule which solves the systems of linear equations by using the determinant.
One of the most important concepts in linear algebra is eigenvalues and eigenvectors. Eigenvalues have almost all important informations by components from an object with
components. Eigenvalues are not only important in theoretical perspective but also applicable to almost all areas related to matrix, such as, finding the solutions of differential equations, computing the power of given matrix, Google search, and image compression, etc. In the last section of this chapter, we compute eigenvalues by using the determinant.
4.1 Definition of Determinant
Reference video: https://youtu.be/MeVCFwg1Eq0 (http://youtu.be/DM-q2ZuQtI0)
Practice site: http://matrix.skku.ac.kr/knou-knowls/CLA-Week-5-Sec-4-1.html
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In this section, we introduce a determinant function which assigns any square matrix |
Definition |
[Permutation] |
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For a set of natural numbers
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We simply denote a permutation as
. As a permutation
is an one to one correspondence, the range
is simply a rearrangement of
. Hence, there are
permutations on
: fixing the position of first element leaves
possibilities to permute the rest. We denote the set of all permutations of set
by
.
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[Remark] |
Inversion |
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In permutation
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Number of inversions for
: after (
)-th index, the number of indexes which is smaller than
-th index
is called the number of inversions for
. In the above example, the number of inversion for
is
. Number of inversions for a permutation
is the total sum of each number of inversions for
,
.
Definition |
[Even permutation and odd permutation] |
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If number of inversions for a permutation is even than it is called an even permutation, If the number is odd than it is called an odd permutation. |
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Determine whether the permutation is even or odd by computing the inversion numbers for a permutation in
.
The number of inversions for 5 is 4. The number of inversions for 1 is 0, for 2 is 0, for 4 is 1, and 3 is the last index. Hence, the total sum is
, and it is an odd permutation. □
● http://matrix.skku.ac.kr/RPG_English/4-TF-Permutation.html
[[0, 1], [0, 2], [0, 3], [0, 4], [3, 4]] # Note!! Index starts from 0
5 # Number of inversions
False # Permutation([5,1,2,4,3]) is not an even permutation ■
Definition |
[Signature function] |
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A signature function
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Classify the permutations of to either even or odd permutation.
permutations |
number of inversions |
class |
sign |
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even |
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even |
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even |
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odd |
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odd |
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odd |
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In permutation, if two numbers switch the location then the signature is changed
Theorem |
4.1.1 |
Let |
Definition |
[Determinant] [Leibniz formula] |
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Let
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By definition,
matrix
has it's determinant as
.
Each term
in the determinant is from the matrix
, by choosing a row and a column, without any overlapping, then multiplying them and assigning a corresponding signature.
Find the , where
.
As is
matrix,
. Since
,
, we have
.
https://en.wikipedia.org/wiki/Rule_of_Sarrus ■
Find the , where
.
As is
matrix,
.
Since ,
,
,
,
,
,
by substituting them into the definition of the determinant, we have
■
Compute the determinant of the following matrices.
,
.
.
. □
● http://matrix.skku.ac.kr/RPG_English/4-B1-Det-matrix.html
240 ■
[Remark] |
Sarrus' method cannot be applied to the case of degree 4 or higher. |
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Hence, the determinant with degree 4 or higher should be computed by the definition. But in that case, there are too many terms and signs to be determined. (Indeed, for degree 4 case, there are |
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Properties of the determinant
Theorem |
4.1.2 |
A square matrix |
● https://www.projectrhea.org/rhea/index.php/Determinant_Transpose_Proof
In ,
, and
. Since
we have . □
240 ■
The properties of the determinant regarding to rows also work to columns.
Theorem |
4.1.3 |
Let |
Let
be a matrix obtained by replacing
th and
th row of
,
. This means
,
and
if
.
(by definition)
(by theorem 4.1.1)
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Let . Since
and
,
. ■
Theorem |
4.1.4 |
If a square matrix |
● http://www.math.vanderbilt.edu/~msapir/msapir/jan29.html
Let which has identical first and third rows. Note
0 ■
Theorem |
4.1.5 |
If a square matrix |
Let which has identical zeros in the third row. Observe
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Theorem |
4.1.6 |
Let |
Let . Note that
■
Theorem |
4.1.7 |
If a square matrix |
Theorem |
4.1.8 |
Let |
Let
be a new matrix whose second row is obtained by adding
times of the first row of
to
.
