BECC
104: MATHEMATICAL
METHODS
IN ECONOMICS
Tutor Marked Assignments
Course Code: BECC 104
Assignment Code: ASST BECC 104/TMA/ July 2023 and January 2024
Total Marks: 100
Assignment A
Answer the following Long Category questions in about 500 words each. Each question
carries 20 marks. Word limit will not apply in the case of numerical questions.
2X20=40
(1)
Find
I
8]
—
A
—I
Consider the following two matrices
[l
1.
the rank of ‘A’ and
(i)
Show that (AB)' =B
(iii)
Show that (A)y' = A
(iv)
Show that (By' =B
‘B’
A"
2. An individual consumer consumes two commodities X; & X,.
The utility function is
U=Xxx3°
The price of commodity one is P; = Rs.3.00, the price of commodity two is P> = Rs.4.00,
the individual’s
income per period is Rs.108.
Determine the utility maximizing level of
X1 & X5 and derive the demand curves for the two commodities.
Assignment B
Answer the following Middle Category questions in about 250 words each. Each question
carries 10 marks. Word limit will not apply in the case of numerical questions.
3X10=30
3. Let Z = f(x.y) = 3x°-5y-225x + 70y + 23.
(1)
Find the stationary points of z.
(i1)
Determine if at these points the function is at a relative maximum, relative
minimum, infixion point, or saddle point.
4. Solve the following differential equation
2
d 572Q+10y=0,
dx
dx
given Y(0)=4
dy
— (0
() =1
5.
T Z=1(x,y) =xy
Find the maximum value for f(x,y) if x and y are constrained to sum to 1 (That is,
X +y = 1). Solve the problem in two ways: by substitution and by using the Lagrangian
multiplier method.
Assignment C
Answer the following Short Category questions in about 100 words each
5X6=30
6. Detine
a.
b.
c.
Adjugate of a matrix
Decomposable matrix
Singular matrix
7. Evaluate [(7x-2)\3x+2 dx
8. Explain the concept of maximum value function.
9. Let the production function by Q = AL°K" . Find the elasticity of production with respect to
labour (L).
10. Denote by a, b and ¢ the column vectors
1
-2
a=|2|b=|1
3
-2
J|ec=|-1
-3
|
Calculate
2a - 5b, 2a-5b +c, a’.b,
BECC
104: MATHEMATICAL METHODS
ECONOMICS
IN
Course Code: BECC
Assignment Code: ASST BECC
104
104/TMA/ July 2023 and January 2024
Total Marks: 100
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Assignment A
Answer the following Long Category questions in about 500 words each. Each
question carries 20 marks. Word limit will not apply in the case of numerical
questions.
1. Consider the following two matrices
A
(i) Find the rank of ‘A’ and ‘B’
To find the rank of matrices A and B, we need to perform row operations to reduce each
matrix to its row-echelon form (also known as reduced row-echelon form) and then count
the number of nonzero rows in the resulting matrices.
Let's start by performing row operations on matrix A to bring it to row-echelon form:
Matrix A:
Step 1: Add Row | to Row 2
Step 3: Divide Row 3 by 4
Step 4: Subtract Row 2 from Row 3
Step 5: Add Row 3 to Row 1
Matrix A 18 now in row-echelon form, and it has two nonzero rows. Therefore, the rank of
matrix A is 2.
Next, let's perform row operations on matrix B:
Matrix B:
Step 1: Add Row 1 to Row 2
Step 2: Subtract Row | from Row 3
Step 3: Add Row 2 to Row 3
Matrix B is now in row-echelon form, and it has two nonzero rows. Therefore, the rank
of matrix B is 2.
In conclusion, the rank of matrix A 1s 2, and the rank of matrix B 1s also 2.
(ii) Show that (AB)'=B'A !
To show that (AB)'=B7'47", we need to prove that the inverse of the product of matrices
AB 1s equal to the product of the inverses of matrices B and 4 in that order.
