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System of Linear
Equations
Linear Algebraic Equation
Linear algebra is one of the
corner stones of modern
computational mathematics.
Almost all numerical schemes
such as the finite element
method and finite difference
method are in act techniques
that transform, assemble,
reduce, rearrange, and/or
approximate the differential,
integral, or other types of
equations to systems of linear
algebraic equations.
A system of linear algebraic equations can
be expressed as
11 12 1 1 1
21 22 2 2 2
1 2
where a and b are constants,
i= 1,2,..., , 1,2,..., .
n
n
m m nm m m
ij i
a a a x b
a a a x b
a a a x b
m j n
     
     
     
     
     
     

L
L
M M O M M M
L
Solution of Linear Algebraic
Equation
Solving a system with a
coefficient matrix is
equivalent to finding the
intersection point(s) of
all m surfaces (lines) in
an n dimensional space.
If all m surfaces happen
to pass through a single
point then the solution
is unique
• If the intersected part is a line or
a surface, there are an infinite
number of solutions, usually
expressed by a particular
solution added to a linear
combination of typically n-m
vectors. Otherwise, the solution
does not exist.
• In this part, we deal with the case
of determining the values x1,
x2,…,xn that simultaneously
satisfy a set of equations.
mxnA
Cramer’s Rule
Cramer’s Rule
Cramer’s rule is
another technique that
is best suited to small
numbers of equations.
This rule states that
each unknown in a
system of linear
algebraic equations
may be expressed as a
fraction of two
determinants with
denominator D and
with the numerator
obtained from D by
replacing the column of
coefficients of the
unknown in question
by the constants b1, b2,
… ,bn.
1 12 13
2 22 23
3 32 33
1
b a a
b a a
b a a
x
D

For example, x1 would be
computed as
Example
Solution:
We begin by setting up
and evaluating the three
determinants:
Use Cramer’s Rule to solve the system:
5x – 4y = 2
6x – 5y = 1
     1 1
2 2
5 4
5 5 6 4 25 4 1
6 5
a b
D
a b
   
             
  
     1 1
2 2
2 4
2 5 1 4 10 4 6
1 5
c b
Dx
c b
   
             
  
     1 1
2 2
5 4
5 1 6 2 5 12 7
6 1
a c
Dy
a c
   
          
  
From Cramer’s Rule, we have:
6
6
1
Dx
x
D

  

and
7
7
1
Dy
x
D

  

The solution is (6,7)
Cramer’s Rule does not apply if D=0. When D=0 , the system is either
inconsistent or dependent. Another method must be used to solve it.
Example

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system of linear equations

  • 2. Linear Algebraic Equation Linear algebra is one of the corner stones of modern computational mathematics. Almost all numerical schemes such as the finite element method and finite difference method are in act techniques that transform, assemble, reduce, rearrange, and/or approximate the differential, integral, or other types of equations to systems of linear algebraic equations. A system of linear algebraic equations can be expressed as 11 12 1 1 1 21 22 2 2 2 1 2 where a and b are constants, i= 1,2,..., , 1,2,..., . n n m m nm m m ij i a a a x b a a a x b a a a x b m j n                                      L L M M O M M M L
  • 3. Solution of Linear Algebraic Equation Solving a system with a coefficient matrix is equivalent to finding the intersection point(s) of all m surfaces (lines) in an n dimensional space. If all m surfaces happen to pass through a single point then the solution is unique • If the intersected part is a line or a surface, there are an infinite number of solutions, usually expressed by a particular solution added to a linear combination of typically n-m vectors. Otherwise, the solution does not exist. • In this part, we deal with the case of determining the values x1, x2,…,xn that simultaneously satisfy a set of equations. mxnA
  • 5. Cramer’s Rule Cramer’s rule is another technique that is best suited to small numbers of equations. This rule states that each unknown in a system of linear algebraic equations may be expressed as a fraction of two determinants with denominator D and with the numerator obtained from D by replacing the column of coefficients of the unknown in question by the constants b1, b2, … ,bn. 1 12 13 2 22 23 3 32 33 1 b a a b a a b a a x D  For example, x1 would be computed as
  • 6. Example Solution: We begin by setting up and evaluating the three determinants: Use Cramer’s Rule to solve the system: 5x – 4y = 2 6x – 5y = 1      1 1 2 2 5 4 5 5 6 4 25 4 1 6 5 a b D a b                           1 1 2 2 2 4 2 5 1 4 10 4 6 1 5 c b Dx c b                           1 1 2 2 5 4 5 1 6 2 5 12 7 6 1 a c Dy a c                  
  • 7. From Cramer’s Rule, we have: 6 6 1 Dx x D      and 7 7 1 Dy x D      The solution is (6,7) Cramer’s Rule does not apply if D=0. When D=0 , the system is either inconsistent or dependent. Another method must be used to solve it. Example