132
Chapter 3
CHAPTER 3
LINEAR SYSTEMS AND MATRICES
SECTION 3.1
INTRODUCTION TO LINEAR SYSTEMS
This initial section takes account of the fact that some students remember only hazily the method of
elimination for
2
2
×
and
3
3
×
systems.
Moreover, high school algebra courses generally
emphasize only the case in which a unique solution exists.
Here we treat on an equal footing the
other two cases — in which either no solution exists or infinitely many solutions exist.
1.
Subtraction of twice the first equation from the second equation gives
5
10,
y
−
= −
so
y
= 2,
and it follows that
x
= 3.
2.
Subtraction of three times the second equation from the first equation gives
5
15,
y
= −
so
y
= –3,
and it follows that
x
= 5.
3.
Subtraction of
3/2
times the first equation from the second equation gives
1
3
,
2
2
y
=
so
y
= 3,
and it follows that
x
= –4.
4.
Subtraction of
6/5
times the first equation from the second equation gives
11
44
,
5
5
y
=
so
y
= 4, and it follows that
x
= 5.
5.
Subtraction of twice the first equation from the second equation gives
0
1,
=
so no
solution exists.
6.
Subtraction of
3/2
times the first equation from the second equation gives
0
1,
=
so no
solution exists.
7.
The second equation is
–2 times the first equation, so we can choose
y
=
t
arbitrarily.
The first equation then gives
10
4 .
x
t
= −
+
8.
The second equation is
2/3
times the first equation, so we can choose
y
=
t
arbitrarily.
The first equation then gives
4
2 .
x
t
=
+
9.
Subtraction of twice the first equation from the second equation gives
9
4
3.
y
z
−
−
= −
Subtraction of the first equation from the third equation gives
2
1.
y
z
+
=
Solution of

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Section 3.1
133
these latter two equations gives
1,
3.
y
z
= −
=
Finally substitution in the first equation
gives
x
= 4.
10.
Subtraction of twice the first equation from the second equation gives
3
5.
y
z
+
= −
Subtraction of twice the first equation from the third equation gives
2
3.
y
z
−
−
=
Solution of these latter two equations gives
1,
2.
y
z
=
= −
Finally substitution in the
first equation gives
x
= 3.
11.
First we interchange the first and second equations.
Then subtraction of twice the new
first equation from the new second equation gives
7,
y
z
−
=
and subtraction of three
times the new first equation from the third equation gives
2
3
18.
y
z
−
+
= −
Solution of
these latter two equations gives
3,
4.
y
z
=
= −
Finally substitution in the (new) first
equation gives
x
= 1.
12.
First we interchange the first and third equations.
Then subtraction of twice the new first
equation from the second equation gives
7
3
36,
y
z
−
−
= −
and subtraction of twice the
new first equation from the new third equation gives
16
7
83.
y
z
−
−
= −
Solution of these
latter two equations gives
3,
5.
y
z
=
=
Finally substitution in the (new) first equation
gives
x
= 1.

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