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EC220. Open Economy Macroeconomics
University of Vermont
Department of Economics
System of Equations and Matrix Algebra
This is an aid guide for the math that we will be covering in this particular course. If
you are interested in more details, I strongly recommend the following textbook.
Chiang, Alpha and Kevin, Wainwright (2005) Fundamental Methods of Mathematical
Economics. 4th Edition. McGraw Hill.
In order to solve a system of three linear equations, the basic concepts that you need to
understand and be able to work with are:
a. Linear equations and system of equations
b. Matrix and determinants
c. Cramer’s rule
a. Linear Equations and System of Equations
The general form of a linear equation can be written as:
ay  bx  cz  0
(1)
where y, x, and z are variables and a, b, and c are called coefficients or parameters
because their values are predetermined or given. Variables can be exogenous or
endogenous. An endogenous variable is one whose value(s) is(are) determined within
the system while an exogenous variable is one whose value is determined outside the
system. For instance, in some simple Economic models, exogenous variables might be
the money supply, government spending, and taxes. Output, the price level, and the
real interest are endogenous variables.
A familiar linear equation is the one for a straight line, such as:
y  a  bx
For example:
y  3  2x
(2)
A system of linear equations is a set of two or more linear equations. For example: the
system the demand and supply form. These equations form a system because there are
two equations, there are two unknown variables (Price and Quantity) and each
equation has variables that we assumed are exogenous. Another example is the SavingInvestment model. Saving and investment depend on the real interest rate. The
equilibrium condition requires that Saving=Investment (loanable funds) and the real
interest rate adjust to make them equal. In other words, the two unknown variables are
“the funds” (S=I) and the real interest rate. The exogenous variables are (T-G) and
“exogenous” investment (Io) (investor preferences, income, productivity, investment
tax credits, and so on).
More precisely, a general form of a system of equations can be written as:
ax  by  cz  d
ex  fy  gz  h
(3)
ix  jy  kz  l
For example:
5x  3 y  2 z  0
4y  z  3
(4)
3x  2 y  4 z  7
This is a system of 3 equations and 3 unknown variables: x, y, and z.
b. Matrices and Determinants
Suppose you have two matrices A and B and a column vector V given by:
a

A = e
 i
b
f
j
c
g  ;
k 
m n o p 
B =  q r s t  ;
 u v w  
 
V =  
 
Notice that matrix A has rank 3x3 because it has 3 rows and 3 columns. B’s rank is 3x4
because B is a matrix with 3 rows and 4 columns, and finally V is a column vector with
3 rows and 1 column or 3x1.
Two vectors can be multiplied if and only if the number of columns of a ROW vector
equals the number of rows of a COLUMN vector (both vectors are compatible). By the
same token, two matrices A and B can be multiplied if and only if the number of
columns of A equals the number of rows of B (Both matrices are compatible).
Using matrices A and B and the column vector V, we see that A has 3 columns, B has 3
rows, and V has 3 rows. According to the previous paragraph, multiplication is possible
between matrices A and B and between matrix A and column vector V because A & B
and A & V are compatible. By the same logic, the product between B and V is not
feasible or not compatible.
Let W be a column vector with 3 rows and 1 column or 3x1. Therefore, W3x1 = A3x3V3x1
a

W = e
 i
b
f
j
c
g 
k 
 
 
  = a  b  c e  f  g i  j  k 
 
(5)
As you can see, W has 3 elements: a  b  c , e  f  g , and i  j  k . Each one
of these elements of W is the sum of the products between the corresponding elements
of a row of matrix A and the column vector V.
For example:
1 2 3


A =  4 5 6 ;
7 8 9
12 11 10 9
B =  8 7 6 5 ;
 4 3 2 1
2
V =  3 
 4 
Then, W = AV
1 2 3  2 

 
W = 4 5 6  3  = 2  6  12 8  15  24 14  24  36 = 20 47 74
7 8 9  4 
If we apply the same logic, we can compute the product between A and B. Let C = AB.
Notice that C is going to be 3x4.
a

C = e
 i
b
f
j
c
g 
k 
m n o p   am  bq  cu an  br  cu ao  bs  cw ap  bt  c 
q r s t  


