ECONOMICS MODEL
Linear Models and Matrix Algebra
Mathematics For EconomicsLectures:
Dr. Ahmad Zaki, M.Si.
Baharuddin, S.Pd., M.Pd.
By:
Group 4
Farnesya Putri 220101512005
Rahmawati 220101510006
Muh. Zainal Pratama 220101511004
CLASS CICP MATHEMATICS EDUCATION STUDY PROGRAM MATHEMATICS DEPARTMENT
MAKASSAR STATE UNIVERSITY 2024
FOREWORD
Praise our gratitude to God Almighty, because for the abundance of grace the authors can complete this paper on time without any significant obstacles and in accordance with expectations.
Our gratitude goes to Dr. Ahmad Zaki, M.Si. and Mr. Baharuddin, S.Pd., M.Pd. as lecturers of the Economic Mathematics course who have helped provide direction and understanding in the preparation of this paper.
We realize that in the preparation of this paper there are still many shortcomings due to our limitations. Therefore, the compiler really hopes for criticism and suggestions to improve this paper. Hopefully what is written can be useful for all parties in need.
Makassar, 10 September 2024
Group 4
TABLE OF CONTENTS
Title Page. ...
Foreword …...
Table of Contents ………...
CHAPTER I INTRODUCTION ………...
1.1 Backround...
1.2 Problem Formulation....
...
1.3 Purpose of
Writing... ...
CHAPTER II DISCUSSION ...
2.1 Conditions for Nonsingularity of a Matrix...
2.2 Test of Nonsingularity by Use of Determinant...
2.3 Basic Properties of
Determinants...
2.4 Fiding the Inverse Matrix...
2.5 Cramer’s Rule...
2.6 Application to Market and National-Income Models...
2.7 Leontif Input-Ouput Models...
2.8 Limitations of Static Analysis...
CHAPTER III CONCLUSION ...
3.1 Conclusion ...
3.2 Suggestions...
BIBLIOGRAPHY ...
i ii iii 1 1 1 2
CHAPTER 1 INTRODUCTION 1.1 Backround
Economic mathematics is a very relevant branch of science in the context of a modern economy. Mathematics plays an important role in analyzing complex economic phenomena, developing mathematical models to understand market behavior, predicting economic trends, and managing economic resources to make rational decisions. A very useful tool in mathematical economics are linear models and matrix algebra.
Linear models and matrix algebra are very important foundations in analysis modern economics because it allows us to describe and understand relationships complex between various economic variables. Within the scope of this paper, they will be explored why linear models and matrix algebra have become important tools in mathematics economist
First, linear models allow economists to establish mathematical relationships between various variables in an economic context. In economics, we often faced with many variables such as price, output, demand, and many other factors that influence economic decisions. The linear model allows us to express the relationship between these variables in a simple way, so that facilitates analysis and predictions.
Second, matrix algebra plays an important role in managing economic systems complex. In this case the matrix is used to represent economic data, such as an input-output table that explains the relationship between various economic sectors. With Using matrix algebra, we can calculate the impact of a change in a sector to other sectors, which is very important in economic planning and policy.
In addition, linear models and matrix algebra are also used in optimization methods in economic field. For example, in the problem of allocating limited resources to produce diversified products, linear models
and matrix algebra are used to optimize the allocation of these resources to produce the greatest economic impact.
In this paper, we will discuss the basic concepts of linear models and Matrix algebra and how it is applied in an economic context. Besides, it will several examples of practical applications of linear models and deep matrix algebra are shown economic analysis. By understanding this concept, you will understand how mathematics works economics can be an important basis for making more economic decisions smart and effective.
