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Linear Algebra with

Contents

Chapter 1. Vectors

*1.1 Vectors in n-space

*1.2 Inner product and Orthogonality

1.3 Vector equations of lines and planes

1.4 Excercise

Chapter 2. Linear system of equations

2.1 Linear system of equations

2.2 Gaussian elimination and Gauss-Jordan elimination

2.3 Exercise

Chapter 3. Matrix and Matrix Algebra

3.1 Matrix operation

3.2 Inverse matrix

3.3 Elementary matrix

3.4 Subspace and linear independence

3.5 Solution set of a linear system and matrix

3.6 Special matrices

*3.7 LU-decomposition

3.8 Excercise

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

4.6 Excercise

Chapter 5. Matrix Model

5.1 Lights out Game

5.2 Power Method

Chapter 6. Linear Transformations

6.1 Matrix as a Function (Transformation)

6.2 Geometric Meaning of Linear Transformations

6.3 Kernel and Range

6.4 Composition of Linear Transformations and Invertibility

*6.5 Computer Graphics with Sage

6.6 Exercises

Chapter 7. Dimension and Subspaces

7.1 Properties of bases and dimensions

7.2 Basic spaces of matrix

7.3 Rank-Nullity theorem

7.4 Rank theorem

7.5 Projection theorem

*7.6 Least square solution

7.7 Gram-Schmidt orthonomalization process

7.8 QR-Decomposition; Householder transformations

7.9 Coordinate vectors

7.10 Exercises

Chapter 8. Diagonalization

8.1 Matrix Representation of LT

8.2 Similarity and Diagonalization

8.3 Diagonalization with orthogonal matrix, *Function of matrix

8.6 SVD and Pseudo-Inverse

8.7 Complex eigenvalues and eigenvectors

8.8 Hermitian, Unitary, Normal Matrices

*8.9 Linear system of differential equations

8.10 Exercises

Chapter 9. General Vector Spaces

9.1 Axioms of Vector Space

9.2 Inner product spaces; *Fourier Series

9.3 Isomorphism

9.4 Exercises

Chapter 10. Jordan Canonical Form

10.1 Finding the Jordan Canonical Form with a Dot Diagram

*10.2 Jordan Canonical Form and Generalized Eigenvectors

10.3 Jordan Canonical Form and CAS

10.4 Exercises

Appendix

*1.1 Vectors in n-space

*1.2 Inner product and orthogonality

1.3 Vector equations of lines and planes

1.4 Excercise

Linear algebra is the branch of mathematics concerning vectors and mappings. Linear algebra is central to both pure and applied mathematics. Combined with calculus, linear algebra facilitates the solution of linear systems of differential equations. Techniques from linear algebra are also used in analytic geometry, engineering, physics, natural sciences, computer science, computer animation, and the social sciences (particularly in economics). A geometric quantity described by a magnitude and a direction is called a vector. In this chapter, we begin with studying basic properties of vectors starting from 3-dimensional vectors and extending these properties to -dimensional vectors. We will also discuss the notion of the dot product (or inner product) of vectors and vector equations of lines and planes.

Introduction : http://youtu.be/Vx1r-y_xIeM (http://youtu.be/Mxp1e2Zzg-A)

 For , , , , , express the vectors , , in component form.    , ,                  and are equivalent.   Copy the following code into http://sage.skku.edu                 or http://mathlab.knou.ac.kr:8080/ to practice.                                                                        o=vector([0, 0])#creates a vector, x=vector([component , component ]) p1=vector([0, -4]) p2=vector([-3, 1]) q=vector([2, 3]) q1=vector([2, -1]) q2=vector([-1, 4]) print "vector OQ=", q-o              # subtract print "vector P1Q1=", q1-p1          # subtract print "vector P2Q2=", q2-p2          # subtract print "vector OQ = vector P1Q1= vector P2Q2"                                                                      vector OQ= (2, 3) vector P1Q1= (2, 3) vector P2Q2= (2, 3) vector vector OQ = P1Q1= vector P2Q2                          ■

 For vectors , in , find , , and .                                                          Copy the following code into http://sage.skku.edu or http://mathlab.knou.ac.kr:8080/ to practice.                                                                       x=vector([1, 2])  #creates a vector, x=vector([component , component ]) y=vector([-2, 4]) print "x+y=", x+y           # adds vectors print "x-y=", x-y           # subtract vectors print "-2*x=", -2*x        # multiplies vectors by scalar, (you must                               include '*' when multiplying)                                                                      x+y=(-1, 6) x-y=(3, -2) -2*x=(-2, -4)                                                    ■

 For , , , , , , express the vectors , , in component form.   , , and are equivalent.

