Optional Mathematics Grade 10 PDF

Optional Mathematics Grade 10 Government of Nepal Ministry of Education, Science and Technology Curriculum Development Centre Sanothimi, Bhaktpur Nepal  Optional Mathematics Grade – 10 Government o...

Optional Mathematics Grade 10 Government of Nepal Ministry of Education, Science and Technology Curriculum Development Centre Sanothimi, Bhaktpur Nepal  Optional Mathematics Grade – 10 Government of Nepal Ministry of Education, Science and Technology Curriculum Development Centre Sanothimi, Bhaktapur 2076 Publisher: Government of Nepal Ministry of Education, Science and Technology Curriculum Development Centre Sanothimi, Bhaktapur © Publisher First Edition: 2076 B.S.  Preface The curriculum and curricular materials have been developed and revised on a regular basis with the aim of making education objective-oriented, practical, relevant and job oriented. It is necessary to instill the feelings of nationalism, national integrity and democratic spirit in students and equip them with morality, discipline and self- reliance, creativity and thoughtfulness. It is essential to develop in them the linguistic and mathematical skills, knowledge of science, information and communication technology, environment, health and population and life skills. it is also necessary to bring in them the feeling of preserving and promoting arts and aesthetics, humanistic norms, values and ideals. It has become the need of the present time to make them aware of respect for ethnicity, gender, disabilities, languages, religions, cultures, regional diversity, human rights and social values so as to make them capable of playing the role of responsible citizens. This textbook for grade nine students as an optional mathematics has been developed in line with the Secondary Level Optional Mathematics Curriculum, 2074 so as to strengthen mathematical knowledge, skill and thinking on the students. It is finalized by incorporating recommendations and feedback obtained through workshops, seminars and interaction programmes. The textbook is written by Prof. Dr. Siddhi Prasad Koirala, Mr. Ramesh Prasad Awasthi and Mr. Shakti Prasad Acharya. In Bringing out the textbook in this form, the contribution of the Director General of CDC Dr. Lekha Nath Poudel is highly acknowledged. Similarly, the contribution of Prof. Dr. Ram Man Shrestha, Mr. Laxmi Narayan Yadav, Mr. Baikuntha Prasad Khanal, Mr. Krishna Prasad Pokharel, Mr. Anirudra Prasad Neupane, Ms. Goma Shrestha, Mr. Rajkumar Mathema is also remarkable. The subject matter of the book was edited by Dr. Dipendra Gurung, Mr. Jagannath Adhikari and Mr. Nara Hari Acharya. The language of the book was edited by Mr. Nabin Kumar Khadka. The layout of this book was designed by Mr. Jayaram Kuikel. CDC extends sincere thanks to all those who have contributed to developing this textbook. This book contains various mathematical concepts and exercises which will help the learners to achieve the competency and learning outcomes set in the curriculum. Efforts have been made to make this textbook as activity-oriented, interesting and learner centered as possible. The teachers, students and all other stakeholders are expected to make constructive comments and suggestions to make it a more useful textbook. 2076 Curriculum Development Centre Sanothimi, Bhaktapur  Table of Content Unit 1 Algebra 1 1.1 Function 1 1.2 Polynomials 16 1.3 Sequence and Series 29 1.4 Linear Programming 68 1.5 Quadratic Equations and Graphs 81 Unit 2 Continuity 92 Unit 3 Matrix 103 Unit 4 Coordinate Geometry 121 Unit 5 Trigonometry 148 Unit 6 Vector 207 Unit 7 Transformation 237 Unit 8 Statistics 253 Answer 276 Unit 1 Algebra 1.1 Function 1.1.0 Review Study the square given and answer the following questions: (a) What is the area function 'f' in terms of side (l)? (b) What is the perimeter function ‘g’ in terms of side (l)? (c) Are functions 'f' and ‘g’ one to one? (d) Are functions 'f' and 'g' onto? Study the following information and relate them with function. (a) To each person, there corresponds an annual income. (b) To each square, there corresponds an area. (c) To