# 6.4.2 Solutions of Linear Systems Using ```6.4.2 Solutions of Linear Systems Using
Matrices - Part II
In 6.4.1, we wrote the system:
2x + 4y + 4z = 4
x + 3y + z = 4
-x + 3y + 2z = -1
as the augmented matrix:
2 4 4 4
1 3 1 4


 1 3 2  1
and then demonstrated (forward) Gaussian elimination,
transforming the augmented matrix into row-echelon form:
1 2 2 2 
0 1  1 2 


0 0 1  1
We then rewrote the system as:
x + 2y + 2z = 2
y - z = 2
z = -1
and used back-substitution to finish the solution.
Instead of back-substitution, we can complete the process
on the system in matrix form using backward elimination.
6.4.2-1
Backward elimination
 start from the bottom, and work on one row at a time to
the top (that’s why it’s called backward elimination)
 turning the entries above the leading 1's to 0's (row op 3)
The resulting matrix will be:
1 0 0 2 
0 1 0 1 


0 0 1 1
This matrix is in reduced row-echelon form (RREF):
 every non-zero row has a leading 1
 below and above the leading 1’s we find only 0’s
 the leading 1's &quot;march&quot; from left to right
 rows of all zeros are at the bottom (not shown here)
Rewriting the matrix as a system, the solution is apparent:
x
y
z
= 2
= 1
= -1
I will illustrate backward elimination with our example
system at the board.
6.4.2-2
What else can “go wrong”?
A matrix in RREF form may have a row of 0’s:
1 0 5 5 
0 1 2 3


0 0 0 0
 this indicates a system with  many solutions
 a dependent system
Q: How do we state &quot;the solution&quot; when there are infinitely
many solutions?
A: We do so parametrically.
The above reduced form corresponds to the system:
x
y
+ 5z
+ 2z
=5
=3
Here’s how to state the solution parametrically:
1. Solve each equation for its first variable:
x = 5 - 5z
y = 3 - 2z
The variables x and y do not appear on the right. z can be
chosen arbitrarily to be any real number (z is referred to as
a parameter), and will determine the corresponding
numerical values for x and y.
2. State your (infinitely many) solutions parametrically:
(5 - 5z, 3 - 2z, z)
z = 0  (5, 3, 0) is a solution
z = 1  (0, 1, 1) is a solution, etc. for  many solutions
6.4.2-3
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