Math 1090-001
Midterm 2
10/23/2014
Instructions: Follow the given instruction for every exercise. You have
60 minutes to complete the exam. Show proof of your work. The space
left for each question should be sufficient, but possibly not necessary, to
answer the question. You can use scratch paper, but I appreciate if you
turn in only the sheets of the test, so copy your answers here, including
convincing proof of your work. The number in the square next to the
problem deignates the number of points that problem is worth. The test
will be graded out of 100, but the exercises sum to 105 points, this means
5 points are extra credit.
First and Last Name:
UnID:
1
1. Solve the following quadratic equations for x, using a different technique for each of
them. Declare the technique you’re using.
10
(a) x2 − 2x − 35 = 0;
10
(b) −4x2 + 10x + 1 = 0.
Solution: The first one can be factored: it is equal to (x − 7)(x + 5) = 0, hence
the roots are 7 and −5. For the second, use the square root technique: write it as
9
2x2 = 78 + 41 = 98 , which yields x2 = 16
, hence the roots are 34 and − 34 . We can use
the quadratic formula to deal with the last one: identify the coefficients a = −4,
b/2 = 5, and c = 1. Then the roots are
p
√
−b/2 + (b/2)2 − ac
−5 + 29
x1 =
=
(0.1)
a
−4
p
√
−b/2 − (b/2)2 − ac
5 + 29
=
.
(0.2)
x2 =
a
4
Page 2 of 7
2. A toxic waste storage facility has space for 11, 000 tons of total waste. They store
three types of toxic waste, based on the toxicity. They’re labeled L (low), M (middle)
and H (high). The cost per ton to store is $4/ton for type L, $5/ton for type M ,
and $7/ton for the last type. The total cost of storage is $63, 000. Profits need to be
$25, 000. How much of each type of waste can be stored at this site if the profit is
$1/ton for type L,$2/ton for type M and $4/ton for type H?
10
(a) You have informations about the total number of tons, the cost and the profit.
Use these to write down linear equations in the variables x, y and z, which denote
respectively the amount of waste of type L, M and H stored.
15
(b) Solve the linear system you obtained with the technique you prefer.
Solution: We know that the total space available is 11, 000 tons, hence if l, m and
h denote respectively the amount of waste of type L, M and H stored we have
l + m + h = 11, 000. Write down the total cost of storage: 63, 000 = 4l + 5m + 7h.
Likewise the total profit will be given by 25, 000 = l + 2m + 4h. Write down the
augmented matrix corresponding to the system:


1 1 1 11, 000


(0.3)
4 5 7 63, 000
1 2 4 25, 000
Our task is to manipulate this matrix to
row with R2 − 4R1 :

1 1 1

0 1 3
1 2 4
get to the identity. replace the second

11, 000

19, 000
25, 000
Replace the third row with R3 − R1 = (0, 1, 3, 14, 000):


1 0 0 2, 000


0 1 3 19, 000
0 1 3 14, 000
Subtract the third row once from the

1 0

0 1
0 0
(0.4)
(0.5)
second:

0 2, 000

3 19, 000
0 5, 000
(0.6)
This system does not have a solution, in particular the equation from the third
line reads 0 = 5, 000, which is always false.
Page 3 of 7






1 −1 0
1
0 1
16 0
8






3. Consider the matrices B =  0
1 3 and C =  3 −3 3 and A =  18 −2 15.
−1 2 1
−1 0 1
−2 2
1
10
(a) Compute (B + C)A;
10
(b) Compute B t C.




1 0 −1
18 −2
7




Solution: We have B t = −1 1 2 , and CA =  96 12 −18, hence
0 3 1
−18 2
−7


19 −2
6


t
B + CA =  95 13 −16.
−18 5
−6


2 −1 −1


We have B + C =  3 −2 6 . Computing the product we get
−2 2
2


16 0 0


(B + C)A =  0 16 0 .
0
0
16
Page 4 of 7
5
10
5
4. Consider the following matrices and compute their inverses, when it’s possible. Otherwise, say why it is not.


1 −1


(a)  1
3
−1 1
!
1 −5
(b) 2
2
3


1 0 0


(c) 0 1 0
0 0 1
Solution: The first matrix is not invertible, since it’s not square.
For the second we have a formula,
a b
c d
1
us that the inverse matrix is
2 + 10
3
2
− 23
!−1
The third matrix is the identity. We can
Otherwise, define the augmented matrix

1 0 0 1

0 1 0 0
0 0 1 0
1
=
ac − bd
!
5
=
1
3
8
− 18
15
16
3
16
!
d −b
, which gives
−c a
!
.
notice that II = I, hence I = I −1 .

0 0

1 0
0 1
(0.7)
. We want to reduce the second half of the matrix to look like the identity, but it
already does. Hence, I −1 = I.
Page 5 of 7
20
5. Solve the following linear system using the elementary row operations on the
augmented matrix. You may stop when the matrix is upper triangular (i.e. when
all the entries below the diagonal are zeros) and then plug-in the solution backwards,
if you want. (Hint: start subtracting the third row from the second one, and replacing
the second row with the result).


−3z + x = 6
−2y + 7x = 3z − 16 .


7x = 14 + 3y + 5z
Solution: Write down the augmented

1 0

7 −2
7 −3
matrix corresponding to the system:

−3
6

(0.8)
−3 −16
−5 14
Our task is to manipulate this matrix to get to the identity. replace the second
row with the difference between the third and second row:


1 0 −3
6


(0.9)
2 −30
0 1
7 −3 −5 14
Now consider the last row, and subtract from it 7R1 . Sum to the result 3R2 (the
last column will have entry 14 − 6(7) − 3(30) = 14 − 42 − 90 = −28 − 90 = 118).
The result is


1 0 −3
6


(0.10)
−30 
0 1 2
0 0
22
−118
divide the last row by 22:


1 0 −3
6


0 1 2 −30
59
0 0 1 − 11
(0.11)
Subtract twice the last row from the second, and sum three times the last row to
the first. You get


1 0 0 − 111
11


(0.12)
0 1 0 − 212
11 
59
0 0 1 − 11
Read the last column, that’s the solution of the system.
Page 6 of 7
Question
Points
1
20
2
25
3
20
4
20
5
20
Total:
105
Page 7 of 7
Score