=> . (by Theorem 4.1.4) ■
Let and 2 times of the second row is added to the first row and name it as matrix
. Then
.
Note that . ■
Theorem |
4.1.9 |
If
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From the previous theorem, . ■
[Remark] |
How to compute the determinant |
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1. Use elementary row operations to make many zeros to a certain row (column). 2. Multiply the diagonal elements.
※ Note that during the elementary row operations, if you multiply k times a row (column) or switch two rows (columns), do not forget to multiply 1/k and -1. |
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Find the determinant of a matrix , where
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Theorem |
4.1.10 |
Let |
[Remark] |
The determinant of an elementary matrix |
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1. If 2. If 3. If 4. If |
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Equivalent conditions for invertible matrix
Theorem |
4.1.11 |
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Theorem |
4.1.12 |
For any two |
Verify the above theorem with matrices .
Since
, and
. □
det(AB)= -10
det(A)*det(B)= -10 ■
Theorem |
4.1.13 |
If a square matrix |
Verify the above Theorem with a matrix .
is invertible with
. Observe
and
. □
A=matrix(QQ, 2, 2, [1, 2, 3, 4])
Ai=A.inverse()
print "det(A)=", A.det()
print "det(A^(-1))=", Ai.det()
det(A)= -2
det(A^(-1))= -1/2 ■
4.2 Cofactor Expansion and Applications of the Determinants
Reference Video: https://youtu.be/w6eTWgw-JZs http://youtu.be/XPCD0ZYoH5I
Practice Site: http://matrix.skku.ac.kr/knou-knowls/CLA-Week-5-Sec-4-2.html
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In this section, we introduce a method which is convenient to compute the determinant as well as important in theory. Moreover, by applying this method, we introduce an easier formula to compute the inverse of a matrix. |
Definition |
[Minor and cofactor] |
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We denote a submatrix, by removing the |
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For given matrix , find the minor and cofactor of
for
.
The minor of
for
is
and the cofactor of A for
is
. ■
Definition |
[Adjoint matrix] |
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Let
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Find of the following matrix.
Note the cofactor of for each element is as follows.
Therefore, □
● http://matrix.skku.ac.kr/RPG_English/4-MA-adjoint.html
[-18 -6 -10]
[ 17 -10 -1]
[ -6 -2 28] ■
Cofactor expansion
The determinant of matrix
can be expanded as follows.
(Expand along the first column)
This is known as a (Laplace) cofactor expansion of along the first column.
Cofactor expansion works for any column and any row.
For any
matrix
, the following identity holds. That is,
.
Which shows .
Read: http://nptel.ac.in/courses/122104018/node29.html
For the previous example ,
det(A)= -94
A*adj(A)=
[-94 0 0]
[ 0 -94 0]
[ 0 0 –94] ■
Therefore the following holds.
Theorem |
4.2.1 [Cofactor expansion] |
Let
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When computing the determinant, it is advantageous to use the cofactor
expansion along the row (column) which has many zeros.
Find the determinant of a given matrix by using the cofactor expansion.
Multiply (-2) to the 2nd row and add it to the 3rd row. Multiply (-3) to the 2nd row and add it to the both 1st and 4th row. Then
.
Now we cofactor expand along the first column,
. ■
Theorem |
4.2.2 [Finding inverse matrix with adjoint matrix] |
Let |
From , find the inverse matrix of
.
http://sage.skku.edu or http://mathlab.knou.ac.kr:8080
(1/dA)*adjA=
[ 9/47 3/47 5/47]
[-17/94 5/47 1/94]
[ 3/47 1/47 -14/47]
A^(-1)=
[ 9/47 3/47 5/47]
[-17/94 5/47 1/94]
[ 3/47 1/47 -14/47]
[ 3/47 1/47 –14/47] ■
4.3 Cramer's Rule
Reference video: https://youtu.be/ESE9AWVI2Dw http://youtu.be/OImrmmWXuvU
Practice site: http://matrix.skku.ac.kr/knou-knowls/CLA-Week-6-Sec-4-3.html
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In this section, we introduce Cramer’s rule which is very useful tool for solving a system of linear equations. |
Cramer's rule can be applied to systems of linear equations with the same number of unknowns and the equations.