 Copy the following code into http://sage.skku.edu                 or http://mathlab.knou.ac.kr:8080/ to practice.                                                                       o=vector([0, 0, 0])    #creates a vector p1=vector([0, -4, 2]) p2=vector([-3, 1, 0]) q=vector([2, 3, 4]) q1=vector([2, -1, 6]) q2=vector([-1, 4, 4]) print "vector OQ=", q-o              # subtract print "vector P1Q1=", q1-p1          # subtract print "vector P2Q2=", q2-p2          # subtract print "vector OQ = vector P1Q1= vector P2Q2"                                                                      vector OQ= (2, 3, 4) vector P1Q1= (2, 3, 4) vector P2Q2= (2, 3, 4) vector OQ = vector P1Q1= vector P2Q2                          ■

 Find , , when and in . ,,                         □     Copy the following code into http://sage.skku.edu to practice.                                                                       x=vector([1, 2, -3, 4])  y=vector([-2, 4, 1, 0])           print "x+y=", x+y                # adds vectors print "x-y=", x-y                # subtracts vectors print "-2*x=", -2*x                                                                      x+y=(-1, 6, -2, 4) x-y=(3, -2, -4, 4) -2*x=(-2, -4, 6, -8)                                              ■

 Find , when , and in .              Copy the following code into http://sage.skku.edu or              http://mathlab.knou.ac.kr:8080/ to practice.                                                                       x=vector([1, 2, -3, 4])  y=vector([-2, 4, 1, 0]) z=vector([5, -2, 3, -7]) print "2*x-3*y+z=", 2*x-3*y+z             # linear combination                                                                     2*x-3*y+z=(13, -10, -6, 1)                                        ■

 The above can also be done in Sage as follows. First, we build the relevant vectors and the command for a linear combination of many vectors. Then, we can combine all into one line, as follows.              Copy the following code into http://sage.skku.edu or              http://mathlab.knou.ac.kr:8080/ to practice.                                                                       x=vector(QQ,  [1, 2, -3, 4])  # computations with quotient numbers in Q y=vector(QQ,  [-2, 4, 1, 0]) z=vector(QQ,  [5, -2, 3, -7]) print "2*x-3*y+z=", 2*x-3*y+z             # linear combination vectors = [x, y, z] scalars = [2, -3, 1] multiples = [scalars[i]*vectors[i] for i in range(3)] print "a*x+b*y+c*z=", sum(multiples)      # linear combination                                                                      2*x-3*y+z = (13, -10, -6, 1)  a*x+b*y+c*z = (13, -10, -6, 1)                                   ■

 In this section, we will discuss the concepts of vector length, distance, and how to calculate the angle between two vectors, as well as vector parallelism and orthogonality in .

 For the vectors  , in , we have the following.              .                                            □                 Copy the following code into http://sage.skku.edu to practice.                                                                       x=vector([2, -1, 3, 2])            y=vector([3, 2, 1, -4]) print x.norm()                 # calculate the norm of x print y.norm()                 # calculate the norm of y print (x-y).norm()             # calculate distance                                                                     3*sqrt(2)                         # sqrt(2) means sqrt(30) 5*sqrt(2)                                                        ■

 Using the vectors and in , calculate .   .           □   Copy the following code into http://sage.skku.edu to practice.                                                                       x=vector([2, -1, 3, 2])            y=vector([3, 2, 1, -4]) print x.inner_product(y)         # find the dot product                                                                     -1                                                               ■

 For two vectors and in , establish orthogonality.              □   Copy the following code into http://sage.skku.edu to practice.                                                                       x=vector([1, 0, 1, 1])            y=vector([-1, 0, 0, 1]) print x.inner_product(y)                                                                        0 #orthogonal                                                   ■

 Using the vectors and from , verify that the triangle inequality holds.   , , ,  and . Hence . So, .                      ■

 In this section, we will derive vector equations of lines and planes in , and we will examine shortest distance problems related to these equations.

