College of Business and Economics, Department of Accounting and Finance
Chapter Two
Matrix Algebra and Its Applications
A matrix: - is a rectangular array of real numbers arranged in m rows & n columns.
- It is enclosed by a bracket and
- It is written in the form of
a11
a12      a1n 

a
In which all the elements in the array are real numbers.
A  
a22      a 2n 
21
a m1
am2      amn 
Each number appearing in the array is said to be an elements or components of the matrix.
Elements of a matrix are subscripted as a ij, to give the row & column location of the element
within the array. The first subscript always refers to the row location of the element; the second
subscript always refers to its column location.
The number of rows (m) & the number of columns (n) of the array give its order of matrices or
its dimension; m x n (reads ““m
m”” by ““nn””)).. The following are examples of matrices
A=
1
5
4
7
3
2
element
this is 3 x 2 matrix
a 12 = 7
a21 = 5
a32= 2
Types of Matrices
There are different types of matrices. These are
1. Vector matrix
2. Square matrix
3. Zero matrix
4. Diagonal Matrix
5. Identity matrix
6. Scalar matrix
1. Vector matrix: –– is a matrix, which consists only one row or one column. It is an m x 1 or
1 x n matrix.

 Row vector: is a 1 x n matrix i.e. a matrix with 1 row
eg.
W = -1
0 6
1x3
 Column vector: is an m x 1 matrix i.e. a matrix with one column only.
eg.
0
Z=
20
5
3x1
2. Square matrix: - It is an n by n matrix a matrix that has the same number of rows &
columns.
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College of Business and Economics, Department of Accounting and Finance
E.g.
1 2 3
A= 2 4 5
7 8 4
3x3
3. Null or zero matrix: - is a matrix that has zero for every entry.
0000
A=
0000
0000
4. Diagonal matrix- It is a square matrix having all zero elements except on its main diagonal.
e.g.
1 0 0
A= 0 4 0
0 0 4 3x3
5. Identity (unit) matrix: - a square matrix with 1 as the element in each position on the main diagonal
(upper left to lower right) and 0 has the element in all other positions and it is denoted by I. It is a scalar
matrix of one or it has one on the main diagonal.
e.g.
1 0 0
A= 0 1 0
0 0 1
3x3
N.B. Each identity matrix is a square matrix
* Primary diagonal represents:
a11, a22, a33, a44 -------- ann entries element
 The most important properties of identity matrix is illustrated by the statements A x I = A &
I x A = A that is, the product of any given matrix & the identity matrix is the given matrix
itself. Thus, the identity matrix behaves in a matrix multiplication like number 1 in an
ordinary arithmetic.
7. Scalar matrix: - is a square matrix where elements on the primary or main diagonal are the
same. ““A
Ann identity matrix is a scalar matrix but a scalar matrix may not be an identity
matrix”.
e.g. 3 0 0 0
A= 0 3 0 0
0 0 3 0
0 0 0 3 4x4
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College of Business and Economics, Department of Accounting and Finance
Operation of Matrix (Addition, Subtraction, Multiplication)
Matrix Addition/ Subtraction
Addition and Subtraction of Matrices
If two matrices have the same number of rows and columns, we can add the matrices by adding
their corresponding entries.
If matrix A and matrix B are of the same order and have elements aij and bij and respectively,
then their sum A + B is a matrix C whose elements Cij = aij +bij are for all i and j. That is,
Cij = aij +bij
for i = 1, 2 --------m
C11 = a11 + b11
C12 = a12 + b12
j = 1, 2 ---------n
C22 = a22 + b22 etc
8
12
eg. 1
3
7
9
––66
=
10
2
4
+ 8
-10
The above matrices
dimension.
––22
eg.
4
are comfortable for addition and subtraction because they have the same
7
6
2
9
+
8
6
7
4
The above matrices are not comfortable for addition and subtraction because they do not have
same dimension. Two matrices of the same dimension are said to be CONFORMABLE FOR
ADDITION AND SUBSTRACTION. Adding/Subtracting corresponding elements from the two
matrices & entering the result in the same raw -column position of a new matrix perform the
addition and subtraction.
Matrix Multiplication
a) Scalar multiplication
It is multiplying each component in the matrix by a constant. The result is a new matrix of the
same dimension as the original matrix.

