7Matrices and determinants

7.1 MATRICES

We start with an introductory example.

Example 7.1 Assume that a firm uses three raw materials denoted by R1, R2 and R3 to produce four intermediate products S1, S2, S3 and S4. These intermediate products are partially also used to produce two final products F1 and F2. The numbers of required units of the intermediate products are independent of the use of intermediate products as input for the two final products. Table 7.1 gives the number of units of each raw material which are required for the production of one unit of each of the intermediate products. Table 7.2 gives the number of units of each intermediate product necessary to produce one unit of each of the final products.

The firm intends to produce 80 units of S1, 60 units of S2, 100 units of S3 and 50 units of S4 as well as 70 units of F1 and 120 units of F2. The question is: how many units of the raw materials are necessary to produce the required numbers of the intermediate and final products?

Table 7.1 Raw material requirements for the intermediate products

Table 7.2 Intermediate product requirements for the final products

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To produce the required units of the intermediate products, we need

2 · 80 + 1 · 60 + 4 · 100 + 0 · 50 = 620

units of raw material R1. Similarly, we need

3 · 80 + 2 · 60 + 1 · 100 + 3 · 50 = 610

units of raw material R2 and

1 · 80 + 4 · 60 + 0 · 100 + 5 · 50 = 570

units of raw material R3. Summarizing the above considerations, the vector yS of the required units of raw materials for the production of the intermediate products is given by

620

yS = 610 , 570

where the kth component gives the number of units of Rk required for the production of the intermediate products.

Next, we calculate how many units of each raw material are required for the production of the final products. Since the intermediate products are used for the production of the final products (see Table 7.2), we find that for the production of one unit of final product F1 the required amount of raw material R1 is

2 · 2 + 1 · 4 + 4 · 1 + 0 · 3 = 12.

Similarly, to produce one unit of final product F2 requires

2 · 3 + 1 · 0 + 4 · 4 + 0 · 1 = 22

units of R1. To produce one unit of final product F1 requires

3 · 2 + 2 · 4 + 1 · 1 + 3 · 3 = 24

units of R2. Continuing in this way, we get Table 7.3, describing how many units of each raw material are required for the production of each of the final products.

Table 7.3 Raw material requirements for the final products

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Therefore, for the production of the final products, there are

12 · 70 + 22 · 120 = 3, 480

units of raw material R1,

24 · 70 + 16 · 120 = 3, 600

units of raw material R2 and finally

33 · 70 + 8 · 120 = 3, 270

units of raw material R3 required. The vector yF containing as components the units of each raw material required for the production of the final products is then given by

3, 480

yF = 3, 600 . 3, 270

So the amount of the individual raw materials required for the total production of the intermediate and final products is obtained as the sum of the vectors yS and yF . Denoting this sum vector by y, we obtain

The question is whether we can simplify the above computations by introducing some formal apparatus. In the following, we use matrices and define operations such as addition or multiplication in an appropriate way.

Definition 7.1 A matrix A is a rectangular array of elements (numbers or other mathematical objects, e.g. functions) aij of the form

Any element (or entry) aij has two indices, a row index i and a column index j. The matrix A is said to have the order or dimension m × n (read: m by n).

If m = n, matrix A is called a square matrix.

For a matrix A of order m × n, we also write A = A(m,n), or A = (aij )(m,n), or simply

A = (aij ).

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Definition 7.2 Let a matrix A of order m × n be given. The transpose AT of matrix A is obtained by interchanging the rows and columns of A, i.e. the first column becomes the first row, the first row becomes the first column and so on. Thus:

Obviously, matrix AT in Definition 7.2 is of order n × m.

Example 7.2 Let

Since matrix A is of order 2 × 4, matrix AT is of order 4 × 2, and we get

27

14

is a special matrix with one column (i.e. a matrix of order m × 1). Analogously, a transposed vector aT = (a1, a2, . . . , am) is a special matrix consisting only of one row.

So only for matrices of the same order m × n can we decide whether both matrices are equal.

Definition 7.4 A matrix A of order n × n is called symmetric if A = AT, i.e. equality aij = aji holds for 1 ≤ i, j ≤ n. Matrix A is called antisymmetric if A = −AT, i.e. aij = −aji for 1 ≤ i, j ≤ n.

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Definition 7.7 A matrix O = (oij ) of order m × n with

·· · 00 0

is called a zero matrix.