 Find vector, parametric and symmetric equations of the line that passes through the point and is parallel to the vector .   (1) The vector equation of the line is given by             ,   or             . (2) The parametric equation is given by  (). (3) The symmetric equation is given by .   ■

 Find parametric equations for the line that passes through the points and .   Two points and with position vectors and forms a vector           and the vector equation can be written  as                      ,   .     Thus, the parametric equations are:   , , ()            ■

 Find vector and parametric equations of the plane that passes through the three points:  , , and .     Let , , , and . Then we have two vectors that parallel to the plane as   , .   Then, from our above definitions, we have  ,           which is a vector equation of the plane.   If we further simplify the above expression, we have    .    In particular, , , .      is the parametric equations of the plane.                         ■

 For vectors , , find (the projection of onto ) and (the component of orthogonal to ).   Since , we have    □
 Copy the following code into http://sage.skku.edu to practice.                                                                       a=vector([2, -1, 3])            b=vector([4, -1, 2]) ab=a.inner_product(b) aa=a.inner_product(a) p=ab/aa*a;w=b-p           print "p=", p print "w=", w                                                                     p= (15/7, -15/14, 45/14) w= (13/7, 1/14, -17/14)                                         ■

 Find the distance from the point to the plane .     Figure 12 tells us that . Here, ,                       where , so                          .              □   Copy the following code into http://sage.skku.edu to practice.                                                                       n=vector([1, 3, -2]) v=vector([3, -1, 2]);d=-6 vn=v.inner_product(n) nn=n.norm() Dist=abs(vn+d)/nn print Dist                                                                     5/7*sqrt(14)         #                            ■

For points , find the vector .

Sol)

Checked by Sage)  http://math1.skku.ac.kr/home/pub/2505/ by 변희성

INPUT

P1=vector([5,-2,1])

P2=vector([2,4,2])

vec=P2-P1

print vec

OUTPUT

(-3, 6, 1)      :　OK

What is the initial point of the vector with terminal point  ?

Sol)    Initial point

Checked by Sage)  http://math1.skku.ac.kr/home/pub/2506/  by 변희성

INPUT

x=vector([1,1,3])

B=vector([-1,-1,2])

point=B-x

print point

OUTPUT

(-2, -2, -1)  :　OK

((NEW )

What is the initial point of the vector  3 2 1  with terminal point

2 3 8 ?

Sol)

■

Checked by Sage) http://math3.skku.ac.kr/home/pub/23  by 김민기

INPUT

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

B = vector([2, 3, 8])

print "vector xB ", B-x

OUTPUT

vector xB  (-1, 1, 7)          :   OK   ■

For vectors , , and

, compute the following:

Sol)

■

Checked by Sage)  http://math1.skku.ac.kr/home/pub/2507/  by 변희성

INPUT

u=vector([-3,1,2,4,4])

v=vector([4,0,-8,1,2])

w=vector([6,-1,-4,3,-5])

ans=(2*u-7*w)-(8*v+u)

print ans

OUTPUT

(-77, 8, 94, -25, 23)   :　OK

Using the same vectors and from above, find the vector that

satisfies the following:

Sol) Let

⇒

⇒

Checked by Sage)  http://math1.skku.ac.kr/home/pub/2447  by 김원경

INPUT

u=vector([-3,1,2,4,4])

v=vector([4,0,-8,1,2])

w=vector([6,-1,-4,3,-5])

x=(2*u-v-w)/6

print x

OUTPUT

(-8/3, 1/2, 8/3, 2/3, 11/6)  :　OK

(New)

For vectors , , and

. Find the vector that satisfies the following:

Sol)    =>

=>

=>

=>

∴         ■

Checked by Sage)  http://math1.skku.ac.kr/home/pub/2427     by 배성준

INPUT

u=vector([-3,1,2,4,4])

v=vector([4,0,-8,1,2])

w=vector([6,-1,-4,3,-5])

x=(2*u-3*v-w)/8

print"x=",x

OUTPUT

x= (-3, 3/8, 4, 1/4, 7/8)  :　OK

For vectors calculate , where is

the angle between x and y.

Sol)

.

.

Note : ∴

Checked by Sage)  http://math1.skku.ac.kr/home/pub/2508/ by 임상훈

INPUT

x=vector([-1,-2,3])

y=vector([3,-2,-1])

print "cos(theta)= ", x.inner_product(y)/(x.norm()*y.norm())

OUTPUT

cos(theta)=  -1/7    :　OK

Find the distance between the two points and .

