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) 
■
.
is a 
matrix, and 
. 
is a 
matrix, and 
, and 
are 
matrix respectively. The main diagonal entries of 
are 
, and 
is also written as 
. 
Let 
be the coefficient matrix, 
the unknown, and 
the constant, then
to the second equation.
to get
to the third equation.
.
in the second equation, we get 
. Substituting
and 
in the first equation,
. Hence the solution is 
do not satisfy the above properties (1), (2), (3) respectively, they are not 
Its
and its REF by EROs is 
.
. 
. 
From 
, we can express a general solution as a 
.
. 
, and its RREF is 
.
, 
, 
and
, 
, 
to free variables. We have the following.
, 
, 
,
Answer the questions for the following linear system.
.
.
, 
, 
.
.
■
Find a linear system with its augmented matrix.
.)
be the coefficient matrix,
= 
, 
= 
<=> 
=
Find the number of leading variables and
, 
= 
.
= RREF
Leading variables are 
, 
, 
, 
, 
.
Which matrices are 
, 
.
, 
.
row’s first number(except 0) is not zero. So it is not REF and also RREF too.
row’s first number(except 0)is in the same column with 
row. So it is not REF and RREF.
,
.
,
. 
Solve the system using 
, 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.
, 
.
In the following circuit, write a linear system to find current 
.
where 
is the current 
where 
is the voltage 
, 
(From ⓵)
, 
(From ⓶)
, 
, 
, 
,
where 
, 
, 
coefficient matrix 
of 
)
Current 
: 
, 
, 
, 
. ■
In general, we are given a linear system with 
equations and 
unknowns. 
free variables, what is the number of leading variables?
and 
, we can make equations.
⋯①
⋯②
is 2, there are 2 linearly independent equations.
⋯①+②
■
(문제변형) Suppose that three points 
pass through the parabola 
.
by solving 
. 
, 
, 
pass through the parabola)
, where 
, 
, 
, 
■
Write a linear system with four unknowns and four equations satisfying each condition below.
is an example.
. By using a Sage, RREF(A) is 
.
. Therefore, this example is a solution set with one unknown.
is an example.
. By using a Sage, RREF(B) is 
.
. Therefore, this example is a solution set with two unknowns.
Double checked by Sage> 