7.2 MATRIX OPERATIONS

In the following, we discuss matrix operations such as addition and multiplication and their properties.

Definition 7.8 Let A = (aij ) and B = (bij ) be two matrices of order m × n. The sum A + B is defined as the m × n matrix (aij + bij ), i.e.

A + B = (aij )(m,n) + (bij )(m,n) = (aij + bij )(m,n).

Thus, the sum of two matrices of the same order is obtained when corresponding elements at the same position in both matrices are added. The zero matrix O is the neutral element with respect to matrix addition, i.e. we have

A + O = O + A = A,

where matrix O has the same order as matrix A.

Definition 7.9 Let A = (aij ) be an m × n matrix and λ R. The product of the scalar λ and the matrix A is the m × n matrix λA = (λaij ), i.e. any element of matrix A is

multiplied by the scalar λ. The operation of multiplying a matrix by a scalar is called scalar multiplication.

Using Definitions 7.8 and 7.9, we can define the difference of two matrices as follows.

Definition 7.10 Let A = (aij ) and B = (bij ) be matrices of order m × n. Then the difference of matrices A and B is defined as

A − B = A + (−1)B.

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Consequently, matrix A − B is given by the m × n matrix (aij − bij ), i.e.

A − B = (aij )(m,n) − (bij )(m,n) = (aij − bij )(m,n).

Next, we give some rules for adding two matrices A and B of the same order and for multiplying a matrix by some real number (scalar multiplication). Let A, B, C be matrices of order m × n and λ, µ R.

Rules for matrix addition and scalar multiplication

(1)A + B = B + A; (commutative law)

(2)(A + B) + C = A + (B + C); (associative law)

(3) λ(A + B) = λA + λB; (λ + µ)A = λA + µA. (distributive laws)

We have already introduced the notion of a vector space in Chapter 6. Using matrix addition and scalar multiplication as introduced in Definitions 7.8 and 7.9, we can extend the rules presented above and get the following result.

THEOREM 7.1 The set of all matrices of order m × n constitutes a vector space.

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Next, we introduce the multiplication of two matrices of specific orders.

Definition 7.11 Let A = (aij ) be a matrix of order m × p and B = (bij ) be a matrix of order p × n. The product AB is a matrix of order m × n which is defined by

Notice that the product AB is defined only when the number of columns of matrix A is equal to the number of rows of matrix B. For calculating the product A(m,p)B(p,n), we can use Falk’s scheme which is as follows:

k=1

From the above scheme we again see that element cij is obtained as the scalar product of the ith row vector of matrix A and the jth column vector of matrix B.

If we have to perform more than one matrix multiplication, we can successively apply Falk’s scheme. Assuming that the corresponding products of n matrices are defined, we can (due to the validity of the associative law) perform the multiplications either starting from the left or from the right. In the former case, we obtain C = A1A2A3 · · · An according to C = [(A1A2)A3] · · · An, i.e. by using Falk’s scheme repeatedly we obtain

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Next, we discuss some properties of matrix multiplication.

(1)Matrix multiplication is not commutative, i.e. in general we have AB = BA. It may even

happen that only one of the possible products of two matrices is defined but not the other. For instance, let A be a matrix of order 2 × 4 and B be a matrix of order 4 × 3. Then the product AB is defined and gives a product matrix of order 2 × 3. However, the product

BA is not defined since matrix B has three columns but matrix A has only two rows.

(2)For matrices A(m,p), B( p,r) and C(r,n), we have A(BC) = (AB)C, i.e. matrix multiplication is associative provided that the corresponding products are defined.

(3)For matrices A(m,p), B(p,n) and C(p,n), we have A(B + C) = AB + AC, i.e. the distributive law holds provided that B and C have the same order and the product of matrices A and B + C is defined.

(4)The identity matrix I of order n × n is the neutral element of matrix multiplication of square matrices of order n × n, i.e.

AI = IA = A.

Let A be a square matrix. Then we write AA = A2, and in general An = AA . . . A, where factor A occurs n times, is known as the nth power of matrix A.

be given. The product BA is not defined since matrix B has two columns but matrix A has three rows. The product AB is defined according to Definition 7.11, and the resulting product matrix C = AB is of the order 3 × 2. Applying Falk’s scheme, we obtain

Example 7.5 Three firms 1, 2 and 3 share a market for a certain product. Currently, firm 1 has 25 per cent of the market, firm 2 has 55 per cent and firm 3 has 20 per cent of the market.