Theorem |
4.3.1 [Cramer's Rule] |
For a system of linear equations,
let
where |
Since
,
is invertible. Hence the system of linear equations
has a unique solution
. Since
, we have
.
Observe the th component of
is
. Since
,
if we denote as a matrix
with the
th column replaced by the vector
, then we have
. ■
Solve the following system of linear equations by Cramer's rule.
Let
be the coefficient matrix. Then
, and hence
,
,
. ■
-2
-4
-6
-8
x = 2
y = 3
z = 4 ■
Solve the following system of linear equations by Cramer's rule.
From
, and each matrix
has zeros column,
. Hence, the solution is z
■
Theorem |
4.3.2 [Invertible Matrix Theorem] |
For an (1) RREF (2) (3) (4) (5) (6) The columns of (7) The rows of (8) |
Note that there are more equivalent statements for the above theorem. For more equivalent statements, refer Theorem 7.4.9 in section 7.4.
4.4 *Application of Determinant
Reference video: http://youtu.be/OImrmmWXuvU, http://youtu.be/KtkOH5M3_Lc
Practice site: http://matrix.skku.ac.kr/knou-knowls/CLA-Week-6-Sec-4-4.html
The concept of determinant was first introduced by Japanese Takakazu Seki Kowa in 1683. The term determinant originated from the meaning of determining the existence of roots. It was Cauchy who used the term in modern concept in 1815. In this section, we introduce some geometric and algebraic applications among many other applications of the determinant.
By using a determinant, we can easily find areas, volumes, equations of lines, equations of elliptic curves, or equations of plane. Also, the determinant of Vandermonde matrix connects between discrete data, which arise from statistical data and experimental labs, etc.
Show that the equation of a line, which passes through two distinct points , is as follows.
Note that the above equation is degree 1 for both and
. As the equation holds by substituting
and
into the equation, the equation must be the equation of the line which passes through two given points.
■
[Remark] |
Computer simulation |
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[An equation of a line which passes through two distinct points] http://www.geogebratube.org/student/m9504
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Similar to
, an equation of a plane, which passes through three distinct points
, is as follows:
.
[Remark] |
Computer simulation |
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[An equation of a plane which passes through three distinct points] http://www.geogebratube.org/student/m56430
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Consider an arbitrary non-singular square matrix
. Let
be the
th column and
.
For the case is a parallelogram, and for the case
is a generalized parallelepiped.
Parallelogram can be expressed by adding two vectors as follows.
The area of the above parallelogram is which is the same as the absolute value of
. Similarly, a parallelepiped is generated by three vectors which do not lie on the same plane. Let matrix
's columns consist by these three vectors. Then the volume of the parallelepiped is absolute value of
.
[Remark] |
Computer simulation |
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[Volume of parallelepiped] http://www.geogebratube.org/student/m57553
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Theorem |
4.4.1 |
(1) Let (2) Let (3) The area of parallelogram generated by two vectors |
We will prove only (3).
Note that .
Also, the area of parallelogram is . Now, the determinant
(square of base times height)
makes the square of the area of the parallelogram generated by .■
Show that the area of a triangle generated by three points ,
,
is as follows.
As the area is not changing by parallel translation, the area of triangle generated by three given vectors are the same as half of the area of parallelogram generated by and
. Hence, by Theorem 4.1.1, we have
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Vandermonde matrix and the determinants
If there are
distinct points in the
-plane with mutually distinct
coordinates, then there exist a unique polynomial which passes through all given points with degree
. Let's find the polynomial.
Let
be
distinct points in the
-plane with mutually distinct
coordinates. We want to find a polynomial of degree
which passes through all given points. Let
be such a polynomial.
Since these points satisfy the given polynomial, we have
.
Moreover, as are mutually distinct, the coefficient matrix has
.
This coefficient matrix is called Vandermonde matrix with degree
. Now we introduce how to compute the determinant of Vandermonde matrix. For the case
,
.
