Tanzania Institute of Accountancy.

Module Title: Business Mathematics and Statistics

Code: GSU 07201

BAC I & BPLM I

3 x 3 Matrices
What is a matrix?

A matrix is a rectangular pattern of collection of numbers arranged into a fixed

number of rows and columns. We usually enclose the numbers with brackets.

So, for example, the following are all matrices.

( )
5 8 1
2

( ) ( )
a b 12 3 −4 0 3
4
c d ( 2 0 9 −1 ) 0 7 2 11 0 −6

Note that in each case we have a rectangular pattern of numbers. These

numbers can be any numbers we choose - positive, negative, zero, fractions,
decimals, and so on. To refer briefly to a specific matrix we might label it,
usually with a capital letter, so we might write.

A=
( ac bd ) R=( 2 0 9 −1 ) (
Q= 12 3 −4
0 7 2 )
( )
5 0 0
2
B= 0 3
0
0 0 −6

Clearly, all these matrices have different sizes. When we want to refer to the size of
a matrix we state its number of rows and number of columns, in that order. Matrix A
has two rows and two columns; we write that it is a 2 × 2 matrix and say that it is a
‘two by two’ matrix.

Similarly we observe R is 1 × 4, Q is 3 × 2 and B is 3 × 3.

Each number in a matrix is referred to as an element of the matrix.

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Some special types of matrix.

Some types of matrix occur quite frequently, have special properties or are particu-
larly important. We give these matrices special names.

A square matrix, as the name suggests, has the same number of rows as col-
umns. So the matrices A and B above are square.

A diagonal matrix is a square matrix with zeros everywhere except possibly on

the diagonal which runs from the top left to the bottom right. This diagonal is called
the leading diagonal. Matrix B is a diagonal matrix.

An identity matrix, sometimes called a unit matrix, is a diagonal matrix with all its
diagonal elements equal to 1. The following are identity matrices.

( )
1 0 0
I 3= 0 1 0
( )
I 2= 1 0
0 1 0 0 1
The symbol I is usually reserved for labelling identity matrices.

Identity matrices have a special and very important property. We shall see in a later
leaflet, when we consider multiplication of matrices, that multiplying a matrix by an
identity matrix, leaves that matrix unchanged.

Zero Matrix.

The matrix is said to a Zero Matrix if all its elements are zero.

( )
0 0 0

( 00 00 )
Z= 0 0 0
Z=
0 0 0
This matrix has zero property items of matrix operations.

An equal Matrixes.

Two Matrixes are said to be equal if and only if corresponding elements in the matri-
ces are equal.

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Operations of a 3 × 3 matrix
Addition, subtraction and scalar multiplication of matrices

This leaflet will look at the condition necessary to be able to add or subtract two
matrices, and when this condition is satisfied, how to do this. It will also cover how
to multiply a matrix by a number.

Compatible matrices.

Two matrices are said to be compatible when they have the same size, that is, the
same number of rows and the same number of columns. When two matrices are
compatible they can be added (or subtracted).

B=¿¿ ( )
2 8 4
C=¿ 5 −6 0 ¿ ¿ ¿¿
A= (
4 −1
3 6 ) ¿

( )
1 7
D= −5 3
8 0

First consider the size of each of these matrices:

A is a 2 × 2 matrix, B is 3 × 2, C is 3 × 3 and D is 3 × 2.

From these four matrices only B and D are compatible. This means we can calculate
B+D, B−D, D − B, but we cannot add or subtract any other pair of these matrices.

Adding and subtracting matrices.

When two matrices are compatible we add (or subtract) them by adding
(or subtracting) the elements in corresponding positions.
We have seen that B and D have the same size; they are both 3 × 2 matrices. This
means we can add them by adding elements in the corresponding (that is, the
same) positions:

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( )( ) ( )( )
4 6 1 7 4 +1 6+7 5 13
B+D= 1 0 + −5 3 = 1+ (−5 ) 0+3 = −4 3
−2 3 8 0 −2+8 3+0 6 3

Note that the answer we get is another matrix of the same size as the ones we
started with.
Similarly, we can find B − D by subtracting elements in the corresponding positions.

