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1
Quizz 3 (Notes, books, and calculators are not authorized)
Show all your work in the blank space you are given on the exam sheet. Always justify your
answer. Answers with no justification will not be graded.
1 a
3
Question 1: Consider the system
X=
.
a 25
b
(a) For which values of a does the system have a unique solution?
The system has a unique solution iff
1
a
a
= 25 − a2 6= 0
25
This is equivalent to a 6= ±5. The admissible values are a ∈ R\{−5, +5}.
(b) Find those values of (a, b) for which the system has no solution.
The augmented matrix is
M∼
1
a
a 3
1
∼
25 b
0
a
25 − a2
3
.
b − 3a
The system has no solution iff the second pivot is zero and b − 3a 6= 0. This means a = ±5 and
b 6= ±15. In conclusion the above system has no solution iif either a = 5 and b 6= 15, or a = −5
and b 6= −15.
−6x2 + 2x1 + 7x3 = 1
Question 2: Consider the following linear system
3x3 + 4x2 = 8
2x3 = 4
(a) Write the augmented matrix.
The system can be rewritten into the following form:
2x1 − 6x2 + 7x3 = 1
4x2 + 3x3 = 8
2x3 = 4
and the augmented matrix is

2 −6
0 4
0 0

7 1
3 8
2 4
(b) Determine the pivot and free variables.
There is no free variable. There are three pivots, which are 2, 4 and 2 in the first, second and third
column, respectively.
2
Quizz 3, September 20, 2010

2
Question 3: (a) Compute the echelon form of the augmented matrix M = 4
8

2
M = 4
8
2
4
8
−1
1
−1
 
6 4
2
10 13 ∼ 0
26 23
0
2
0
0
−1
3
3
 
4
2 2
5 ∼ 0 0
7
0 0
6
−2
2
−1
3
0
2
4
8

6 4
10 13
26 23
−1
1
−1

4
5
2
6
−2
4
(b) Compute the reduced echelon form of M .

2
M ∼ 0
0
2
0
0
−1
3
0
6
−2
1
 
4
2
5  ∼ 0
1
0
2
2
0
0
−1
3
0
0
0
1
 
1
2
6  ∼ 0
1
0
2
2
0
0
−1
1
0
0
0
1
 
1
2
2  ∼ 0
1
0
2
2
0
0
0
1
0
0
0
1

3
2
1
2
The reduced echelon form of M is

1 1 0 0
M ∼ 0 0 1 0
0 0 0 1
3
2

2
1
2
Question 4: Let A be m × n matrix and B a n × p matrix. (a) Show that if B has a zero
column then AB has a zero column.
Let C1 , . . . , Cp be the columns of B. The definition AB is
AB = [AC1 AC2 . . . ACp ],
in other words, the columns of AB are AC1 , AC2 , . . . ACp . If column Cj is zero then ACj is also
zero, meaning that the jth column of AB is zero.
(b) Show that if A has a zero row then AB has a zero row. (Work with B T AT and use (a).)
T
Let R1 , . . . Rm the rows of A. Then the columns of AT are R1T , . . . , Rm
. If RjT is zero then
from (a) we know that the jth column of B T AT is also zero. This also mean that the jth row of
(B T AT )T = AB is zero.