 If K is any real number & A is an m x n matrix, then the product KA is defined to be the
matrix whose components are given by K times the corresponding component of A; i.e.
KA = Kaij
(m x n)
eg. X = 6 5 7,
then 2X =
2X =
( 2 x 6)
12
(2 x 5)
10
( 2 x 7)
14
b) Matrix by matrix multiplication
If A & B are two matrices, the product AB is defined if and only if the number of columns in A
is equal to the number of rows in B, i.e. if A is an m by n matrix, B should be an n by b matrix.
If this requirement is met A is said to be conformable to B for multiplication. The matrix
resulting from the multiplication has dimension equivalent to the number of rows in A & the
number columns in B.
Lecture Notes on Business Mathematics, Chapter II
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College of Business and Economics, Department of Accounting and Finance
If A is a matrix of dimension n x m (which has m columns) & B is a matrix of dimension p x q
(which has p rows) and if m and p are not the same product A.B is not defined. That is,
multiplication of matrices is possible only if the number of columns of the first equals the
number of rows of the second. If A is of dimension n x m & if B is of dimension m x p, then the
product A.B is of dimension n x p
Given an m by n matrix A and an n by p matrix B, the matrix product AB is an m by p matrix C,
with the ij entry of C given by the formula
cij = ai1b1j + ai2b2j + …
……
……
… + anbn
E.g. ABC Industries must choose a supplier for the raw materials that it uses in its two
manufacturing divisions at Clarion and Brooks. Each division uses different unit amounts of
steel, wood, and plastic as shown in the table below.
Clarion
Brooks
Steel Wood
20
30
22 25
plastic
8
15
The two supply companies being considered, Western and Coastal, can each supply all of these
materials, but at different prices per unit, as described in the following table.
Steel
Wood
Plastic
Western
300
100
145
Coastal
290
90
180
Use matrix multiplication to decide which supplier should be chosen to supply?
(a) The Clarion division, (b) The Brooks division, and (c) Both divisions
Solution
The different amounts of products needed can be placed in matrix A.
Steel wood plastic
A= 20
30
8 clarion
22
25
15 brooks
The prices charged by the suppliers are given in matrix B.
Western coastal
290
steal
300
B= 100
90
wood
180
plastic
145
Lecture Notes on Business Mathematics, Chapter II
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College of Business and Economics, Department of Accounting and Finance
The price from each supplier for each division is found from the product AB.
AB =
20
22
30
25
8
15 x
300
100
145
= 20 x 300 + 30 x 100 + 8 x 145
22 x 300 + 25 x 100 + 15 x 145
=
10,160
290
90
180
20 x 290 + 30 x 90 + 8 x 180
22 x 290 + 25 x 90 x 15 x180
9,940
11,275 11,330
a) The price for the Clarion division for supplies from Western is in the first row, first
column and is $10,160; the price for Clarion for supplies from Coastal is in the first row,
second column and is $9940. Thus the best price for Clarion is from Coastal.
b) The price for the Brooks division for supplies from Western is in the second row, first
column and is $11,275; the price for Brooks for supplies from Coastal is in the second
row, second column and is $11,330. Thus the best price for Brooks is from Western.
c) The sum of the first column is $21,435, the price for supplies from Western for both
divisions; the sum of the second column is $21,270, the price for supplies from Coastal
for both divisions. Thus the best price is from Coastal if one supplier is used for both
divisions.
Note- When the columns and rows of matrix A are interchanged to create a matrix B, and vice
versa, we say that A and B are transposes of each other and write AT = B and BT = A.
Inverse of Matrix
The inverse of a matrix A is commonly denoted by A-1 or inv A.




The inverse of an n x n matrix A is the matrix A-1 such that AA-1 = I = A-1A
A matrix which has an inverse is called nonsingular
A matrix that does not have an inverse is singular.
An inverse of 2 x 2 matrix A exists if ad-bc≠ 0
If the product of A and B is the identity matrix, I, we say that B is the inverse of A and A is the
inverse of B. The matrix B is called the inverse matrix of A, denoted A-1.
We can find the inverse of a matrix by using elementary row operations. These operations are
i.
ii.
Interchange two rows.
