L1058 APPLIED MATHEMATICS FOR ECONOMISTS (SOLUTIONS) 2019-20
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L1058
BSc EXAMINATION 2019-20
APPLIED MATHEMATICS FOR ECONOMISTS
(SOLUTIONS)
1. (a) Define formally what it means for vectors to be linearly independent. Explain
intuitively how this notion is related to the rank of a matrix.
[3 marks] Vectors x1 , . . . , xn are linearly independent if α 1 x1 + . . . + αn xn = 0 if and only if α 1 = . . . = αn . The rank of a matrix is the maximal num- ber of linearly independent (row or column) vectors in the matrix.
Suppose that A is a 3 × 4 matrix given by
A =
.
Find the rank of the matrix A. Is the rank full?
[9 marks] The rank of the matrix is 3 . One can easily check that the first three vectors are linearly independent. Since the rank is equal to the num- ber of rows in A , the rank is full.
(c) Consider the following system of linear equations
1x1 + 3x2 + 4x3 = 3;
2x1 + 7x2 + 3x3 = −7;
2x1 + 8x2 + 6x3 = −4.
Does it have a solution? Does it have more than one solution? Make sure to
precisely motivate your answer.
[7 marks] From the previous part of this question, we know that vectors (1, 2, 2) , (3, 7, 8), and (4, 3, 6) are linearly independent. Therefore, the rank of the matrix of coefficients corresponding to the above system of equa- tions is full. In particular, vector (3, −7, −4) must be a linear com- bination of the above vectors, which suffices to conclude that the sys- tem of equations has at least one solution. Since there are only three variables, there is exactly one solution.
(d) Consider a system of equations
1x1 + 3x2 + 4x3 + 3x4 = b1 ;
2x1 + 7x2 + 3x3 − 7x4 = b2 ;
2x1 + 8x2 + 6x3 − 4x4 = b3 .
for some arbitrary values of b1 , b2 , b3 . Depending on the values of the numbers b1 , b2 , b3 on the right hand side, does the system of equations always have a solution? Can it have more than one solution? Motivate your answer.
[6 marks] The matrix of coefficient corresponding to the above system of equa- tions is equivalent to the matrix A. We know that the rank of A is full, so the above system has a solution for any b1 , b2 , and b3 . How- ever, since there are 4 variables and 3 rows, there are infinitely many solutions.
2. (a) Explain precisely what it means for a matrix to be invertible.
[3 marks] The inverse of a matrix A is another matrix, usually denoted by A− 1 , such that the inner product A · A− 1 is equal to the unit matrix. A matrix is invertible if its inverse exist.
Suppose that A is a 3 × 3 matrix given by
A =
0 1 2 .
(b) Find the inverse of the matrix A (if it exists).
[8 marks] The method of finding the inverse of a matrix that was discussed in the lecture and seminars was based on elementary transformations. In other words, one can first transform matrix A to a unit matrix. Then, apply the same transformations to the unit matrix in order to obtain the inverse of A . The solution would be
A− 1 = 1
2 − 1 0 2 .
(c) Solve the following system of linear equations.
2x2 + 2x3 = 2;
x1 + x2 + x3 = 1;
x2 + 2x3 = 0.
How many solutions does it have? Make sure to motivate your answer.
[6 marks] Since the matrix of coefficients in the above system is equal to the matrix A, it suffices to apply the same list of transformations used in question (b) to the vector (2, 1, 0) . The resulting vector (0, 2, − 1) is the unique solution to the system of equations. Alternatively, one could note that the matrix A is invertible, or has the full rank, so the corresponding system of equations has exactly one solution.
(d) Show that vector (x1 , x2 , x3 ) given by
= A− 1 · ,
is a solution to the system of linear equations in part (c), where A− 1 denotes the inverse of the matrix A.
[3 marks] Indeed, vector (0, 2, − 1) is a solution to the above system of equations.
(e) “If an ℓ × ℓ matrix is invertible then its rank must be full.” Give a sketch of an
argument supporting this claim.
[5 marks]
The easiest way to show this is to notice that whenever matrix A is invertible, then equation A ·x = b has exactly one solution, for any vector b . In particular, we have x = A− 1 ·b . However, this must im- ply that the matrix A is of full rank.
3. Consider the function
f (x, y, z) = x3 + 3y2 + 2xz3 − z3 y − 1.
(a) Compute the Jacobian of f and discuss whether the function is continuously
differentiable.
[4 marks]
The Jacobian is a vector
Df (x, y, z) = = 2y2
Clearly, each partial derivative exists and is continuous, so func- tion f is continuously differentiable.
(b) Verify that the Hessian matrix of the function f is symmetric. Is this true for
any function for which the Hessian exists? Explain precisely.
[5 marks]
The Hessian is a 3 × 3 matrix
D2 f (x, y) = 0(6x)
6z2
0
6
−3z2
12xz − 6zy ,
which is symmetric. This follows from Schwarz’s theorem (or Clairaut’s theorem on equality of mixed partials). However, this need not al- ways hold. This is always true as long as the cross-partial deriva- tives are continuous.
Next, consider the function
g(x, y) =
(c) Sketch a level graph of the function g, with the horizontal axis corresponding to variable x and the vertical axis corresponding to variable y .
[4 marks] The level curves will be hyporbolas in the positive orthant and all the rest will be flat and equal to 0.