each number, there corresponds its cube. Each (a), (b) and (c) are examples of function. Take an example of function and represent it in different forms. Discuss about the graph among your friends. Graphs provide a means of displaying, interpreting and analyzing data in a visual form. To graph an equation is to make a drawing that represents the solutions of that equation. When we draw the graph of an equation, we must remember the following points: (a) Calculate solutions and list the ordered pairs in a table. (b) Use graph paper and scale the axes. (c) Plot the ordered pairs, look for patterns and complete the graph with the equation being graphed. 1.1.1.(a): Algebraic function and it's graph Study the following functions and think about their graphical representation: f(x) = x + 5, f(x) = x2, f(x) = x3 The algebraic equation consisting of the least one variable is called the algebraic function. A function f is linear function if it can be written as f(x) = mx + c, where m and c are constant. If m = 0, the function is constant function and written as f(x) = c. If m = 1 and c = 0, the function is the identity function written as f(x) = x. 1  Linear Fuction Identity function Constant function functions like f(x) = ax2 + bx + c, a ≠ 0 and f(x) = ax3 + bx2 + cx + d, a ≠ 0 are examples of Non-linear functions. f(x) = ax2 + bx + c, a ≠ 0 is quadratic function where a, b, and c are real numbers. The graph of quadratic function is a parabola. The vertex form of f(x) = ax2 + bx + c, a ≠ 0 is f(x) = a(x-h)2 + k For example, f(x) = x2 + 2x + 2 = (x-(-1))2 + 1. The vertex of the parabola is (h, k) = (-1, 1). When, x = 0, the curve meets y-axis at (0, 2). x = h is the axis of parabola about which the curve is symmetric. If a > 0, the curve turns upward from vertex and it turns downwards for a < 0. 2 The function defined as f(x) = ax3 + bx2 + cx + d, a ≠ 0 is called cubic function, where a, b, c and d are real numbers. The nature of curve in graph is as shown at the right. When b, c, d are non-zero, the curve meets X-axis at three different points. o o o If each of b, c, d is zero, then the curve passes through the origin. Example 1 (a) When we draw the graph of y = 2x2, we can take different values of x and find their corresponding y-values. Representing the ordered pairs (x, y) in graph, we can find the shape of curve in graph. (b) When we draw the graph of y = -2x2, we follow the same steps as we do in (a) and give the shape of curve in graph. 3 Exercise: 1.1.1 (a) 1. (a) Define linear function with example. (b) What is the coordinates of vertex of f(x) = ax2 + bx + c, a ≠ 0 (c) Identify the identity function: f(x) = 5 and f(x) = x. 2. Study the following graphs and identify their nature as identity, constant, quadratic and cubic function. 4 3. Draw the graph of the following function (a) y = x + 2 (b) y = 6 (c) y = x2 (d) y = -x2 (e) y = x3 4. Pemba estimates the minimum ideal weight of a woman, in pounds is to multiply her height, in inches by 4 and subtract 130. Let y = minimum ideal weight and x = height. (a) Express y as a linear function of x. (b) Find the minimum ideal weight of a woman whose height is 62 inches. (c) Draw the graph of height and weight. 5. Investigate the nature of graph showing linear, quadratic and cubic function in our daily life. Make a report and present it in classroom. 1.1.1 (b): Trigonometric function Discuss about trigonometric ratios of any angle in your classroom. Also, tell any nine relations among trigonometric ratios of any magnitude. The function which relates angles and their measurement to the real number is called trigonometric function. It associates an angle with the definite real number. We know sin (x + 2) = sinx, cos (x + 2) = cosx and tan(x + ) = tanx. So, if a function f(x + k) = f(x) for positive value of k. k is called period for f(x). In case of sinx and cosx, the period is taken as 2, but for tanx, it is taken as  (Why?) Trigonometric functions are said to be non-algebraic or transcendental functions. (i) Graph of sinx or sine – graph: Let us take the values of x differing 90° and corresponding values of y for y = sinx. The