Mathematics 1090
2005
1. Solve
 the following

x − 3y + 4z
a) 2x + 2y + z


3x − 4y + 2z
Using matrices:
Then
 the solution

x − y + z
b) 3x + 2z


x − 4y + 2z
Using matrices:
PRACTICE EXAM II-Solution
linear systems of equations:
=0
=1
=9

1
2
3

1
0
0

1
0
0

1
0
0

1
0
0

1
0
0

−3 4 | 0
2 1 | 1 R2 ← R2 − 2R1
−4 2 | 9 R3 ← R3 − 3R1

−3
4
| 0
8
−7 | 1 R2 ← R2 × 81
5 −10 | 9

−3
4
| 0
R1 ← R1 + 3R2

1 −7/8 | 1/8
5
−10 | 9
R3 ← R3 − 5R2

0 −11/8 | 3/8
1 −7/8 | 1/8 
0 −45/8 | 67/8 R3 ← R3 × −8
45

0 −11/8 |
3/8
R1 ← R1 − 11
8 R3
1 −7/8 |
1/8 
R2 ← R2 + 87 R3
0
1
| −67/45

0 0 | 109/45
1 0 | −53/45 
0 1 | −67/45
of the system is x = 109/45, y = −53/45, z = −67/45.
=3
=7
=5

1
3
1

1
0
0

1
0
0

1
0
0

−1 1 | 3
0 2 | 7 R2 ← R2 − 3R1
−4 2 | 5 R3 ← R3 − R1

−1 1 | 3
3 −1 | −2 R2 ← R2 × 31
−3 1 | 2

−1
1
|
3
R1 ← R1 + R2

1 −1/3 | −2/3
−3
1
|
2
R3 ← R3 − 3R2

0 2/3 | 7/3
1 −1/3 | −2/3
0
0
|
0
Spring
This is the reduced form. The corresponding system (we omit the last equation
because it is useless) is
(
(
x = − 32 z + 73
x + 23 z = 37
⇐⇒
y − 13 z = − 32
y = 31 z − 23
This system has an infinite number of solutions given by x = − 23 z + 73 , y = 31 z − 23
where z is considered as a parameter (for each different value of z you have a different
solution to the system).
2. A bank lent a company $135500 for the developpement of 3 new products, A, B and
C. If the loan for product A is $15500 more than for products B and C altogether,
and the amount for product B was twice as much as the amount for product C, how
much was lent for each product?
We note x the amount lent for product A, y the amount lent for product B,z the
amount lent for product C. Then they must satisfy the following equations:


x + y + z = 135500 The total loan is the sum of the loan for each product
x = 15500 + y + z
Loan for A is 15500 plus loan for B plus loan for C


y = 2z
amount lent for B is twice the amount lent for C
Then you have to solve the following system:


x + y + z = 135500
x − y − z = 15500


y − 2z
=0
The solution is x = 75, 500, y = 40, 000, z = 20, 000.
3. Find
inverse of the following matrices (if it exists):
the 1 2
a)
2 3
This is a 2x2 matrix, then we calculate ad − bc = 1 × 3 − 2 × 2 = −1 6= 0 then the
−1
1
3 −2
−3 2
1 2
=
=
matrix has an inverse given by:
2 −1
2 3
−1 −2 1
2
2
b) 3
2 6
2
Same thing we calculate ad − bc = × 6 − 2 × 2 = 0 then the matrix has no inverse.
3


0 1 0
c) 1 1 0
0 1 1

0
1
0

1
0
0

1
0
0

1 0 | 1 0 0
R2 ↔ R1
1 0 | 0 1 0
1 1 | 0 0 1

1 0 | 0 1 0 R1 ← R1 − R2
1 0 | 1 0 0
1 1 | 0 0 1 R3 ← R3 − R2

0 0 | −1 1 0
1 0 | 1 0 0
0 1 | −1 0 1
−1 

0 1 0
−1 1 0
then 1 1 0 =  1 0 0
0 1 1
−1 0 1

3.bis (you can use the result of exercise 3)(
=1
.
= 45
1 2
x
1
This system can be rewritten in matrix form as
=
then by multi2 3
y
45
−1 1 2
−3 2
plying both sides by
we obtain
=
2 3
2 −1
87
−3 × 1 + 2 × 45
1
−3 2
x
=
=
=
−43
2 × 1 + (−1) × 45
45
2 −1
y
a) Solve the following linear system:
x + 2y
2x + 3y
then x = 87, y = −43.

