Dr. Neal, WKU
MATH 307
€
Row Echelon Form
Consider a system of n equations, n unknowns: AX = B . From det( A ), we can
determine how many solutions the system has. But how do we find them? One
method, called Gauss-Jordan elimination, is to perform row operations on the augmented
matrix A B to convert it to a system C D , where C is upper-triangular. If in fact
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det(A) ≠ 0 , then row operations can produce the matrix In S , where In is the n × n
identity matrix and S is the unique solution.
All of this work is greatly simplified €
with TI’s built-in rref( command in the
€MATRIX MATH menu. We will
€ use this command throughout the course. Essentially,
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we will give a
rref ( A B ) = In S whenever det(A) ≠ 0 . Before we use this command,
€
general idea of row operations.
2x − y + 2z = 3
Consider the system AX = B :
x + y − 2z = 6
€
−3x + 2y − z = 3
 2 −1 2 


A =  1 1 −2


−3 2 −1
€  x
 
X =  y
 
 z
 3
 
€ B = 6
 
 3
 2 −1 2 3 


A B =  1 1 −2 6 
 −3 2 −1 3 


Then det( A ) = 9 ≠ 0, so there is exactly one solution to the system.
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x + c12 y + c13 z = d1
We first€want to convert the system to the form:
y + c 23 z = d2
z = d3
Allowable Rules: (i) We can switch rows. (ii) We can multiply or divide any row
(i.e., equation) by a non-zero scalar. (iii) We can add a non-zero multiple of any row
€
(equation) to any other row (equation).
The variable symbols x , y , and z are not important, so we can do these row
operations on the rows of the augmented matrix A B .
1. We want the leading
of the top equation to be 1. So we have two choices
€ € coefficient
in this case. (i) We can
divide€the original top equation by 2, or (ii) We can switch the
first and second rows (i.e., equations). €
 2 −1 2 3 
→


A B =  1 1 −2 6 
 −3 2 −1 3  Switch R1 and R2


 1 1 −2 6 


 2 −1 2 3 
−3 2 −1 3 


2. We now use the leading 1 in the top row to eliminate the values below it using Rule
€
(iii). Afterwards, we can conveniently
divide Row 2 by –3.
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 1 1 −2 6 
1 1 −2 6 
1 1 −2 6 
→
→






 2 −1 2 3  −2R1 + R2 0 −3 6 −9  R2 ÷ (−3) 0 1 −2 3 
−3 2 −1 3  3R1 + R3 0 5 −7 21 
0 5 −7 21 






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Dr. Neal, WKU
3. Now use the leading 1 in the second row to eliminate the value below it using Rule
(iii). Conveniently, we can then divide Row 3 by 3.
1 1 −2 6 
1 1 −2 6 
→
→




0 1 −2 3  −5R2 + R3 0 1 −2 3  R3 ÷ 3
0 5 −7 21 
0 0 3 6 




€
1 1 −2 6 


0 1 −2 3 
0 0 1 2 


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€
At this point, the last augmented matrix can be read as
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x + y − 2z = 6
y − 2z = 3
z=2
So, z = 2 . From the second equation, y = 3 + 2z = 3 + 2(2) = 7 . Then from the first
equation, x = 6 − y + 2z = 6 − 7 + 2(2) = 3 . So the unique solution is (x, y,z) = (3, 7, 2).
Solving the system this way is called Gaussian elimination€with back substitution.
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But we also can continue the €row operations to create I3 on the left side of the
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augmented matrix.
4. Now use the 1 in the last row to eliminate the values above it. Then use the 1 in the
€
second row to eliminate the value above it.
1 1 −2 6 
→


0 1 −2 3  2R3 + R1
0 0 1 2  2R3 + R2


€
 1 1 0 10 


0 1 0 7 
0 0 1 2 


→
−1R2 + R1
1 0 0 3 


0 1 0 7 
0 0 1 2 


The system is now in the form In S where S gives the unique solution (3, 7, 2).
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Completing these
last row operations is called
Gauss-Jordan elimination and the final
€
€ matrix is the reduced
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row echelon form of the augmented matrix
A B.
In this case, rref ( A B ) = In S because
€ det( A ) ≠ 0. We will see more examples later
in the cases when det( A ) = 0, and when we have m equations and n unknowns with
m ≠ n . In these cases, we may not obtain In on the left side of the last reduced matrix.
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how to use the rref( command
on your €
calculator. This command
∗ You must learn
€ row echelon form of A B . ∗
will directly produce the reduced
€
2x − y + 2z = 3
Given x + y − 2z = 6
−3x + 2y − z = 3
€
 2 −1 2 3 


€ €
Enter the augmented 3 × 4 matrix  1 1 −2 6  as [C] (or


−3 2 −1 3 
€
use some other matrix name.) Then enter rref[[C])
to obtain
€
€
1 0 0 3 


0 1 0 7  .


0 0 1 2 