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Review for Midterm Exam 1
General stuff
:
The exam will cover the following sections in the book: 1.1, 1.2, 1.3, 1.4, 1.5, 1.7 and 1.8.
:
The questions are going to be similar to hw problems and to problems done in class.
They might be phrased as multiplechoice questions, though.
:
No textbooks, notebooks, calculators, laptops, cellphones (or any other communication
devices) will be allowed.
:
You will have 45 min for the exam (the exam will start on 10:05am promptly).
:
Do not cheat.
NOTE:
I might have made simple mistakes (such as 2+3 = 6) in the solutions. Please recheck
all the computations (please email me if you find any).
On the other hand, the methods of
solutions are correct.
Course:
M340L
Instructor:
G.
ˇ
Zitkovi´
c
Semester:
Fall, 2005
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Section 1.1
You should:
:
know what
a solution
to a system of linear equations means.
:
know the terminology (you will not be asked to define terms, but they will appear in
problems).
:
be familiar with consistency (existence) and uniqueness issues.
Problem 1.
Here is a system of linear equations:
3
x
1
+
2
x
2
+
2
x
4

6
x
5
=
0
2
x
1
+
8
x
2

2
x
3
+
3
x
4

2
x
5
+
x
6
=
5

4
x
1
+
16
x
2
+
x
4
+
7
x
5
+
2
x
6
=
0
32
x
1
+
3
x
2

x
4
+
27
x
5

7
x
6
=
0
(a) Find
a
solution to this system.
(b) Are there any other solutions?
Solution:
(a) Inspection of the system above reveals that
x
1
=
x
2
=
x
4
=
x
5
=
x
6
= 0,
x
3
=

5
/
2 is
a solution.
(b) Once we know (from (a)) that the system admits at least one solutions, the fact that the
reduced echelon form will necessarily have pivotless columns (more columns than rows)
implies that the description of the solution set will involve free variables. Presence of
free variables in a consistent system indicates that there are infinitely many solutions.
Therefore, the answer is yes.
Problem 2.
For what values of
h
and
k
does the following system have infinitely many solutions?
4
x
+
6
y
=
h

2
x

3
y
=
k
Solution:
First we reduce the augmented matrix of the system to echelon form (no need for
reduced echelon form):
4
6
h

2

3
k
∼
4
6
h
0
0
k
+
1
2
h
The system will be consistent only if there are no rows of the form [0 0
. . .
0
a
],
a
= 0. In this
case, if and only if
k
+
1
2
h
= 0. In that case,
x
2
is a free variable and the system has infinitely
many solutions. (When
k
+
1
2
h
= 0, then the system has no solutions. For no combination of
h
and
k
will the system have a unique solution.)
Course:
M340L
Instructor:
G.
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 Spring '09
 KOCH
 Linear Algebra, Vector Space, G. Zitkovi´

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