Add a multiple of one row to another row.
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College of Business and Economics, Department of Accounting and Finance
iii.
Multiply a row by a nonzero constant.
If A= a b
then A-1= 1
d
-b
this formula works only for2x2 matrix. And ad –bc
Finding the inverse of a square matrix
Example
Procedure
To find the inverse of the square matrix A:
Find the inverse of matrix A = 1 2
1 1
1. Form the augmented matrix [A / I ],
1 2 1 0
where A is the n x n matrix and I is the n x n
1 1 0 1
identity matrix.
R2 = -1xR1 + R2 1 2
1 0
2. Perform elementary row operations on [A
0 -1 -1 1
/ I] until we have an augmented matrix of
R2 = -R2
1 2
1 0
the form [I / B]—that is, until the matrix A
0 1
1 -1
on the left is transformed into the identity
R1 = (-2) R2 + R1
1 0
-1 2
matrix.
0 1
1 -1
The inverse of A is B = -1 2
1 -1
3. The matrix B (on the right) is the inverse
of matrix A.
ad – bc
c d
-c a
must have to be different from zero.
If the inverse exists for a square matrix A, we find A-1 as follows.
Find the inverse of matrix A, below.
2
A=
1
1
5
4
4 3
-3 -2
Solution- To find the inverse of matrix A, we induce A in [A / I] (the matrix A augmented with
the identity matrix I)
2 5 4
1 0 0
1 4 3
0 1 0
1 -3 -2
0 0 1
When the left side becomes an identity matrix, the inverse is in the augment.
Switch R1 and R2
1 4 3
2 5 4
1 -3 -2
0 1 0
1 0 0
0 0 1
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College of Business and Economics, Department of Accounting and Finance
R2 = -2R1 + R2
R3 = -R1 + R3
R1 = -4R2 + R1
0
1
0
1/3
2/3
1
R1 = -1/3R3 + R1
R2 = -2/3R1 + R2
A =
0
1
0
-1/3 2/3 0
0
-1 1
1 0 1/3
0 1
2/3
0 0 -1/7
R3 = 7R2 + R3
-1
0 1 0
1 -2 0
0 -1 1
1 4 3
0 1 2/3
0 -7 -5
R2 = -R2/3
1
0
R3 = -3R3 0
1 4 3
0 -3 -2
0 -7 -5
4/3 -5/3
-1/3 2/3
7
-11
1 0
0 1
0 0
4/3
-1/3
-7/3
-5/3 0
2/3
0
11/3 1
0
0
-3
0
0
1
-1 2 1
-5 8
2
7 -11 -3
-1 2
1
-5 8
2
7 -11 -3
Find the inverse of the matrix
A=
1 2 1
0 1 1
1 2 2
Application of Matrix
I.
Solving Systems of Equations
We can use matrices to solve systems of linear equations by performing the same operations
on the rows of a matrix to reduce it as we do on equations in a linear system.
The three different operations we can use to reduce the matrix are called elementary row
operations.
There are two well recognized methods for solving linear equations:
1) Solution with Inverse Matrices method
Steps 1. Change the system of linear equation into matrix form. The result will be 3
different matrices constructed using coefficient of the variables, unknown values
and right hand side (constant) values
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College of Business and Economics, Department of Accounting and Finance
2. Find the inverse of the coefficient matrix
3. Multiply the inverse of coefficient matrix with the vector of constant, and the
resulting values are the values of the unknown matrix.
eg.
2X + 3Y = 4
X + 2Y = 2
Given this system of linear equation applying
inverse method we can find the unknown values.
Step 1. Change it into matrix form
Using coefficient construct one matrix i.e. coefficient matrix
2
3 = Coefficient matrix
1
2
Using the unknown variables construct unknown matrix & it is a column vector (a matrix
which has one column)
X = vector of unknown
Y
-Using the constant values again construct vector of constant
4
= vector of constant
2
Step 2. Find inverse of the coefficient matrix
Now we are familiar how to find an inverse for any square matrix. Assuming once first method
find the inverse for matrix 2
3
2
1
Its inverse become
2
-1
-3
2
Step 3. Multiply the coefficient inverse with the vector of constant
2
-3
4
=
2
-1
2
2
0
Therefore the resulting matrix that is 2 is;
0
 The value for the unknown variables i.e.