Similarly, as we illustrated in the above case, the determinant of Vandermonde matrix
with degree
is product of
(with
). That is,
.
[Reference] http://www.proofwiki.org/wiki/Vandermonde_Determinant
Find a polynomial that passes through the points (39, 34), (99, 47), (38, 58) by using a Vandermonde matrix.
V=
[ 1.0 39.0 1521.0]
[ 1.0 99.0 9801.0]
[ 1.0 38.0 1444.0]
x= (1558.34590164, -54.568579235, 0.396994535519)
p=0.396994535519*x^2 -54.568579235*x + 1558.34590164
plot(p, (x, -20, 120)) # plot the graph
■
[Remark] |
Computer Simulation |
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[Curve Fitting] http://www.geogebratube.org/student/m9911 [Area of parallelogram] http://www.geogebratube.org/student/m113
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4.5 Eigenvalues and Eigenvectors
Reference video: https://youtu.be/U7KG5jFR0q4 http://youtu.be/OImrmmWXuvU,
Practice site: http://matrix.skku.ac.kr/knou-knowls/CLA-Week-6-Sec-4-5.html
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For an |
Definition |
[Eigenvalues and Eigenvectors] |
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Let |
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Let and
. Then
=
. Hence 3 is an eigenvalue of
, and
is an eigenvector corresponding to 3. ■
Since , for any
, the only eigenvalue of identity matrix
is
. Also, any nonzero vector
is an eigenvector of
corresponding to 1. ■
If
is an eigenvector of
corresponding to
, then
is also an eigenvector of
corresponding to
for any nonzero scalar
.
.
General method to find eigenvalues
Since
and , the system of linear equations
should have nonzero solution. Therefore, the characteristic equation,
should hold.
is called the characteristic polynomial.
Theorem |
4.5.1 |
Let
(1) (2) (3) System of linear equations |
Find all eigenvalues and corresponding eigenvectors of
.
If satisfies
. Then,
(1)
However, as mentioned above, this system of linear equations should have nontrivial (nonzero) solution. Hence,
① Let’s find an eigenvector corresponding to .
From (1),
② Let’s find an eigenvector corresponding to .
From (1),
■
[Remark] |
Computer simulation |
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[Visualize the eigenvalues and eigenvectors] http://www.geogebratube.org/student/b73259#material/11114
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Do eigenvalues exist for any square matrix?
Theorem |
4.5.2 [Fundamental Theorem of Algebra] |
For any real (or complex) coefficient polynomial with degree
has |
That is, a real square matrix
with degree
always has
eigenvalues in complex domain. However, in this textbook we have limited the scalar as real numbers, and hence there is no eigenvalues means there is no real eigenvalues.
Find eigenvalues and eigenvectors of a matrix .
● http://matrix.skku.ac.kr/RPG_English/4-BN-char_ploy.html
① Characteristic polynomial of
x^2 - 2*x – 8
② Hence the eigenvalues are as follows.
[x == -2, x == 4]
③ We can find the eigenvalues directly by using the built in command.
[4, -2]
④ In order to find eigenvector for , solve
.
[ 1 -1]
[ 0 0]
=>
⑤ In order to find eigenvector for , solve
.
[ 1 1]
[ 0 0]
Hence,
⑥ We can find the eigenvectors directly by using the built in command.
[(4, [(1, -1)], 1), (-2, [(1, 1)], 1)] # [eigenvalues, eigenvectors(it may appear in different form), multiplicity] ■
Find eigenvalues and eigenvectors of a matrix .
● http://matrix.skku.ac.kr/RPG_English/4-VT-eigenvalues.html
① Characteristic polynomial of
x^3 - 3*x^2 - 9*x + 27
② Hence the eigenvalues are as follows.
[x == -3, x == 3]
③ We can find the eigenvalues directly by using the built in command.
[-3, 3, 3] # shows root with multiplicity 2
④ In order to find eigenvector for , solve
.
[ 1 0 2/3]
[ 0 1 -1/3]
[ 0 0 0]
Hence,
⑤ In order to find eigenvector for , solve
.
[ 1 -1 -1]
[ 0 0 0]
[ 0 0 0]
Hence,
( and
are real numbers not simultaneously become zero)
⑥ We can find the eigenvectors directly by using the built in command.