1.2 Problem Formulation
1. What are the conditions needed for a matrix to have an invers e(nonsingular)?
2. Hw do you calculate the determinant of a matrix, expecally for matrices of order 1, 2 and 3?
3. How does the exchange between rows and columns in a matrix affect the value of its determinant?
4. How to find the inverse matrix?
5. How is Cramer's rule used in solving systems of linear equations?
6. How are linear and matrix models applied to market and national income models?
7. How to apply the leontive input-output model related to life daily?
8. What are the limitations of static analysis?
1.3 Purpose of Writing
1. To understand the necessary and sufficient conditions for a matrix to have an inverse
2. To understand how to calculate the determinant of order 1, 2, and 3 matrices usung the correct method, including the product addition rule for the determinant of a 3×3 matrix.
3. To understand that the determinant value of a matrix does not change when the rows and columns of the matrix are exchanged (transposed) , so
¿A∨¿∨A '∨¿
4. To know how to find the inverse matrix.
5. To find out the use of Cramer's rule in solving the system of linear equations system of linear equations.
6. To To find out the application of linear and matrix models to market models and national income.
7. To find out the application of the leontive input-output model related to everyday life.
8. To find out the limitations of static analysis.
CHAPTER II DISCUSSION 2.1 Conditions for Nonsingularity of a Matrix
A given coefficient matrix A can have an inverse (i.c., can be "nonsingular") only if it is square. As was pointed out earlier, however, the squareness condition is necessary but not sufficient for the existence of the inverse A−1. A matrix can be square, but singular (with- out an inverse) nonetheless.
Necessary versus Sufficient Conditions
The concepts of "necessary condition" and "sufficient condition" are used frequently in economics. It is important that we understand their precise meanings before proceeding further.
A necessary condition is in the nature of a prerequisite: Suppose that a statement p is true only if another statement q is true; then q constitutes a necessary condition of p. Sym- bolically, we express this as follows:
p → q
Which is read as "p only if q," or alternatively, "ilp, then q." It is also logically correct to interpret (5.1) to mean "p implies q." It may happen, of course, that we also have p⇒ w at the same time. Then both q and w are necessary conditions for p.
Example 1
If we let p be the statement "a person is a father" and q be the statement "a person is male," then the logical statement p → q applies. A person' is a father only if he is male, and to be male is a necessary condition for fatherhood, Note,
however, that the converse is not true: fatherhood is not a necessary condition for maleness.
A different type of situation is one in which a statement p is true if q is true, but p can also be true when q is not true. In this case, q is said to be a sufficient condition for p. The truth of a suffices to establish the truth of p, but it is not a necessary condition for p. This case is expressed symbolically by
p ← q
which is read: "p if q" (without the word only)-or alternatively, "if q, then p," as if read- ing
p ← q backward. It can also be interpreted to mean "q implies p."
Example 2
If we let p be the statement "one can get to Europe" and a be the statement "one takes a
plane to Europe, then p ← q Flying can serve to get one to Europe, but since ocean trans- portation is also feasible, is not a prerequisite. We can write p ← q, but not p → q
In a third possible situation, y is both necessary and sufficient for p. In such an event. we write
p ↔ q
which is read: "p if and only ifq" (also written as "p ↔ q"). The double-headed arrow is really a combination of the two types of arrow in p ← q and p → q, hence the joint use of the two terms "if" and "only if." Note that p ↔ q states not only that p implics q but also thatq implies p.
2.2 Test of Nonsingularity by Use of Determinant Determinants and Nonsingularity
The determinant of a square matrix A, denoted by [A], is a uniquely defined scalar (num- ber) associated with that matrix. Determinants are defined only for square matrices. The smallest possible matrix is, of course, the 1×1 matrix A =
[
a11]
Bydefinition, its deter- minant is equal to the single element a11 itself: [A] =
[
a11]
=a11 The symbol
[
a11]
here must not be confused with the look-alike symbol for the absolute value of a number. In the absolute-value context, we have, for instance, not only [5] = 5 but also [−5] = 5 because the absolute value of a number is its numerical value without regard to the algebraic sign. In contrast, the determinant symbol preserves the sign of the element, so while [8] = 8 (a positive number), we have [−8] = −8 (a negative number). This distinction proves to be crucial in the later discussion when we apply determinantal tests whose results depend critically on the signs of determinants of various dimensions, including 1×1 ones, such as[
a11]
= a11For a 2×2 matrix A=
[
aa1121aa1222]
, its determinant is defined to be the sum of two terms of follows:[A]=
[
aa2111aa1222]
=a11a12−a21a22 [¿a scalar]which is obtained by multiplying the two elements in the principal diagonal of A and then subtracting the product of the two remaining elements. In view of the dimension of matrix A, the determinant [A] given in (5.5) is called a second-order determinant.
