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Name:
Section Registered In:
Math 125 – Exam 2 – Version 1 March 27, 2006
- Let A =
α^2 2 4
, B =
^2
, and C =
−^2
(a) (1pt) Compute AB.
Solution:
AB =
α^2 2 4
^2
α^ ·^ 2 + 2^ ·^3 α^ ·^ 1 + 2^ ·^2 2 · 2 + 4 · 3 2 · 1 + 4 · 2
2 α^ + 6^ α^ + 4 16 10
(b) (1pt) Compute AC.
Solution:
AC =
α^2 2 4
−^2
α^ · −2 + 2^ ·^5 α^ ·^ 7 + 2^ · −^1 2 · −2 + 4 · 5 2 · 7 + 4 · − 1
−^2 α^ + 10^7 α^ −^2 16 10
(c) (2pt) Solve for α such that AB = AC.
Solution: Recall that for two matrices to be equal, we need the entries in each matrix to be the same. For AB = AC, we need
2 α + 6 = − 2 α + 10 α + 4 = 7 α − 2 16 = 16 10 = 10
The third and fourth equations are obviously satisfied. Solving for α in the first equation, we see that for equality in the a 11 position, α = 1. Solving the second equation for α, we get
the same solution: α = 1. Therefore, AB = AC when α = 1.
(d) (1pts) Note that here AB = AC but B 6 = C. What property is A lacking that AB = AC does not require that B = C? Briefly explain your answer.
Solution: When A is invertible, we can left multiply the equation AB = AC by A−^1. This results in the equation B = C.
- A small town has three primary industries: a copper mine, a railroad, and an electric utility. To produce $1 of copper, the copper mine uses $0.20 of copper, $0.10 of transporta- tion, and $0.20 of electric power. To provide a $1 of transportation, the railroad uses $0. of copper, $0.10 of transportation, and $0.40 of electric power. To provide a $1 of electricity, the electric utility uses $0.20 of copper, $0.20 of transportation, and $0.30 of electric power. (a) (2pts) Form the consumption matrix for the above economic model.
Solution: copper railroad electric
C =
copper used railroad used electricity used (You did not have to label the matrix for full-credit.)
(b) (2pts) Define what it means for an economy to be productive.
Solution: An economy is productive if given any external demand ( D~), there is a pro- duction schedule ( X~) that can meet the demand. Mathematically, there is an X~ such that X^ ~ = [I −C]−^1 D~ has all positive entries for any demand vector D~. Either answer is acceptable.
(c) (3pts) Determine if the small town’s economy is productive.
Solution: There are multiple solution to this question. First, we can note that every industry is profitable (since every column sums to less than 1), hence the economy is productive. Another solution would be to recognize that every row sums to less than 1, hence the economy is productive. Lastly, we can show that [I − C]−^1 is non-negative.
[I − C]−^1 =
Since no entry in the inverse is negative, the economy is productive.
(d) (3pts) Suppose that during the year there is an outside demand of 1.2 million dollars for copper, 0.8 million dollars for transportation, and 1.5 million dollars for electric power. How much should each industry produce to satisfy the demands?
Solution: We are given that D~ =
(where the units on each entry is in millions of
dollars). We are asked to solve for X~ such that X~ = [I − C]−^1 D~. Using the calculator,
X^ ~ ≈
. Therefore the copper industry should produce 2.81 million dollars of copper,
the railroad should produce 2.12 million dollars of transportation, and the electricity indus- try should produce 4.16 million dollars in electricity.
- Let A be an invertible n × n matrix. Determine if the following are always true. Fully justify your answer. (a) (3pts) (I + A)(I + A−^1 ) = 2I + A + A−^1
Solution: This is always true. Using the properties of matrix arithmetic:
(I + A)(I + A−^1 ) = II + IA−^1 + AI + AA−^1 = I + A−^1 + A + AA−^1 by properties of I = I + A−^1 + A + I by properties of the inverse = 2 I + A−^1 + A since addition is commutative
(b) (2pts) If A^2 is invertible then (A^2 )−^1 = (A−^1 )^2.