(d) Evaluate the Jacobian of g at any point (x, y) such that x 0 and y 0.
[5 marks]
The Jacobian is a vector
Dg(x, y) = ( ) = ( ) ,
for x > 0 and y > 0, and Dg(x, y) = (0, 0) otherwise.
(e) By applying the formal definition of the partial derivative, compute the partial
derivatives of function g at (0, 0). Is this function continuously differentiable? [Hint: Check how the partial derivatives change as x and y approach 0.]
[7 marks]
Using the formal definition
(x, y) = lim = lim = 0.
4. (a) Define formally the notion of a local maximum and local minimum of a function.
How are these notions different from global maxima and global minima?
[3 marks]
A local minimum of a function f , is an argument such that f (x) ≤ f (y) , for all y is some open ball around x . We define a local maximum anal- ogously. Global minima/maxima satisfy the above condition for any y in the domain.
(b) Find all extremal points of the function
f (x, y) := x4 + x2 − 6xy + 3y2
and classify each one as a local maximum, local minimum, or saddle point. [9 marks]
The Jacobian of the function is
Df (x, y) = ) ,
which is equal to 0 at (0, 0) , (1, 1), and ( − 1, − 1) . The Hessian is
D2 f (x, y) = ) .
Hence, matrices D2 f (1, 1) and D2 f ( − 1, − 1) are positively semi-definite, so (1, 1) and ( − 1, − 1) are local minima. The matrix D2 f (0, 0) is non-definite, so (0, 0) is a saddle point.
(c) Define formally what it means for a function to be concave. What does it mean for a function to be strictly concave?
[3 marks]
Function f is concave if for any x , y and α ∈ [0, 1] , we have
f (αx + (1 − α)y) ≥ αf (x) + (1 − α)f (y).
The function is strictly concave if the above inequality is strict for all α ∈ (0, 1) .
(d) Suppose that functions f : R → R and g : R → R are concave. Show, without assuming that the functions are differentiable, that the function
h(x) = min {f (x), g(x) },
is also concave. [By min{y, z}, we mean the lower of the two numbers. That is, if y < z, then min{y, z} = y. Otherwise,we have min{y, z} = z .]
[10 marks] Take any x , x′ . If f (x) ≥ g(x) and f (x′ ) ≥ g(x′ ) , the concavity con- dition holds trivially. Let f (x) ≥ g(x) and f (x′ ) < g(x′ ) . Take any α ∈ [0, 1] and suppose that h(αx + (1 − α)x′ ) = g(αx + (1 − α)x′ ) . Then
αh(x) + (1 − α)h(x′ ) = αg(x) + (1 − α)f (x′ )
≤ αg(x) + (1 − α)g(x′ )
≤ g(αx + (1 − α)x′ )
= h(αx + (1 − α)x′ ).
If h(αx + (1 − α)x′ ) = f (αx + (1 − α)x′ ) , we prove this analogously.
5. An individual maximises her preferences represented by the utility function u(x, y), where
u(x, y) := √x + 2 √y ,
with respect to the amounts x, y of goods 1 and 2, respectively. We assume that both goods can be consumed only in positive amounts. Moreover, the consumer faces the following budget constraint
x + 2y ≤ 24.
Since consuming each good takes time, she also faces a time constraint 2x + y ≤ 24.
(a) Draw a graph of the set of constraints that the consumer is facing in this opti-
misation problem. Determine whether the set is compact and/or convex.
[3 marks]
y
12
8
12
The set is bounded and closed, hence, compact. It is also convex.
(b) Determine whether the objective function is continuous over the set of con-
straints. Is the function (strictly) concave? Determine if the consumer’s optimi- sation problem has a solution. Is it unique? Precisely motivate your answer.
[3 marks]
The objective function is well-defined and continuous function over the set of constraints. One can show that it is also strictly con- cave. Given that the set of constraints is compact and convex, this is sufficient to conclude that there is a unique solution to the above optimisation problem.
(c) Write down the Lagrange function corresponding to the optimisation problem and write down the Kuhn-Tucker conditions for this problem. [Hint: You may neglect the non-negativity constraints for x and y without loss.]
[5 marks]
The Lagrange function is given by
√x + 2 √y + λ[24 − x − 2y] + µ[24 − 2x − y],
where λ and µ are Lagrange multipliers. We neglected the non-negativity constraints. The Kuhn-Tucker conditions are given by
− λ − 2µ = 0;
− 2λ − µ = 0;
λ[24 − x − 2y] + µ[24 − 2x − y] = 0,
where λ, µ ≥ 0 .
(d) By examining each possible combination of binding constraints, derive all so- lutions to the Kuhn-Tucker conditions, clearly stating the corresponding values of x and y, and the Lagrange multipliers. Determine which solutions to the Kuhn-Tucker conditions are solutions to the original optimisation problem.
[9 marks] The conditions are satisfied for (x, y) = (8, 8) . However, this requires for µ = 0 and λ = 1/(2√8) , even though both constraints are bind- ing.
(e) What is the interpretation of the Lagrange multipliers? Would the consumer
prefer to have more money or more time?
[5 marks] Lagrange multipliers are the marginal increase in the maximised value function of the consumer once we relax the constraint. Therefore, the consumer would be better off if money increased marginally (λ > 0)
but would not benefit from a marginal increase in time (µ = 0).
2022-01-12