Example 1
Given A=
[
10 48 5]
and B=[
03 5−1]
, their determinants are [A]=[
10 48 5]
=10(5)−8(4)=18, and[B]=
[
0−13 5]
=3(−1)−0(5)=−3While a determinant (enclosed brackets) is by definition a scalar, a matrix as such does not have a numerical value. In other words, a determinant is reducible to a number, but a matrix is, in contrast, a whole block of numbers. It should also be emphasized that a determinant is defined only for a square matrix, whereas a matrix as such does not have to be square.
Even at this early stage of discussion, it is possible to have an inkling of the relationship berween the linear dependence of the rows in a matrix A, on the one hand, and its determi- nant A, on the other. The two matrices
C=
[
cc1'2']
=[
3 83 8]
=¿D=[
dd1'2']
=[
8 2426]
bothave linearly because c1'=c2
'∧d1
'=d2
'=4d1'their determinants also turn out to be equal to zero:
[C]=
[
3 83 8]
=3(8)−3(8)=0[D]=
[
8 2426]
=2(24)−8(6)=0This result strongly suggests that a "vanishing" determinant (a zero-value determinant) may have something to do with linear dependence. We shall see that this is indeed the case. Furthermore, the value of a determinant ¿A∨¿ can serve not only as a criterion for testing the linear independence of the rows (hence the nonsingularity) of matrix A, but also as an input in the calculation of the inverse
A−1 if it exists.
First, however, we must widen our vista by a discussion of higher-order determinants.
Evaluating a Third-Order Determinant
A determinant of order 3 is associated with a 3 * 3 matrix. Given
[A]=
[
aaa112131aaa122232aaa132333]
its determinant has the value
[A]=
[
aaa112131aaa122232aaa132333]
=a11[
aa2232aa2333]
−a12[
aa2131aa2333]
+a13[
aa2131aa2232]
¿a11a22a33−a11a23a32+a12a23a31−a12a21a33+a13a21a32−a13a22a31[¿a scalar]
Looking first at the lower line of on, we see the value of ¿A∨¿ expressed as a sum of six product terms, three of which are prefixed by minus signs and three by plus signs. Complicated as this sum may appear, there is nonetheless a very easy way of
"catching" all these six terms from a given third-order determinant. This is best explained diagram- matically (Fig. 5.1). In the determinant shown in Fig. 5.1, each element in the top row has been linked with two other elements via two solid arrows as follows: a11→ a22→ a33, a12→ a23→ a31,∧a13→ a32→ a21. Each triplet of elements so linked can be multiplied out, and their product taken as one of the six product terms in (5.6). The solid-arrow product terms are to be prefixed with plus signs.
On the other hand, each top-row element has also been connected with two other cle- ments via two broken arrows as follows:
a11→ a32→ a23, a12→ a21→ a33,∧a13→ a22→ a31.Each triplet of elements so connected can also be multiplied out, and their product taken as one of the six terms in (5.6). Such products are prefixed by minus signs. The sum of all the six products will then be the valuc of the determinant.
Example 2
[
4 5 67 8 9213]
=(2) (5) (9)+(1) (6) (7)+(3) (8) (4)−(2) (8) (6)−(1) (4) (9)−(3) (5) (7)=−92.3 Basic Properties of Determinants
We can now discuss some properties of determinants which will enable us to
"discover" the connection between linear dependence among the rows of a square matrix and the vanishing of the determinant of that matrix.
