Table 15.1
Cars required
Compact Intermediate Large
People carrier Luxury limousine
Week 1 Week 2 Week 3
4 7 2 3 5 5 12 9 5 2 1 3 1 1 2
The total car hire bill for each week can then be calculated by multiplying the number of cars to be hired in each category by the corresponding price.
A matrix is deﬁned as an array of numbers (or algebraic symbols) set out in rows and columns. Therefore, the car hire requirements for the 3-week period in this example can be set out as the matrix
4 7 2 3 5 5
A = 12 9 5
2 1 3 1 1 2
where each row corresponds to a size of car and each column corresponds to a week. The usual notation system is to denote matrices by a capital letter in bold type, as for matrix A above, and to enclose the elements of a matrix in a set of squared brackets, i.e. [ ].
Matrices may also be speciﬁed with algebraic terms instead of numbers. Each entry is usually known as an ‘element’. The elements in each matrix must form a complete rectangle, without any blank spaces. For example, if there are 5 rows and 3 columns there must be 3 elements in each row and 5 elements in each column. An element may be zero though.
The size of a matrix is called its ‘order’. The order is speciﬁed as:
(number of rows)×(number of columns)
For example, the matrix A above has 5 rows and 3 columns and so its order is 5 ×3. Matriceswithonlyonecolumnorrowareknownasvectors. Theseareusuallyrepresented
by lower case letters, in bold. For example, the set of car rental prices we started this chapter with can be speciﬁed (in £) as the 1 ×5 row vector
p = 139 160 205 340 430
and the car hire requirements in week 1 can be speciﬁed as the 5 ×1 column vector
4 3
q = 12
2 1
Matrix addition and subtraction
Matrices that have the same order can be added together, or subtracted. The addition, or subtraction, is performed on each of the corresponding elements.
© 1993, 2003 Mike Rosser
Example 15.1
A retailer sells two products, Q and R, in two shops A and B. The number of items sold for the last 4 weeks in each shop are shown in the two matrices A and B below, where the columns represent weeks and the rows correspond to products Q and R, respectively.
A = 10
4 12 7 12 9 14
and B = 8
9 3 4 18 21 5
Derive a matrix for total sales for this retailer for these two products over the last 4 weeks.
Solution
Total sales for each week will simply be the sum of the corresponding elements in matrices A and B. For example, in week 1 the total sales of product Q will be 5 plus 8. Total combined sales for Q and R can therefore be represented by the matrix
T = A +B = 10 12 5 +8 4 +9
10 +8 12 +18
12 7 8 9 3 4
9 14 8 18 21 5
12 +3 7 +4 13 13 15 11
9 +21 14 +5 18 30 30 19
An element of a matrix can be a negative number, as in the solution to the example below.
Example 15.2
If A = 12
30 15
and B = 4
35 what is A −B?
Solution
A −B = 12
30 7 35 12 −7 15 4 8 8 −4
30 −35 5 −5 15 −8 4 7
Scalar multiplication
There are two forms of multiplication that can be performed on matrices. A matrix can be multiplied by a speciﬁc value, such as a number (scalar multiplication) or by another matrix (matrix multiplication). Scalar multiplication simply involves the multiplication of each element in a matrix by the scalar value, as in Example 15.3. Matrix multiplication is rather more complex and is explained later, in Section 15.2.
© 1993, 2003 Mike Rosser
Example 15.3
The number of units of a product sold by a retailer for the last 2 weeks are shown in matrix A below, where the columns represent weeks and the rows correspond to the two different
shop units that sold them. 12 30
8 15
If each item sells for £4, derive a matrix for total sales revenue for this retailer for these two shop units over this two-week period.
Solution
Total revenue is calculated by multiplying each element in matrix of sales quantities A by the scalar value 4, the price that each unit is sold at. Thus total revenue can be represented
(in £) by the matrix
4 ×12 4 ×30 48 120
4 ×8 4 ×15 32 60
Thescalarvaluethatamatrixismultipliedbymaybeanalgebraictermratherthanaspeciﬁc number value. For example, if the product price in Example 15.3 above was speciﬁed as p instead of £4 then the total revenue matrix would become
12p 30p 8p 15p
Scalardivisionworksinthesamewayasscalarmultiplication, butwitheachelementdivided by the relevant scalar value.
