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   

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.

 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

In this section, we introduce a determinant function which assigns any square matrix to a real number . In order to define the determinant function, we first introduce permutation. Then we review the properties of the determinant function.

 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 .

[Remark]		Inversion
	In permutation , an inversion is the case when a bigger natural number placed on the left hand side of a smaller natural number. For example, in a permutation , is placed on the left hand side of , and hence is an inversion. Similarly, is an inversion.

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 , .

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 

Permutation([5,1,2,4,3]).inversions()           # inversions

[[0, 1], [0, 2], [0, 3], [0, 4], [3, 4]]      # Note!! Index starts from 0

Permutation([5,1,2,4,3]).number_of_inversions()

# Number of inversions

5

Permutation([5,1,2,4,3]).is_even() 

# check whether it is even permutation

False    ■

Classify the permutations of to either even or odd permutation.

permutations	number of inversions	class	sign
		even
		even
		even
		odd
		odd
		odd

       ■

 In permutation, if two numbers switch the location then the signature is changed

 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 

B=matrix(QQ, 3, 3, [1, 2, 3, -4, 5, 6, 7, -8, 9])

print B.det()                   # compute the determinant

240                                                              ■

[Remark]		Sarrus' method cannot be applied to the case of degree 4 or higher.
	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 terms, and for degree 10, there are terms to compute). Therefore, it is more effective to study the properties of the determinant and find the determinant by using these properties. (We will skip the proofs but will verify them by examples).

Properties of the determinant

● https://www.projectrhea.org/rhea/index.php/Determinant_Transpose_Proof 

In , , and . Since

we have .      □

B=matrix(QQ, 3, 3, [1, 2, 3, -4, 5, 6, 7, -8, 9])

print B.transpose().det()    

240                      ■

The properties of the determinant regarding to rows also work to columns.

  Let ​ be a matrix obtained by replacing th and th row of  , . This means  , ​​ and if .

     (by definition)

          (by theorem 4.1.1)

                                                                  ■

Let . Since and ,  .              ■

● http://www.math.vanderbilt.edu/~msapir/msapir/jan29.html

Let which has identical first and third rows. Note

A=matrix(QQ, 3, 3, [1, 2, 3, -1, 0, 7, 1, 2, 3])

print A.det()                           # compute the determinant

0               ■

Theorem	4.1.5
If a square matrix has a row (column) with identical zeros then .

Let which has identical zeros in the third row. Observe

          ■

Let . Note that

    ■

Theorem	4.1.7
If a square matrix has two proportional rows then .

  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 .              ■

From the previous theorem, .  ■

[Remark]		How to compute the determinant
	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.

Find the determinant of a matrix , where

              ■

[Remark]		The determinant of an elementary matrix
	1. If is obtained by multiplying to a row of , 2. If is obtained by switching two rows of , 3. If is obtained by multiplying times a row and adding it to another row of , 4. If is an matrix and is an elementary matrix,     .

Equivalent conditions for invertible matrix

Theorem	4.1.11
is invertible if and only if .

Theorem	4.1.12
For any two matrices and ,  .

Verify the above theorem with matrices .

 Since , and

 .                                          □

                                                                    A=matrix(QQ, 2, 2, [1, 2, 3, 4])

B=matrix(QQ, 2, 2, [2, -1, 1, 2])

C=A*B

print "det(AB)=", C.det()

print "det(A)*det(B)=", A.det()*B.det()

det(AB)= -10

det(A)*det(B)= -10               ■

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                  ■

 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

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]
	We denote a submatrix, by removing the th row and th column of a given square matrix ,  as . We call its determinant as minor of for . We also call as cofactor of for .

For given matrix , find the minor and cofactor of for .

  The minor of for is and the cofactor of A for is .         ■

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 

A=matrix(QQ, 3, 3, [3, -2, 1, 5, 6, 2, 1, 0, -3])

print A.adjoint()                                  # adjoint matrix

[-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 ,

 http://sage.skku.edu or http://mathlab.knou.ac.kr:8080

A=matrix(QQ, 3, 3, [3, -2, 1, 5, 6, 2, 1, 0, -3])

print "det(A)=", A.det()

print "A*adj(A)="

print A*A.adjoint()  

det(A)= -94

A*adj(A)=

[-94   0   0]

[  0 -94   0]

[  0   0 –94]                                                            ■

Therefore the following holds.