Five basic properties will be discussed here. These are properties comnion to determinants of all orders, although we shall illustrate mostly with second-order determinants:
Property I The interchange of rows and columns does not affect the value of a determinant. In other words, the determinant of a matrixA has the same value as that of its transposc A ' that is, ¿A∨¿∨A '∨¿
Example 1
[
4 35 6]
=[
4 53 6]
=9Example 2
[
a bc d]
=[
b da c]
=ad−bc2.4 Fiding the Inverse Matrix
If the matrix A in the linear-equation system Ax=d is non singular, then A−1 exists, and the solution of the system will be x=A−1d. We have learned to test the nonsingularity of A by the quote. ¿a∨≠0.The next question is, How can we find the inverse A−1if A docs pass that test?
2.4.1 Expansions of a Determinant by Alien Cofactors
Before answering this question, let's discuss another important property of determinants.
Property I Expansion of the determinant with different cofactors (cofactors of the 'wrong' row or column) always results in zero.
Example :
If we expand the determinant
|
4 1 25 2 11 0 3|
by using the elements of the first rowelements but the cofactors of the second row elements We geta11
|
C21|
+a12|
C22|
+a13|
C23|
=4(−3)+1(10)+2(1)=0More generally, applying the same type of expansion by alien cofactors as
described in Example I to the determinant |A|=
|
aaa112131 aaa122232 aaa132333|
will yield a zero sum of products as follows:The reason lies in the fact that the sum of the product in (5.12) can be thought of as a fixed expansion of the second row of the other determinant
|A|=
|
aaa112131 aaa122232 aaa132333|
which is only different from |A| in its second row while the first row is the same. As an exercise, write down the second row cofactor of¿A∨¿and prove that this is exactly the cofactor of the cofactor arising in (5.12) and with the correct sign . Because |A|= 0,Since the two rows are the same, the expansion with different cofactors as shown in (5.12) must also yield zero.
Property VI applies to determinants of all orders and is used if a determinant is expanded by different cofactors of any row or column. Thus, we can state in general that for a determinant of any order the following holds:
[xpansion with i-th row and cofactor with i-th row']
[expansion with jth row and cofactor with jth row']
Carefully compare (5.13) with (5.8). And finally (the general Laplace expansion), the signs ij of aijand ¿Cij∨¿must be identical in any multiplication of terms in the sum. On the other hand, in expansions by different cofactors as in (5.13), one of the two signs (the chosen value of i' or j') is inevitably out of place.
2.4.2 Matrix Inversion
Property VI, as summarized in (5.13) helps directly in developing a method for matrix inversion, namely in finding the inverse of a matrix. Given a non- singular matrixA , n × n
Since each element of A has a cofactor |Cij|, we can form a cofactor matrix by swapping each element aij in (5.14) with its cofactor |Cij|. Such a cofactor matrix, written as C = [ ¿Cij| ] , must also be n×n. However, for present purposes we are more interested in the transpose of C. The transpose of C' is usually denoted as the adjoint of A and uses the symbol adj A. The adjoint is written in the form
The matrices A and C' are suitable for multiplication, and the product AC' is another n×n matrix where each element is the sum of the products. Using the Laplace expansion formula and the VI property of determinants, the product AC' can be expressed as follows:
Since the determinant |A| is a nonzero scalar, it is permissible to divide both sides of the equation AC' = |A|I by |A|. The result is
Multiply the front part of both sides of the last equation by A−1, and using the result that A−1A=I, we obtain C '
¿A∨¿=A−1¿or A−1= 1
¿A∨¿=adj A¿ [by (5.15]
Now we have found a way to invert the matrix A!
Thus, the general procedure for finding the inverse of a square matrix A will follow these steps: (1) find ¿A∨¿[we proceed to the next step if and only if
¿A∨≠0, since if | A∨¿0, the inverse in (5.16) is undetermined]; (2) Compute the cofactors for all elements of A and arrange them as a matrix C=¿; (3) Use the transpose of C to obtain adj A; and (4) Divide adj A by the determinant
¿A∨¿. The result is the A−1desired inverse.
