Example
Find the solution set of the following system of linear equations:
.
0
8
2
6
=
−
+
=
+
+
=
+
+
z
y
x
z
y
x
z
y
x
Rb
Ra
↔
cRa
Ra
→
Solution
Since the system of linear equations is already in standard form we can write down its augmented
matrix immediately. We then perform a sequence of elementary row operations until a row
equivalent matrix is obtained from which we can readily determine the solution for the given system
of equations. The following notation will be used to indicate the row operation(s) being performed on
the matrices at each stage of the process:
means interchange rows a and b.
means replace row a by c
×
row a.
8
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cRb
Ra
Ra
+
→
means replace row a by the sum of row a and c
×
row b.
⎯
⎯
⎯
→
⎯
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
⎯
⎯
⎯
→
⎯
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
⎯
⎯
⎯
→
⎯
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
→
→
→
R2

R1
R1
R1

R3
R3
R1

R2
R2
6

2

0
0
2
0
1
0
6
1
1
1
0
1

1
1
2
0
1
0
6
1
1
1
0
1
1
1
8
1
2
1
6
1
1
1
⎥
⎥
⎥
⎦
⎤
⎡
⎤
⎡
⎤
⎡
1
0
0
1
4
1
0
1
4
1
0
1
⎢
⎢
⎢
⎣
⎯
⎯
⎯
→
⎯
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎣
⎯
⎯
⎯
⎯
→
⎯
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎣
−
−
→
⎟
⎠
⎞
⎜
⎝
⎛
→
3
1
0
0
2
0
1
0
3
1
0
0
2
0
1
0
6
2
0
0
2
0
1
0
R3

R1
R1
R3
2
1
R3
The system of equations that corresponds to this last matrix is
3
2
1
=
=
=
z
y
x
.
Therefore, the solution to the
original system of linear equations is the ordered triple (1, 2, 3).
Example
Use elementary row operations on the augmented matrix to solve the following system of linear
equations:
9
4
3
2
6
3
2
=
+
+
=
+
+
z
y
x
z
y
x
3
=
+
+
z
y
x
⎥
⎤
⎢
⎡
−
⎥
⎤
⎢
⎡
⎥
⎤
⎢
⎡
−
→
−
→
−
→
−
→
0
1
0
1
3
1
1
1
3
1
1
1
R2
R3
R3
R2
R1
R1
2R1
R3
R3
R1
R2
R2
Solution
⎥
⎥
⎦
⎢
⎢
⎣
⎯
⎯
⎯
⎯
→
⎯
⎥
⎥
⎦
⎢
⎢
⎣
⎯
⎯
⎯
⎯
→
⎯
⎥
⎥
⎦
⎢
⎢
⎣
0
0
0
0
3
2
1
0
3
2
1
0
3
2
1
0
9
4
3
2
6
3
2
1
0
(1)
2
3
(2)
x
z
y
z
.
The equations corresponding to this last matrix are
0
0
(3)
−
=
+
=
=
.
=
These equations may be rewritten as
0
0
=
2
3
−
=
z
y
z
x
.
We see that
x
and
y
depend on
z
and can
write the solution as (
z
, 3 – 2
z
,
z
) where z is arbitrary. This system has infinitely many solutions,
each value of
z
produces a new solution.
Example
Use elementary row operations on the augmented matrix to solve the following system of linear
equations.
10
4
3
2
6
3
2
3
=
+
+
=
+
+
+
+
z
y
x
z
y
x
z
y
x
=
Vector Geometry and Linear Algebra
MATH 1300
Unit 3
9
Solution
.
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
⎯
⎯
⎯
⎯
→
⎯
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
⎯
⎯
⎯
⎯
→
⎯
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
→
−
→
−
→
1
0
0
0
3
2
1
0
3
1
1
1
4
2
1
0
3
2
1
0
3
1
1
1
10
4
3
2
6
3
2
1
3
1
1
1
R2
R3
R3
2R1
R3
R3
R1
R2
R2
3
0
2
3
0
0
0
1
x
y
z
x
y
z
The equations corresponding to this last matrix are
x
y
z
+
+
=
+
+
=
+
+
=
.
Consider the last of these three equations. Clearly there are no values for
x
,
y
and
z
that will satisfy
this last equation and so the system has no solution.
3.4 Problems
1.
Use elementary row operations on the augmented matrices to find the solution set for each of
the following systems of linear equations.
a.
b.
1
=
−
y
x
5
=
+
y
x
1
3
2
2
=
+
+
y
x
=
y
x
2
c.
4
2
2
=
−
−
y
x
=
y
x
1
=
−
y
x
2
6
=
−
+
=
z
y
x
6
5
3
3
=
+
+
d.
12
3
2
=
+
y
x
e.
f.
2
0
=
+
−
z
y
x
+
+
z
y
x
7
6
4
3
=
+
+
=
+
+
z
y
x
z
y
x
2
3
6
1
z
x
y
z
x
y
+
z
y
x
3
=
+
+
z
y
x
+
=
g.
2
2
x
y
z
−
+
=
+
−
=
h.
x
2
=
+
z
x
1
=
+
−
z
y
3.5 The GaussJordan elimination procedure
In this section a systematic procedure for solving systems of linear equations is introduced. This
procedure is called the GaussJordan elimination procedure. The system of linear equations is