Math 129A MIDTERM EXAM I with solutions 1 (20 pts) Find all h and

```Math 129A MIDTERM EXAM I with solutions
1 (20 pts) Find all h and k such that the system
x1 + hx2
= 2
4x1 + 8x2 = k
(a) has no solution,
(b) has a unique solution,
(c) has many solutions.
Solution. After applying the row operation R2 7→ R − 4R1 to the augmented matrix [A b]
of the system we obtain the matrix:
&quot;
1
h
2
#
0 8 − 4h k − 8
Therefore the answers are as follows.
(a) h = 2 and k 6= 8 (because then the matrix has a row [0 0 k − 8], k − 8 6= 0, and thus
the system is inconsistent);
(b) h 6= 2 (because then there is a pivot in each column of A, and thus the system has a
unique solution);
(c) h = 2 and k = 8 (because then there is a pivot-free column, and thus there is a free
variable);


3
 

2 (20 pt) Determine if b = 
−7 is a linear combination of the vectors formed from the
−3

1

columns of the matrix A = 
0
−2

−4 2
1

2
1
5
Solution. We first apply the row operations R3 7→ R3 + 2R1 , R3 7→ R3 + 7R2 , and then
R3 7→
1
R
23 3
to the matrix [A b]. As a result we obtain the following matrix:


1 −4 2 3


0 1 2 −7 .


0 0 1 −2
Since there is a pivot in every column of A, the answer is ”Yes”, i.e. b is a linear combination of the columns of A. Note that it is not necessary to write down this combination,
i.e. to solve the corresponding linear system.




 
−1
5
 
 
 

 
 
3 (20 pts) Let v1 = 
 0 , v2 = −1, and v3 = 3. Does the set {v1 , v2 , v3 } span
−1
1
1
3
1
Solution. After applying the row operation R3 7→ R3 + R1 to the matrix [v1 v2 v3 ] we
obtain the following matrix:


1 −1 5


0 −1 3 .


0 0 6
Since there is a pivot in every row, the vectors {v1 , v2 , v3 } does span R3 .
4 (20 pts) Describe and compare the solution sets of the following two systems:
(1) 2x1 − x2 + 3x3 = 0, and
(2) 2x1 − x2 + 3x3 = 2
h
i
Solution. The system (1) has augmented matrix 1 −1 3 0 . This matrix has two pivot
free columns, namely the second and the third one. The solution then is
1
3
x2 − x3
2
2
= free
x1 =
x2
x3 = free
We can write the solution in the vector form as well:
 
 


1
1
3
− 23
r − 2s
2
2

 
 


 
 
x=
 r  = r 1 + s  0  .
s
0
1
For the system (2) with similar reasoning we find that the solution is
3
1
x 1 = 1 + x2 − x3
2
2
x2 = free
x3 = free
In vector form we have

1
r
2
x=

 
  
 
1
− 32 s
1
−3
  
2
 2
 
 
 
r 
 = 0 + r  1  + s  0  .
s
0
0
1
The solutions of (1) and (2) are represented by two parallel planes in R3 : one of the planes
(corresponding to the first system) passes through the origin (0, 0, 0), and the other (corresponding to the first system) passes through (1, 0, 0).
Note. We have to solve and compare two systems with one equation each. We don’t have
one system with two equations!

 
2
 

5 (20 pts) Find the value(s) of h for which the vectors 
−4,
1
independent.

−6
 
 7 ,
 
−3
 
h
 
 2  are linearly
 
4
Solution. We first apply the row operation R1 ↔ R3 , and then apply R2 7→ R2 + 4R1 and
R3 7→ R3 − 2R1 to the matrix A whose columns are the three given vectors we obtain the
matrix:


1 −3
4


.
B=
0
−5
18


0 0 h−8
The three vectors in question are linearly independent if and only if there is a pivot in each
column of B (or equivalently, of A). Therefore the answer is h 6= 8.
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