Sol)

Checked by Sage)  http://math1.skku.ac.kr/home/pub/2509  by 정양헌

INPUT

P=vector([-1,2,1])

Q=vector([-3,-4,5])

print "P-Q =",(P-Q).norm()

OUTPUT

P-Q = 2*sqrt(14)  :　OK

For vectors , , find the real number

that satisfy .

Sol)

■

Checked by Sage)  http://math1.skku.ac.kr/home/pub/2510/  by 정양현

INPUT

a=var('a')

x=vector([a,2,-1,a])

y=vector([-a,-1,3,6])

xy=[x[i]*y[i] for i in range(4)]

solve (sum(xy), a)

OUTPUT

[a == 1, a == 5] :　OK

(New)

For vectors x     , y   2  2 , find the real number

that such that x ⋅y .

Sol)

Checked by Sage) http://math3.skku.ac.kr/home/pub/41 by 박시현

INPUT

a=var('a')

x=vector([a,2,-1,a])

y=vector([a,2,3,2])

xy=[x[i]*y[i] for i in range(4)]

print xy

solve ([a^2+2*a+1],a)

OUTPUT

[a^2, 4, -3, 2*a]

[a == -1]               :　OK

Find a vector equation of the line between the two points .

Sol)   A equation of line passes through the point

either parallel to a vector

= ()

∴ , ()  ■

Check by Sage) http://math1.skku.ac.kr/home/pub/2488 by 주민규

INPUT

p=vector([-5,1,3])

q=vector([2,-3,4])

t=real

x=q-p

print x

OUTPUT

(7, -4, 1)       :　OK

NOTE)

We can use this vector to find the equation of line by adding a vector to equation multiplied by a real number .

(New)

Find a vector equation of the line between the two points

and .

Sol)

The vector equation pass through two points and :

or

∴

Check by Sage) http://math3.skku.ac.kr/home/pub/126 by 주민규

INPUT

var('x,y,z')

var('t')

P = vector([5,0,1])

Q = vector([0,2,6])

R= Q-P

print "vector equation= ", P+R*t

OUTPUT

vector equation=  (-5*t + 5, 2*t, 5*t + 1)   :  OK      ■

Find a normal vector perpendicular to the plane .

Sol) Normal vector : Length of vector is 1.

.

(perpendicular vector to the plane)

.

■

[Projection] For and , find the scalar projection and vector projection of onto .

Sol)

Checked by Sage)  http://math1.skku.ac.kr/home/pub/2511/  by 정양현

INPUT

a=vector([2, -1, 3])

b=vector([4, -1, 2])

ab=a.inner_product(b)

aa=a.inner_product(a)

p=ab/aa*a;w=b-p

print "p=", p

print "w=", w

OUTPUT

p= (15/7, -15/14, 45/14)

w= (13/7, 1/14, -17/14)              :OK

[Discussion] Vectors with the same magnitude and direction are considered to be equivalent.

However, in a vector space,  discuss the relationship between vectors with the same slope but expressed with different equations.

Ans)

Equivalence

When each components of two vectors are same, then we say

that and are equivalent (or equal) and we write .

Conclusively, if the two lines' direction and the gradient are each the same, then a line moved parallely and the length changed, they can be the same vector(line).

Sol)

Check :  and

=> : orthonormal vectors

[Find which is orthonormal to both and .]

=>

=> , .

We know that for some .

In addition, must be a normal vector which means that =>

[Solution] Section 1-1  http://youtu.be/fbCMyh-iDCQ

Section 1-2  http://youtu.be/sEFj_7t_bqc

Section 1-3  http://youtu.be/avVJfeEoeVs and http://youtu.be/tys3taO5IHs

2

2.1 Linear system of equations

2.2 Gaussian elimination and Gauss-Jordan elimination

2.3 Exercise

A system of linear equations and its solution is one of the most important problems in Linear Algebra. A linear system with thousands of variables occurs in natural and social sciences, engineering, as well as traffic problems, weather forecasting, decision-making, etc. Even differential equations concerning derivatives such as velocity and acceleration can be solved by transforming them into a linear system.