( )( )( )( )
4 6 1 7 4−1 6−7 3 −1
B−D= 1 0 − −5 3 = 1− (−5 ) 0−3 = 6 −3
−2 3 8 0 −2−8 3−0 −10 3

Since any matrix is the same size as itself we can always add a matrix to itself. If we
do this for matrix A we find

( )( )(
A+ A= 4 −1 + 4 −1 = 4 +4 −1+−1 = 8 −2
3 6 3 6 3+3 6+6 6 12 )( )
Another way to write A + A is 2A. Therefore we have

(
A+ A=2 A=2 4 −1 = 8 −2
3 6 6 12 )( )
We see that the elements of 2A are each twice the elements of A. This illustrates
how to multiply a matrix by a number, and leads us to the topic of scalar multipli-
cation:

Scalar multiplication
When working with matrices there are two kinds of multiplication: scalar multiplica-
tion and matrix multiplication. Scalar multiplication is where a matrix is multiplied
by a single number. Matrix multiplication is where a matrix is multiplied by another
matrix.

To multiply a matrix by a scalar (that is, a single number), we simply multi-

ply each element in the matrix by this number.

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( )( )( )
1 7 3×1 3×7 3 21
3 D=3 −5 3 = 3×−5 3×3 = −15 9
8 0 3×8 3×0 24 0

−2 C=−2 (25 = )(
8 4 −2×2 −2×8 −2×4
=
−4 −16 −8
−6 0 −2×5 −2×−6 −2×0 −10 12 0 )( )

( )( )
1 1 1
×4 ×−1 2 −
1
2
A=
(
1 4
2 3
−1
6
=
)
2
1
×3
2
1
×6
=
3
3
2

2 2 2

Multiplication
One of the most important operations carried out with matrices is matrix multiplica-
tion. At first sight this is done in a rather strange way. The reason for this only be-
comes apparent when matrices are used to solve equations.

So, if the first matrix has size p × q, that is, it has p rows and q columns,
and the second has size r × s, that is, it has r rows and s columns, we can
only multiply them together if q = r. When this is so, the result of multiply-
ing them together is a p × s matrix.

Two matrices can only ever be multiplied together if the number of columns
in the first is the same as the number of rows in the second.

To multiply A and B,

No. Columns in A = No. Rows in B

Then A ¿ B = C
(1 ¿ 3) (3 ¿ 2) = (1 ¿ 2)

( ) ( )
a b c r s t
A= d e f B= u v w
g h i x y z

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( )( )( )
a b c r s t ar+bu+cx as+bv +cy at+bw+cz
C= d e f u v w = dr+eu +fx ds+ev+fy dt +ew +fz
g h i x y z gr+hu +ix gs+hv +iy gt+hw +iz

Example: Find

( )( )
1 2 −1 2 −1 3
AB= 3 4 0 1 −2 1
1 5 −2 0 3 −2

Matrix transpose.
Transpose of m × n matrix A, denoted AT is n × m matrix.

( )(
0 4 T
7 0 =
3 1
0 7 3
4 0 1 )
Example: . Transpose converts row vectors to column vec-
tors, vice versa

DISCUSION QUESTIONS

Problem 1

Suppose Fatima has two factories at places A and B. Each factory produces sport
shoes for boys and girls in three different price categories labelled 1, 2 and 3. The
quantities produced by each factory are represented as matrices given below:

Factory A Factory B
Boys Girls Boys Girls

[ ] [ ]
1 80 60 1 90 50
2 75 65 2 70 55
3 90 85 3 75 75

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Suppose Fatima wants to know the total production of sport shoes in each price cat-
egory.

Problem 2

a) Suppose Neema and Lucy are two friends. Neema wants to buy 2 pens and 5
story books, while Lucy needs 8 pens and 10 story books. They both go to a
shop to enquire about the rates which are quoted as follows: Pen – Tshs 5
each, story book – Tshs 50 each. How much money does each need to spend?
b) Suppose that they enquire about the rates from another shop, quoted as fol -
lows: pen – Tshs 4 each, story book – Tshs 40 each. How much money does
each need to spend in each shop?