  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,

      .                        ■

From , find the inverse matrix of .

 http://sage.skku.edu or http://mathlab.knou.ac.kr:8080

A=matrix(QQ, 3, 3, [3, -2, 1, 5, 6, 2, 1, 0, -3])

dA=A.det()                         # compute the determinant

adjA=A.adjoint()                   # compute adjoint matrix

print "(1/dA)*adjA="

print (1/dA)*adjA                 # compute inverse matrix

print 

print "A^(-1)="

print A.inverse()           # compare the results of inverse matrix

(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]                                         ■

 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

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. 

 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

,,  . ■ 

A = matrix(3,3,[-2,3,-1,1,2,-1,-2,-1,1]);

A1 = matrix(3,3,[1, 3, -1,4, 2,-1,-3,-1,1]);

A2 = matrix(3,3,[-2,1, -1, 1,4,-1,-2,-3,1]);

A3 = matrix(3,3,[-2,3,1, 1, 2, 4, -2,-1,-3]);

print A.det()

print A1.det()

print A2.det()

print A3.det()

print "x =",  A1.det()/A.det()

print "y =",  A2.det()/A.det()

print "z =",  A3.det()/A.det()

-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      ■

 Note that there are more equivalent statements for the above theorem. For more equivalent statements, refer Theorem 7.4.9 in section 7.4. 

 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
	[An equation of a line which passes through two distinct points]   http://www.geogebratube.org/student/m9504

 Similar to , an equation of a plane, which passes through three distinct points  , is as follows:

.

[Remark]		Computer simulation
	[An equation of a plane which passes through three distinct points]   http://www.geogebratube.org/student/m56430

 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
	[Volume of parallelepiped]   http://www.geogebratube.org/student/m57553

 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

                            .      ■

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.

def Vandermonde_matrix(x_list):          # generate Vandermonde matrix

    n=len(x_list)

    A=matrix(RDF, n, n, [[z^i for i in range(n)] for z in x_list])

    return A

x_list=[39, 99, 38]                          # x coordinate

V=Vandermonde_matrix(x_list)

y_list=vector([34, 47, 58])                    # y coordinate

print "V="

print V

print

print "x=", V.solve_right(y_list)

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
	[Curve Fitting]   http://www.geogebratube.org/student/m9911 [Area of parallelogram]  http://www.geogebratube.org/student/m113

 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

For an matrix and a vector , is a vector in . One of the important questions in applied problems is “Is there any nonzero vector , which makes both and parallel?” If such a vector exists, then it is called an eigenvector and it plays many important roles in linear transformation. In this section, we introduce eigenvectors and eigenvalues.

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.

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
	[Visualize the eigenvalues and eigenvectors]   http://www.geogebratube.org/student/b73259#material/11114

 Do eigenvalues exist for any square matrix?

 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

A=matrix(QQ, 2, 2, [1, -3, -3, 1])    # input A

print A.charpoly()                  # characteristic equation of A

x^2 - 2*x – 8

② Hence the eigenvalues are as follows.

solve(x^2 - 2*x - 8==0, x)

[x == -2, x == 4]

③ We can find the eigenvalues directly by using the built in command.

A.eigenvalues()               # eigenvalues of A

[4, -2]

④ In order to find eigenvector for , solve .

(-2*identity_matrix(2)-A).echelon_form() # consider only coefficient matrix

[ 1 -1]

[ 0  0]

=>     

⑤ In order to find eigenvector for , solve .

(4*identity_matrix(2)-A).echelon_form()  # consider only coefficient matrix

[ 1  1]

[ 0  0]

Hence,    

⑥ We can find the eigenvectors directly by using the built in command.

A.eigenvectors_right() 

[(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  

A=matrix(QQ, 3, 3, [1, 2, 2, 1, 2, -1, 3, -3, 0])  # input A

print A.charpoly()                    # characteristic equation of A

x^3 - 3*x^2 - 9*x + 27

② Hence the eigenvalues are as follows.

solve(x^3 - 3*x^2 - 9*x + 27==0, x)

[x == -3, x == 3]

③ We can find the eigenvalues directly by using the built in command. 

A.eigenvalues()               # eigenvalues of A

[-3, 3, 3]                      # shows root with multiplicity 2

④ In order to find eigenvector for , solve .

(-3*identity_matrix(3)-A).echelon_form() # consider only coefficient matrix

[   1    0  2/3]

[   0    1 -1/3]

[   0    0    0]

Hence,       

⑤ In order to find eigenvector for , solve .

(3*identity_matrix(3)-A).echelon_form()  # consider only coefficient matrix

[ 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.

