TOPICS

Unit-3(Examples)

Matrices

**Example-1 :-** Consider the following information regarding the number of men and women workers in three factories I, II and III
Represent the above information in the form of a 3 × 2 matrix. What does the entry in the third row and second column represent?

The information is represented in the form of a 3 × 2 matrix as follows: The entry in the third row and second column represents the number of women workers in factory III.

**Example-2 :-** If a matrix has 8 elements, what are the possible orders it can have?

We know that if a matrix is of order m × n, it has mn elements. Thus, to find all possible orders of a matrix with 8 elements, we will find all ordered pairs of natural numbers, whose product is 8. Thus, all possible ordered pairs are (1, 8), (8, 1), (4, 2), (2, 4) Hence, possible orders are 1 × 8, 8 × 1, 4 × 2, 2 × 4.

**Example-3 :-** Construct a 3 × 2 matrix whose elements are given by aᵢⱼ = 1/2|i - 3j|.

In general a 3 × 2 matrix is given by Now, aᵢⱼ = 1/2|i - 3j|, i = 1,2,3 and j = 1,2 a₁₁ = 1/2|1 - 3x1| = 1/2|-2| = 1 a₁₂ = 1/2|1 - 3x2| = 1/2|-5| = 5/2 a₂₁ = 1/2|2 - 3x1| = 1/2|-1| = 1/2 a₂₂ = 1/2|2 - 3x2| = 1/2|-4| = 2 a₃₁ = 1/2|3 - 3x1| = 1/2|0| = 0 a₃₂ = 1/2|3 - 3x2| = 1/2|3 - 6| = 3/2 Hence the required matrix is given by

**Example-4 :-**
Find the values of a, b, c, x, y and z.

As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get x + 3 = 0; x = -3, z + 4 = 6; z = -4, 2y – 7 = 3y – 2; 3y - 2y = -7 + 2; y = -5, a – 1 = – 3; a = -3 + 1; a = -2, 0 = 2c + 2; 2c = -2; c = -1, b – 3 = 2b + 4; 2b - b = -3 - 4; b = -7, we get a = – 2, b = – 7, c = – 1, x = – 3, y = –5, z = 2

**Example-5 :-** Find the values of a, b, c, and d from the following equation:

By equality of two matrices, equating the corresponding elements, we get 2a + b = 4 ......(i), a – 2b = –3 ......(ii) By solving eq. (i)and (ii) 2a + b = 4 (a - 2b = -3) x 2 2a - 4b = -6 2a + b - 2a + 4b = 4 + 6 5b = 10; b = 2 Put in eq. (i), we get 2a + 2 = 4; 2a = 2; a = 1 Now, 5c – d = 11 ......(iii), 4c + 3d = 24......(iv) (5c - d = 11) x 3 15c - 3d = 33 4c + 3d = 24 15c - 3d + 4c + 3d = 33 + 24 19c = 57; c = 3 Put in eq. (iii), we get 5x3 - d = 11; 15 - d = 11; -d = -4; d = 4 Therefore, a = 1, b = 2, c = 3 and d = 4

**Example-6 :-**
Given
Find A + B.

Since A, B are of the same order 2 × 3. Therefore, addition of A and B is defined and is given by

**Example-7 :-**
then find 2A - B.

**Example-8 :-**
then find the matrices of X, such that 2A + 3X = 5B.

We have 2A + 3X = 5B or 2A + 3X – 2A = 5B – 2A or 2A – 2A + 3X = 5B – 2A (Matrix addition is commutative) or O + 3X = 5B – 2A (– 2A is the additive inverse of 2A) or 3X = 5B – 2A (O is the additive identity) or X = 1/3(5B – 2A)

**Example-9 :-**
Find X and Y,

**Example-10 :-**
Find the values of x and y from the following equation:

**Example-11 :-**
Two farmers Ramkishan and Gurcharan Singh cultivates only three varieties of rice namely Basmati, Permal and Naura.
The sale (in Rupees) of these varieties of rice by both the farmers in the month of September and October are given by the following matrices A and B.
(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.

(i) Combined sales in September and October for each farmer in each variety is given by

(ii) Change in sales from September to October is given by

(iii) 2% of B = 2B/100 = 0.02 × B Thus, in October Ramkishan receives ₹ 100, ₹ 200 and ₹ 120 as profit in the sale of each variety of rice, respectively, and Grucharan Singh receives profit of ₹ 400, ₹ 200 and ₹ 200 in the sale of each variety of rice, respectively.

**Example-12 :-**
Find AB,

The matrix A has 2 columns which is equal to the number of rows of B. Hence AB is defined. Now

**Example-13 :-**
then find AB, BA. Show that AB ≠ BA.

Since A is a 2 × 3 matrix and B is 3 × 2 matrix. Hence AB and BA are both defined and are matrices of order 2 × 2 and 3 × 3, respectively. Clearly AB ≠ BA.

**Example-14 :-**

Thus matrix multiplication is not commutative.

**Example-15 :-**
Find AB,

Thus, if the product of two matrices is a zero matrix, it is not necessary that one of the matrices is a zero matrix.

**Example-16 :-**
Find A(BC), (AB)C and show that (AB)C = A(BC).

Clearly, (AB) C = A (BC)

**Example-17 :-**
Calculate AC, BC and (A + B)C. Also, verify that (A + B)C = AC + BC.

Clearly, (A + B) C = AC + BC

**Example-18 :-**
then show that A³ – 23A – 40 I = O

**Example-19 :-**
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 paise) 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.

So the total amount spent by the group in the two cities is 340,000 paise and 720,000 paise, i.e., ₹ 3400 and ₹ 7200, respectively.

**Example-20 :-**
verify that

(i) (A′)′ = A,

(ii) (A + B)′ = A′ + B′,

(iii) (kB)′ = kB′, where k is any constant.

(i) Clearly, (A′)′ = A

(ii) Clearly, (A + B)′ = A′ + B′

(iii) Clearly, (kB)′ = kB′

**Example-21 :-**
verify that (AB)′ = B′A′

Clearly, (AB)′ = B′A′

**Example-22 :-**
Express the matrix
as the sum of a symmetric and a skew symmetric matrix.

Thus, B is represented as the sum of a symmetric and a skew symmetric matrix.

**Example-23 :-**
By using elementary operations, find the inverse of the matrix

In order to use elementary row operations we may write A = IA.

In order to use elementary column operations we may write A = AI.

**Example-24 :-**
Obtain the inverse of the following matrix using elementary operations

In order to use elementary row operations we may write A = IA.

**Example-25 :-**
Find P^{-1}, if it exists, given

In order to use elementary row operations we may write P = IP. We have all zeros in the second row of the left hand side matrix of the above equation. Therefore, P^{-1}does not exist.

CLASSES