Problem 3

In a legislative assembly election, a political group hired a public relations firm to

promote its candidate in three ways: telephone, house calls, and letters. The cost
per contact (in Tshs) is given in matrix A as

The number of contacts of each type made in two cities X and Y is given by

Find the total amount spent by the group in the two cities X and Y.

Problem 4

A manufacturer produces three products x, y, z which he sells in two markets. An-

nual sales are indicated below:

(a) If unit sale prices of x, y and z are Tshs 2.50, Tshs 1.50 and Tshs 1.00, respec -
tively, find the total revenue in each market with the help of matrix algebra.

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(b) If the unit costs of the above three commodities are Tshs 2.00, Tshs 1.00 and
Tshs 50 respectively. Find the gross profit.

Problem 5

Find the values of x and y from the following equation:

2
[ 7x 5
+
][
3 −4
y−3 1 2
=
][
7 6
15 14 ]

Problem 6

Two farmers Mabula and Masanja cultivates only three varieties of rice namely
Maize, Cassava and Beans. The sale (in Tshs) of these varieties of crops by both the
farmers in the month of September and October are given by the following matrices
A and B.

September sales ( in Tshs)

Maize Cassava Beans

A= [
10 , 000 20 , 000 30 , 000 ⇒Mabula
50 , 000 30 , 000 10 , 000 ⇒Masanja ]
October sales (in Tshs)

Maize Cassava Beans

B= [
5,000 10 , 000 6 , 000 ⇒ Mabula
20 ,000 10 , 000 10 ,000 ⇒ Masanja]
i). Find the combined sales in September and October for each farmer in each
variety.
ii). Find the decrease in sales from September to October.
iii).If both farmers receive 2% profit on gross sales, compute the profit for each
farmer and for each variety sold in October.

Problem 7
A manufacturer produces three products x, y, z which he sells in two markets.
Annual sales are indicated below:
Market Products
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I 10,000 2,000 18,000

II 6,000 20,000 8,000
(a) If unit sale prices of x, y and z are Tshs 4.00 , Tshs 3.00 and Tshs 2.00, respec-
tively, find the total revenue in each market with the help of matrix algebra.
(b) If the unit costs of the above three commodities are Tshs 3.00, Tshs 2.00 and
Tshs 1.00 respectively. Find the gross profit.

The determinant of a 3 × 3 matrix

Notation

Given the matrix A, its determinant is abbreviated as det (A) or |A|

The determinant of the matrix can be used to solve systems of equations, but first
we need to discuss how to find the determinant of a matrix. Here we will be learning
how to find the determinant of a 3×3 matrix.
Note: Determinant can be evaluated by using any row or any column

Cofactor method
Example : Find the determinant of the 3×3 matrix below.

[ ]
2 -3 1
A= 2 0 -1
1 4 5
The set-up below will help you find the correspondence between the generic ele-
ments of the formula and the elements of the actual problem.
Applying the formula,

[ ]
a b c
det d e f =a⋅det
g h i
e f
h i [ ]
−b⋅det
d f
g i
+c⋅det[ ]
d e
g h [ ]

[ ]
2 −3 1
det 2 0 −1 =2×det
1 4 5
4 5[ ]
0 -1
−(−3 ) ×det [ ]
2 -1
1 5 [ ]
+1×det
2 0
1 4
=2 [ 0−(−4 ) ] +3 [ 10−(−1 ) ] +1 [ 8−0 ]
=2 ( 0−4 ) +3 ( 10+1 ) +1 ( 8 )

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=2 ( 4 ) +3 ( 11 ) +8
=8+33+8
=49

The presence of zero (0) in the first row should make our computation much easier.
Remember, those elements in the first row, act as scalar multipliers. Therefore, zero
multiplied to anything will result in the entire expression to disappear.

Here’s the setup again to show the corresponding numerical value of each variable
in the formula.