[(-3, [(1, -1/2, -3/2)], 1), (3, [ (1, 0, 1), (0, 1, -1) ], 2)] # [eigenvalues, eigenvectors (it may appear in different form), multiplicity]
For a triangular matrix
with degree
, the main diagonal components of
are
(
). Therefore, the characteristic polynomial of
is
, and hence the eigenvalues of the triangular matrix
are its main diagonal components,
.
Find the characteristic polynomial and all the eigenvalues of triangular matrix .
As ,
's eigenvalues are
. ■
Definition |
[Eigenspace] |
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Let |
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That is, an eigenspace of
corresponding to
is the set of all eigenvectors of
corresponding to
and the zero vector, which is a subspace of
.
From the given matrix in
, find eigenspaces of
corresponding to each eigenvalue
.
From the result of ,
① if , by solving
, we have
② When , by solving
, we have
(
,
)
■
● http://matrix.skku.ac.kr/LA-Lab/index.htm
● http://matrix.skku.ac.kr/knou-knowls/cla-sage-reference.htm
[Solution] Section 4-1 http://youtu.be/Fne4gaZtE_Q
Section 4-3 http://youtu.be/Ygu4_7I4fGQ
Is permutation
of
even or odd?
Sol)
The number of inversions for 2 is 2. The number of inversions for 0 is 0, for 1 is 0, for 3 is 0, for 5 is 0, for 10 is 2, for 8 is 1 and 7 is the last index.
Hence, the total sum is
Answer)
It is an odd permutation. ■
Double Checked by Sage)
[[0, 1], [0, 2], [5, 6], [5, 7], [6, 7]] (5 inversions) ■
(New) Find the following determinants.
(1).
Sol)
(
)
(
)
(
)
(By Theorem 4.1.9)
So =-1.
Double checked by Sage : http://math3.skku.ac.kr/home/pub/19 (by SC.Kim)
det(A)= -1 ■
[ Note. 문제에서는 5차 정사각행렬이 주어졌기 때문에 어느 정도의 과정을 거치면 금방 determinant를 구할 수 있지만 차수가 점점 올라갈수록 복잡해진다. Sage를 이용한다면 어떠한 행렬에 관하여도 위와 같은 풀이를 이용하여 해결할 수 있다는 것을 알 수 있다. ]
Let
be
matrix and
, find the followings.
(1) (2)
(3)
(4)
Sol)
(1)
(2)
(3)
(4)
Double checked by Sage : http://math1.skku.ac.kr/home/pub/2503 (by MS.Kang)
[ 1 0 0]
[ 0 1 0]
[ 0 0 -4]
-4
16 #
-1/4 #
-32 #
-1/32 # ■
[ Note : Determinant의 성질을 알게 되면 다양한 변형된 행렬의 Determinant를 복잡한 계산 없이 아주 손쉽게 풀 수 있다는 것을 보여주는 대표적인 문제이다. ]
For given matrices.
,
(1) show .
(2) show .
Sol) (1) ,
(2)
Double Checked by Sage : http://math1.skku.ac.kr/home/pub/2513/ (by LG.Kim)
(1)
det(A)= -10
det(B)= -10
(2)
det(AB)= 100
det(A)*det(B)= 100 ■
[ Note 1 :
인 특수한 행렬에 대해서
를 구해보면,
,
또한, 위의 문제 (2) 에서 행렬 는 행렬
의 2번째-3번째 열을 바꾸고, 그 다음 2번째-3번째 행을 바꾼 형태이므로 Theorem 4.1.3 에 의해
이 성립한다.
실제로, Sage 를 통해 확인해보면; http://math1.skku.ac.kr/home/pub/2514 (by LG.Kim)
-10
-10
성립한다는 것을 확인할 수 있다. ]
[ Note 2 :
Determinant 의 성질이 True 임을 알 수 있는 대표적인 문제이다. 실제로 matrix 일 때에도 성립한다는 것을 보일 때 Determinant 의 성질이 완벽히 참이라는 것이 증명이 되지만 여기서는 그 경우까지 고려하면 약간 복잡해지기 때문에 이 페이지에 싣지는 않을 것이다. 그렇지만 일반적인 matrix 에 대하여
가 성립하며 이는 upper triangular matrix 또는 diagonal matrix 등 특수한 matrix 의 성질을 이용하여 증명할 수 있다. ]
For which
and
, the given matrix is invertible?
