Download Systems of Linear Equations

MATH 250
Linear Equations and Matrices
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Topics
•
•
•
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Preliminaries
Systems of Linear Equations
Matrices
Algebraic Properties of Matrix Operations
Special Types of Matrices and Partitioned
Matrices
• Matrix Transformations
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Systems of Linear Equations
System of m equations in n unknowns
a11 x1  a12 x2 
a21 x1  a22 x2 
 a1n xn  b1
 a2n xn  b2
am1 x1  am2 x2 
 amn xn  bm
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Systems of Linear Equations
Comments
• If a system has a solution, call it consistent
• If a system doesn’t have a solution, call it
inconsistent
• If b1  b2   bm  0 , the system is called
homogeneous. A homogeneous system always
has the trivial solution x1  x2   xn  0
• If two systems have the same solution, then they
are called equivalent. The solution strategy for
linear systems is to transform the system through
a series of equivalent systems until the solution is
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obvious
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Matrices
Systems of Equations
• Consider a11 x1  a12 x2   a1n xn  b1
• Define
a21 x1  a22 x2 
 a2n xn  b2
am1 x1  am2 x2 
 amn xn  bm
 a11

a21

A

 am1
a12
a22
am 2
a1n 
a2n 


amn 
• Express system as AX  B
 x1 
 
x2 

X 
 
 xn 
 b1 
 
b2 

B 
 
bm 
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Matrices
Systems of Equations
• Since the solution of the system involves the
a and b values only, will often work with the
augmented matrix
a
 11
 a21


 am1
a12
a22
am2
a1n b1 

a2n b2 


amn bm 
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Systems of Linear Equations
Elementary Operations on Systems
1) Switch two equations
2) Multiply an equation by nonzero constant
3) Add multiple of one equation to another
The application of any combination of
elementary
operations to a linear system yields a new
linear system
that is equivalent to the first
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Inverting a Matrix
• Usually not a good idea to compute x=A-1b
– Inefficient
– Prone to roundoff error
• In fact, compute inverse using linear solver
– Solve Axi=bi where bi are columns of identity,
xi are columns of inverse
– Many solvers can solve several R.H.S. at
once
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Solving Linear Systems Using
Gaussian Elimination
• Write the augmented matrix for the system
• Use matrix row operations to simplify the
matrix to one with 1s down the diagonal
from upper left to lower right, and 0s below
the 1s
• Write the system of linear equations
corresponding to the matrix in step 2, and
use back-substitution to find the system’s
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solutions
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Theorem 1: Equivalent systems and equivalent
matrices:
If the augmented coefficient matrices of two
linear systems are row equivalent, then the
two systems have the same solution set.
Definition: Echelon Matrix
The matrix E is called an echelon matrix provided it has the
following two properties:
1. Every row of E that consists entirely of zeros (if any)
lies beneath every row that contains a nonzero element.
2. In each row of E that contains a nonzero element, the
nonzero element lies strictly to the right of the first
nonzero element in the preceding row (if there is a
preceding row).
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Example
Use matrices to solve the system
3x
+
y
+
2z
=
31
x
+
y
+
2z
=
19
x
+
3y
+
2z
=
25
Solution
Step 1 Write the augmented matrix for the system.
Linear System
3x + y + 2z = 31
x + y + 2z = 19
x + 3y + 2z = 25
Augmented Matrix
3
1
2
31
1
1
2
19
1
3
2
25
Example cont.
Solution
Step 2 Use matrix row operations to simplify the matrix to one with 1s
down the diagonal from upper left to lower right, and 0s below the 1s.
Our goal is to obtain a matrix of the form
1
0
0
a
1
0
b
d
1
c
e
f
.
Our first step in achieving this goal is to get 1 in the top position of the first
column.
We want 1 in
this position.
3
1
1
1
1
3
2
2
2
31
19
25
To get 1 in this position, we interchange rows 1 and 2. (We could also
interchange rows 1 and 3 to attain our goal.)
1
3
1
1
1
3
2
2
2
19
31
25
This was row 2; now it’s row 1.
This was row 1; now it’s row 2.
Example cont.
Solution
Now we want to get 0s below the 1 in the first column.
We want 0 in
these positions.
1
3
1
1
1
3
2
2
2
19
31
25
Let’s first get a 0 where there is now a 3. If we multiply the top row of
numbers by –3 and add these products to the second row of numbers, we
will get 0 in this position. The top row of numbers multiplied by –3 gives
-3(1) or –3, -3(1) or –3,
-3(2) or –6,
-3(19) or –57.
Now add these products to the corresponding numbers in row 2. Notice that
although we use row 1 to find the products, row 1 does not change.
1
1
2
19
3 + (-3) 1 + (-3) 2 + (-3) 31 + (-3)
1
3
2
25
=
1
0
1
We want 0 in
this position.
1
-2
3
2 19
-4 -26
2 25
Example cont.
Solution
We are not yet done with the first column. If we multiply the top row of
numbers by –1 and add these products to the third row of numbers, we will
get 0 in this position. The top row of numbers multiplied by –1 gives
-1(1) or –1, -1(1) or –1,
-1(2) or –2,
-1(19) or –19.
Now add these products to the corresponding numbers in row 3.
1
1
2
19
0
-2
-4
-26
1 + (-1) 3 + (-1) 2 + (-2) 25+(-19)
=
1
0
0
1
-2
2
2 19
-4 -26
0 6
We move on to the second column. We want 1 in the second row, second
column.
We want 1 in
this position.
1
0
0
1
-2
2
2 19
-4 -26
0 6
Example cont.
Solution
To get 1 in the desired position, we multiply –2 by its reciprocal, -1/2.
Therefore, we multiply all the numbers in the second row by –1/2 to get
1
› (0)
0
1
› (-2)
2
2
› (-4)
0
19
› (-26)
6
=
1
0
0
1
1
2
2
2
0
19
13
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We want 0 in
this position.
We are not yet done with the second column. If we multiply the top row of
numbers by –2 and add these products to the third row of numbers, we will
get 0 in this position. The second row of numbers multiplied by –2 gives
-2(0) or 0, -2(1) or –2,
-2(2) or –4,
-2(13) or –26.
Now add these products to the corresponding numbers in row 3.
1
0
0+0
1
2
19
1
2
13
2 + (-2) 0 + (-4) 6+(-26)
=
1
0
0
1
1
0
2 19
2 13
-4 -20
Example cont.
Solution
We move on to the third column. We want 1 in the third row, third column.
1
0
0
We want 1 in
this position.
1
1
0
2 19
2 13
-4 -20
To get 1 in the desired position, we multiply –4 by its reciprocal, -1/4.
Therefore, we multiply all the numbers in the third row by –1/4 to get
1
1
2
19
0
1
2
13
-1/4(0) -1/4(0) -1/4(-4) -1/4(-20)
=
1
0
0
1
1
0
2
2
1
19
13
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We now have the desired matrix with 1s down the diagonal and 0s below the 1s.
Step 3 Write the system of linear equations corresponding to the
matrix in step 2, and use back-substitution to find the system’s solution.
The system represented by the matrix in step 2 is
Example cont.
Solution
3
1
1
1
1
3
2
2
2
31
19
25
x + y + 2z = 19
y + 2z = 13
z = 5
We immediately see that the value for z is 5. To find y, we back-substitute 5
for z in the second equation.
y + 2z = 13
y + 2(5) = 13
y=3
Equation 2
Substitute 5 for x.
Solve for y.
Finally, back-substitute 3 for y and 5 for z in the first equation:
x + y + 2z = 19
Equation 1
x + 3 + 2(5) = 19
Substitute 3 for y and 5 for x.
x + 13 = 19
Multiply and add.
x=6
Subtract 13 from both sides.
The solution set for the original system is {(6, 3, 5)}.
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Gaussian Elimination
Definition
• A matrix is in echelon form if
• Any rows consisting entirely of zeros are
grouped at the bottom of the matrix.
• The first nonzero element of each row is 1.
This element is called a leading 1.
• The leading 1 of each row after the first is
positioned to the right of the leading 1 of the
previous row.
• (This implies that all the elements below a
leading 1 are zero.)
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Example 1
Solving the following system of linear equations using the
method of Gaussian elimination.
x1  2 x2  3 x3  2 x4  1
 x1  2 x2  2 x3  x4  2
2 x1  4 x2  8 x3  12 x4  4
Solution
Starting with the augmented matrix, create zeros below the pivot
in the first column.
2
3 2  1