First, let's find the inverses of matrices 4 and B:
3|
Matrix A:
Matrix B:
To find the inverse of a matrix, we will use the Gaussian elimination method:
For matrix A4:
Performing row operations:
¢«
Row
1+ Row2->Row?2
e
Rowl-2*Row3->Row3
Divide Row 2 by 2:
Subtract 4 * Row 2 from Row 3:
[AlI]l]=[-1
1
2]1
[0O
O
O]
[0
O
O]O
0
0]
0.5
0.5
-2
0]
1]
Divide Row 3 by -2:
[A|lIl=[-1
1
2|1
0
[O
O
O]
0.5
[0
O
O]
O
0]
0.5
1
0]
-0.5]
Add 2 * Row 2 to Row 1 and subtract Row 2 from Row 3:
[AlIl=[-11
[0 O
[0 O
2]2
1 0]
0] 0.5 0.5 0]
0]0
O -0.5]
Divide Row 1 by -1:
[A|I]l]=[1
-1
-2
|
-2
-1
[0O
O
0]
0.5
[0O
O
O]
O
0]
0.5
O
0]
-0.5]
Divide Row 2 by 0.5:
[AlI]l]=0[1
-1
-2
| -2
-1
0]
[0
0O
O0]1
1
0]
[0
O
O0]0
O
-0.5]
Divide Row 3 by -0.5:
(Al
I]l]=0[1
-1
-2
]|
-2
-1
0]
[0
O
O]
1
1
0]
[0O
O
O]
O
O
1]
The left part of the augmented matrix is now the identity matrix, which means that the
inverse of matrix A 1s:
For matrix B:
Performing row operations:
e«
Rowl+Row?2->Row?2
e«
Rowl-Row3->Row?3
Divide Row 3 by 1:
Subtract Row 2 from Row
Subtract Row
6|
1:
1 from Row 3:
This augmented matrix does not lead to the identity matrix on the left side, which means
that matrix B does not have an inverse. Therefore, B! does not exist.
Now, let's check if (4B)
'=B 14
(AB) '=Not possible
B '4 '=Not possible
Since matrix B does not have an inverse, we cannot proceed with the computation of
(AB)—", and therefore we cannot compare it with B7'47!.
In conclusion, the relation ((AB)"{-1
(iii) Show that (A1)
1=A
To show that (4 ') '=4, we need to prove that the inverse of the inverse of matrix A is
equal to matrix A itself.
We have already found the inverse of matrix 4 in a previous response:
Now, let's find the inverse of 4!
The inverse of A™' is simply matrix A itself, as shown above. Therefore,
(A1 =4,
In conclusion, we have successfully shown that (47" 1=4.
(iv) Show that (B')'=B
To show that (B™')"'=B, we need to prove that the inverse of the inverse of matrix B is
equal to matrix B itself.
Given matrix B:
Let's first find the inverse of matrix B:
We want to find the inverse matrix B™' such that BB '=/, where I is the identity matrix.
Let's set up the augmented matrix [B|/] and perform row operations to find the inverse:
Performing row operations:
«
Row
1+ Row?2->Row?2
«
Rowl-Row3->Row3
Add Row 2 to Row 3:
This augmented matrix does not lead to the identity matrix on the left side, which means
that matrix B does not have an inverse. Therefore, B! does not exist.
Since B ! does not exist, (B ') ! is also undefined.
In conclusion, we cannot show that (B7')"'=B because the inverse of matrix B does not
exist.
2. An individual consumer consumes two commodities X; & X;. The utility function
is
U=
X|fl.4 XZO.()
The price of commodity one is P; = Rs.3.00, the price of commodity two is P; =
Rs.4.00, the individual’s income per period is Rs.108. Determine the utility
maximizing level of X; & X: and derive the demand curves for the two commodities.