A.eigenvectors_right() 

[(-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 .   ■

 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)

http://math1.skku.ac.kr/home/pub/2470

Permutation([2,0,1,3,5,10,8,7]).inversions()

[[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)

A=matrix(5,5,[1,0,1,0,1,1,1,1,1,1,1,1,0,-1,1,0,1,0,1,1,1,0,1,-1,1])

print "det(A)=",; print A.det()                                                                           

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)

A=matrix(QQ,3,3,[1,0,0,0,1,0,0,0,-4])    

print A

print A.det()                       

print (A*A).det()                           # (1)

print (A.inverse()).det()                      # (2)

print (2*A).det()                           # (3)

print ((2*A).inverse()).det()           # (4)

[ 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)

A=matrix(QQ, 3, 3, [1, 0, 2, 0, 5, 0, 3, 0, 4])

B=matrix(QQ, 3, 3, [1, 0, 3, 0, 5, 0, 2, 0, 4])

print "det(A)=", A.det()

print "det(B)=", B.det()

det(A)= -10

det(B)= -10

(2)

A=matrix(QQ, 3, 3, [1, 0, 2, 0, 5, 0, 3, 0, 4])

B=matrix(QQ, 3, 3, [1, 2, 0, 3, 4, 0, 0, 0, 5])

C=A*B

print "det(AB)=", C.det()

print "det(A)*det(B)=", A.det()*B.det()

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)

A=matrix(QQ, 3, 3, [1, 0, 2, 0, 5, 0, 3, 0, 4])

B=matrix(QQ, 3, 3, [1, 2, 0, 3, 4, 0, 0, 0, 5])

print A.det()

print B.det()

-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)

A=matrix(QQ, 3, 3, [0, -1, -2, 1, 4, 1, 2, 0, -4])

B=matrix(QQ, 3, 3, [0, 1, 2, -1, 4, 0, -2, 1, -4])

print A.det()

print B.det()

10

10

(2)

A=matrix(QQ, 3, 3, [0, -1, -2, 1, 4, 1, 2, 0, -4])

B=matrix(QQ, 3, 3, [3, 6, 0, 0, 4, 1, 0, 1, 5])

C=A*B

print A.det()*B.det()

print C.det()

570

570

(3)

A=matrix(QQ, 3, 3, [0, -1, -2, 1, 4, 1, 2, 0, -4])

B=A.inverse()

print 1/A.det()

print B.det()

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)

A=matrix(QQ, 3, 3, [1, 1, 5, 3, -6, 9, 2, 6, 2])

B=A.adjoint()

C=B.transpose()

print C

[-66  12  30]

[ 28  -8  -4]

[ 39   6  -9]

(2)

A=matrix(QQ, 5, 5, [1, 2, 3, 4, 5, 0, 1, 0, 1, 0, -1, 1, -1, 1, -1, 0, 1, 2, 3, 4, -4, 2, -3, 1, 5])

B=A.adjoint()

C=B.transpose()

print C

[-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 3^rd column .

⇒

The answer is .

Double checked by Sage : http://math3.skku.ac.kr/home/pub/9 (by SC.Kim)

A=matrix(QQ, 4, 4, [4, 1, 0, 2, -1, 3, 5, 1, 0, 1, 4, 0, 2, -1, -1, 1])

print "det(A)=", A.det()

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)

        A=matrix(QQ,4,4,[4,1,0,2,-1,3,5,1,0,1,4,0,2,-1,-1,1]);

        print A.adjoint()

        [ 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)

A=matrix([[1, 0, 1], [-1, 3, 0], [1, 0, 2]]) dA=A.det() adjA=A.adjoint() print "inverse of A = (1/dA)*adjA ="  print (1/dA)*adjA

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)

A=matrix(5,5,[1,0,1,3,5,-1,3,0,7,2,1,0,2,1,8,2,-4,0,0,3,-8,9,2,5,4]) dA=A.det() adjA=A.adjoint() print "inverse of A = (1/dA)*adjA =" print (1/dA)*adjA

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) 

A=matrix(3, 3, [3, -3, -2, -1, -4, 2, 5, 4, 1])

b=vector([3, 2, 1])

print "x=", A.inverse()*b  

print

print "x=", A.solve_right(b)  

x= (60/101, -55/101, 21/101)

(2) 

A=matrix(4, 4, [1, -1, -1, -1, -1, -1, 1, 1, 1, 1, -1, 1, 1, 1, 1, 1])

b=vector([0, 2, 1, 1])

print "x=", A.inverse()*b  

print

print "x=", A.solve_right(b)  

x= (1/2, -1, 0, 3/2)