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We can summarize this in a so-called market share vector s, where component si is a real number between zero and one giving the current percentage of firm i as a decimal so that the sum of all components is equal to one. In this example, the corresponding market share vector s = (s1, s2, s3)T is given by

0.25

s = 0.55 . 0.20

In the course of one year, the following changes occur.

(1)Firm 1 keeps 80 per cent of its customers, while losing 5 per cent to firm 2 and 15 per cent to firm 3.

(2)Firm 2 keeps 65 per cent of its customers, while losing 15 per cent to firm 1 and 20 per cent to firm 3.

(3)Firm 3 keeps 75 per cent of its customers, while losing 15 per cent to firm 1 and 10 per cent to firm 2.

We compute the market share vector s after the above changes. To do this, we introduce a matrix T = (tij ), where tij is the percentage (as a decimal) of customers of firm j who become a customer of firm i within the next year. Matrix T is called a transition matrix. In this example, matrix T is as follows:

0.800.15 0.15

0.05 0.65 0.10 .

0.150.20 0.75

To get the percentage of customers of firm 1 after the course of the year, we have to compute

s1 = 0.80s1 + 0.15s2 + 0.15s3.

Similarly, we can compute the values s2 and s3, and we find that vector s is obtained as the product of matrix T and vector s:

Hence, after one year, firm 1 has 31.25 per cent of the customers, firm 2 has 39 per cent and firm 3 has 29.75 per cent.

Example 7.6 Consider again the data given in Example 7.1. Introducing matrix RS(3,4) as the matrix giving the raw material requirements for the intermediate products as in Table 7.1 and matrix S(F4,2) as the matrix of the intermediate product requirements for the final products as in Table 7.2, we get the raw material requirements for the final products described by matrix RF(3,2) by matrix multiplication:

RF(3,2) = RS(3,4) · S(F4,2).

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Let vectors x(S4,1) and x(F2,1) give the number of units of each of the intermediate and final products, respectively, where the ith component refers to the ith product. Then we obtain the

vector y of the total raw material requirements as follows:

y(3,1) = y(S3,1) + y(F3,1)

=RS(3,4) · x(S4,1) + RF(3,2) · x(F2,1)

=RS(3,4) · x(S4,1) + RS(3,4) · S(F4,2) · x(F2,1).

The indicated orders of the matrices confirm that all the products and sums are defined.

We now return to transposes of matrices and summarize the following rules, where A and B are m × n matrices, C is an n × p matrix and λ R.

Rules for transposes of matrices

(3)(λA)T = λAT;

(4)(AC)T = CTAT.

Definition 7.12 A matrix A of order n × n is said to be orthogonal if ATA = I .

As a consequence of Definition 7.12, we find that in an orthogonal matrix A, the scalar product of the ith row vector and the jth column vector with i = j is equal to zero, i.e. these vectors are orthogonal (cf. Chapte 6.2).

7.3 DETERMINANTS

Determinants can be used to answer the question of whether the inverse of a matrix exists and to find such an inverse matrix. They can be used e.g. as a tool for solving systems of

264 Matrices and determinants

linear equations (this topic is discussed in detail in Chapter 8) or for finding eigenvalues (see Chapter 10).

Let

be a square matrix and Aij denote the submatrix obtained from A by deleting the ith row and jth column. It is clear that Aij is a square matrix of order (n − 1) × (n − 1).

Definition 7.13 The determinant of a matrix A of order n × n with numbers as elements is a number, assigned to matrix A by the following rule:

n

det A = |A| = (−1) j+1 · a1j · |A1j |.

j=1

For n = 1, we define |A| = a11.

Whereas a matrix of order m × n is a rectangular array of m · n elements, determinants are defined only for square matrices and in contrast to matrices, a determinant is a number provided that the elements of the matrix are numbers as well. According to Definition 7.13, a determinant of a matrix of order n × n can be found by means of n determinants of matrices of order (n − 1) × (n − 1). The rule given in Definition 7.13 can also be applied when the elements of the matrices are e.g. functions or mathematical terms.

For n = 2 and matrix

we get

|A| = a11a22 − a12a21.

For n = 3 and matrix

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=a11(a22a33 − a32a23) − a12(a21a33 − a31a23) + a13(a21a32 − a31a22)

=a11a22a33 + a12a23a31 + a13a21a32 − a11a23a32 − a12a21a33 − a13a22a31.