[ ]
-5 0 −1
det 1 2 −1 =−5×det
-3 4 1
2 -1
4 1 [ ]
−( 0 )×det
1 -1
-3 1 [ ]
+ (−1 )×det
1 2
-3 4 [ ]
=−5 [ 2−(−4 ) ]−0 [ 1− (3 ) ] −1 [ 4−(−6 ) ]
=−5 ( 2+4 )−0−1 ( 4+6 )
=−5 ( 6 )−1 ( 10 )
=−30−10
=−40

Diagonals method
Example : Find the determinant of
−3 3 2
| 5 4 −1|
2 1 4
Step 1: Rewrite the first two columns of the matrix
−3 3 2 −3 3 2 −3 3
| 5 4 −1|→| 5 4 −1| 5 4
2 1 4 2 1 4 2 1
Step 2: Multiply diagonally downward and diagonally upward.

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Step 3: Add the downward numbers together.

−48+ (−6 ) +10=−44
Step 4: Add the upward numbers together
16+3+ 60=79
Step 5: Subtract the upward sum from the downward sum to get the
determinant.
−44−79=−123

The determinant of matrix is –123.

Matrix inversion of a 3 × 3 matrix

To find inverse of a matrix, it is necessary first to check for singularity of the
matrix, i.e. to see whether determinant of that matrix is zero;
i) If yes, then inverse does not be exists.
ii) If no, then proceed as follows.

Minors
Each element in a square matrix has its own minor. The minor is the value of the
determinant of the matrix that results from crossing out the row and column of the
element under consideration.

For the element 7 in matrix A, since this element is in the first row and first column,

we cross out the first row and column of A to leave the 2 × 2 matrix
( 34 −1
−2 ) .
We now evaluate the determinant of this matrix:

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3 −1
| |=( 3×−2 )−( 4×−1 )=(−6 )−(−4 )=−6+4=−2
4 −2
The minor of the element 7 is thus −2.

For the element 4, since this element is in the third row and second column, we

cross out the third row and second column of A to leave the 2 × 2 matrix
( 7 1
0 −1 )
. We now evaluate the determinant of this matrix:
7 1
| |=( 7×−1 )−( 0×1 )=(−7 )−( 0 )=−7+0=−7
0 −1
The minor of the element 4 is thus −7.

We could continue in this way to find the minor of every element.

Place signs
Each element in a square matrix has a place sign. The place sign of the element in
the first row and first column is always ‘+’. The place signs then alternate from ‘+’
to ‘−’ as we move across the rows and down the columns. So, for a 3 × 3 matrix the
place signs are:

( )
+ − +
− + −
+ − +
Cofactors
Each element in a square matrix has its own cofactor. The cofactor is the product
of the element’s place sign and minor.

The element 7 in matrix A has place sign + and minor −2 so its cofactor is +(−2) =
−2.
The element 4 in matrix A has place sign − and minor −7 so its cofactor is −(−7) =
7.
Proceeding in this way we can find all the cofactors.
The original matrix, its matrix of minors and its matrix of cofactors are:

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( ) ( ) ( )
7 2 1 −2 −3 9 −2 3 9
A= 0 3 −1 Minor= −8 −11 34 Cofactor= 8 −11 −34
−3 4 −2 −5 −7 21 −5 7 21

Adjoint
The adjoint and inverse of a matrix
Before you work through this leaflet, you will need to know how to find the determi-
nant and cofactors of a 3 × 3 matrix. If necessary you should refer to previous
leaflets in this series which cover these topics.
Here is the matrix A that we saw in the leaflet on finding cofactors and determi-
nants. Alongside, we have assembled the matrix of cofactors of A.

( )
−2 3 9
Cofactor= 8 −11 −34
−5 7 21
In order to find the inverse of A, we first need to use the matrix of cofactors, C, to
create the adjoint of matrix A. The adjoint of A, denoted adj(A), is the transpose
of the matrix of cofactors:

adj( A ) = CT
Remember that to find the transpose, the rows and columns are interchanged, so
that

( )
−2 8 −5
T
adj( A ) = C = 3 −11 7
9 −34 21
Then the formula for the inverse matrix is

Given a matrix A, its inverse is given by

-1 1
A = adj( A )
det( A )

where det(A) is the determinant of A, and adj(A) is the ad-

joint of A.