 = em  fq  gu en  fr  gv eo  fs  gw ep  ft  g 
 u v w    im  jq  ku in  jr  kv io  js  kw ip  jt  k 
(6)
For example:
1 2 3 12 11 10 9  12  16  12 11  14  9 10  12  6 9  10  3 


 

C = 4 5 6  8 7 6 5 = 48  40  24 44  35  18 40  30  12 36  25  6
7 8 9  4 3 2 1 84  64  36 77  56  27 70  48  18 63  40  9
 40 34 28 22 


C = 112 97 82 67 
184 160 136 112
A determinant of a matrix |D| is a number associated with a squared matrix and
equals the sum of the products of the elements along the main diagonals minus the sum
of the products of the elements along the secondary diagonals.
Let D be a squared matrix.
a

D = e
 i
b
f
j
c
g 
k 
Then, |D| = (afk + bgi + ejc) – (cfi + bek + jga)
For example:
2 1 3
D = 4 0 1 then |D| = (0 + 2 + 36) – (0 + 6 + 20) = 38 – 26 = 12
2 3 5
|D| = 12
Now, turning the attention to the system of equations and by making use of matrix
multiplication properties, system (3) above can be written in matrix form as:
a
e

 i
b
f
j
c   x  d 
g   y    h 
k   z   l 
(7)
notice that if you multiply the matrix of the coefficients by the column vector of the
variables you get the left hand side of the system above.
Example: consider the system of equations in (4)
5x  3 y  2 z  0
4y  z  3
3x  2 y  4 z  7
Write this system in matrix form:
5  3  2  x  0
0 4
1   y   3

3 2  4  z  7
As you can see, if you multiply the matrix of the coefficients (3x3) by the vector of
unknown variables (3x1) you will get a COLUMN vector (3x1) with three elements.
Each element is the corresponding left hand side of each equation in system (4)
c. Cramer’s Rule
In brief, Cramer’s rule is an approach that allows us to solve a system of linear
equations. It consists of the following steps:
1. Write the system in matrix form
a
e

 i
b
f
j
c   x  d 
g   y    h 
k   z   l 
2. Compute the determinant for the unknown variables coefficient matrix.
|D| = (afk + bgi + ejc) – (cfi + bek + jga)
3. Substitute the left hand side column of the coefficient matrix with the right hand
side vector of values on the column of the coefficient matrix that correspond to
the order of the variable in the vector of unknown variables. For example, if you
want to find x, then the substitution is for the first column. For y would be the
second column, and so on.
For x:
d

Dx =  h
 l
For y:
b
f
j
c
g 
k 
Dy =
For z:
a d
e h

 i l
c
g 
k 
a

Dz =  e
 i
b
f
j
d
h 
l 
4. Compute the determinants of Dx, Dy, and Dz.
|Dx| = (dfk + bgl + hjc) – (cfl + jgd + hbk)
|Dy| = (ahk + dgi + elc) – (chi + edk + lga)
|Dz| = (afl + bhi + ejd) – (dfi + jha + ebl)
5. Find the value of each variable: divide each variable associated determinant
(obtained in 4) by the determinant of the coefficient matrix |D|.
x
| Dx |
|D|
y
| Dy |
|D|
z
| Dz |
|D|
Example:
Let’s follow the instructions step by step:
1. Write the system of equations in matrix form:
5  3  2  x  0
0 4
1   y   3

3 2  4  z  7
2. Compute the determinant of the coefficient matrix
|D| = (–80 – 9 + 0) – (–24 + 10 + 0) = – 89 + 24 – 10 = -75
3. Substitute the left hand side column of the coefficient matrix with the right
hand side vector of values on the column of the coefficient matrix that
correspond to the order of the variable in vector of unknown variables.
For x:
For y:
For z:
0  3  2

1 
Dx =  3 4
7 2  4
5 0  2 
0 3 1 
Dy = 

3 7  4
5  3 0 

4 3
Dz = 0
3 2 7
4. Compute their determinants
|Dx| = (0 – 21 – 12) – (– 56 + 36 + 0) = – 33 +20 = -13
|Dy| = (-60 + 0 + 0) – (–18 + 35 + 0) = -77
|Dz| = 140 – 27 + 0 – (0 + 0 +30) = 83
5.
Find the value of each variable.
x = -13/-75 = 0.1733;
y = -77/-75 = 1.0267;
z = 83/-75 = -1.1067