Download Row-Echelon Form and Reduced Row

CHAPTER 8.1
Matrices and Systems of
Equations
Matrix- a streamlined technique for solving systems of linear
equations that involves the use of a rectangular array of numbers
 a11 a12 a13 


a
a
a
21
22
23 

a a a 
 31 32 33 
N columns
M rows
MxN
Order of Matrices
 2

1 3 0
2 4
3 6


6 
0 
 
 5 
1 1
1
2 
1 4
2 2
31
system
augmented
matrix
coefficient
matrix
 2 x  4 y  6 z  10

  x  6 y  2 z  6
4 x
 2z  2

2
 1

 0
3
2
3
3 4
4 3 
1 5 
 4 2 3 
2 5 1


 0 3 2 
Writing an Augmented Matrix
1. Begin by writing the linear system and aligning the variables.
2. Next, use the coefficients and constant terms as the matrix
entries. Include zeroes for each missing coefficients.
 x  4y  8

 y  2 z  3
 x  5z  0

1
0

1
4
0 8
1 2 3
0 5 0 
Elementary Row Operations
1. Interchange two rows.
2. Multiply a row by a nonzero constant.
3. Add a multiple of a row to another row.
Associated Augmented Matrix
 1 2 3 9 
 1 3 0  4 

 R1  R2
 2 5 5 17 
1
0

 2
2
1
5
3 9
3 5 
5 17  2R1  R3
1 2 3 9 
0 1

3
5


0 1 1  1 R2  R3
1
0

0
2
1
0
1
0

0
2 3
1 3
0 1
3 9
3 5  1
R3
2 4 
2
9
5
2 
Use back-substitution to find the solution
x 1
y  1
z2
Row-Echelon Form and Reduced Row-Echelon Form
1. All rows consisting of zeroes occur at the bottom of the
matrix.
2. For each row that does not consist of zeroes, the first
nonzero entry is 1(called a leading 1).
3. For two successive (nonzero) rows, the leading 1 in the
higher row is farther to the left than the leading 1 in the lower
row.
Reduced row-echelon form- if every column that has a
leading 1 has zeroes in every position above and below its
leading 1.
For example:
Row-Echelon Form
1 2 1 4 
0 1 0 3 


0 0 1 2 
1 5
0 0

0 0

0 0
2 1 3 
1 3 2 
0 1 4

0 0 1
Reduced row-echelon form
1
0

0

0
0
1
0
0
0
0

0
1
0
0
0
0
1
0
0
1
0
1
2 
3

0
5
3 
0 
Solve the system
y  z  2 w  3


x  2y  z  2


2 x  4 y  z  3w  2

 x  4 y  7 z  w  19
Switch row 1 with row 2
1 2 1 0 2 
0 1

1

2

3


2 4
1 3 2  2R1  R3

 R  R
1

4

7

1

19
4

 1
1 2 1 0 2 
0 1 1 2 3 


0 0 3 3 6 


0

6

6

1

21

 6R2  R4
1
0

0

0
2 1 0
1 1 2
0 3 3
0 0 13
2 
3 
1
6  R3
 3
39 
1
0

0

0
2
1
0
0
1
1
1
0
1
0

0

0
2
1
0
0
1 0 2 
1 2 3
1 1 2 

0
1 3
0
2 
2 3 
1 2 
  1 R
13 39 
4
13
 x  2y  z  2
 y  z  2 w  3


z  w  2


w3
Use back-substitution
x  1
y2
z 1
w3
Gaussian Elimination with Back-Substitution
1. Write the augmented matrix of the system of linear
equations.
2. Use elementary row operations to rewrite the
augmented matrix in row-echelon form.
3. Write the system of linear equations corresponding to
the matrix in row-echelon form and use backsubstitution to find the solution.
Solve the system
1 1 2
1 0 1

 2 3 5

 3 2 1

1
0

2

3

1
0

0

3
1 2
1 1
3 5
2 1
1 2
1 1
1 1
2 1
4
6   R1  R2
4

1

4
2

4 2R1  R3

1

4
2

4

1  3R1  R4

1 1 2
0 1 1

0 1 1

0 5 7

1
0

0

0

4 
2 

4  R2  R3

11
1 2
1 1
0
0
5 7
In row 3,

4 
2 

2 

11
0  2
inconsistent, no solution
1 2 3
0 1 3

0 0 1
9  2R2  R1
5 
2
Apply additional elementary row operations until you obtain a matrix in
reduced row-echelon form.

1 0 9
0 1 3

0 0 1

19 9 R3  R1
5  3R3  R2

2 

1 0 0
0 1 0

0 0 1

1
1

2 
x 1
y  1
z2
Solve the system
 x  5z  2

 y  3 z  1
 2 4 2
3 5 0

1
R1
0
2
1 
1 2 1
3 5 0

0
1  3R1  R2
1 2 1
0 1 3

0
1   R2
1 2 1
0 1 3

0
1 2R  R
1 0 5
0 1 3

2
1
x  5 z  2
y  1  3 z
Let z  a where a is
2
1
A real number, then the
solution set has the
form
(5a  2, 3a  1, a)
=infinite number of solutions