(3)

A=matrix(3, 3, [1, 2, 1, 2, 2, 1, 1, 2, 3])

b=vector([5, 6, 9])

print "x=", A.inverse()*b  

print

print "x=", A.solve_right(b)  

x= (1, 1, 2)

(4)

A=matrix(4, 4, [1, 1, 0, 0, 0, 1, 1, -2, 1, 0, 2, 1, 1, 1, 0, 1])

b=vector([4, 1, 0, 0])

print "x=", A.inverse()*b  

print

print "x=", A.solve_right(b)  

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)

V=matrix(2,2,[1,-1,1,2])

v=V.inverse()                                   Answer 

y=matrix(2,1,[11,-10])                           [ 4]

a=v*y                                          [-7]

print a

(2)

http://math1.skku.ac.kr/home/pub/2475 (by SY.Lee)

V=matrix(3,3,[1,1,1,1,2,4,1,3,9])                    Answer

v=V.inverse()                                   

y=matrix(3,1,[3,3,5])                             [ 5]

a=v*y                                          [-3]

print a                                          [ 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)

x1=matrix(2,1,[4,3])

x2=matrix(2,1,[7,5])

A=x1.augment(x2)

a=A.det()

print a.abs()

Answer: 1

(2)

http://math1.skku.ac.kr/home/pub/2473 (by SY.Lee)

x1=matrix(3,1,[1,0,4])

x2=matrix(3,1,[0,-2,2])

x3=matrix(3,1,[3,1,-1])

A=x1.augment(x2).augment(x3)

a=A.det()

print a.abs()

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)

A=matrix(QQ, 2, 2, [3, 0, -1, -2])

print A.charpoly()

x^2 - x - 6

solve(x^2 - x - 6==0, x)

[x == 3, x == -2]

print A.eigenvalues()

[3, -2]

(3*identity_matrix(2)-A).echelon_form()

[1 5]

[0 0]

(-2*identity_matrix(2)-A).echelon_form()

[1 0]

[0 0]

A.eigenvectors_right()

[(3, [(1, -1/5)], 1), (-2, [(0, 1)], 1)]

(2)

A=matrix(QQ, 4, 4, [-3, 0, -2, 8, 0, 1, 4, -2, -4, 10, -1, -2, 6, -4, -2, 3])

print A.charpoly()

x^4 - 118*x^2 - 168*x + 1485

solve(x^4 - 118*x^2 - 168*x + 1485==0, x)

[x == 11, x == -9, x == -5, x == 3]

print 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)

A=matrix(QQ,3,3,[1,1,1,1,2,2,1,2,3]);

a=1

b=2

c=3

print A.det()

print a*(b-a)*(c-b)

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

http://math3.skku.ac.kr/home/pub/18

A=matrix(QQ,[[4,1,0,2], [-1,3,5,1], [0,1,4,0],[2,-1,-1,1]]) B=matrix(QQ,[[1,2,3,4,5],[5,4,3,2,1],[2,1,3,5,4],[9,5,7,1,5],[2,1,1,5,1]]) C=matrix(QQ,[[4,1,0,2,6,8],[-1,3,5,1,1,5],[0,1,4,0,1,5],[2,-1,-1,1,1,5],[6,5,1,2,5,1],[1,6,8,8,7,6]]) print "det(adj A)= ",A.adjoint().det() print "(det A)^(n-1)",A.det()^(3) print "det(adj B)= ",B.adjoint().det() print "(det B)^(n-1)",B.det()^(4) print "det(adj C)= ",C.adjoint().det() print "(det C)^(n-1)",C.det()^(5)

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) adjA=matrix(QQ, 4, 4, [2, 0, 0, 0, 0, 2, 1, 0, 0, 4, 3, 2, 0, -2, -1, 2])

print adjA.determinant()                                                             # 8

P=(adjA.determinant())^(1/(4-1))

print P                                                                          # 2

(2) A=P*adjA.inverse()

print A

[ 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)

A=matrix(QQ,3,3,[1,1,1,8,4,2,27,9,3])

b=vector([-2,-2,6])

print A.inverse()*b

(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)

A=matrix(QQ,2,2,[1,2,2,4])

print A.eigenvectors_right()

[(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)

A=matrix(QQ,5,5,[1,1,2,1,1,1,2,3,2,1,2,3,1,2,1,1,2,2,3,1,1,1,1,1,7])

print A.charpoly()

print A.eigenvalues()

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. ]