# Last name: name: 1 Quiz 3 (Notes, books, and calculators are not authorized) ```Last name:
name:
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
a 4
1
Question 1: Consider the system
X=
.
2 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 4
2
2 a = a − 8 6= 0
√
√
√
This is equivalent to a 6= &plusmn;2 2. The admissible values are a ∈ R\{−2 2, +2 2}.
(b) Find those values of (a, b) for which the system has no solution.
√
√
Observe first that it is necessary that a = 2 2 or a = −2 2 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
4
1
a
4
1
a 4 1
∼
M∼
∼
2 a b
0 12 a2 − 4 12 ab − 1
a 12 a2 21 ab
√
The system has no solution iff 12 a2 − 4 = 0 and 21 ab − 1 6= 0. This means a = &plusmn;2 2 and b 6= &plusmn; √12 .
√
In conclusion the above system has no solution if and only if either a = 2 2 and b 6= √12 , or
√
a = −2 2 and b 6= − √12 .
−x3 + 2x1 + 3x2 = 1
x2 + 2x1 = 1
Question 2: Consider the following linear system
2x2 = 2
(a) Write the augmented matrix.
The system can be rewritten into the following form:
2x1 + 3x2 − x3 = 1
2x1 + x2 + 0x3 = 1
0x1 + 2x2 + 0x3 = 2
and the augmented matrix is

2
2
0
3
1
2
−1
0
0

1
1
2
(b) Determine the pivot and free variables.
We reduce the augmented

2 3
2 1
0 2
matrix in echelon form.
 
−1 1
2 3 −1
0 1 ∼ 0 −2 1
0 2
0 2
0
 
1
2
0 ∼ 0
2
0
3
−2
0
−1
1
1

1
0
2
There is no free variable. There are three pivots, which are 2, −2 and 1 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 = 4
8

2
M = 4
8
2
4
8
−1
1
−1
 
6 4
2
10 13 ∼ 0
26 23
0
−1
3
3
2
0
0
6
−2
2
 
4
2 2
5 ∼ 0 0
7
0 0
−1
3
0
 
2
1
6  ∼ 0
1
0
2
0
1
0
−1
1
−1
2
4
8
6
−2
4

6 4
10 13
26 23

4
5
2
(b) Compute the reduced echelon form of M .

2 2
M ∼ 0 0
0 0
−1
3
0
 
2
4
5 ∼ 0
2
0
6
−2
4
2
0
0
−1
3
0
0
0
1
2
0
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

1
Question 4: Let E = −3
0
R2 and R3 are row vectors.
0
1
0
(a)
3
2

2
1
2

 
R1
0
0 and let A = R2  be an arbitrary 3 &times; 3 matrix, where R1 ,
R3
1
Compute EA
E is the matrix on the elementary row operation that consists of replacing the second row of a
matrix by itself minus three times the first row:


R1
EA = −3R1 + R2 
R3
(b) Give the inverse of E
E −1

1
= 3
0
0
1
0

0
0
1
because

1
3
0
0
1
0


0
1
0 EA = 3
1
0
0
1
0

 

0
R1
R1
0 −3R1 + R2  = 3R1 − 3R1 + R2  = A.
1
R3
R3
Using A = I in the above equality means that

1 0
3 1
0 0

0
0 E = I
1
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