To determine the utility-maximizing levels of commodities X and X> and derive the
demand curves, we can use the concept of utility maximization subject to a budget
constraint. The consumer's problem can be formulated as follows:
Maximize U=X104X206 Subject to the budget constraint: Pi.X\+PXo=M
Given: P1=Rs.3.00 P,=Rs.4.00 M=Rs.108.00
Let's solve for the utility-maximizing levels of Xi and Xa:
1.
Set up the Lagrangian function:
L=X104X20.6—A(P1 X1+ P2X2—M)
2. Take the partial derivatives with respect to Xj, X3, and 4, and set them equal to
Zero:
oL
0
X,
oL
0
X,
oL
0
O~
Taking the derivatives:
aL
ax.
_
04X,
9| Pazc
0.6 X" 3063 0—,P1=0
P —
oL
04v.—04_ AP>=0
3 po
ox, _ 0.6.X1"X>
2 PP~
1.
M=0
Solve the system of equations:
From the first equation: 0.4X2"=1P,
From the second equation: 0.6.X,%4=1P>
Equating the two values of A: 0.4X:%=0.6.X,"4
2.
Solve for the ratio of X and X>:
)004:32A5Q6
gy
AYJ_
e
3.
L
:
114y
.
AX-/_)
:
Substitute the budget constraint: P1.X1+P2Xo=M
3X1+4X>=108
4.
Substitute the expression for X from step 4 into the budget constraint:
— 108
3(3)7 XX,
5.
Solve for Xz:
!
Los
T
4”3(%)11
]
§
Substitute the value of X into the expression for X, from step 4:
n=7 s
Now we have the utility-maximizing levels of X| and X>. To derive the demand curves for
the two commodities, we can solve for X; and X: as functions of their respective prices
while keeping utility constant.
The demand curve for X can be derived by solving the budget constraint for Xi:
_ M-P, X,
=
Th,
Similarly, the demand curve for Xz can be derived by solving the budget constraint for Xa:
M-Py X,
2
Pz
Substitute the values of M, Py, and P: as given in the problem, and then plug in the values
of X1 and X> as derived earlier.
Finally, the utility-maximizing levels of X1 and X and their corresponding demand
curves can be presented as the solutions to the problem.
Assignment B
Answer the following Middle Category questions in about 250 words each. Each
question carries 10 marks. Word limit will not apply in the case of numerical
questions.
3. Let Z. = f(x,y) = 3x> -5y? -225x + 70y + 23.
(i) Find the stationary points of z.
To find the stationary points of the function f{x,y)=3x3—5y2—225x+70y+23, we need to
find the critical points where both partial derivatives are equal to zero. In other words, we
.
0z
0z
need to find the points (x,y) where Py 0 and 3y
Let's calculate the partial derivatives of f{x,y) with respect to x and y:
Ox0z=9x2—225
oyoz—10y+70
Now, we'll set these partial derivatives equal to zero and solve for x and y:
—10y+70=0 10
10y=70
=17
So, we have two potential stationary points:
1. (x)=(5,7)
2.
(x»)=(=5,7)
These are the points where the partial derivatives are both equal to zero, which are the
potential stationary points of the function z=f{x,y). To determine whether these points are
local maxima, local minima, or saddle points, we would need to perform the second
derivative test or analyze the behavior of the function around these points.
(ii) Determine if at these points the function is at a relative maximum, relative
minimum, infixion point, or saddlec point.
To determine the nature of the stationary points of the function
z=fx,y)=3x*—5y?=225x+70y+23 (which we found in the previous part to be (5,7)(5,7) and
(—5,7)(—5,7)), we can use the second derivative test. The second derivative test helps us
classify these points as relative maxima, relative minima, or saddle points.
First, let's calculate the second partial derivatives of f(x,y):
5 = 0 (Since the mixed partial derivatives are equal)
Now, let's evaluate the second partial derivatives at the stationary points:
1. For the point (5,7)(5,7):
)
)
)
The determinant of the Hessian matrix D =
0%z
9%z
0x?
dy?
9%z
—( Py ay)z =(90)(-10) -
(0)>=—900<0.
