Lesson 10.2 – Row Operations and Solving Systems
1. Solve the following system of equations.
 x  3y  5z  8

2x  5y  4z  8

3x  5y  4z  5
2. Write an augmented matrix from the system in problem 1 and follow the row operations indicated.









R2  2r1  r2 

R3  3r1  r3







R3  4r2  r3 





The resulting matrix is what we call Row Echelon Form and has the general form

1 a b c 
0 1 d

0 0 1

1 
R3  r3 
13






e 
f 
3. Write a system of equations form the resulting augmented matrix from problem 2 and solve.
4. Write the augmented matrix from the following system and convert to row echelon form by the steps indicated.

 x  3y  4z  4


2x  5y  6z  7 matrix 


3x  3y  4z  18











1 
R3  r3 
4


5. Solve the resulting system from problem 4.

system







R2  2r1  r2 

R3  3r1  r3








R3  6r2  r3 






6. Write the augmented matrix from the following system. Then convert to row echelon form by the steps indicated.

2 x  3 y  z  2


matrix 
 x yz 8
3x  2 y  9 z  9




R2  R3 


R1  r3  r1 
R2  12r3  r2 



R3  5r2  r3 






R2  2r1  r2 
R3  3r1  r3 




1 
 R3  57 r3 






 R1  r2  r1 



R1  R2

















The resulting matrix is what we call Row Reduced Echelon Form and has the general form.











1 0 0 a 
0 1 0 b 


0 0 1 c 
This form is nice because you can clearly see the solution of the initial system: x = _________ , y = _________, and z = _________.
7. Write the augmented matrix from the following system and convert to row reduced echelon form by the steps indicated.

 x  3y  z  6


2x  5y  z  2 matrix 


 x  y  2z  7

1 
R3   r3 
3






R2  2r1  r2 
R3  r1  r3 



 R1  r

3  r1
 R  3r  r 
3
2
 2







R3  2r2  r3 






 R1  3r2  r1 


Tips For Writing in Row Reduced Echelon Form








x
y
z

1) Make the number in thefirst row first column (top left) a 1 by switching rows or multiplying it by a nonzero multiple

2) Use this 1 (and row operations) to make everything beneath it a zero
3) Repeat steps 1 and 2 with second row second column number, etc to get in row echelon form
4) Use the 1 in the last row (and row operations) to make everything above it zero
5) Repeat step 4 as you work back up.
Step 1
Step 2
Step 5
a
1
 1

b



1 c 
Step 3
Step 4
Note: This example is with three
equations with three variables. We
will not always work with these. We
may have more or less.