X = 2
Y
0
Then X = 2 and Y = 0 that is unique solution
Under this method the number of equations has to be equal with the number of variables. Let as
consider this example
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College of Business and Economics, Department of Accounting and Finance
2x + 5y + 4z = 4
x + 4y + 3z = 1
x - 3y - 2z = 5
find the value of x, y & z
The first step under this method is to separate the equation.
2 5 4
X
1 4 3 x Y
1 -3 -2
Z
=
4
1
5
Let the first matrix be called A the second X and the last one B. Then
AxX=B this means AX=B, X=B/A or X=1/A. B which is X=A-1B
For the above example X= A-1B
Which is:
A-1 =
Then
X=
X=3
-1 2
1
-5 8
2
and
7 -11 -3 3x3
-1 2
1
-5 8
2 x
7 -11 -3
Y = -2
the solution of A inverse is solved in the above example.
4
B= 1
5 3x1
4
1
5
3
which will give -2
2
Z=2
2) Gauss Jordan method
This method is very general because it is used on circumstances where the numbers of equations
and the number of variables are equal, or different.
It is developed by a mathematician Karl F. Gauss (1777-1855). It helps to solve systems of linear
equations with different solution approaches i.e. unique solution, No solution and infinite
solution cases.
“n” by “n” systems
Step: 1. Change the system of linear equation into a matrix form
2. Augment the coefficient matrix with the vector of constant.
3. Change the coefficient matrix into identity form by applying elementary row
Operation and apply the same on the vector of constant.
1. The resulting values of the vector of constant will be the solution or the value of the
Unknown A/B -------- I/S
N=m
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College of Business and Economics, Department of Accounting and Finance
N<m
where N = number of equations and m = number of variables.
N>m
We can have 3 possible forms of solution.
0 Unique solution
1 An infinite solution
2 No solution
In solving systems with the left-to-right elimination method used in “Solutions of Systems of
Linear Equations,” we operated on the coefficients of the variables x, y, and z and on the
constants. If we keep the coefficients of the variables x, y, and z in distinctive columns, we do not
need to write the equations. In solving a system of linear equations with matrices, we first write
the coefficients and constants from the system (on the left) in the augmented matrix (on the
right).
e.g. 1
x+y=2
2x + 2y = 4
Step one: separate
1 1
2 2
Step two: augment
1 1
2 2
Step three: ERO
x = 2
y
4
2
4
1 1
R2 = -2R1+R2
0 0
e.g.
2x + y = 2
2x + 2y = 3
Step one: separate
1 1
2 2
Step two: augment
1 1
2 2
Step three: ERO
R2 = -2R1+R2
2
0
Thus x + y = 2 and 0(y) = 0 where y have infinite solution
x = 2
y
3
2
3
1 1
0 0
2
-1
thus x + y = 2 and 0(y) = -1 which have no solution for y.
E.g.1
x + 2y + 3z = 6
x
-z = 0
x -y -z= 4
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College of Business and Economics, Department of Accounting and Finance
step 0ne
1 2 3
1 0 -1
1 -1 -1
Step two 1 2 3
1 0 -1
1 -1 -1
X
Y
Z
6
= 0
4
6
0
4
Step three implement ERO then it will give you
1
0
0
0
1
0
0 -0.5
0 4
1 -0.5
this have unique solution X= -0.5 Y= 4
Z= -0.5
Use matrices to solve the system
2x + 5y + 4z = 4
x + 4y + 3z = 1
x - 3y - 2z = 5
first step
2 5 4
1 4 3
1 -3 -2
Second step 2 5 4
1 4 3
1 -3 -2
Third step
X
Y
Z
4
= 1
5
4
1
5
ERO
Gauss-Jordan Elimination Method
Goal (for Each Step)
Row Operation
1. Get a 1 in row 1, column Interchange row 1 and row 2.
1.
2. Add multiples of row 1
only to the other rows to get
zeros in other entries of
column 1.
3. Use rows below row 1 to
get a 1 in row 2, column 2.
4. Add multiples of row 2
only to the other rows to get
zeros as the other entries in
column 2.
5. Use rows below row 2 to
get a
1 in row 3, column 3.