62
Since D <0 and fi
> (), this implies that the point (5,7) is a saddle point.
For the point (—5,7)(—5,7):
9°Z
_ 18 . (-5)=-90
dx?2
2
)
)
)
The determinant of the Hessian matrix D =
0%z
9%z
0x2
dy?
9%z
—( Py ay)z =(-90)(-10) -
(0)°=900>0.
2z
Since D > 0 and Py < 0, this implies that the point (-5,7) 1s a relative maximum.
In summary, the function f{x,))=3x*-5y*-225x+70y+23 has a relative maximum at the
point (=5,7)(—5,7) and a saddle point at the point (5,7)(5,7).
4. Solve the following differential equation
given Y(0) =4
To solve the given second-order linear homogeneous differential equation:
d*y
dy
mf2g+
_
10y =0
with the initial conditions 1(0)=4 and Z—z(O) =1, we can follow these steps:
1.
Write down the characteristic equation associated with the differential equation:
?=2r+10=0
. Solve the characteristic equation for its roots r1 and r:
ra—2r+10=(r—(1+30))(r—(1-3i)) So,
r=1+3i and r=1-3i.
3.
The general solution of the differential equation is given by:
v(x)=e"x(c1cos(3x)+c2sin(3x))
4.
Use the initial conditions to solve for the constants ¢ and c2:
o«
y(0)=4 gives: c1=4
.
3—1(0) =1 gives: 3¢>=1, s0 3¢2=31
Therefore, the particular solution to the given differential equation is:
y(x)=ex(4cos(3x)+31sin(3x))
5. If Z=A1(x,y) =xy
Find the maximum
value for f(x,y) if x and y are constrained to sum to 1 (That is, x
+y = 1). Solve the problem in two ways: by substitution and by using the Lagrangian
multiplier method.
To tind the maximum value of the function f{x,y)=xy subject to the constraint x+y=1, we
can solve the problem using both the substitution method and the Lagrangian multiplier
method.
Substitution Method:
We have the constraint x+y=1, so we can solve for one variable in terms of the other and
substitute it into the function f{x,y)=xy.
From the constraint, y=1-x.
Substitute y into the function f{x,y):
Ax)=x(1—x)=x—x2
Now, we want to find the maximum value of f{x). To do this, we'll take the derivative of
Ax) with respect to x and set it equal to zero to find critical points:
f'(x)=1-2x=0 Solving for x:
2x=1=x=21
So, the critical point is x= "2 . Plug this back into the original function f{x,y)=xy:
1
1
1
Q=z-U-27327%
1
Therefore, the maximum value of f{(x,))=xy subject to the constraint x+y=11s Y4 .
14 |
Lagrangian Multiplier Method:
Let's set up the Lagrangian function:
L(xy,A)=xy—A(x+y—1)
Now, we'll find the partial derivatives of L with respect to x, y, and 4, and set them equal
to zero to find the critical points:
aL_
ax_y
=0
B
From the first two equations, we have y=4 and x=4. Substituting these into the third
equation gives A+A—1=0, which implies 1= "% .
Substitute A= "> back into x and y to find the critical points: x="
and y=".
Finally, plug x and y into the original function f(x,y)=xy:
Again, we find that the maximum value of f(x,y)=xy subject to the constraint x+y=1 is % ,
consistent with the result from the substitution method.
Assignment C
Answer the following Short Category questions in about 100 words each.
6. Define
a. Adjugate of a matrix
The adjugate of a square matrix is a fundamental concept in linecar algebra. Also known
as the adjoint matrix, it is obtained by taking the transpose of the cofactor matrix of the
given matrix. The cofactor of an element in a matrix is the determinant of the submatrix
obtained by removing the row and column containing that clement. The adjugate matrix
is used in various mathematical operations, particularly in finding the inverse of a matrix.
Specifically, if the determinant of the original matrix is non-zero, the inverse can be
computed by multiplying the adjugate matrix by the reciprocal of the determinant.
