MATHEMATICAL ECONOMICS
Linear Algebra
Cramer’s Rule
Any system of linear equations can be solved
by using Cramer’s Rule
Solutions are expressed as ratios of
determinants.
Example
Y = C*+c(1-t)Y+I*-bR+G*+NX*-mY
Y=(A*-bR)
Y=A*-bR
MS=kY-hr
In Matrix Form
1

Y  bR  A *
kY  hr  MS
1 b  Y   A * 

    

k  h  R   MS 
Cramer’s Rule
For any variable i, it’s value is |A(i)|/|A|
where:
|A| is the determinant of the matrix of
coefficients; and
|A(i)| is the coefficient matrix with the i’th
column replaced with the vector of constants.
Solution
1 b
A
 h  bk
k h
A* b
A(1) 
  A * h  bMS
MS  h
1 A*
A(2) 
 MS  A * k
k MS
Y* 
A * h  bMS
 MS  A * k
R* 
h  bk
h  bk
Expansion by Cofactors
How to calculate determinants for matrices
with dimensions greater than 2  2 using
expansion by cofactors.
Expansion by cofactors works by expanding a
single row or column by its associated
cofactors.
Each matrix is made up of a number of
elements. Each element’s position in the array
is described by the numerical subscripts.
These refer to first, the row and second, the
column.
Example:
1 0 
A

2
1


The element a11 = 1, the element a12 = 0, the
element a21 = 2 and the element a22 = 1.
Any matrix can also be subdivided into smaller
matrices.
These submatrices also have determinants.
The determinant of a submatrix is termed a
cofactor.
Each element in a matrix has an associated
cofactor. The cofactor is the determinant of
the matrix formed by omitting the row and
column of the associated element.
Example
1 0 2 
A   2 1 3 
 1 3  2 .
1
The cofactor C11 = 3
0
C21 = 3
2
3
 2 or -11.
0 2
 2 or -6. C31 = 1 3 or -2.
EXPANSION
The determinant of A can be found by
expanding the first column with its associated
cofactors.
o In fact, any single row or column can be
used.
The formula is:
| A |  (1) i  j aij Cij
In this example it is:
(-1)1+1  -1  (-11) + (-1)2+1  -2  (-6) + (-1)3+1
 1  (-2)
= (11) + (-12) + (-2)
= -3
Multivariate Calculus
A function f(x,y) may be differentiated to
produce f´(x,y).
Setting all partial derivatives equal to zero can
simultaneously generate all optimal solutions.
Example
C  2 E 2  78 E  2 X 2  66 X  2 EX
C
 0  4 X  66  2 E
X
C
 0  4 E  78  2 X
E
A solution can be found by applying
Cramer’s Rule
|A|=12,
|A(1)|= 108, |A(2)|= 180.
X*=9, E*=15
Hessians
A maximum (minimum) can only be
proved with a Hessian matrix.
o A Hessian (H) is a matrix of all the
partial derivatives.
Cross partial derivatives have the same
value (Young’s Theorem) if the
endogenous variables are the same.
Partial Derivatives
 2C
 2C
 4
 4
2
2
X
E
Cross Partial Derivatives
 2C
 2C

 2
XE EX
Hessian Matrix
C XX
H
C XE
C EX    4  2


C EE    2  4
Principle Minor
Each Hessian is made up of principal
minors H1,H2,…,Hn=H.
The principal minors above are:
H1    4
H 1  4
  4  2
H 2   H   

  2  4
H 2  12
Second Order Condition for a maximum
All principal minors alternate in sign, the
first |H1| being negative
Second Order Condition for a minimum
All principal minors are positive
Hence, the solutions above prove the optimal
values are associated with a maximum point.
Implicit Function Theorem
(One Equation)
Consider the function f ( x, y )  0.
Take the total differential of the equation:
f ( x, y )
f ( x, y )
dx 
dy  0
x
y
Then rearrange to get:
dy
f / x

dx
f / y
…or the implicit function theorem.
Example
Ax y   U  U  Ax y   f ( x, y)  0
f ( x, y )
  Ax  1 y 
x
f ( x, y )
  Ax  y  1
y
dy
 Ax  1 y 
 y