Solution
1 4 3 1
2 5 4 4
1 -3 - 2 5
Add -2 times row 1 to row 2; 1 4 3 1
put the result in row 2. Add -1 0 -3 - 2 2
times row 1 to row 3; put the 0 -7 - 5 4
result in row 3.
1
Divide row 2 with -3 and put 1 4 3
0
1
2/3
-2/3
the result in row 2.
0 -7 -5
4
Add -4 times row 2 to row 1; 1 0 1
11/3
put the result in row 1. Add 7 0 1 2/3 -2/3
times row 2 to row 3; put the 0 0 -1/3 -2/3
result in row 3.
Multiply row 3 by -3; put the 1 0 1/3 11/3
0 1 2/3 -2/3
result in row 3.
0 0 1
2
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AAU, College of Business and Economics, Department of Accounting and Finance
3
6. Add multiples of row 3 Add -1/3 times row 3 to row 1 0 0
0
1
0
-2
only to the other rows to get 1; put the result in row 1. Add
2
zeros as the other entries in -2/3 times row 3 to row 2; put 0 0 1
the result in row 2.
column 3.
7. Repeat the process until it All rows have been used. The
matrix is in reduced form.
cannot be continued
Thus the solution is X= 3 Y= -2 Z= 2 and the solution has unique solution
When a system of linear equations has a unique solution, the coefficient part of the reduced
augmented matrix will be an identity matrix. Note that for this to occur, the coefficient matrix
must be square; that is, the number of equations must equal the number of variables. Recall that
the Gauss-Jordan elimination method (outlined in Example 1) may be used with systems of any
size.
All the systems considered so far had unique solutions, but it is also possible for a system of
linear equations to have an infinite number of solutions or no solution at all.
One of the three things will result:
1. An m by n identifying matrix above m – n bottom rows that are all zeros, giving the
unique solution:
1
0
0
3
3X1 + 2X2 + X3 = 23
0
1
0
-5
X1 + 3X2 + 2X3 = 26
0
0
1
4
2X1 + X2 + 2X3 = 10
3, 5, 4
0
0
0
0
4X1 + 5X2 + 3X3 = 49
2. A row that is m – n bottom raw is all zeros except in the constant column, indicating that there
are no solutions
eg.
1
0
0
0
0
1
0
0
0
0
1
0
3
-5
7
1
2X1 + X2 = 30
X1 + 2X2 = 24
4X1 + 5X2 = 72
3. A matrix in a form diffe rent from (1) & (2), indicating that there are an unlimited
number of solutions
eg.
II.
1
0
0
0
0
1
0
0
2
3
0
0
-4
8
0
0
3X1 + 2X2 + X3 = 6
6X1 + 4X2 + 3X3 = 12
9X1 + 6X2 + 3X3 = 18
15X1 + 10X2 + 5X3 = 30
Markov Chain Analysis
This model is a forecasting model. It is probabilistic (stochastic) model. A Russian
Mathematician called Andrew Markov around 1907 develops this model.
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College of Business and Economics, Department of Accounting and Finance
Markov chains are models, which are useful in studying the evolution of certain systems over
repeated trials. These repeated trials are often successive time periods where the state (outcome
condition) of the systems in any particular time period can’t be determined with certainty.
Therefore, a set of transition probabilities is used to describe the manner in which the system
makes transition from one period to the next. Hence,. We can predict the probabilities of the
system being in a particular state at a given time period. We can also talk about the long run or
equilibrium or steady state.
The necessary assumptions of the chain:
1. The system condition (outcome) state in any given period depends on its state in the
Preceding period & on the transition probabilities
2. The transition probabilities are constant overtime
3. Change in the system will occur once & only once each period
eg. If it’s a week, it’s only once in a week
4. The transition period occurs with regularities
* if we start with days, we use the day until we reach our end.
Information flow in the analysis
The Markov model is based on two sets of input data

 The set of transition probabilities

 The existing or initial or current conditions or states
The Markov process, therefore, describes the movement of a system from a certain state in the
current state/time period to one of n possible states in the next stage. The system makes in an
uncertain environment, all that is known is the probability associated with any possible move or
transition. This probability is known as transition probability, symbolized by Pij. It is the
likelihood that the system which is currently in state i will move to state j in the next period.