15|
b. Decomposable matrix
A decomposable matrix is a square matrix that can be expressed as the sum of two or
more matrices, often simpler in structure. In other words, a matrix 1s decomposable 1f it
can be broken down into the sum of two or more non-zero matrices. Decomposability is a
useful concept in matrix analysis and can aid in understanding properties and behaviors
of matrices. Decomposing matrices into simpler components can provide insights into
their eigenvalues, determinant, rank, and other algebraic properties. Decomposable
matrices play a role in various mathematical applications, including linear
transformations, system of lincar equations, and matrix factorization.
c. Singular matrix
A singular matrix, also known as a degenerate matrix, is a squarce matrix that does not
have an inverse. In other words, a matrix is singular if its determinant is zero.
Geometrically, a singular matrix represents a linear transformation that collapses the
space in at least one dimension, causing a loss of information. This implies that the
columns (or rows) of a singular matrix are linearly dependent, making it impossible to
find a unique solution to a system of lincar equations represented by the matrix. Singular
matrices have important implications in various fields such as linear algebra, physics,
engineering, and statistics. They are used to analyze critical points, stability, and behavior
of systems described by matrices.
7. Evaluate
j(7x
- 2}V3x
+ 2 dx
To evaluate the integral J(7x - 2N(3x + 2)) dx, we will perform the integration step by
step.
J(7x - 2N(3x +2)) dx
First, let's integrate the term 7x with respect to x:
[(7x) dx = (7/2)x*2 + C1
Next, let's integrate the term -2V(3x + 2) with respect to x:
Letu=3x+2
Then, du/dx = 3
dx = du/3
Now we substitute the values of u and dx back into the intcgral:
16|
[(-27u) (1/3) du = (-2/3) Nu du
Integrating Vu with respect to u:
(-2/3) * (2/3)
#*u™3/2) + C2 = (-4/9) * u™(3/2) + C2
Substituting back for u:
=(-4/9) * 3x + 2)(3/2) + C2
Now, putting the two results together:
[(7x - 2N(3x + 2)) dx = (7/2)x"2 - (4/9)(3x + 2)"(3/2) + C
So, the evaluated integral 1s:
(7/2)x"2 - (4/9)(3x + 2)(3/2) + C, where C = C1 + C2 1s the constant of integration.
8. Explain the concept of maximum value function.
The concept of a maximum value function is a fundamental idea in mathematics,
particularly in calculus and optimization. A maximum value function, often referred to as
simply a "max function," describes the behavior of a function and identifies the highest
value it reaches within a specified interval or domain. This concept is crucial in various
fields, including economics, physics, engineering, and more, where finding the optimal or
highest point of a process 1s of great importance.
A maximum value function is formally defined as follows: Let f(x) be a function defined
on a specific interval or domain D. The maximum value of the function on D, denoted as
f(max), is the largest value that the function attains within that interval. Mathematically,
this can be expressed as:
f(max) = max {f(x) | x € D}
In simpler terms, f(max) is the highest point on the graph of the function f(x) within the
given interval D.
Understanding and analyzing maximum value functions are essential in a variety of
contexts:
1. Optimization: In real-world applications, such as economics and engineering,
maximum value functions play a vital role in optimization problems. These
problems involve finding the best possible outcome within certain constraints. For
instance, a company might want to maximize its profit given certain production
limitations. Solving such problems involves identifying the maximum value of a
specific function.
2. Calculus: Calculus 1s a branch of mathematics that deals with rates of change and
accumulation. The concept of maximum value functions is closely linked to the
calculus of derivatives and integrals. Finding the maximum value of a function
often involves determining critical points (where the derivative is zero or
undefined) and evaluating the function at those points.
. Graphical Analysis: Visualizing the graph of a function helps in understanding its
behavior. The maximum value function provides insights into the highest point on
the graph and its location within the given domain. This information is valuable
for making informed decisions based on the function's characteristics.