The latter computations for n = 3 can be done as follows. We add the first two columns at the end as fourth and fifth columns. Then we compute the products of the three diagonals from the left top to the right bottom and add them, and from this value we subtract the sum of the products of the three diagonals from the left bottom to the right top. This procedure is known as Sarrus’s rule (see Figure 7.1) and works only for the case n = 3.

Determinants of square submatrices are called minors. The order of a minor is determined by its number of rows (or columns). A minor |Aij | multiplied by (−1)i+j is called a cofactor. The following theorem gives, in addition to Definition 7.13, an alternative way of finding the determinant of a matrix of order n × n.

Figure 7.1 Sarrus’s rule.

THEOREM 7.2 (Laplace’s theorem, cofactor expansion of a determinant) Let A be a matrix of order n × n. Then the determinant of matrix A is equal to the sum of the products of the elements of one row or column with the corresponding cofactors, i.e.

Theorem 7.2 contains Definition 7.13 as a special case. While Definition 7.13 requires cofactor expansion by the first row to evaluate the determinant of matrix A, Theorem 7.2 indicates that we can choose one arbitrary row or column of matrix A to which we apply cofactor expansion. Therefore, computations are simplified if we choose one row or column with many zeroes in matrix A.

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Example 7.8 We evaluate the determinant of matrix

23 5

by applying Theorem 7.2 and performing cofactor expansion by the second column. We get

According to Theorem 7.2, the first determinant of order 2 × 2 on the right-hand side of the first row above is the minor |A12| obtained by crossing out in matrix A the first row and the second column, i.e.

Accordingly, the other minors A22 and A32 are obtained by crossing out the second column as well as the second and third rows, respectively.

We now give some properties of determinants.

THEOREM 7.3 Let A be an n × n matrix. Then |A| = |AT|.

This is a consequence of Theorem 7.2 since we can apply cofactor expansion by the elements of either one row or one column. Therefore, the determinant of matrix A and the determinant of the transpose AT are always equal. In the case of a triangular matrix, we can easily evaluate the determinant, as the following theorem shows.

THEOREM 7.4 Let A be an n × n (lower or upper) triangular matrix. Then

n

|A| = a11 · a22 · . . . · ann = aii .

i=1

As a corollary of Theorem 7.4, we find that the determinant of an identity matrix I is equal to one, i.e. |I | = 1.

If we evaluate a determinant using Theorem 7.2, it is desirable that the determinant has an appropriate structure, e.g. computations are simplified if many elements of one row or of one column are equal to zero. For this reason, we are looking for some rules that allow us to evaluate a determinant in an easier form.

THEOREM 7.5 Let A be an n × n matrix. Then:

(1)If we interchange in A two rows (or two columns), then we get for the resulting matrix

A : |A | = −|A|.

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(2)If we multiply all elements of a row (or all elements of a column) by λ R, then we get for the resulting matrix A : |A | = λ · |A|.

(3)If we add to all elements of a row (or to all elements of a column) λ times the corresponding elements of another row (column), then we get for the resulting matrix A :

|A | = |A|.

COROLLARY 7.1 For the n × n matrix B = λA, we obtain: |B| = |λA| = λn · |A|.

The latter corollary is obtained by a repeated application of part (2) of Theorem 7.5.

THEOREM 7.6 Let A and B be matrices of order n × n. Then |AB| = |A| · |B|.

It is worth noting that in general |A + B| = |A| + |B|. Next, we consider two examples of evaluating determinants.

Example 7.9 We evaluate the determinant of matrix

We apply Theorem 7.5 to generate a determinant having the same value, in which all elements are equal to zero below the diagonal:

In the first transformation step, we have generated zeroes in rows 2 to 4 of the first column. To this end, we have multiplied the first row by 2 and added it to the second row, yielding the new second row. Analogously, we have multiplied the first row by −3 and added it to the third row, and we have multiplied the first row by −4 and added it to the fourth row. In the next transformation step, we have generated zeroes in rows 3 and 4 of the second column. This means we have multiplied the second row (of the second determinant) by −2 and added it to the third row, and we have multiplied the second row by 3 and added it to the fourth row (application of part (3) of Theorem 7.5). Additionally, we have interchanged rows 3 and 4, which changes the sign of the determinant. Finally, we applied Theorem 7.4.