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The inverse has the special property that AA−1 = A−1 A = I (an identity matrix).

( )
0 −3 −2
A= 1 −4 −2
EXAMPLE: Find the inverse of the matrix
-3 4 1
Step 1: check if the matrix A is not a singular matrix, i.e. its determinant ¿
0 which is a condition for inverse to exist.

0 −3 −2
|A|=| 1 −4 −2|=0×det
-3 4 1
(
-4 -2
4 1 )
+3×det
1 -2
-3 1 (
−2×det )
1 −4
-3 4
=1 ( )
Since det(A) ¿ 0, then A-1 exist

Step 2: Replace every entry by its minor

Given an entry in a 3 by 3 matrix, cross out its entire row and column, and take the
determinant of the 2 by 2 matrix that remains (this is called the minor).

In our example, this gives us

[ ][ ]
(−4 )×1−(−2 )×4 1×1−(−2 )× (−3 ) 1×4−(−4 )×(−3 ) 4 −5 −8
M= (−3 )×1−(−2 )×4 0×1−(−2 ) ×(−3 ) 0×4−(−3 )×(−3 ) = 5 −6 −9
(−3 ) ×(−2 )−(−2 )×(−4 ) 0×(−2 )−(−2 )×1 0×(−4 )− (−3 )×1 −2 2 3

Step 3: Change some of the signs

We now change the signs of some of the minors, according to the pattern

( )
+ − +
− + −
+ − +
Thus creating what's called the matrix of cofactors. In our case, this is

( )
4 5 −8
C= -5 −6 9
-2 −2 3

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Step 4: Transpose

We now transpose the matrix of cofactors. In our case, we get

( )
4 −5 −2
t
C = 5 −6 −2
-8 9 3
Step 5: Adjoint

( )
4 −5 −2
T
adj( A ) = C = 5 −6 −2
-8 9 3

Step 6: Divide by the determinant

Finally, we divide by the determinant of the original matrix. In our case, the deter-
minant is det(A) = 1

Given a matrix A, its inverse is given by

-1 1
A = adj( A )
det( A )

So the inverse is simply

( )
4 −5 −2
-1
A = 5 −6 −2
-8 9 3
You should verify this is correct by showing that AA−1 = A−1 A = I, the 3 × 3 identity
matrix.

Solving a set of simultaneous equations

We now show how the inverse is used to solve the simultaneous equations:

7 x+2 y−z=21
3 y−z=5
−3 x+4 y−2 z=−1
In matrix form these equations can be written

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( )( ) ( )
7 2 1 x 21
0 3 −1 y = 5
−3 4 −2 z −1
Recall that when AX = B, then X = A−1B so

()( )( ) ( )( )
x −2 8 −5 21 −42+40+5 3
y = 3 −11 7 5 = 63−55−7 = 1
z 9 −34 21 −1 189−170−21 −2
So x = 3, y = 1 and z = −2.
These values should be checked by substituting them back into the original equa-
tions.

Finally, note that if the determinant of the coefficient matrix A is zero, then it will be
impossible to find the inverse of A, and this method will not be applicable.

Cramer’s Rule
Cramer’s rule is a method for solving linear simultaneous equations. It makes use of
determinants and so knowledge of these is necessary before proceeding.

Cramer’s rule for three equations

For the case of three equations in three unknowns: If
a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3

|a1 b1 c1 ¿| a2 b2 c2 ¿|¿¿¿
¿
then x, y and z can be found from

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|d1 b1 c1¿|d2 b2 c2 |¿ |a1 d1 c1¿|a2 d2 c2 |¿ |a1 b1 d1¿|a2 b2 d2 |¿

x= ¿ ¿ y= ¿ ¿ z= ¿ ¿
Example:
¿ ¿ ¿
A company produces three products every day. Their total production on a certain
day is 45 tons. It is found that the production of the third product exceeds
the production of the first product by 8 tons while the total combined production of
the first and third product is twice that of the second product. Determine
the production level of each product using Cramer’s rule.