Example:
Find the inverse of A=
[
3 21 0]
.Since |A|=−2≠0, the inverse A−1exists. The cofactor of each element is in this case a 1x1determinant, which is simply defined as the scalar element of that determinant itself (that is, ¿aij∨≡ aij).Thus, we have
c=¿
Observe the minus signs attached to 1 and 2, as required for cofactors.
Transposing th cofactor matrix yields
adj A=
[
−10 −32]
So the inverse A−1can be written as A=−1
2
[
−10 −32]
=[
012 −2−23]
2.5 Cramer’s Rule
Cramer's rule is generally used to find solutions to systems of linear equations.
2.5.1 Deification of the rule
Given an equation system Ax=d, where Ais n × n, the solution can be written as
x¿=A d= 1
|A|(adj A)d
Where 𝐴 1 is defined as the matrix 𝐴 whose first column is replaced by the elements of the matrix d. So the solution to the system of equations Ax=dcan be expressed as
(5.18)
The result in (5.18) is the statement of Crmer's rule. Note that, when the matrix inversion method yields the solution values f all the endogeneus variables at
once ( x¿is a vector), Cramer's rule can give us the solution value of only a single endogeneus variable at a time ( x¿, is a scalar).
Example:
Find the solution of the equation system
The relevant determinants ¿A∨¿and ¿Aj∨¿are found to be
Thus the solution values of the variables are
Notice in each of these examples we find ¿A∨≠0.This is a necessary condition for the application of Cramers rule, as for the existence of the inverse A−1. 2.5.2 Note on Homogeneous-Equation System
The system of equations 𝐴𝑥 = 𝑑 can have any constant in the vector d.
However, if 𝑑 = 0, namely if 𝑑 1 = 𝑑 2 =. . . = 𝑑𝑛 = 0, then the system of equations becomes 𝐴𝑥 = 0 (vector 0), then the equation This is called a homogeneous vector. If this happens, the result obtained is a trivial result:
x¿
(n ×1)= A−1 (n ×1)
0
(n ×1)= 0 (n ×1)
Alternatively, this result can be derived from Cramer's rule. The fact that 𝑑 = 0 implies that | Aj| for all j, all columns must contain zeros so the answer becomes:
Surprisingly enough, the only way to obtain a non-trivial answer to a homogeneous system of equations is to use | 𝐴 | = 0, that is, to obtain a singular coefficient matrix A. In this case we obtain:
So in a homogeneous system of equations, Cramer's rule cannot be applied.
2.5.3 Solution Outcomes for a Linear-Equation System
As the first possibility. The system may yield a unique, nontrivial solution. This type of outcome can arise only when we have a non homogeneous system with a non singular coefficient matrix. The second possible outcome is unique, trivial solution, and this is
Vector d Determinants
¿A∨¿
¿A∨≠0 (matrix A is nonsingular)
d ≠0
( nonhomogeneous system)
There exists a unique, non-trivial solutionx¿≠0
d=0 ( homogeneous system)
There exists a unique, trivial solutionx¿=0
|A|=0 (singular matrix A) Dependent
Equations
There exists an infinite number of solutions (not including the trivial one)
There exists an infinite number of solutions
(including the trivial one).
Equations inconsistentnt
No solution exists [Not pissible]
2.6 Application to Market and National-Income Models 2.6.1 Market Model
The following is a market model with two commodities.
The three required determinants have the following values:
Therefore, the price equilibrium must be
The equilibrium quantities can be found, as before, by setting P1=P1¿and P2=P2¿in the demand or supply function.
Example:
If it is known that the demand function for a good is P = 15 – Q, and the supply function is P = 3 + 0.5Q. Calculate what equilibrium price and quantity are created in the market?
Solution:
Market equilibrium is created when demand = supply
After obtaining the equilibrium price in the market of Rp. 7, the next step is to enter this value into one of the functions.
Qd=15−p Qs=−6+2p
Qd=15−7 Qs=−6+2(7) Qd=8 Qs=8
So market equilibrium is created at an equilibrium price of 7 and an equilibrium quantity of 8 units.