. Real-World Applications: Maximum value functions are used to model and solve
real-world problems. For example, in physics, they can describe the maximum
height a projectile reaches, while in finance, they can represent the highest stock
price over a given period.
To determine the maximum value of a function, various methods can be employed:
a. Graphical Analysis: Plotting the function and identifying its highest point visually.
b. Analytical Techniques: Calculating critical points and endpoints of the interval, then
evaluating the function at these points to find the maximum value.
¢. Second Derivative Test: If the function is twice differentiable, analyzing the concavity
of the function and using the second derivative test to determine whether a critical point
is a maximum or minimum.
In conclusion, the concept of a maximum value function is a fundamental idea in
mathematics that has widespread applications in various fields. It helps us understand and
optimize processes, make informed decisions, and model real-world scenarios. Analyzing
and finding the maximum value of a function require a combination of graphical,
analytical, and calculus-based approaches, contributing to the rich and versatile toolkit of
mathematical problem-solving.
9. Let the production funetion by Q = AL? K®. Find the clasticity of production with
respect to labour (L).
The elasticity of production with respect to labor (L) measures the responsiveness of the
quantity of output (Q) to a change in the amount of labor input (L), while holding other
inputs constant. Mathematically, the elasticity of production (Ep) with respect to labor is
given by the following formula:
Ep =(2Q/dL) * (L/Q)
In the given production function, Q = A * L"a * K*b, where Q represents the quantity of
output, L is the amount of labor input, A is a constant, K is another input (capital), and a
and b are exponents representing the elasticities of output with respect to labor and
capital, respectively.
Let's find the partial derivative of Q with respect to L:
0Q/cL = O(A * L a * K"b)/0L
=A*a*LMa-1)*K"
Now, we can substitute this value into the clasticity formula:
Ep=(A*a*LMNa-1)*K"b)*(L/(A*L"a*K"b))
Ep=a*LMa-1)*K*
*(L/(L"a* K"b))
Ep=a* (La-1)/L")*
(K"b/K"b)
Simplifying the terms in the parentheses:
Ep=a*(1/L)*1
Finally, the elasticity of production with respect to labor (L) is:
Ep=a/L
In this context, the value of the exponent "a" in the production function determines the
elasticity of production with respect to labor. If "a" is greater than 1, the elasticity is
positive, indicating that an increase in labor input will lead to a proportionally larger
increase in output. If "a" is less than 1, the elasticity is less than 1, indicating that an
increase 1n labor input will lead to a proportionally smaller increase in output. If "a"
cquals 1, the elasticity is equal to 1, implying a constant proportional change in output
with respect to changes in labor input.
It's important to note that the ¢lasticity of production can vary based on the specific
values of the parameters and the production function itself. The above analysis provides a
general framework for understanding how changes 1n labor input affect the quantity of
output in a production process.
10. Denote by a, b and c¢ the column vectors
1
-2
-2
a=|2(,b=|1|,c=]-1
3
-3
1
Calculate 2a - 5b, 2a-5b +¢c,a. b,
19 |
Let's perform the calculations step by step using the given column vectors a, b, and c:
Given vectors:
a=[1,23]"T
b=[-2,1,-3]"T
c=[-2,-1,11"T
Calculate 2a - 5b:
2a-5b=2*[1,2,3]"T-5*[-2,1,-3]"T
=[2, 4, 6]"T +[10, -5, 15]"T
=[12,-1,21]"T
So,2a-5b=[12,-1,21"T.
Calculate 2a - Sb +c¢:
2a-5b+c=[12,-1,21)"T +[-2, -1, 1|*"T
=[10, -2, 22]"T
So,2a-5b+c¢=[10,-2,22]"T.
Calculate the dot product (inner product) of vectors a and b:
a.b=(1*2)+Q2*1)+(3*-3)
=-2+2-9
So, the dot product of vectors a and b, denoted as a . b, is -9.
To summarize:
2a-5b=[12,-1,21"T
2a-5b+c¢=[10,-2, 22]"T
The dot product of vectors aand b, a . b, is -9.