maximum and minimum values of sinx are 1 and -1 respectively. x: -360° -270° -180° -90° 0° 90° 180° 270° 360° sinx: 0 1 0 -1 0 1 0 -1 0 y = sinx Domain: -360° < x° < 360° -2 < x < 2 Range: -1 < y < 1 Period: 2 5 (ii) Graph of cosx or cosine – graph Let us take the values of x differing 90° and corresponding values of y for y = cosx. The maximum and minimum values of cosx are 1 and -1 respectively. x -360° -270° -180° -90° 0° 90° 180° 270° 360° cosx 1 0 -1 0 1 0 -1 0 1 y = cosx Domain: -360° < x° < 360° -2 < x < 2 Range: -1 < y < 1 Period: 2  6 (iii) Graph of tanx or tangent - graph Let us take the values of x differing 90° and the corresponding values of y. For y = tanx. As x passes through – 360° to 360°, the values of tanx suddenly changes from very large positive and negative values. The line parallel to the y-axis corresponding to the odd multiples of 90° are continuously approached by the graph on either side, but never actually meet. Such lines are called asymptote to the curve. x -360° -270° -180° -90° 0° 90° 180° 270° 360° tanx 0 Not 0 Not 0 Not 0 No 0 defined defined defined defined y = tanx range: - 1 Sn can also be written as follows: 𝑎(𝑟 𝑛 −1) 𝑠𝑛 = 𝑟−1 𝑎𝑟 𝑛 −𝑎 = 𝑟−1 𝑎𝑟 𝑛−1 × 𝑟−𝑎 = 𝑟−1 𝑙𝑟−𝑎 𝑠𝑛 = 𝑟−1 [where, l = ar n−1 ] 𝑎−𝑙𝑟 And 𝑠𝑛 = if r < 1 1−𝑟 62 Note: to solve the problem easily, we can suppose 𝑎 i) Three terms in GP be , a, ar 𝑟 𝑎 𝑎 ii) Four terms in GP be , , ar, ar3 𝑟3 𝑟 𝑎 𝑎 iii) Five terms in GP be 2 , , a, ar, ar2 𝑟 𝑟 Example 1 Find the sum of the following series: 36 + 12 + 4 + …… up to 6 terms. Solution: Here, The given series is 36 + 12 + 4 + … to 6 terms First term (a) = 36 12 1 Common ratio (r) = 36 = 3  It is geometric series no. of terms (n) = 6 Sum of 6 terms (s6) = ? By formula, 𝑎(1−𝑟 𝑛 ) 𝑠𝑛 = 1−𝑟 r 0, ax + by + c < 0, or ax + by + c ≥ 0 or ax + by + c ≤ 0 is known as the linear inequality in two variables x and y. Let us consider one example; x ≥ 3. i) How many variable are there in the above inequality? ii) What do you mean by sign >? The associated equation of x ≥ 3 is x = 3 is a straight line (say MN) parallel to y-axis at a distance of 3 units from x-axis. From the graph, we observe that the line x = 3 i.e. MN divides the whole plane into two parts, one on the right side of MN each point of which x-coordinate is greater than 3 and other on the left of MN each point of which x-coordinate less than 3. Each point on MN has x-coordinate 3. The line x = 3 is said to be the boundary line. Hence, the graph of x ≥ 3 is shown in the plane region on the right of MN including MN. Hence, MN must be drawn by the solid line because the line MN is also included on the graph. Let’s take another inequality x 4  0 > 4 which is false. Hence, the graph of x-2y ≥ 4 is the plain region does not contain the origin. Similarly, taking testing point (0, 0) in 2x+y ≤ 8, Put x = 0, y = 0, then 2 x 0 + 0 ≤ 8 = 0 ≤ 8 which is true. Hence, the graph of 2x+y ≤ 8 is the plane region containing the origin.  The intersection part of the shaded region gives the required solution set of the given system of inequalities. 1.4.3 Linear programming Most of the business and economic activities may have various problem of planning due to limited resources. In order to achieve the business goal i.e. minimizing the cost of production and maximize the profit from the optimum use of available limited resources, linear programming is used. Linear programming is a mathematical technique of finding the maximum or minimum value of the objective function satisfying the given condition. The problem which has object of finding maximum or minimum value satisfying all the given condition is called linear programming (L.P) problem. To define linear programming, we need some basic definitions: (i) Decision variables: The non-negative independent variables involving in the L.P problem are called decision variables. For example: in 2x + 3y = 7, x and y are decision variable. (ii) Objective function: The linear function whose value is to be maximized or minimized (optimized) is called an objective function. 