Sol)
invertible =>
.
The matrix is a lower triangular matrix. By Theorem 4.1.9,
.
∴ Answer ; ■
For given matrices,
(1) show .
(2) show .
(3) show .
Sol)
(1)
(2)
(3)
Double checked by Sage : http://math1.skku.ac.kr/home/pub/2515/ (by LG.Kim)
(1)
10
10
(2)
570
570
(3)
1/10
1/10 ■
Find all cofactors of the following matrices.
(1) (2)
Sol)
(1) ,
,
,
,
,
,
(2) (Using Sage)
Double checked by Sage : http://math1.skku.ac.kr/home/pub/2516 (by LG.Kim)
(1)
[-66 12 30]
[ 28 -8 -4]
[ 39 6 -9]
(2)
[-18 14 22 -14 -4]
[ 36 -28 -44 28 8]
[-18 14 22 -14 -4]
[ 18 -14 -22 14 4]
[ 0 0 0 0 0] ■
[ Note :
위 문제에서와 같이 (1)번 같은 경우는 손으로 쉽게 해결할 수 있지만, (2)번과 같이 행렬의 차원이 점점 커질 때 손으로 풀기 힘들어지고 복잡해진다. 그래서 Sage Code를 이용하여 문제를 쉽게 풀 수 있었다. ]
Find the determinant of the matrix by cofactor expansion.
Sol)
Use cofactor expansion on the 1 column
.
⇒ .
Use cofactor expansion on the 3rd column .
⇒
The answer is .
Double checked by Sage : http://math3.skku.ac.kr/home/pub/9 (by SC.Kim)
det(A)= 30 ■
Find the adjoint matrix
of the matrix
form (Problem 8).
Sol) Checked by Sage : http://math1.skku.ac.kr/home/pub/2498/ (by MS.Kang)
[ 10 -10 10 -10]
[ 12 0 -6 -24]
[ -3 0 9 6]
[-11 20 -17 32] ■
Find the inverse matrix of the given matrix by cofactor expansion.
(1) (2)
Solution) (1) by cofactor expansion
(2) by cofactor expansion
=
,
=
=
Double checked by Sage)
(1) http://math1.skku.ac.kr/home/pub/2483 (by JW.Baek)
inverse of A = (1/dA)*adjA = [ 2 0 -1 ]
[ 2/3 1/3 -1/3 ]
[ -1 0 1 ]
(2) http://math3.skku.ac.kr/home/pub/25 (by SY.Lee)
inverse of A = (1/dA)*adjA =
[ -18/133 23/133 2/7 -48/133 -29/133]
[ -30/19 13/19 1 -4/19 -4/19]
[ 999/133 -412/133 -27/7 -129/133 80/133]
[ 164/133 -47/133 -5/7 -6/133 13/133]
[-268/133 106/133 8/7 39/133 -18/133]
Solve the systems of linear equations by using the Cramer's rule.
(1)
(2)
(3)
(4)
Sol)
(1)
Let ,
,
=-101,
=-60,
=55,
=-21
=
,
=
,
=
(
Cramer's Rule)
=
=
,
=
=
,
=
=
(2)
Let ,
,
=8,
=4,
=-8 ,
=0,
=12
=
,
=
,
=
,
=
=
=
,
=
=-1,
=
=0,
=
=
(3)
Let ,
,
=-4,
=-4,
=-4,
=-8
=
,
=
,
=
=
=1,
=
=1,
=
=2
(4)
Let ,
,
=3,
=26,
=-14 ,
=-7,
=-12
=
,
=
,
=
,
=
=
=
,
=
=
,
=
=
,
=
=-4
Double check by Sage)
http://math1.skku.ac.kr/home/pub/2484 (by JW.Baek)
(1)
x= (60/101, -55/101, 21/101)
(2)
x= (1/2, -1, 0, 3/2)
(3)
x= (1, 1, 2)
(4)
x= (26/3, -14/3, -7/3, -4) ■
Solve the following problems by using the determinant of Vandermonde matrix.
