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1
Quiz 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.
a 9
3
Question 1: Consider the system
X=
.
1 a
b
(a) For which values of a does the system have a unique solution?
The system has a unique solution if and only if the determinant is non zero:
a
1
9
= a2 − 9 6= 0
a
This is equivalent to a 6= ±3. The admissible values are a ∈ R\{−3, +3}.
(b) Find those values of (a, b) for which the system has no solution.
Observe first that it is necessary that a = 3 or a = −3 for the above system not to have a solution
(otherwise the determinant is nonzero). In particular, a cannot be zero. Then, the augmented
matrix is reduced in echelon form as follows:
a 9 3
a 9 3
a
9
3
M∼
∼
∼
.
1 a b
a a2 ab
0 a2 − 9 ab − 3
The system has no solution iff a2 − 9 = 0 and ab − 3 6= 0. This means a = ±3 and b 6= ±1. In
conclusion the above system has no solution if and only if either a = 3 and b 6= 1, or a = −3 and
b 6= −1.
−6x3 + 2x1 + 2x2 = 1
3x2 + 4x1 = 8
Question 2: Consider the following linear system
2x2 = 4
(a) Write the augmented matrix.
The system can be rewritten into the following form:
2x1 + 2x2 − 6x3 = 1
4x1 + 3x2 + 0x3 = 8
0x1 + 2x2 + 0x3 = 4
and the augmented matrix is

2
4
0
2
3
2
−6
0
0

1
8
4
(b) Determine the pivot and free variables.
We reduce the augmented

2 2
4 3
0 2
matrix in echelon form.
 
−6 1
2 2 −6
0 8 ∼ 0 −1 12
0 4
0 2
0
 
1
2
6 ∼ 0
4
0
2
−1
0
−6
12
24

1
6
16
There is no free variable. There are three pivots, which are 2, −1 and 24 in the first, second and
third column, respectively.
2
Quiz 3, September 18, 2011

2
Question 3: (a) Compute the echelon form of the augmented matrix M = 2
6

2
M = 2
6
2
2
6
1
3
4
1
5
6
 
3
2
9  ∼ 0
13
0
2
0
0
1
2
1
 
3
2
6 ∼ 0
4
0
1
4
3
2
0
0
1
2
0
1
4
1
2
2
6
1
3
4
1
5
6

3
9
13

3
6
1
(b) Compute the reduced echelon form of M .

2
M ∼ 0
0
2
0
0
1
2
0
1
4
1
 
3
2
6 ∼ 0
1
0
2
0
0
1
1
0
1
2
1
 
3
2
3 ∼ 0
1
0
2
0
0
0
1
0
−1
2
1
 
0
2
3 ∼ 0
1
0
2
0
0
0
1
0
0
0
1

1
1
1
The reduced echelon form of M is

1 1 0 0
M ∼ 0 0 1 0
0 0 0 1
1
2

1
1
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, i.e., B = [C1 C2 . . . Cp ]. The definition of the product AB
implies that
AB = [AC1 AC2 . . . ACp ],
in other words, the columns of the matrix 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 be the rows of A. Then the columns of AT are R1T , . . . , Rm
. If RjT is zero then
T T
from (a) we know that the jth column of B A is also zero. This also mean that the jth row of
(B T AT )T = AB is zero.