In Linear Algebra, a solution of a linear system is obtained by Gauss elimination method or with determinants. In Chapter 2, we consider a geometric meaning of the solution of a linear system and its solution, and investigate some applications of a linear system.

 The theory of linear systems is the basis and a fundamental part of linear algebra, a subject which is used in most parts of modern mathematics. Computational algorithms for finding the solutions are an important part of numerical linear algebra, and play a prominent role in engineering, physics, chemistry, computer science, and economics. In this section, we study the process of finding solutions of linear system of equations and its geometric meanings.

 Equations , can be written as , and they are linear. But  , are not linear.

 Describe all possible solution sets in of a linear system with three equations and three unknowns.     One can show that there are three possibilities by a geometric method. Let us denote each equation by a plane respectively.   ① It has a unique solution. Three planes meets in a unique point. [Ex]

 ②It has infinitely many solutions. [Ex] (1)     (2)     (3)                 ③ It has no solution. (It is called 'inconsistent').   [Ex] (1)            (2)       (3)             (4)                                                                                                                                       ■

 Solve the following linear system. Since there are five unknowns and three equations, assign to the any two unknowns arbitrary real numbers. Rearranging each equation, we get     Substitute , ( are arbitrary real numbers) to get     Therefore, the solution of a given linear system is ( are arbitrary real numbers).   The solution set is    . Thus this system has infinitely many solutions.                  ■

 Consider matrices                  .    is a matrix, and . is a matrix, and , and are matrix respectively. The main diagonal entries of are , and is also written as  .            ■

 Find the augmented matrix of the following linear system of equations.    Let be the coefficient matrix, the unknown, and the constant, then     Hence we have          Its augmented matrix is                                                                     □                                                                         A=matrix(3, 3, [1,1,2,2,4,-3,3,6,-5])            # 3x3 matrix b=vector([9,1,0])                                # constant vector print A.augment(b,subdivide=True)               # augmented matrix                                                                     [ 1  1  2 |  9 ] [ 2  4 –3 |  1 ] [ 3  6 –5 |  0 ]                                                  ■

 Gaussian elimination (also known as row reduction) is an algorithm for solving systems of linear equations. It is usually understood as a sequence of operations performed on the associated matrix of coefficients. Using row operations to convert a matrix into reduced row echelon form is sometimes called Gauss–Jordan elimination. Linear system of equations can be easily solved by using Gauss–Jordan elimination.

 In the following procedure, the left side shows solving a linear system directly, and the right side shows solving it using an augmented matrix.                         Add the multiplication of the first equation by to the second equation.                  Add the multiplication of the first equation by -3 to the third equation.                      Multiply the second equation by to get                        Add the multiplication of the second equation by to the third equation.                  Multiply the third equation by .                      Thus the system reduces to  Now substituting in the second equation, we get . Substituting and in the first equation, we get . Hence the solution is                          ■

 The following are all in REF.

 Consider matrices Since matrices do not satisfy the above properties (1), (2), (3) respectively, they are not in REF.

 The following are all in RREF.

 The following are equivalent.           ■

 Find the RREF of .                                                                                      A=matrix(3, 5, [1,1,1,4,4,2,3,4,9,16,-2,0,3,-7,11])      # 3x5 matrix input print A.rref()                                     # A's RREF                                                                     [ 1  0  0  2 –1] [ 0  1  0  3  2] [ 0  0  1 –1  3]                                                 ■

 Solve the following by the Gauss elimination.                                                              Its augmented matrix is and its REF by EROs is  . Therefore, since the corresponding linear system of the above augmented matrix is      ,i.e,    The solution is .                              ■

Gauss-Jordan elimination: This is a method to transform the augmented matrix of a linear system into RREF.

 Solve the following system using the Gauss-Jordan elimination.   We will use Sage to solve this.                          □                                                                         A=matrix([[1,3,-2,0,2,0],[2,6,-5,-2,4,-3],[0,0,5,10,0,15], [2,6,0,8,4,18]]) b=vector([0,-1,5,6])                       print A.augment(b).rref()                                                          [  1   3   0   4   2   0   0] [  0   0   1   2   0   0   0] [  0   0   0   0   0   1 1/3] [  0   0   0   0   0   0   0]   Its corresponding linear system is     By letting ( are any real), its solution is   . ■

From  , we can express a general solution as a vector form with any .