72 (iii) Constraints: - The conditions satisfied by the decision variables are called constraints. For example: if x and y be the number of first two kinds of articles produced, then x + y ≥ 1000; x ≥ 0 and y ≥ 0 are the constraints (iv) Feasible region (convex polygonal region): A closed plane region bounded by the intersection of finite number of boundary lines is known as feasible region. (v) Feasible solution: The values of decision variables x and y involved in objective function satisfying all the given condition is known as feasible solution. Maximum or the minimum value of an objective function will always occur at the vertex of feasible region. Example 3 Find the maximum and the minimum values of the objective function (F) = 4x – y subject to 2x + 3y ≥ 6, 2x – 3y ≤ 6 and y ≤ 2 Solution: - Here, The given constraints are 2x + 3y ≥ 6, 2x – 3y ≤ 6 and y ≤ 2 The objective function (F) = 4x – y To find: The maximum and the minimum value. The corresponding equation of given constraints are 2x+3y = 6 …………….. (i) 2x-3y = 6 ……………… (ii) and y = 2 ……………... (iii) From equation (i), 3y = 6 – 2x 6−2𝑥 or, y = 3 Table from equation (i) x 0 3 6 y 2 0 -2 Similarly, from equation (ii) 2x = 6+3y 6+2𝑦 or, x = 3 73 Table form equation (ii) x 3 0 -3 y 0 -2 -4 The Boundary line (i) passes through the point (0, 2), (3, 0) and (6, -2). Draw a solid line through these points in graph. Similarly, another boundary line (ii) passes through the points (3, 0), (0, -2) and (-3, -4). Draw a solid line through these points in graph. Taking common testing point (0, 0) in both inequality 2x + 3y ≥ 6 and 2x – 3y ≤ 6 Now, x = 0, and y = 0 then 2x + 3y ≥ 6 = 2 x 0 + 3 x 0 ≥ 6 = 0 ≥ 6 which is false. So, the graph of 2x + 3y ≥ 6 is the plane region does not contain the origin. Similarly, taking x = 0 and y = 0 in (ii) then 2x – 3y ≤ 6  2 x 0 – 3 x 0 ≤ 6 = 0 < 6 which is true. So the graph of 2x-3y ≤ 6 is the plain region containing the origin. From equation (iii), y = 2 is a straight line parallel to x-axis lying at a distance of 2 units above from x-axis. C B(6, 2 The graph of y ≤ 2 is the lower half plane from y = 2 including the A boundary line. From the graph, ∆ABC is the feasible region. The vertices of the feasible region are A(3, 0), B(6, 2) and C(0, 2) Now, the feasible solution is Vertices Objective function Remarks F = 4x – y A (3, 0) F = 4 x 3 – 0 = 12 B (6, 2) F = 4 x 6 – 2 = 22 22 (max) C (0, 2) F = 4 x 0 – 2 = –2 -2 (min)  Maximum value of F = 22 at the vertex B(6, 2) i.e. when x = 6 and y = 2 and the minimum value of F = -2 at the vertex C(0, 2) i.e. when x = 0 and y = 2 74 Example 4 Maximize and minimize Z = 2x + y under the constraints, x + y ≤ 6, x – y ≤ 4, x ≥ 0, y ≥ 0 Solution: Here, The given constraints are, x + y ≤ 6, x-y ≤ 4, x ≥ 0, and y ≥ 0 The objective function (Z) = 2x+y To find: maximum and minimum value The corresponding equation of the given constraints are: x + y = 6 …………… (i) x – y = 4 ……………. (ii) x = 0 ……………… (iii) y = 0 ………………. (iv) From the equation (i) y = 6 – x Table from equation (i) x 0 6 3 y 6 0 3 From equation (ii) y = x – 4 Table from equation (ii) x 0 4 2 y -4 0 -2 The boundary line (i) passes through (0, 6), (6, 0) and (3, 3). Plot the points and join them by the solid line. Similarly, the boundary line (ii) passes through (0, -4), (4, 0) and (2, -2). Plot the points and join them by the solid line. Taking common testing point (1, 1) in the given inequalities Now, x = 1 and y = 1 in x + y < 6 1 + 1 ≤ 6 = 2 ≤ 6 which is true. So the graph of x + y ≤ 6 in the plane region containing the testing point (1, 1). Similarly, taking x = 1 and y = 1 in x – y ≤ 4 75  1-1 ≤ 4 = 0 ≤ 4 which is true. So the graph of x – y ≤ 4 in the plane region containing the origin. From the equation (iii) and (iv) x = 0 and y = 0 are the y-axis and x-axis respectively. x ≥ 0 is the right half plane containing the y-axis and y ≥ 0 is the upper half plane containing the x-axis. From the graph quadrilateral OABC is the feasible region. The vertices of the feasible region are O(0, 0), A(4, 0), B(5, 1) and C(0, 6). Hence the feasible solution is Vertices Objective function Remarks Z = 2x + y O (0, 0) Z=2x0+0=0 0 (min) A (4, 0) Z=2x4+0=8 B (5, 1) Z = 2 x 5 + 1 = 11 11 (Max) C (0, 6) Z=2x0+6=6  Maximum value of Z = 11 at the vertex B(5, 1) and minimum value of Z = 0 at the vertex O (0, 0) Example 5 In the given diagram the coordinates of A, B, and C are (2, 0), (6, 0) and (1, 4) respectively. The shaded region inside the ∆ABC is represented by inequalities. Write down the equations of these inequalities and also calculate the minimum value of 2x+3y from the values which satisfy all the three inequalities. Solution: Here, ∆ABC is a feasible region and their coordinates are A(2, 0), B(6, 0) and C(1, 4) The objective function F = 2x + 3y To find: (i) Equation of inequalities (ii) The minimum value 76 For the line BC, The equation of line BC passing through B(6, 0) and C(1, 4) 𝑦2 −𝑦1 𝑦 − 𝑦1 = (𝑥 − 𝑥1 ) 𝑥2 −𝑥1 4−0 Or, 𝑦 − 0 = (𝑥 − 6) 1−6 4 Or, 𝑦 = (𝑥 − 6) −5 Or, 4𝑥 − 24 = −5𝑦 Or, 4𝑥 + 5𝑦 = 24 Since, the half plane with the boundary line BC contains origin. So the inequality of BC is 4x + 5y ≤ 24 Similarly, for AC The equation of line AC Passing through A(2, 0) and C(1,4) is 𝑦 −𝑦 𝑦 − 𝑦1 = 𝑥2−𝑥1 (𝑥 − 𝑥1 ) 2 1 4−0 Or, 𝑦 − 0 = 1−2 (𝑥 − 2) 4 Or, 𝑦 − 0 = −1 (𝑥 − 2) Or, 𝑦 − 0 = −4𝑥 + 8 Or, 4𝑥 + 𝑦 = 8 Since the half plane with the boundary line AC doesn’t contain the origin, the inequality of AC is 4x + y ≥ 8. The equation of AB means the equation of x-axis is y = 0 and the equation of y-axis is x = 0. The shaded region lies in the first quadrant only. So the inequality of AB is y > 0 and inequality of y-axis is x ≥ 0 The inequality equations are 4x + 5y ≤ 24, 4x + y > 8, x > 0, y > 0 77  Again for the minimum value, Vertices Objective function Remarks (F) = 2x+3y A (2, 0) F= 2 x 2 +3 x 0 = 4 4 (Min) B (6, 0) F = 2 x 6 + 3 x 0 = 12 C (1, 4) F = 2 x 1 + 3 x 4 = 14  Minimum value of F = 4 at the vertex A(2, 0) Exercise 1.4.1 1. a) Define boundary line with an example. b) In which condition the boundary line is dotted line? 2. a) Define constraints with an example. b) What do you mean by objective function? Also write an example. 3. a) In an objective function (F) = 5x – 2y, one vertex of feasible region is (5, 2) then find the value of objective function. b) From the given inequality 4x + 3y ≥ 10, Write the boundary line equation. 4. a) From the adjoining figure, i) What is called the shaded region ∆ABC? ii) Find the equation of AB. c) Draw the graph of x ≥ 0. d) Draw the graph of y ≤ 0. Y' 5. Where does the solution set lies for the inequalities x ≥ 0 and y ≥ 0? 6. a) Draw the graph of the following inequalities. i) x ≥ y ii) x + 2y ≤ 8 iii) x ≥ -5 iv) y ≥ 2x 78  b) Draw the graph of the following inequalities and shade the common solution set. i) 2x + 2y ≥ 6 and y ≥ 0 ii) 2x + y ≥ 6 and x ≥ 2 iii) x + y ≤ 2 and x ≤ 0 iv) x – 3y ≤ 6 and y ≤ 3 7. a) Find the maximum and minimum value of the function Z = 3x + 5y for each of the following feasible region, i) ii) Y' Y' 8. Draw the graphs of the following inequalities and find the feasible region. Also find the vertices of the feasible region. i) x + y ≤ 3 ii) x – 2y > 4 iii) 2x + y ≥ 4 iv) 2y ≥ x – 1 x≥2 2x + y ≤ 8 3x + 4y ≤ 12 x+y≤4 y≤1 y ≥ -1 x ≥ 0, y ≥ 0 x ≥ 0, y ≥ 0 9. Find the maximum values of the following objective functions with the given constraints. i) p = 14x + 16y subject to ii) z = 6x + 10y + 20 subject to 3x + 2y ≤ 12 3x +5y ≤ 15 7x + 5y ≤ 28 5x + 2y ≤ 10 x ≥ 0, y ≥ 0 x ≥ 0 and y ≥ 0 iii) F = 6x + 5y subject to iv) Q = 3x + 2y subject to x+y≤6 x+y≥0 x – y ≥ -2 x–y≤0 x ≥ 0, y ≥ 0 y ≤ 2, x ≥ -1 79 10. Find the minimum values of the following objective functions with the given constraints. i) 2x + 4y < 8 ii) Objective function (M) = x + y subject to 3x + y ≤ 3 3x + 4y ≤ 21 x ≥ 0, y ≥ 0 2x + y ≥ 4 objective function (F) = 5x + 4y x ≥ 0, y ≥ 0 iii) Objective function (L) = 3x + 5y iv) Objective function (F) = 6x + 9y subject to 2x + y ≤ 6 Subject to x+y≥3 2x + y ≤ 9 x ≥ 0, y ≥ 0 y ≥ x, x ≥ 1 11. In the adjoining figure, the coordinates of A, B, C are (-3, 0), (2, 0) and (0, 2) respectively. Find the inequalities represented by the shaded region and also calculate the maximum value of 3x + 4y which satisfy all three inequalities. Y' 12. Study and discuss, in the adjoining figure, find the