2.1 Linear system of equations
2.2 Gaussian elimination and Gauss-Jordan elimination
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.
②It has infinitely many solutions.
[Ex] (1) (2) (3)
③ It has no solution. (It is called 'inconsistent').
[Ex] (1) (2)
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.
(Find augmented matrix from a given LSE.)
Let be the coefficient matrix, the unknown, and the constant, then
Hence we have
Its augmented matrix is
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.
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 RREF of a given matrix.) (RREF를 구하시오.)
Find the RREF of .
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
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. □
[ 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
, , , , , .
Sec 2.2 Example 8. Solve LSE by using RREF (Gauss 소거법).
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.
(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 ■
Find a linear system with its augmented matrix.
(Put the unknowns as .)
Let be the coefficient matrix, the unknown, and b the constant.
then = , =
Hence we have
∴ Therefore, linear system is
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 , = .
Leading variables are , , , and free variables are , .
Double checked by Sage > http://math3.skku.ac.kr/home/pub/69 by 주영은
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.
The augmented matrix of the system can be expressed as below.
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.
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 , ,
Current : , , , . ■
Double checked by Sage http://math1.skku.ac.kr/home/math2013/343/
[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.
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.
Because dimension of is 2, there are 2 linearly independent equations.
<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 .
(, , 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.
(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.
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