Solution
Let the production level of the three products be x, y and z respectively.
Therefore, we will have the following equations

x + y +z=45
−x +z=8
x−2 y +z=0
Therefore, we have, using (1), (2) and (3)

( )( ) ( )
1 1 1 x 45
−1 0 1 y = 8
1 −2 1 z 0
Which gives us
1 1 1
Δ=|−1 0 1|=6
1 −2 1
Since ∆ ≠ 0, there is a solution.

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45 1 1 1 45 1
Δ x =| 8 0 1|=66 Δ y =|−1 8 1|=90
0 −2 1 1 0 1
1 1 45
Δz =|−1 0 8 |=114
1 −2 0
Therefore,
Δx 66 Δy 90
x= = =11 y= = y= =15
Δ ,6 Δ 6 and
Δ 114
z= z =z= =19
Δ 6
Hence, the production levels of the products are as follows:
First product - 11 tons
Second product - 15 tons
Third product- 19 tons

Example.
A salesman has the following record of sales during three months for three items A,
B and C which have different rates of commission.

Months Sales of Units Total commission drawn in Tsh

A B C
January 90 100 20 800
February 130 50 40 900
March 60 100 30 850

Find out the rates of commission on the items A ,B and C by using Cramer’s rule.

Solution.
Let x, y and z be the rates of commission in Rs. per unit for A, B and C items re -
spectively.

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Hence the rate of commision for A, B and C are Tsh2, Tsh 4 and Tsh 11 respectively.

Example:
A n a u t o m o b i l e c o m p a n y u s e s t h r e e t y p e s o f s t e e l s1, s2 and s3 for pro-
ducing three types of cars c 1, c2 and c3. The steel requirement (in tons) for each
type of car is given below:

CARS
C1 C2 C3
S1 2 3 4
Steel S2 1 1 2
S3 3 2 1

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Determine the number of cars of each type which can be produced using
29, 13 and 16 tons of steel of the three types respectively

Solution
Let x, y and z denote the number of cars that can be produced of each type. Then
we have

2 x+3 y+4 z=29

x+ y+2 z=13
3 x+2 y+z=16
The above information can be represented using the matrix method, as under.

[ ][ ] [ ]
2 3 4 x 29
1 1 2 y = 13
3 2 1 z 16

The above equation can be solved using inverse method.

Therefore, z = 4. Substituting y = 3 and z = 4 in (1), we get x = 2. Hence the solu-
tion is x = 2, y = 3 and z = 4

Example:
The following system of simultaneous equations represent the profit and loss of a
certain Oil Company of in one the smallest towns in Tanzania for the business of
petrol (x), diesel (y) and kerosene (z) in (Tsh.“000,000”) per day.

REQUIRED
Use Cramer's rule to find loss or profit for this company as follows:
i). Loss or profit of each individual type of oil (i.e. petrol, diesel, and kerosene)
per day
ii). Total business profit/(loss) per day
Solution:

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( )( ) ( )
3 1 1 x 3
2 2 5 y = −1
Matrix form
1 −3 −4 z 2

3 1 1
Δ=|2 2 5 |=3(-8 + 15) - 1(-8 - 5 ) + 1(-6 - 2)= 21 + 13 - 8 = 26
1 −3 −4
Here Determinant is not equal to zero, Hence the Cramer's rule can be applicable.
Δx Δy Δz
x= y= z=
Δ , Δ and Δ

3 1 1
Δx =|−1 2 5 |=3( -8 + 15) - 1(4 - 10) + 1(3 - 4 ) = 21 + 6 - 1 = 26
2 −3 −4

3 3 1
Δ y =|2 −1 5 |=3(4 - 10) - 3(-8 - 5 ) + 1(4 + 1) = -18 + 39 + 5 = 26
1 2 −4

3 1 3
Δz =|2 2 −1 |=3(4 - 3) - 1(4 + 1) + 3(-6 -2) = 3 - 5 - 24 = -26
1 −3 2

By using Cramer's rule:

Δx 26 Δy 26
x= = =1 y= = y= =1
Δ 26 Δ 26
, and
Δ 26
z= z =z=− =−1
Δ 26

So the Unknown variables are x = 1, y = 1 & z = -1 (In millions of money)

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Therefore
i). x = Sh.1,000,000
y = Sh.1,000,000
z = (Sh.1,000,000)

ii). Total profit = Sh.1,000,000 + Sh.1,000,000 - Sh.1,000,000

= Sh.1,000,000

DISCUSSION QUESTIONS –APPLICATION FOR CRAMER’S RULE

QUESTION 1:

a) Use matrix to solve the following system of equations:

x - y + z = 19

-3x + 2y - 3z = -6

2x – 5y + 4z = 5

b) Three chemicals are combined to form three grades of fertilizer. One unit of
grade one fertilizer requires 10kg. of chemical A,30 of B, and 60 of C. One
unit of grade two requires 20kg of A, 30 of B, and 50 of C. One unit of grade
three requires 50kg. of A, and 50 of C. If 1600kg. of A, 1200 of B, and 3200 of
C are available, how many units of three grades should be produced to use all
available supplies?

QUESTION 2:
a) What is meant by these terms
i) Matrix?
ii) Singular matrix?
iii) Transpose of matrix?

b) If the arrangement of items in one of super-markets at Mlimani City in Dar-es-

Salaam is in number and matrix form as.
i) Find determinant
ii) Find the inverse of the matrix

QUESTION: 3
The Mselebwende Rent Truck Company (MRTC) plans to spend Tsh 5 million on 200
new vehicles. Each van will cost Tsh 20,000, each small truck Tsh 25,000, and each

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large truck Tsh 35,000. Past experience shows that MRTC needs twice as many vans
as small trucks. How many of each kind of vehicle can the company buy?

QUESTION 4:
John receives an inheritance of Tsh 50,000. His financial advisor suggests that he in-
vest this in three mutual funds: a money-market fund, a blue-chip stock fund, and a
high-tech stock fund. The advisor estimates that the money-market fund will return
5% over the next year, the blue-chip fund 9%, and the high-tech fund 16%. John
wants a total first-year return of Tsh 4000. To avoid excessive risk, he decides to in -
vest three times as much in the money-market fund as in the high-tech stock fund.
How much should he invest in each fund?

QUESTION 5.
An animal feed is to be made from corn, soybeans, and cottonseed. Determine how
many units of each ingredient are needed to make a feed that supplies 1800 units
of fiber, 2800 units of fat, and 2200 units of protein, given that 1 unit of each ingre -
dient provides the numbers of units shown in the table below.
Corn Soybeans Cottonseed
Units of Fiber 10 20 30
Units of Fat 30 20 40
Units of protein 20 40 25

QUESTION 6
3
Mkubwa Mwamasso invested different amounts at 8%, 8 4 % and 9%, all at sim-
ple interest. Altogether he invested Tsh 40,000 and earns Tsh3,455 per year. How
much does he have invested at each rate if he has Tsh4,000 more invested at 9%
than at 8%?, Solve by using Cramer’s rule.

QUESTION 7

Production Scheduling SIDO wishes to produce three types of bolts: types A, B, and
C. To manufacture a type-A bolts requires 2 minutes on machine I, 1 minute on ma -
chine II, and 2 minutes on machine III. A type-B bolts requires 1 minute on machine
I, 3 minutes on machine II, and 1 minute on machine III. A type-C bolts requires 1
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minute on machine I and 2 minutes each on machines II and III. There are 3 hours
available on machine I, 5 hours available on machine II, and 4 hours available on
machine III for processing the order. How many bolts of each type should SIDO
make in order to use all of the available time? Use Cramer’s rule.

SOLUTION: The given information may be tabulated as follows:

We have to determine the number of each of three types of bolts to be made. So,
let x, y, and z denote the respective numbers of type-A, type-B, and type-C bolts s
to be made.