 1
 1 2 3 2  1
 1  2  2 1 2 R 2  R1 0 0 1 3
1
 2
4
8 12 4 R3  (2) R1 0 0 2 8 6

At this stage, we create a zero only below the pivot.
 1 2 3 2  1   1 2 3 2  1

0 0 1 3
1 1 0 0 1 3
1
R3 
R3  (2) R 2 


2
0 0 0 2 4 
0 0 0 1 2 
Echelon form
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We have arrived at the echelon form.
The corresponding system of equation is
x1  2 x2  3x3  2 x4  1
x3  3x4  1
x4  2
We get
x3  3(2)  1
x3  5
Substituting x4 = 2 and x3 = 5 into the first equation,
x1  2 x2  3(5)  2(2)  1
x1  2 x2  10
x1  2 x2  10
Let x2 = r. The system has many solutions. The solutions are
x1  2r  10, x2  r , x3  5, x4  2
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Example 2
Solving the following system of linear equations using the
method of Gaussian elimination, performing back substitution
using matrices.
x1  2 x2  3 x3  2 x4  1
 x1  2 x2  2 x3  x4  2
2 x1  4 x2  8 x3  12 x4  4
Solution
We arrive at the echelon form as in the previous example.
2
3 2  1
 1
 1 2 3 2  1
  1  2  2 1 2
   0 0 1 3
1
 2
0 0 0 1 2 
4
8 12 4



Echelon form
This marks the end of the forward elimination of variables from
equations. We now commence the back substitution using
matrices.
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 1 2 3 2  1
0 0 1 3
1
0 0 0 1 2 



R1  (2) R 2
R 2  (3) R3
 1 2 3 0  5
0 0 1 0  5
0 0 0 1
2


 1 2 0 0 10
R1  (3) R 3 0 0 1 0  5
0 0 0 1
2

This matrix is the reduced echelon form of the original
augmented matrix. The corresponding system of equations is
x1  2 x2  10
x3  5
x4  2
Let x2 = r. We get same solution as previously,
x1  2r  10, x2  r , x3  5, x4  2
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