2.6.2 National Income Model
The simultaneous equations of the national income model are:
This equation can be rearranged into the form
Now the coefficient matrix is in the form of
[
−b1 −11]
and the constant column vector[
I0+aG0]
By Cramer's rule, we obtain:Example:
The consumption function of a country's society is 𝐶 = 100 + 0.8 𝑌𝑑 and investment is 100. Government spending is 250 and taxes collected are 250. The government provides a subsidy of 50, while exports amount to 300 and imports amount to 200. Determine the amount of equilibrium national income?
Solution:
Y=C+I+G+(X−M)
Where :
Y = national income
C = consumption, expressed as C = a+bY I = a government investment
G = government spending
(XM) = net exports (exports minus imports)
Y=100+0,8(Y−250+50)+100+250+(300−200) 0,2Y=390
Y=1950
2.6.3 IS-LM Model: Closed Economy
IS-LM analysis is one of the basic analyses in macroeconomics to see the balance that occurs in the goods market (Investment-Saving/IS) and in the money market (Liquidity Money/LM). Where both show the relationship between national income and interest rates.
The IS curve equation can be formed by equating the investment equation to the savings equation.
For example: (I0 – S0)/s = Yb; and; p/s = b, then the form of the IS equation can be written: Y = f (i) = Yb – bi
The LM equation can be formed by equating the money demand equation (liquidity preference) with the money supply equation.
For example: (M0 - L0)/k = Yu; and; h/k = u, then the form of the LM curve equation can be written as:
Y=f(i)=Yu+ui Example
If it is known that the consumption function of a country is C = 500 + 0.80Y, and the investment function is I = 2000 – 5000i. Then the amount of money in circulation (money supply) is 9,000, and the function of money demand by the community is L = 10,000 + 0.4Y – 20,000i. Make the IS-LM equation and its balance?
Solution:
IS Equation
C=500+0,80Y , maka S=−500+0,20Y I=2000–5000i I=S
2000–5000i=−500+0,20Y 2500–5000i=0,20Y
Y=12.500–25.000i
IS-LM equation IS=LM
12.500–25.000i=−2.500+50.000i
15.000=75.000i i=0,20
After getting i = 0.20, then enter it into one of the IS or LM equations.
IS →Y=12.500–25.000i
Y=12.500–25.000(0,20) Y=7.500
LM →Y=−2.500+50.000i
Y=−2.500+50.000(0,20) Y=7.50 0
2.7 Leontif Input-Ouput Models
In its static version, the input-output analysis of Professor Wassily Leontief, a Nobel Prize winner, poses this particular question: “What level of output should each of the *n* industries in the economy produce in order to satisfy the total demand for that product?” In brief, input-output analysis is not a form of general equilibrium analysis. Although the interdependence of various industries is emphasized, the correct level of output considered is the level that satisfies technical input-output relationships rather than market equilibrium conditions.
Nevertheless, the problem faced by input-output analysis also means solving a system of simultaneous equations, and once again, matrix algebra can be used.
Example
Suppose the economy consists of 3 industries: agriculture, mining, and manufacturing.
To produce one unit of agricultural output, Rp 0.2 of its own output, Rp 0.3 of mining output, and Rp 0.2 of manufacturing output are required.
To produce one unit of mining output, Rp 0.4 of its own output, Rp 0.1 of agricultural output, and Rp 0.2 of manufacturing output are required.
To produce one unit of manufacturing output, Rp 0.1 of its own output, Rp 0.3 of agricultural output, and Rp 0.2 of mining output are required.
The final consumer demand is Rp20.000 for agriculture, Rp10.000 for mining, and Rp40.000 for manufacturing.
Calculate the equilibrium output quantities for the three industrial sectors!