.

 Using the Gauss-Jordan elimination, express the solution of the following homogeneous equation as a vector form.   Its augmented matrix is , and its RREF is . Thus, leading entry 1's correspond to leading variables , , and the rest variables , , to free variables. We have the following.   , ,   Now let free variables be , , , then   , , , , , .       Therefore                         ■

Answer the questions for the following linear system.

(1) Find the coefficient matrix.

(2)Express the linear system in the form .

(3) Find it’s augmented matrix.

Solution.

(1) The coefficient matrix of the following linear system is .

(2) Let , , .

So the following linear system can be expressed by .

(3) It’s augmented matrix is

http://math3.skku.ac.kr/home/pub/15

Find a linear system with its augmented matrix.

(Put the unknowns as .)

Solution.

Let be the coefficient matrix, the unknown, and b the constant.

then = ,   =

Hence we have

<=> =

Therefore, linear system is

Note:

This problem does not need to use Sage.  ■

Find the number of leading variables and

free variables in the solution set of the following system.

SOL> Let , = .

=  RREF

Leading variables are , , , and free variables are , .

Double checked by Sage > http://math3.skku.ac.kr/home/pub/69 by 주영은

okay. ■

Which matrices are in REF or RREF? If one is not in RREF, transform it to a form in RREF.

, .

Sol) Let the matrices , .

At the matix row’s first number(except 0) is not zero. So it is not REF and also RREF too.

At the matix B row’s first number(except 0)is in the same column with row. So it is not REF and RREF.

For the matrix ,

,

For the matrix B,

.

Answer : There are nor REF and RREF.

For each matrix RREF is ,.

Solve the system using Gaussian elimination.

Sol)

The augmented matrix of the system can be expressed as below.

Find

However, at , the last equation in the system is and it is impossible. Therefore, there’s no answer for the system           ■

Solve the system using Gauss-Jordan elimination.

Sol) The forth row is the double of the second row, so we now ignore it in this system of linear equations.

,

So .

We can also solve this problem by using Sage   http://math1.skku.ac.kr/home/pub/2430/    ■

In the following circuit, write a linear system to find current .

[Kirchhoff's Law : ⓵ where is the current 1)

⓶ where is the voltage 2)]

, (From ⓵)

, (From ⓶)

, , , ,

⇒ where , ,

coefficient matrix

of )

Current : , , , .    ■

Double checked by Sage http://math1.skku.ac.kr/home/math2013/343/

Aug =matrix([[1,-1,0,0,0],[0,1,-1,-1,0],[0,1,3,0,9],[0,1,0,6,9]])

print Aug.rref()

[1 0 0 0 3]

[0 1 0 0 3]

[0 0 1 0 2]

[0 0 0 1 1]

In general, we are given a linear system with equations and unknowns.

If there are free variables, what is the number of leading variables?

From this, think about the relation among the numbers of free variables, leading variables, and unknowns.

Sol.

In a homogeneous linear system with n unknowns, if the RREF of the augmented matrix has k leading 1’s, the solution set has n-k free variables.

From this, we can conclude that the sum of the number of free variables and that of leading variables is a whole number of unknowns.

This problem does not need to use Sage.  ■

2015.10.11. Final OK by SGLee

Using and , we can make equations.

⋯①

⋯②

Because dimension of is 2, there are 2 linearly independent equations.

⋯①+②

∴

■

<Chapter 2 p3 NEW>

(문제변형) Suppose that three points pass through the parabola .

By plugging in these points, obtain three linear equations. Find coefficients by solving

Sol)

(, , pass through the parabola)

, where ,

, ,

Double checked by Sage : http://math1.skku.ac.kr/home/pub/2487

(by 2009313441 김동욱)    ■

Write a linear system with four unknowns and four equations satisfying each condition below.

(a) A solution set with one unknown.

(b) A solution set with two unknowns.

Sol.

(a) is an example.

Let A be the matrix . By using a Sage, RREF(A) is .

Free variable is only . Therefore, this example is a solution set with one unknown.

(b)

is an example.

Let B be the matrix . By using a Sage, RREF(B) is .

Free variables are . Therefore, this example is a solution set with two unknowns.

Double checked by Sage> http://math3.skku.ac.kr/home/pub/70

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 http://matrix.skku.ac.kr/LA-Lab/Solution/