The total amount of time that machine I is used is given by

2 x+ y+z =180
The total amount of time that machine II is used is given by
x +3 y+2 z=300
The total amount of time that machine III is used is given by
2 x+ y+2 z=240

QUESTION 8: A farmer has 200 acres of land suitable for cultivating crops A, B,
and C. The cost per acre of cultivating crop A, crop B, and crop C is Tshs 40, Tshs
60, and Tshs 80, respectively. The farmer has Tshs 12,600 available for land cultiva -
tion. Each acre of crop A requires 20 labour-hours, each acre of crop B requires 25
labour-hours, and each acre of crop C requires 40 labour-hours. The farmer has a
maximum of 5950 labour hours available. If she wishes to use all of her cultivatable
land, the entire budget, and all the labour available, how many acres of each crop
should she plant?

SOLUTION: Let x, y, and z denote the number of acres of crop A, crop B, and crop
C, respectively, to be cultivated. Then, the condition that all the cultivatable land be
used translates into the equation

x + y + z=200
Next, the total cost incurred in cultivating all three crops is:-
40 x +60 y +80 z=12 , 600

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 Tanzania Institute of Accountancy.

Finally, the amount of labour required to cultivate all three crops is

20 x+25 y+40 z=5950
Thus, the solution is found by solving the above system of linear equations

QUESTION 9

The annual returns on MASAI’s three investments amounted to Tshs 21,600: 6% on

a savings account, 8% on mutual funds, and 12% on bonds. The amount of MASAI’s
investment in bonds was twice the amount of his investment in the savings account,
and the interest earned from his investment in bonds was equal to the dividends he
received from his investment in mutual funds. Find how much money he placed in
each type of investment.

QUESTION 10

A private investment club has Tshs 200,000 earmarked for investment in stocks. To
arrive at an acceptable overall level of risk, the stocks that management is consid-
ering have been classified into three categories: high risk, medium risk, and low
risk. Management estimates that high-risk stocks will have a rate of return of 15%/
year; medium-risk stocks, 10%/year; and low-risk stocks, 6%/year. The members
have decided that the investment in low-risk stocks should be equal to the sum of
the investments in the stocks of the other two categories. Determine how much the
club should invest in each type of stock if the investment goal is to have a return of
Tshs 20,000 per year on the total investment.

QUESTION 11

A theater has a seating capacity of 900 and charges Tshs 4 for children, Tshs 6 for
students, and Tshs 8 for adults. At a certain screening with full attendance, there
were half as many adults as children and students combined. The receipts totaled
Tshs 5600. How many children attended the show?

QUESTION 12

Mr. and Mrs. Joseph have a total of Tshs 100,000 to be invested in stocks, bonds,
and a money market account. The stocks have a rate of return of 12%/year, while
the bonds and the money market account pay 8%/year and 4%/year, respectively.
The Joseph have stipulated that the amount invested in stocks should be equal to
the sum of the amount invested in bonds and 3 times the amount invested in the
money market account. How should the Joseph allocate their resources if they re-
quire an annual income of Tshs 10,000 from their investments?

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QUESTION 13

( )
6 7 −1
3 x 5
a) Find the value of x if the matrix
9 11 x has no inverse.

( )
1 3 4
−1
A =3 2 2
b) If
1 1 1 ,find matrix A

c) Find the values of x and y given that

(
A= 4 x 9 6
6 2 2x ) ,
(
B= − y 0 7
4 −8 y )
and

(
A+B= 8 9 13
10 −6 10 )
( ) ( )
0 1 2 −1 1 −1
A= 1 2 3 p −6 q
d) If
3 1 1 and adj(A) is
−5 3 −1 ,
i) Find the values of p and q

ii) Determine ( A−1 )T

( ) ( )
1 2 3 1 −3 2
A= 2 4 5 B= −3 3 −1
e) If
3 5 6 and
2 p q , find the value of P and Q such
that
B= A−1 .

( )
2 0 4
A= 5 −1 1
f) If
9 7 8
i) Show that A is non singular,
ii) Compute A−1 .

QUESTION 14.

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[ ]
2 1 3
A= 4 0 2
A matrix A is defined as
0 −1 −3 , given that f(A) = A2 – 2A + 10. Find
the value of f(A).

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