inequalities which represent the boundaries of shaded region as solution set and find the maximum and minimum value of Z = 3x + 5y. Prepare a report and present it in your class room. Y' 80 1.5 Quadratic equations and graph 1.5.0 Review Let us consider, an equation and discuss on it y = 2x + 3 (i) What are x and y called? (ii) What are the maximum degree of x and y? (iii) What has the degree of constant term? (iv) Write the name of given equation. Can you draw the graph of the given equation? If it is, let’s discuss on its shape. Again, let us consider another equation x2 + 3x + 2 = 0 (i) Write the degree of the equation. (ii) What is the variable in that equation? (iii) Can you write the roots of the equation x2 + 3x + 2 = 0? Discuss above questions in different groups and write the conclusion 1.5.1 Graph of quadratic function The general form of quadratic function is y = ax2 + bx + c where, a, b and c are called coefficient of x2, coefficient of x and constant term respectively. The graph of quadratic function is called parabola. a) Graph of the quadratic function (y = ax2) i) y = ax2 where a = 1 to draw the graph, let’s find some values of x and y x 0 ±1 ±2 ±3 ±4 y 0 1 4 9 16 From the table, plotting the pair of points in a graph and joined them freely ii) y = x2 where a = -1 To draw the graph, let’s find some values of x and y x 0 ±1 ±2 ±3 ±4 0 -1 -4 -9 -16 y Plotting the pair of points in a graph and joined them freely. 81 From graphs, we can say that, for different values of a, we get different curves of the same nature known as the parabola with same turning point origin known as the vertex of parabola. Also, we can find the following information: i) The parabola turns upward for a > 0 and turns downward for a < 0. ii) Each parabola is symmetrical about y-axis i.e. y-axis divides each parabola into two identical parts. iii) Greater value of ‘a’ numerically narrower will be the faces of parabola and lesser the value of ‘a’ numerically wider will be the faces of parabola. b) Graph of the quadratic function y = ax2 + bx + c the given quadratic function is Or, 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 𝑏 𝑐 Or, 𝑦 = 𝑎 (𝑥 2 + 𝑥 + ) [ Taking a common] 𝑎 𝑎 𝑏 𝑏 2 𝑏 2 𝑐 Or, 𝑦 = 𝑎 {(𝑥)2 + 2. 𝑥. +( ) −( ) + } 2𝑎 2𝑎 2𝑎 𝑎 𝑏 2 𝑏2 𝑐 Or, 𝑦 = 𝑎 {(𝑥 + 2𝑎) − 4𝑎2 + 𝑎} 𝑏 2 4𝑎𝑐−𝑏2 Or, 𝑦 = 𝑎 {(𝑥 + 2𝑎) + 4𝑎 2 } 𝑏 2 4𝑎𝑐−𝑏2 Or, 𝑦 = 𝑎 (𝑥 + 2𝑎) + 4𝑎 82  −b 4ac−b2 Which is in the form of y = a (x - h)2 + k Where h = ,k= 2a 4a 2 b 4ac−b  The vertex of parabola = (h, k) = (− , 4a ) 2a When k = 0, the equation of parabola takes the form y = a(x – h)2 where the vertex = (h, 0). Conclusion: i) The graph of y = ax2 + bx + c is called parabola is symmetrical to a line parallel to y-axis. 𝑏 ii) The equation of line of symmetry is x = − 2𝑎 Example 1 Draw the graph of y = x2 +2x – 8. Also find the equation of line of symmetry. Solution: here, P The given quadratic equation is y = x2 +2x -8 …………. (i) Now, comparing equation (i) with y = ax2 + bx + c we get a = 1, b = 2 and c = -8 Now, x - coordinate of the vertex of parabola 𝑏 −(2𝑎) (x) = − = = −1 2𝑎 2𝑥𝑎 y-coordinate of the vertex of parabola (y) = (-1)2 + 2(-1) – 8 Q = 1 -2 - 8 = -9  Vertex or turning point of the parabola (h, k) = (-1, -9) Table from equation (i) x -1 0 1 2 -2 3 -3 -4 -5 Y -9 -8 -5 0 -8 7 -5 0 7 Plotting the pair of points (-1, -9), (0, -8), (1, -5), (2, 0), (-2, -8), (3, 7), (-3,-5), (-4,0), (-5,7). Join these points freely. The Parabola meets the x-axis at two points (2, 0) and (-4, 0). Again, PQ is the line of symmetry and its equation is x = -1. Note: For the quadratic function y = ax2 + bx + c the turning point not at the origin. 