Solution: This is always true.
(A^2 )(A−^1 )^2 = (AA)(A−^1 A−^1 ) = AAA−^1 A−^1 = AIA−^1 = AA−^1 = I
Therefore (A^2 )−^1 = (A−^1 )^2. (Technically, we should also do the same calculation for (A−^1 )^2 (A^2 ) = I. But since A^2 is a square matrix, we are guaranteed that, if we have the right inverse, it is the left inverse as well.)
- Consider the linear system
x + 2y + 3z = 5 2 x + 5y + 3z = 3 x + 8z = 17.
(a) (2pts) Write this linear system as a matrix equation of the form A~x = ~b.
Solution:
x y z
(b) (1pt) Find A−^1.
Solution: The calculator says that A−^1 =
(c) (2pts) Use A−^1 to solve for the variable vector. Be sure to show work.
Solution: Since A~x = ~b and A−^1 inverse exist, we can solve for ~x by the equation ~x = A−^1 ~b.
~x = A−^1 ~b =
T 14 =
0.^6000 0.^6000
T 15 =
0.^6000 0.^6000
Hence, it takes 14 years for the market to stabilize.
An alternative way of finding the time period that the market stabilizes is to look for the first power of T such that, when rounding, all of the column vectors are the same vector. From the above matrices, we see that the first time this happens is when t = 14. In the long run, 60% of commuters will be using mass transit.
- (1pt each) Answer the following either True or False and justify your answer.
(a) If A and B are matrices, then both AB and BA are defined if and only if A and B are square matrices.
Solution: False. To multiply matrices, you need the number of columns of the first matrix to equal the number of rows of the second. To have both AB and BA, this requires that A be n × k and B be k × n. Note that if n does not equal k, then A and B are not square but the multiplication is still defined.
(b) Every square matrix is invertible.
Solution: False. We have encountered many square matrices that do not have a multiplica- tive inverse.
(c) If A and B are matrices such that AB = I, then both A and B are invertible.
Solution: False. The definition of multiplicative inverses is that AB = I and BA = I. If A and B are not square, we are not even guaranteed that AB even exists, let alone the identity matrix.
(d) If A is a 3 × 2 matrix and B is a 2 × 4 matrix, than AB is a 3 × 4 matrix.
Solution: True. Since the number of columns of A equal the number of rows of B, we can calculate AB. By construction, the resulting matrix will have the dimensions: the numbers of rows of the first by the number of columns of the second. Here, that gives a 3 × 4 matrix.
(e) The determinant of an n × n matrix can be evaluated using a cofactor expansion along any row.
Solution: True. The cofactor expansion definition of the determinant allows us to expand along any row or down any column.
Name:
Section Registered In:
Math 125 – Exam 2 – Version 2 March 27, 2006
- Consider the linear system
x + 2y + 3z = 6 2 x − 3 y + 2z = 14 3 x + y − z = − 2.
(a) (2pts) Write this linear system as a matrix equation of the form A~x = ~b.
Solution:
x y z
(b) (1pt) Find A−^1.
Solution: The calculator says that A−^1 =
(c) (2pts) Use A−^1 to solve for the variable vector. Be sure to show work.
Solution: Since A~x = ~b and A−^1 inverse exist, we can solve for ~x by the equation ~x = A−^1 ~b.
~x = A−^1 ~b =
(a) (3pts) How is it possible to use the determinant to determine if a matrix is invertible?
Solution: If the determinant of a matrix is nonzero then it is invertible.
(b) (7pts) Find all values of k for which the matrix A =
k −k 3 0 k + 1 1 k − 8 k − 1
is invertible.