Example 15.4
If the set of car rental prices in the vector p = 139 160 205 340 430 includes VAT (Value Added Tax) at 17.5% and your company can claim this tax back, what is the vectorv of prices without this tax?
Solution
First of all we need to ﬁnd the scalar value used to scale down the original vector element values. As the tax rate is 17.5% then the quoted prices will be 117.5% times the basic price. Therefore a quoted price divided by 1.175 will be the basic price and so the vector of prices
(in £) without the tax will be
v = 1.175 = 1.175 139 160 205 340 430 = 1.175 139 1.175 160 1.175 205 1.175 340 1.175 430
= 118.30 136.17 174.47 289.36 365.96
© 1993, 2003 Mike Rosser
Test Yourself, Exercise 15.1
1. A ﬁrm uses 3 different inputs K, L and R to make two ﬁnal products X and Y. Each unit of X produced requires 2 units of K, 8 units of L and 23 units of R. Each unit of Y produced requires 3 units of K, 5 units of L and 26 units of R. Set up these input requirements in matrix format.
2. ‘A vector is a special form of matrix but a matrix is not a special form of vector’. Is this statement true?
3. For the pairs of matrices below say whether it is possible to add them together and
then, where it is possible, derive the matrix C = A +B. (a) A = 2 35 and B = 4 35
(b) A = 5 3 and B = 7 0 2
10 4 3 0
(c) A = 12 and B = 2
6 −9 1 1
4. A company sells 4 products and the sales revenue (in £m.) from each product sold through the company’s three retail outlets in a year are given in the matrix
7 3 1 4 R = 6 3 8 2.5
4 1.2 2 0
If proﬁt earned is always 20% of sales revenue, use scalar multiplication to derive a matrix showing proﬁt on each product for each retail outlet.
15.2 Basic principles of matrix multiplication
If one matrix is multiplied by another matrix, the basic rule is to multiply elements along the rows of the ﬁrst matrix by the corresponding elements down the columns of the second matrix. The easiest way to understand how this operation works is to ﬁrst work through some examples that only involve matrices with one row or column, i.e. vectors.
Returning to our car hire example, consider the two vectors
p = 139 160 205 340 430 and
4 3
q = 12
2 1
The row vector p contains the prices of hire cars in each category and the column vector q containsthequantitiesofcarsineachcategorythatyourcompanywishestohirefortheweek.
© 1993, 2003 Mike Rosser
At the start of this chapter we worked out the total car hire bill as
139 ×4 +160 ×3 +205 ×12 +340 ×2 +430 ×1 = £4,606
In terms of these two vectors, what we have done is multiply the ﬁrst element in the row vector p by the ﬁrst element in the column vector q. Then, going across the row, the second element of p is multiplied by the second element down the column of q. The same procedure is followed for the other elements until we get to the end of the row and the bottom of the column.
Now consider the situation where the car hire prices are still shown by the vector p = 139 160 205 340 430
but there are now three weeks of different car hire requirements, shown by the columns of matrix
4 7 2 3 5 5
A = 12 9 5
2 1 3 1 1 2
To calculate the total car hire bill for each of the three weeks, we need to ﬁnd the vector
t = pA
This should have the order 1 × 3, as there will be one element (i.e. the bill) for each of the three weeks. The ﬁrst element of t is the bill for the ﬁrst week, which we have already found in the example above. The car hire bill for the second week is worked out using the same method, but this time the elements across the row vector p multiply the elements down the second column of matrix A, giving
139 ×7 +160 ×5 +205 ×9 +340 ×1 +430 ×1 = £4,388
The third element is calculated in the same manner, but working down the third column of A. The result of this matrix multiplication exercise is therefore
t = pA = 139 160 205 340
= 4606 4388 3983
4
43012 2
1
7 5 9 1 1
2 5
3 2
The above examples have shown how the basic principle of matrix multiplication involves the elements across a row vector multiplying the elements down the columns of the matrix being multiplied, and then summing all the products obtained. If the ﬁrst matrix has more than one row (i.e. it is not a vector) then the same procedure is followed across each row. This means that the number of rows in the ﬁnal product matrix will correspond to the number of rows in the ﬁrst matrix.
© 1993, 2003 Mike Rosser
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