NCERT Solutions for Class 12 Science Maths Chapter 3 – Matrices
Explore the comprehensive NCERT Solutions for Class 12 Science Mathematics Chapter 3 on Matrices, featuring clear step-by-step explanations. Widely favored by Class 12 Science students, these solutions serve as invaluable resources for completing homework assignments swiftly and preparing for exams effectively. You can access all the questions and answers from Chapter 3 of the NCERT Book for Class 12 Science Mathematics right here, free of charge. Page No 64: Question 1: In the matrix, write: (i) The order of the matrix (ii) The number of elements, (iii) Write the elements a13, a21, a33, a24, a23 ANSWER: (i) In the given matrix, the number of rows is 3 and the number of columns is 4. Therefore, the order of the matrix is 3 × 4. (ii) Since the order of the matrix is 3 × 4, there are 3 × 4 = 12 elements in it. (iii) a13 = 19, a21 = 35, a33 = −5, a24 = 12, a23 = Page No 64: Question 2: If a matrix has 24 elements, what are the possible order it can have? What, if it has 13 elements? ANSWER: We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the ordered pairs of natural numbers whose product is 24. The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), and(6, 4) Hence, the possible orders of a matrix having 24 elements are: 1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, and 6 × 4 (1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13. Hence, the possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1. Page No 64: Question 3: If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements? ANSWER: We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural numbers whose product is 18. The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6,), and (6, 3) Hence, the possible orders of a matrix having 18 elements are: 1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, and 6 × 3 (1, 5) and (5, 1) are the ordered pairs of natural numbers whose product is 5. Hence, the possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1. Page No 64: Question 4: Construct a 2 × 2 matrix,, whose elements are given by: (i) (ii) (iii) ANSWER: In general, a 2 × 2 matrix is given by (i) Therefore, the required matrix is (ii) Therefore, the required matrix is (iii) Therefore, the required matrix is Page No 64: Question 5: Construct a 3 × 4 matrix, whose elements are given by (i) (ii) ANSWER: In general, a 3 × 4 matrix is given by (i) Therefore, the required matrix is (ii) Therefore, the required matrix is Page No 64: Question 6: Find the value of x, y, and z from the following equation: (i) (ii) (iii) ANSWER: (i) As the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get: x = 1, y = 4, and z = 3 (ii) As the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get: x + y = 6, xy = 8, 5 + z = 5 Now, 5 + z = 5 ⇒ z = 0 We know that: (x − y)2 = (x + y)2 − 4xy ⇒ (x − y)2 = 36 − 32 = 4 ⇒ x − y = ±2 Now, when x − y = 2 and x + y = 6, we get x = 4 and y = 2 When x − y = − 2 and x + y = 6, we get x = 2 and y = 4 ∴x = 4, y = 2, and z = 0 or x = 2, y = 4, and z = 0 (iii) As the two matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get: x + y + z = 9 … (1) x + z = 5 … (2) y + z = 7 … (3) From (1) and (2), we have: y + 5 = 9 ⇒ y = 4 Then, from (3), we have: 4 + z = 7 ⇒ z = 3 ∴ x + z = 5 ⇒ x = 2 ∴ x = 2, y = 4, and z = 3 Page No 64: Question 7: Find the value of a, b, c, and d from the equation: ANSWER: As the two matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get: a − b = −1 … (1) 2a − b = 0 … (2) 2a + c = 5 … (3) 3c + d = 13 … (4) From (2), we have: b = 2a Then, from (1), we have: a − 2a = −1 ⇒ a = 1 ⇒ b = 2 Now, from (3), we have: 2 ×1 + c = 5 ⇒ c = 3 From (4) we have: 3 ×3 + d = 13 ⇒ 9 + d = 13 ⇒ d = 4 ∴a = 1, b = 2, c = 3, and d = 4 Page No 65: Question 8: is a square matrix, if (A) m < n (B) m > n (C) m = n (D) None of these ANSWER: The correct answer is C. It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns. Therefore, is a square matrix, if m = n. Page No 65: Question 9: Which of the given values of x and y make the following pair of matrices equal (A) (B) Not possible to find (C) (D) ANSWER: The correct answer is B. It is given that Equating the corresponding elements, we get: We find that on comparing the corresponding elements of the two matrices, we get two different values of x, which is not possible. Hence, it is not possible to find the values of x and y for which the given matrices are equal. Page No 65: Question 10: The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is: (A) 27 (B) 18 (C) 81 (D) 512 ANSWER: The correct answer is D. The given matrix of the order 3 × 3 has 9 elements and each of these elements can be either 0 or 1. Now, each of the 9 elements can be filled in two possible ways. Therefore, by the multiplication principle, the required number of possible matrices is 29 = 512 Page No 80: Question 1: Let Find each of the following (i) (ii) (iii) (iv) (v) ANSWER: (i) (ii) (iii) (iv) Matrix A has 2 columns. This number is equal to the number of rows in matrix B. Therefore, AB is defined as: (v) Matrix B has 2 columns. This number is equal to the number of rows in matrix A. Therefore, BA is defined as: Page No 80: Question 2: Compute …
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