83 1.5.2 Graph of a cubic function The general form of cubic function is defined by y = ax3 + bx2 + cx + d where, a, b, c and d are constants and a ≠ 0. The simplest form of a cubic function passing through origin is y = ax3. Different values of ‘a’ will give different curves passing through the origin with similar nature. Now y = ax3 when a = 1 then discuss on its graph y = x3 …………… (i) Table from equation (i) x 0 1 2 -1 -2 y 0 1 8 -1 -8 Plotting the pair of points (0, 0), (1, 1), (2, 8), (-1, -1), (-2, -8) in a graph and joined them freely. The curve line passes through the points of 1st quadrant, origin and 3rd quadrant points when coefficient of x3 (a) is positive. Example 2 Draw the graph of y = -x3 where a = -1 also, write the nature of graph. Solution: Here, The given cubic function is y = -x3 Table from equation (i) x 0 1 -1 2 -2 y 0 -1 1 -8 8 Plotting the pair of points (0, 0), (1, -1), (-1, 1), (2, -8) and (-2, 8) in a graph and joined them freely Again, the curve line passes through the points of 2nd quadrant, origin and 4th quadrant points when coefficient of x3 (a) is negative 84 Example 3 Draw the graph of y = 2x3, also write the value of ‘a’. Solution: Here, The given cubic equation is y = 2x3 ………………….. (i) Table from the equation x 0 1 -1 2 -2 y 0 2 -2 16 -16 Plotting the pair of points (0, 0), (1, 2), (-1, -2), (2, 16), (-2, -16) in a graph and joined them freely The value of a = 2 Let us discuss on the following equations and draw the graph of (i) 𝑦 = −2𝑥 3 1 (ii) 𝑦 = 𝑥3 2 1 (iii) 𝑦 = − 2 𝑥 3 1.5.3 Solution of quadratic equation and linear equation Let us consider two equations: y = x2 – 5x + 6 and y = 2 Let us discuss on the nature of the graph of both equations and draw the graph Now, y = x2 – 5x + 6 ………… (i) y = 2 ………………….. (ii) Comparing equation (i) with y = ax2 + bx + c, we get a = 1, b = -5, c = 6 −b (−5) 5 X-coordinate of the vertex of parabola (x) = = = = 2.5 2a 2x1 2 Y-coordinate of the vertex of parabola (y) 5 2 5 25 25 25−50+24 −1 = ( ) − 5. + 6 = − +6 = = = – 0.25 2 2 4 2 4 4 5 −1  The vertex of parabola (h, k) = ( 2.5, - 0.25) = (2 , 4 ) 85 Table from equation (i) x 2.5 0 1 2 3 4 5 y -0.25 6 2 0 0 2 6 Plotting the pair of points (2.5, -0.25), (0, 6), (1, 2), (2, 0), (3, 0), (4, 2), (5, 6) in a graph and joined them freely Again, From equation (ii) y = 2 represent a straight line parallel to y- axis which is 2 units above the X-axis. From the graph, the intersection points of the parabola and the straight lines are (1, 2) and (4, 2) The solutions are x = 1, y = 2 and x = 4, y = 2 Hence, x = 1, 4 and y = 2 By substitution method the given equations are y = x2 – 5x + 6 ………………….. (i) and y = 2 ……………………….. (ii) Substituting y = 2 in equation (i), we get or, 2 = x2 – 5x + 6 or, x2 – 5x + 6 – 2 = 0 or, x2 – 5x + 4 = 0 or, x2 – (4 + 1)x + 4 = 0 or, x2 – 4x – x + 4 = 0 or, x (x - 4) – 1 (x - 4) = 0 or, (x – 4) (x – 1) = 0 Either, x – 4 = 0 x=4 or, x – 1 = 0 x=1 x = 4, 1 and y = 2 Note: When vertex of parabola (x-coordinate or y coordinate) is in fraction especially their denominator is 4 then take a scale as 8 small boxes = 1 unit so that it is easy to draw the graph. 86 Example 1 Solve: y = x2 – 3x + 5 and y = 2x + 1 by graphical method and substitution method. Solution: Here, The given equations are: y = x2 – 3x + 5 …………………. (i) y = 2x + 1 ………………………. (ii) Comparing equation (i) with y = ax2 +bx + c, we get a = 1, b = -3, c = 5 −b (−3) 3 Now, x-coordinate of the vertex of parabola (x) = =− = = 1.5 2a 2 2 3 2 3 And y-coordinate of the vertex of parabola (y) = ( ) − 3 ( ) +5 2 2 9 9 9−18+20 11 = − +5 = = = 2.75 4 2 4 4 3 11  The vertex of parabola (h, k) = ( , ) = (1.5, 2. 75) 2 4 Table from equation (i) x 1.5 0 1 -1 2 3 4 y 2.75 5 3 9 3 5 9 Plotting the pair of points (1.5, 2.75), (0, 5), (1, 3), (-1, 9), (2, 3), (3, 5) (4, 9) in a graph and joined them freely Again, From equation (ii) y = 2x + 1 Table for equation (ii) x 0 -1 1 y 1 -1 3 Plotting the pair of points (0, 1), (-1, -1), (1, 3) in the same graph and draw the straight line The intersection points of the parabola and a straight line are (1, 3) and (4, 9) 87   The solutions are x = 1, y = 3, and x = 4, y = 9 Hence, (1, 3) and (4, 9) are solution set By substitution method The given equations are y = x2 – 3x + 5 …………………. (i) y = 2x + 1 ………………………. (ii) Substituting y = 2x + 1 from equation (ii) in equation (i), we get or, 2x + 1 = x2 – 3x + 5 or, x2 - 3x + 5 – 2x -1 = 0 or, x2 – 5x + 4 = 0 or, x2 – (4 + 1)x + 4 = 0 or, x2 – 4x – x + 4 = 0 or, x(x – 4) - 1(x – 4) = 0 or, (x – 4) (x – 1) = 0 Either, x – 4 = 0 x=4 or, x–1=0 x=1 When x = 4 then y = 2 x 4 + 1 = 9 When x = 1 then y = 2 x 1 + 1 = 3 Hence, the solutions are x = 4, y = 9, and x = 1, y = 3 Example 2 Solve the equation graphically: x2 + 2x – 3 = 0. Solution: Here, The given equation is x2 + 2x – 3 = 0 or, x2 = 3 – 2x Let, y = x2 = 3 – 2x Taking, y = x2 …………………. (i) And y = 3 – 2x ……………… (ii) From equation (i) y = x2 is in the form of y = ax2 where (a = 1) so its turning point (vertex) is always origin (0, 0). 