(1) Find the line equation which passes through the two points (-1, 11) and (2, -10).
(2) Find the coefficients a, b, c of parabolic equation which passes
through the three points (1, 3), (2, 3) and (3, 5).
Sol)
(1)
Let the equation of line to be .
The Vadermonde matrix is
.
,
Ans:
(2)
Let
Vardermonde matrix is
,
Ans:
Double check by Sage)
(1)
http://math1.skku.ac.kr/home/pub/2474 (by SY.Lee)
Answer : a= [ 4]
[-7]
(2)
http://math1.skku.ac.kr/home/pub/2475 (by SY.Lee)
Answer : a = [ 5]
[-3]
[ 1]
Solve the following problems by using the determinant.
(1) The area of a parallelogram which is generated by two sides connecting the origin and each point (4, 3) and (7, 5).
(2) The volume of parallelepiped which is generated by three vectors, (1, 0, 4), (0, -2, 2) and (3, 1, -1).
Sol)
(1)
Let
(Area of parallelogram)
(2)
Let
(Area of parallelepiped)
Double check by Sage)
(1)
http://math1.skku.ac.kr/home/pub/2472 (by SY.Lee)
Answer: 24 ■
Find the eigenvalues and eigenvectors of the following matrices.
(1) (2)
Sol)
(1)
,
,
⓵ If , then
(
From
)
, where
.
⓶ If . then
.
, where
.
(2)
By using Sage,
,
⓵ If , then
, where
.
⓶ If , then
, where
.
⓷ If , then
, where
.
⓸ If , then
, where
.
Double checked by Sage : (1) http://math1.skku.ac.kr/home/pub/2518/ (by LG.Kim)
(2) http://math1.skku.ac.kr/home/pub/2517/ (by LG.Kim)
(1)
x^2 - x - 6 # char. poly. of A
[x == 3, x == -2]
[1 5]
[0 0]
[1 0]
[0 0]
[(3, [(1, -1/5)], 1), (-2, [(0, 1)], 1)]
(2)
A.charpoly() = x^4 - 118*x^2 - 168*x + 1485
[x == 11, x == -9, x == -5, x == 3]
A.eigenvalues() = [11, 3, -5, -9]
(3*identity_matrix(4)-A).echelon_form()
[ 1 0 0 -1]
[ 0 1 0 -1]
[ 0 0 1 -1]
[ 0 0 0 0]
(11*identity_matrix(4)-A).echelon_form()
[ 1 0 0 -9/13]
[ 0 1 0 7/13]
[ 0 0 1 11/13]
[ 0 0 0 0]
(-5*identity_matrix(4)-A).echelon_form()
[ 1 0 0 7/5]
[ 0 1 0 7/5]
[ 0 0 1 -13/5]
[ 0 0 0 0]
(-9*identity_matrix(4)-A).echelon_form()
[ 1 0 0 2]
[ 0 1 0 -1]
[ 0 0 1 2]
[ 0 0 0 0]
A.eigenvectors_right() =
[ (11 , [ (1, -7/9, -11/9, 13/9) ], 1),
(3, [ (1, 1, 1, 1) ], 1 ) ,
(-5, [ (1, 1, -13/7, -5/7 ) ], 1),
(-9, [ (1, -1/2, 1, -1/2 ) ], 1) ] ■
[ Note :
만일 2차 정사각행렬 에 대하여
를 만족하는
,
이 존재하면
이 되어야 하므로
.
,
.
,
는 위 이차방정식의 해이므로,
,
. ]
Explain why
for the following matrix
.
Sol)
Hence, has identical rows.
(
Theorem.4.1.4) ■
Show that for two square matrices
and
, if
for an invertible matrix
, then
.
Sol)
■
[ Note :
위 문제는 두 정사각행렬 ,
에 대하여
이 성립한다는 사실을 알고 있다면 쉽게 풀 수 있는 문제 유형이다. ]
(New) Simplify the following determinant.