Solution: We want to use part (a) and take this determinant. To compute the determinant, I will choose to do a cofactor down the first column since there is a zero in it.
det A = k
k + 1 1 − 8 k − 1
∣∣ −^0
−k 3 − 8 k − 1
∣∣ +^ k
−k 3 k + 1 1
= k[(k + 1)(k − 1) + 8] + k[−k − 3(k + 1)] = k[k^2 − 1 + 8] + k[− 4 k − 3] = k^3 + 7k − 4 k^2 − 3 k = k^3 − 4 k^2 + 4k = k(k^2 − 4 k + 4) = k(k − 2)^2
When the determinant is not equal to zero, then A is invertible. We want k(k − 2)^2 6 = 0. Therefore, the matrix is invertible when k 6 = 0 and k 6 = 2.
Lastly, we can show that [I − C]−^1 is non-negative. Since no entry in the inverse is negative, the economy is productive.
(e) (2pts) The current gross production for this economy is $70 million in tourism, $ million in transportation, and $10 million in services. Is the economy self-sustaining in each area? Solution: Here we are given a production schedule and are asked if the is producing enough to sustain its current production.
[I − C]
The economy is not self-sustaining since there is a shortage of 16.5 million dollars of services.
- Let A =
^0
β 2
, B =
^1
, and C =
^2
(a) (1pt) Compute AB.
Solution:
AB =
^0
β 2
^1
0 ·^ 1 + 1^ ·^3 0 ·^ 1 + 1^ ·^4
β · 1 + 2 · 3 β · 1 + 2 · 4
β + 6 β + 8
(b) (1pt) Compute AC.
Solution:
AC =
^0
β 2
^2
0 ·^ 2 + 1^ ·^3 0 ·^ 5 + 1^ ·^4
β · 2 + 2 · 3 β · 5 + 2 · 4
2 β + 6 5 β + 8
(c) (2pts) Solve for β such that AB = AC.
Solution: Recall that for two matrices to be equal, we need the entries in each matrix to be the same. For AB = AC, we need
3 = 3 4 = 4 β + 6 = 2 β + 6 β + 8 = 5 β + 8
The first and the second equations are obviously satisfied. Solving for β in the third equa- tion, we see that for equality in the a 21 position, β = 0. Solving the fourth equation for β, we get the same solution: β = 0. Therefore, AB = AC when β = 0.
(d) (1pts) Note that here AB = AC but B 6 = C. What property is A lacking that AB = AC does not require that B = C? Briefly explain your answer.
Solution: When A is invertible, we can left multiply the equation AB = AC by A−^1. This results in the equation B = C.
- A new mass transit system has just gone into operation. The transit authority has made studies that predict the percentage of commuters who will change to mass transit (M) or continue driving their automobile (A). Data suggests that 30% of mass transit users will switch to their auto next year and 20% of auto commuters will switch to mass transit. (a) (3pts) Construct the transition matrix for this model. Be sure to label your matrix.
Solution:
M A
T =
.^7
to M to A
(b) (2pts) Suppose that the population of the area remains constant and that initially 30% of commuters are using mass transit. What percentage of commuters will be using mass transit after 3 years?
Solution: Writing the initial state as a vector, we get S 0 =
.^3
M
A
S 3 = T 3 S 0 =
.^3875
Mass transit has roughly 38.8% of the market and auto commuters are 61.3% of the market after three years.
(c) (5pts) Find the exact stable vector and determine what percentage of commuters will be using mass transit in the long run.
Solution: Note that [T − I] =
−.^3.^2
(^). Using the stability requirement [T − I]S~ = ~ 0
yields only one equation −. 3 m +. 2 a = 0, where m is the percentage of mass transit users at equilibrium and a the percentage of auto commuters at equilibrium. Using the stochastic requirement for S~, we also have the equation m + a = 1. These two equations form the
matrix equation AS~ = B~ where A =
−.^3.^2
(^) and B~ =
. Solving for the exact
stable vector S~,
S^ ~ = A−^1 B~ =
.^4
In the long run, 40% of commuters will be using the mass transit system.