 
0 1 0
1



b) Let us define A = 1 1 0 and B = 2. Solve the following matrix equation
0 1 1
3
 
x1
for X = x2 :
x3
AX = B
Same as in a),we multiply
 both sides of this (matrix) equation by (the inverse matrix
−1 1 0
of A) A−1 =  1 0 0 then we obtain:
−1 0 1
A−1 AX = A−1 B ⇐⇒ I3 X = A−1 B ⇐⇒ X = A−1 B
then we calculate

  
   
1
1 × (−1) + 2 × 1 + 3 × 0
1
−1 1 0
A−1 B =  1 0 0 × 2 =  1 × 1 + 2 × 0 + 3 × 0  = 1
2
1 × (−1) + 2 × 0 + 3 × 1
3
−1 0 1
that is x1 = 1, x2 = 1, x3 = 2.
4. A small country with a simple economy (only agricultural and manufacturing industries) has the following technology matrix:
0.5 0.5
A=
0.1 0.3
a) Calculate the matrix I2 − A and find its inverse.
0.5 −0.5
0.5 0.5
1 0
=
−
I2 − A =
−0.1 0.7
0.1 0.3
0 1
It is a 2x2 matrix so we calculate 0.5 × 0.7 − (−0.5) × (−0.1) = 0.3 6= 0 then this
matrix has an inverse which is given by
0.5 −0.5
−0.1 0.7
−1
=
1
0.3
0.7 0.5
0.1 0.5
=
7/3 5/3
1/3 5/3
b) Using the correct Leontief model and the corresponding technological equation,
find the gross production needed to have surpluses of 10 units of agricultural products
and 25 of manufacturing products. This economy follows the open Leontief model and
the corresponding technological equation is (I 2 − A)X = D where X is the matrix
representing the gross production and D represents the surplusses. Then the solution
to the technological equation is
7/3 5/3
10
195/3
65
−1
X = (I2 − A) D =
=
=
1/3 5/3
25
135/3
45
Then the gross production must be 65 units of agricultural products and 45 units of
manufactured products.
5. A company selling washers has fixed costs of $100, and variable costs of 100 + 0.2x
per unit.
a) Write the formula for the cost function C(x).
The formula is C(x) = 100 + x(100 + 0.2x) = 100 + 100x + 0.2x 2 .
b) If the selling price for a washer is $112, calculate the profit function P (x) and find
the break-even point(s).
The revenue function is R(x) = 112x and then the profit function is
P (x) = R(x) − C(x) = 112x − 100 − 100x − 0.2x 2 = −0.2x2 + 12x − 100
The break-even point(s) are found by solving P (x) = 0, and you find two solutions
x1 = 10 and x2 = 50. There are then 2 break-even points, one when 10 washers are
produced and sold, the other when 50 are produced and sold.
c) How many washers must the company sell to maximize its profit?
The profit is an opening downward parabola, it then achieves its max. at its vertex
−12
= 30. That is max profit is achieved when producing/selling 30
given by x = 2×(−0.2)
washers.
6. Electricity bill: x is the consumption in kWh, C(x) is the monthly charge for x kWh.
There is a fixed fee for the connection of $10, and the price per kWh is of 10 cents
when you consume less than 100 kWh, 8 cents when you consume between 100 and
200 kWh and 5 cents above 200 kWh. Write the piecewise
How much do you pay if you consume 210 kWh?
We have


10 + 0.1x
C(x) = 10 + 0.1 × 100 + 0.08(x − 100) = 12 + 0.08x


12 + 0.08 × 200 + 0.05(x − 200) = 28 + 0.05x
defined function C(x).
for 0 ≤ x ≤ 100
for 100 < x ≤ 200
for x > 200
For 210 kWh the cost is C(210) = 28 + 0.05 × 210 = $38.5.
7. Compute the following matrix operations, if possible. If not say why.
1 2 0
0 1 −1
+3
2 4 1
1 −2 0


1 −1 0
1
5 6
9
3
8 7 12
1 1 0
0 2 4
1 2
1 0
1
+ 1 2 3
6

T
−1
1

− 2
2
3
4

1
0
1
1 2 0
0 1 −1
1+3×0
2+3×1
0 + 3 × (−1)
1 5 −3
+3
=
=
2 4 1
1 −2 0
2 + 3 × 1 4 + 3 × (−2)
1+3×0
5 −2 1
1 1 0 1
+ 1 2 3 4 is not defined because the two matrices have different
0 2 4 6
orders.

 

1 −1 0
1/3 −1/3 0
1
5 6
9  = 5/3
2
3
3
8 7 12
8/3 7/3 4

 
 
 

T
−1 1
1 1
−1 1
2 0
1 2 1
−  2 0 = 2 0 −  2 0 =  0 0
1 0 2
3 1
1 2
3 1
−2 1