  1
dx
 x
 Ax y
Implicit Function Theorem
(Multiple Equations)
Consider the two-equation system:
f ( x1 , x2 ,  )  0
g ( x1 , x2 ,  )  0
Take the total differentials of the two
equations:
f
f
f
dx1 
dx2 
d  0
x1
x2

g
g
g
dx1 
dx2 
d  0
x1
x2

This system can be written in matrix notation:
 f
 x
 g1

 x1
f 
  f d 
x2   dx1    



g  dx2   g

d 

  
x2 
Use Cramer’s rule to solve for dx1,
f
f
d

x2
g
g

d

x2
dx1 

f
f
x1 x2
g g
x1 x2

dx1


d
f

g

f

g

f
x1
g
x1
f
x2
g
x2
J
f
x2
d
g
x2
f
x2
g
x2
Constrained Optimisation
Many economic problems involve the agent
facing constraints.
The most common approach in mathematical
economics to constrained optimisation is the
Lagrangean technique.
The value of the Lagranegean multiplier is the
shadow price of the constraint.
If we change the constraint by 1 unit, the
Lagrangean multiplier tells us how much
the function will change by.
Thus, the Lagrangean multiplier measures
the marginal utility of income in the case
of a consumer choice problem.
Suppose we have the following production
problem.
max   p(2 ln N  4 ln K )
N ,K
s.t.2 N  3K  720
The first function above is the objective. The
second function is the constraint.
We create a single composite function (the
Lagrangean)- which includes a third variable λ
(the Lagrangean multiplier).
The Lagrangean for the above problem
(assuming that p=$1) is:
L  2 ln N  4 ln K   (720  2 N  3K )
Solution- determine the three first order
conditions, eliminate λ and then solve for N
and K.
The first order conditions are:
L 2
 N  2  0
N
L 4
 K  3  0
K
L
 720  2 N  3K  0

To eliminate the Lagrangean multiplier, we
rearrange the first two first-order conditions
and divide the second into the first.
2
4
N
K

2
3
…then
2K 2

4N 3
Cross-multiply to get:
6K  8N
 N  86 K  0.75 K
Substitute
720  2 N  3K
 720  2(0.75 K )  3K
 4.5 K  720
 K   160, N   120
Bordered Hessians
Hessians cannot be used to derive the second
order conditions for problems of constrained
optimisation. The Hessian will in some
iterations, ignore the effect of the constraints.
The solution is to use a Bordered Hessian
Example
L  u ( x, y )   ( M  p x x  p y y )
The First Order conditions are:
L u ( x, y )

 p x  0
x
x
L u ( x, y )

 p y  0
y
y
L
 M  px x  p y y  0

The second derivatives (not conditions) are:
2L
2L
2L
 U xx
 U yy
0
2
2
2
x
y

2L
2L

 U xy
xy yx
2L
2L

  px
x x
2L
2L

  py
y y
The Bordered Hessian.
 U xx

H   U xy
 p x

U xy
U yy
 py
 px 

 py 
0 
Note that in this case, the size (n) of the
Hessian is 2, even though the matrix is a 3×3.
A border-preserving minor of order r is found
by:
Subtracting (n-r) rows and columns from the
matrix; while
These rows/columns are never the border; then
Calculating the sign of the determinant.
A leading border-preserving minor only
calculates the leading minor of order r.
Example
 4 10  1
H  10 2  2
 1  2 0 
This has two principal minors of order 1-
2 2
H1 
or
2 0
4 1
1 0
The first is the leading principal minor.
The Bordered Hessian has one principal minor
of order 2-
4
10
1
H 2  H  10 2  2
1  2 0
The second order conditions for a maximum
(minimum) are that:
The border-preserving principal minor of
order r of the bordered Hessian has the
sign -1r (<0) for r = 2,…,n.
The value of the Hessian above is 22; hence
satisfy the second order conditions for a
maximum.