From these inputs the model makes two predictions usually expressed as vectors.
1. The probabilities of the system being in any state at any given future time
Period
2. The long run (equilibrium) or steady state probabilities.
This type of analysis is used to predict future proportion. By this it is to mean that many
industries compete in sustaining their market share and are forced to predict their market share
and method of competition.
Example
Currently its known that 80% of customers shop at store 1 & 20% shop at store 2. In reviewing a
past data suppose we find that out of all customer who shopped at store 1 in a given week 90%
remain loyal for the next week (store one again), 10% switch to store 2. On the other hand, out of
all customers who shopped at store 2, in a given week 80% remains loyal for the next week
(store 2 again), 20% switch to store 1. What will be the proportion of customers shopping at
store 1 & 2 in each of the next two weeks?
Solution
Lecture Notes on Business Mathematics, Chapter II
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College of Business and Economics, Department of Accounting and Finance
Let S1 be the proportion of store 1
S2 be the proportion of store 2
Initial state/current state probability matrix for store one and two will be:
V12
=
(0.8
0.2)
To the next weekly shopping period
from one
S1
S2
Transition probability matrix
0.1
week
S1 0.9
is a square matrix such that each
shopping S2 0.2
entry indicates the prob. Of the
0.8
period
system moving from a given state to
another state.
 The sum of rows in the transition matrix should be 1
 We have to be consistent in writing the elements
Markov Chain Formula
a) Vij (n) = Vij (n – 1) x P
where P = Transition matrix
Vij (n) = Vector for period n
Vij (n – 1) = Vector for period n – 1
V12 (0) = (0.8 0.2)
current share
V12 (1) = V12 (1 –1 ) x P
= V12 (0) x P
(0.8 0.2)
0.9
0.1
0.2
0.8
= (0.8 x 0.9) + (0.2 x 0.2)
(0.8 x 0.1) + (0.2 x 0.8)
= 0.72 + 0.04
0.08 + 0.16
= 0.76
0.24
V12 (1) = (0.76
0.24)
V12 (2) = V12 (2 – 1) x P
= V12 (1) x P
(0.76 0.24)
0.9
0.1
0.2
0.8
(0.76x 0.9) + (0.24 x 0.2)
0.76 (0.1) + (0.24 x 0.8)
(0.732
0.268)
b) Long run market share
Assumption
In the log run the share of the systems is assumed to be constant.
Let - the share of store 1 in the long run be V1
- the share of store 2 in the long run be V2
n
p
n+1
(V1 V2)
0.9
0.1
=
(V1 V2)
0.2
0.8
0.9V1 + 0.2V2 = V1
0.1V1 + 0.8V2 = V2
Lecture Notes on Business Mathematics, Chapter II
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College of Business and Economics, Department of Accounting and Finance
-V1 + 0.9V1 + 0.2V2 = 0
0.1V1 + 0.8V2 – V2 = 0
-0.1V1 + 0.2V2 = 0
0.1V1 + (0.2V2) = 0
0.9V1 + 0.2 (1 – V1) = V1
0.9V1 + 0.2 – 0.2V1 = V1
0.7V1 + 0.2 = V1
0.2 = 0.3V1
V1 = 0.2/0.3
V2 = 1 – V1
= 1 – 0.67
V2 = 0.33
In the long run 67% of the customers will shop in store 1 & 33% in shop 2.
For example let us assume that there are 10,000 peoples who use soft drink in A.A city and only
Pepsi and coca are the available soft drinks. The current market share for the city is 20% for
Pepsi and 80% for coca. The management of Pepsi Company planned to increase its market
share with price discount.
The plan shows that 30% 0f users of coca will be forced to use Pepsi and 70% will stay with
coca and 5% users of Pepsi will change to coca where as the others use Pepsi. Required find
Pepsis next week market share and the steady state?
Solution
Pepsi
coca
Pepsi
95%
5%
Coca
30%
70%
Find the next week market share?