Sol)
Double checked by Sage : http://math1.skku.ac.kr/home/pub/2524 (by MS.Kang)
Use matrix . (One example)
1
1
■
[ Note : var{a, b, c}를 첫 줄에 입력하면 실제 미지수에 대해서도 임을 알 수 있다. ]
Solution)
[ (
From Theorem 4.2.2) ]
# multiply
⇒ #
⇒ # determinant each side
⇒ # using
⇒ # using
⇒ # devide
both side
■
Checked by Sage
det(adj A)= 27000
(det A)^(n-1) 27000
det(adj B)= 429981696
(det B)^(n-1) 429981696
det(adj C)= 14808575318291015625
(det C)^(n-1) 14808575318291015625
Let
be a
matrix, and assume that
.
(1) Find . Which relation does this value have with
?
(2) Find .
Sol) [Tip : ,
]
(1) (Using Sage)
, (
From Problem p4)
(2) , then
,
(Using Sage)
Double checked by Sage : http://math1.skku.ac.kr/home/pub/2522 (by LG.Kim)
(1)
determinant(adjA) = 8
(determinant(adjA))^(1/(4-1)) = 2
(2)
[ 1 0 0 0]
[ 0 4 -1 1]
[ 0 -6 2 -2]
[ 0 1 0 1] ■
[ Note : 위 문제는 앞의 문제와 연관이 있는 문제이다.
앞서 문제에서 이 성립한다는 것을 보였기 때문에 이를 응용하여 간단히
를 구할 수 있었다. ]
By using the Cramer's rule,
find the degree 3 polynomial which passes through the following four points.
(0,1), (1,-1), (2,-1), (3,7)
Sol) By substituting each point to the polynomial, we get four equations.
By using Cramer’s rule, (,)
Double checked by Sage : http://math1.skku.ac.kr/home/pub/2521/ (By HK.Seo)
(1, -2, -1)
,
,
■
Let the characteristic polynomial of matrix
be
. Find eigenvalues of matrix
.
Sol)
,
,
(multiple root)
, so
Eigenvalues of matrix
are
,
. ■
[ Note :
차 정사각행렬
의 고윳값을
라고 할 때 (단,
,
,
,
),
인 자연수
에 대하여
의 고윳값이
임을 보이자.
Ⅰ. 일 때
Ⅱ. 일 때
임이 성립한다고 가정하면,
일 때
일 때도 성립한다.
따라서 의 고윳값은
이다. ]
Find the eigenspaces of
, corresponding to each eigenvalue and show that they are orthogonal to each other in the plane.
Sol) ,
,
,
Also,
Hence, eigenspaces are orthogonal to each other in the plane.
Double checked by Sage : http://math1.skku.ac.kr/home/pub/2520 (By HK.Seo)
[(5, [(1, 2)], 1), (0, [(1, -1/2)], 1)] ■
[Note :
e-book을 참고하면 두 vector ,
(단,
,
)에 대해서 내적한 값을 구하였는데 어차피 scalar 값을 고려할 필요가 없기 때문에 eigenvector 들의 내적 값을 구하여
이 되는지만 확인하면 된다. ]
Find the characteristic polynomial of the following matrix. And find the roots of the polynomial by using the Sage.
Sol) The characteristic polynomial of is the determinant of
.
,
,
,
,
Double checked by Sage : http://math1.skku.ac.kr/home/pub/2523/ (By HK.Seo)
x^5 - 14*x^4 + 39*x^3 + 59*x^2 - 121*x + 31
[-1.898437229647213?, 0.3126193021869149?, 1.071200945566419?,
5.280735607688378?, 9.23388137420550?] ■
[ Note : 앞서 계속해서 강조되었듯이 이번 문제를 통하여 Sage 의 힘은 엄청나다는 것을 알 수 있다. 저 특성방정식을 구하고 난 뒤에 일반적으로 풀 수가 없다. 그래서 Sage 가 이 문제를 통하여 엄청 중요한 요소임을 알 수 있다. 저런 복잡한 문제들도 Sage Coding을 통하여 몇 분도 안 되어 풀 수 있다. ]
About the Author
https://www.researchgate.net/profile/Sang_Gu_Lee
https://scholar.google.com/citations?user=FjOjyHIAAAAJ&hl=en&cstart=0&pagesize=20
http://orcid.org/0000-0002-7408-9648
http://www.scopus.com/authid/detail.uri?authorId=35292447100