Solution:
In matrix form, we obtain:
A=
[
0.2 0.1 0.3 0.3 0.4 0.20.2 0.2 0.1
]
, d =[
20.00010.00040.000]
𝑋 = (𝐼 − 𝐴) −1𝑑 (𝐼 − 𝐴) =
[
1 0 00 1 00 0 1]
_[
0.2 0.1 0.30.3 0.4 0.2
0.2 0.2 0.1
]
=[
−−0.20.80.3 −0.1−0.20.6 −0.3−0.90.2]
𝑋 = (𝐼 − 𝐴) −1𝑑 X=
[
0.375 1.66671.25 0.2083 0.50.50.25 0.1667 1.1111
] [
20.00010.00040.000]
X=
[
0.25×0.375×1.25×20.000+20.000+20.000+¿0.1667¿¿0.20831.6667××10.000+¿×10.000+¿10.000+1.1111¿0.5×0.5××40.00040.00040.000]
X=
[
25.0007.5005.000 +16.667+++1.6672.083+¿+¿¿20.00044.44420.000]
X=
[
47.08344.16751.111]
Therefore, the agricultural industry should produce an output of Rp. 51,250, the mining industry Rp. 77,500, and the manufacturing industry Rp. 54,444.
2.8 Limitations of Static Analysis
In the discussion of static equilibrium in the market or in the national income, our primary concern has been to find the equilibrium values of the endogenous variables in the model. A fundamental point that was ignored in such an analysis is the actual process of adjustments.and readjustments of the variables ultimately leading to the equilibrium state (if it is at all attainable). We asked only about where we shall arrive but did not question when or what may happen along the way.
The static type of analysis fails, therefore, to take into account two problems of impor- tance. One is that, since the adjustment process may take a long time to complete, an equi- librium state as determined within a particular frame of static analysis may have lost its relevance before it is even attained, if the exogenous forces in the model have undergone some changes in the meantime.
This is the problem of shifts of the equilibrium state. The second is that, even if the adjustment process is allowed to run its course undisturbed, the equilibrium state envisaged in a static analysis may be altogether unattainable. This would be the case of a so-called unstable equilibrium, which is characterized by the fact that the adjustment process will drive the variables further away from, rather than progressively closer to, that equilibrium state. To disregard the adjustment process, therefore, is to as- sume away the problem of attainability of equilibrium.
The shifts of the equilibrium state (in response to exogenous changes) pertain to a type of analysis called comparative statics, and the question of attainability and stability of equi- librium falls within the realm of dynamic analysis. Each of these clearly serves to fill a sig- nificant gap in the static analysis, and it is thus imperative to inquire into those areas of analysis also. We shall leave the study
of dynamic analysis to Part 5 of the book and shall next turn our attention to the problem of comparative statics.
CHAPTER III CLOSING 3.1 Conclusion
The use of linear models and matrix algebra is very important in life everyday life, especially in the economic sector. For this reason, the definition and concept of various things related to linear models and matrix algebra, as well as various the explanation of the formula needs to be known and understood. Matrix algebra gives us a more concise way of writing any system of linear equations, and also complete the determinant criteria to test the existence of one answer. The matrix method can also provide advantages in calculations, such as if its task is to solve problems at the same time several systems an equation that has an identical coefficient matrix A but with a vector different constants. The goal of input output analysis is to determine how many levels output from each industry that must be produced in an economy, in order can meet total demand for products with certainty.
3.2 Suggestion
We realize that the paper we have written is still far from perfect there are
recommend that you always study other learning sources so you can get more understand the application of the matrix for the sake of correct understanding and/or improvement misunderstanding of concepts that we might convey.
BIBLIOGRAPHY
Chiang, K. (2007). [Alpha C. Chiang, Kevin Wainwright] Fundamental Methods of Mathematical Economics.pdf.
Kalangi, J. B. (2012). Economics and Business Mathematics. Jakarta: Salemba Empat Publishers.
Rahayu, Y., & Nurhadiyono, B. (2012). Implementation of Matrices in Business Mathematics and Economics. Techno.COM, Vol. 11, No. 2, 74-81.
Mohammad Nur Rianto Al Arif, M. (2013). Applied Mathematics for Economics.
Bandung: Syarif Hidayatullah State Islamic University Jakarta.