Worksheet
1. Mandela has a total of br. 30,000 invested in two accounts that yields of 8% and 10%
interest per year respectively. If the interest Mandela receives from the two accounts in a
year is br. 2640, how much does he have invested in each account? ( solve by using both
inverse and Gaussian methods)
2. The total number of passengers riding a certain city bus during the morning shift is 1000.
If the child’s fare is br. 0.50, the adult fare is br. 1.50, and the total revenue from the fares
in the morning shift is br. 1300, how many children and how many adults rode the bus
during the morning shift? ( solve by using both inverse and Gaussian methods)
3. Two sociologists have grant money to study school busing in a particular city. They wish
to conduct an opinion survey using 600 telephone contacts and 400 house contacts.
Survey Company A has personnel to do 30 telephone and 10 house contacts per hour;
survey company B can handle 20 telephone and 20 house contacts per hour. How many
hours should be scheduled for each firm to produce exactly the number of contacts
needed? ( solve by using both inverse and Gaussian methods)
Lecture Notes on Business Mathematics, Chapter II
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College of Business and Economics, Department of Accounting and Finance
4. Kebede carpet co. has an inventory of 1,500 square yards of wool & 1,800 square yards
of nylon to manufacture carpeting. Two grades of carpeting are produced. Each roll of
superior grade carpeting requires 20 sq. yards of wool & 40sq. yards of nylon. Each roll
of quality-grade carpeting requires 30 square yards of wool & 30 square yard of nylon. If
Kebede would like to use all the material in inventory, how many rolls of superior & how
may rolls of quality carpeting should be manufactured? ( solve by using both inverse and
Gaussian methods)
5. A certain manufacturer produces two product P & q. Each unit of product P requires 20
units of row material A & 10 units of row material B. Each unit of product of requires 30
units of raw material A & 50 units of raw material B. there is a limited supply of 1200
units of raw material A & 950 units of raw material B. How many units of P & Q can be
produced if we want to exhaust the supply of raw materials? ( solve by using both inverse
and Gaussian methods)
6. Attendance records indicate that 80,000 South Koreans attended the 2002 world cup at its
opening ceremony. Total ticket receipts were Birr 3,500,000. Admission prices were Birr
37.5 for the second-class and Birr 62.50 for the first class. Determine the number of
South Koreans who attended the football game at first class and second class. ( solve by
using both inverse and Gaussian methods)
7. Getahun invested a total of br. 10,000 in three different saving accounts. The accounts
paid simple interest at an annual rate of 8%, 9% & 7.5% respectively. Total interest
earned for the year was br. 845. The amount in the 9% account was twice the amount
invested in the 7.5% account. How much did Getahun invest in each account? ( solve by
using both inverse and Gaussian methods)
8. A division of the ministry of public health has conducted a simple survey on the public
attitude towards smoking. From the results of the survey the department concluded that
currently only 20% of the population smokes cigarette & every month 10% of nonsmokers become smokers where as 5% of smokers discontinue smoking. Required:a) Write the current & transition matrices
b) b. What will be the proportion of the non-users (non-smokers) & users
(smokers) in the long run?
9. A vigorous television advertising campaign is conducted during the football reason to
promote a well-known brand X shaving cream. For each of several weeks, a survey is
made & it is found that each week 80% of those using brand X continue to use it & 20%
switch. It is also found that those not using brand X, 20% switch to brand X while the
other 80% continue using another brad.
a) Write the transition matrix, assuming the transition percentage continue to hold
for succeeding weeks.
b) b) If 20% of the people are using brand X at the start of the advertising
campaign, what percentage will be brand X 1week later? Two weeks later?
Lecture Notes on Business Mathematics, Chapter II
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College of Business and Economics, Department of Accounting and Finance
10. Populations of 100,000 consumers make the following purchases during a particular
week: 20,000 purchases Brand A, 35,000 Brand B & 45,000 purchase neither Brand.
From a market study, it is estimated that of those who purchase Brand A, 80% will
purchase it again next week, 15% will purchase brand B next week, & 5% will purchase
neither brand. Of those who purchase B, 85% will purchase it again next week, 12% will
purchase brand A next week, & 3% will purchase neither brand. Of those who purchase
neither brand, 20% will purchase A next week, 15% will purchase Brand B next week, &
65% will purchase neither brand next week. If this purchasing pattern continues what will
be the proportion in the next two weeks and in the long run?
Lecture Notes on Business Mathematics, Chapter II
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