88  Table from equation (i) x 0 1 -1 2 -2 3 -3 y 0 1 1 4 4 9 9 Plot the pair of points (0, 0), (1, 1), (-1, 1), (2, 4), (-2, 4), (3, 9) (-3, 9) in a graph and joined them freely Again, from equation (ii) y = 3 – 2x Table from equation (ii) X 0 1 2 Y 3 1 -1 Plotting the pair of points (0, 3), (1, 1), (2, -1) in the same graph and draw the straight line. From the graph, the intersection points of parabola and the straight line are (1, 1) and (-3, 9)  x = 1, -3 Exercise 1.5.1 1. a) Define vertex of parabola. b) In a quadratic equation ax2 + bx + c = 0, what a, b and c are called? 2. a) Define line of symmetry in parabolic curve. b) Write the equation of line of symmetry in the equation y = x2. c) Write the equation of line of symmetry in the equation y = ax2 + bx + c. 89 3. Write the vertex and the equation of line of symmetry of the following graph: Scale: 5 small boxes = 1 unit i) ii) y y x’ O x x’ O x y’ y’ (iii) y x’ O x y’ 90  1 4. a) Write the nature of the graph of 𝑦 = 2 𝑥 2. b) Write the nature of the graph of 𝑦 = − 𝑥 3. 5. a) Draw the graph of the following equation (function): 1 1 i) 𝑦 = −2𝑥 2 ii) 𝑦 = − 𝑥 2 iii) 𝑦 = 𝑥 2 iv) 𝑦 = 3𝑥 2 2 2 b) Draw the graph of following functions: i) 𝑦 = 2𝑥 2 ii) 𝑦 = 3𝑥 3 iii) 𝑦 = −3𝑥 3 6. Find the vertex of the following equations: i) y = 4x2 + 8x + 5 ii) y = x2 – 6x iii) x2 = 2y 2 2 iv) y = x + 3x + 2 v) y = x – 6x + 5 7. Draw the graph of the following function: i) y = x2 + 2x – 5 ii) y = 3x2 – 2 iii) y = x2 + 4x – 1 8. Solve the following equations using graphical method as well as substitution method: i) y = x2 and y = 3 – 2x ii) x2 = 2y and y = x 2 iii) y = x – 2x and y = x – 2 iv) y = x2 + 3x -10 and x = y v) y = x2 + 4x – 7 and y = 2x + 1 vi) y = x2 + 8x – 6 and y = 4 – x 9. Solve the following equations by graphical method: i) x2 + 2x – 3 = 0 ii) x2 – 5x + 6 = 0 iii) x2 – 2x – 15 = 0 iv) 3x2 + 5x + 2 = 0 v) 2x2 – 7x + 3 = 0 vi) x2 + 6x + 5 = 0 10. Observe the adjoining graph, write the vertex of parabola and the equation direction of the opening of parabola and the equation of parabola. Also list the steps to find its equation, prepare a report and present it in your class. 91 Unit 2 Continuity 2. 0 Review Limit of a Function a. Observe the following figures. Here, let n denotes the number of sides of the polygon. If the sides of polygon increases indefinitely i.e. 𝑛 → ∞, then the polygon takes the form of a circle. Thus, we write lim 𝑝𝑜𝑙𝑦𝑔𝑜𝑛 = 𝑐𝑖𝑟𝑐𝑙𝑒. 𝑛→∞ b. For the function 𝑦 = 𝑓(𝑥) = 𝑥 2 , observe the following chart: 𝑥 1.9 1.99 1.999 1.9999 𝑦 = 𝑓(𝑥) = 𝑥 2 3.61 3.9601 3.996001 3.999600011 𝑥 2.0001 2.001 2.01 2.1 𝑦 = 𝑓(𝑥) = 𝑥 2 4.00040001 4.004001 4.0401 4.41 As x approaches to 2 from left, i.e. 𝑥 → 2− then 𝑓(𝑥) = 𝑥 2 approaches 4. We write lim− 𝑓(𝑥) = 4. This is known as left hand limit. Similarly, as x approaches to 𝑥→2 2 from right i.e. 𝑥 → 2+ then 𝑓(𝑥) = 𝑥 2 approaches to 4. We write lim+ 𝑓(𝑥) = 𝑥→2 4. This is known as right hand limit. In such case, we say the limit of the function exists and has the value 4. We write simply, lim 𝑓(𝑥) = lim 𝑥 2 = 4 𝑥→2 𝑥→2 Thus, the existence of limit if any function depends upon the fact that left hand limit and right hand limit of the function exist and is equal. i.e. lim− 𝑓(𝑥) = lim+ 𝑓(𝑥) = 𝑓(𝑎) 𝑥→𝑎 𝑥→𝑎 It can be summarized that let f(x) be defined about 𝑥0 except possibly at 𝑥𝑜 itself. If f(x) gets arbitrarily close to L for all sufficiently close to 𝑥𝑜 , we say that f(x) approaches the limit L as approaches 𝑥𝑜. We write lim 𝑓(𝑥) = 𝐿. 𝑥→𝑥𝑜 92  𝑥 2 −1 c. Consider the function 𝑓(𝑥) =. Finding the limiting value of f(x) as 𝑥 → 1. Here, 𝑥−1 0 x = 1 makes f(x) undefined, since𝑓(𝑥) =. Therefore, we try to find the limiting value 0 of f(x) when x is sufficiently close to 1 but not equal to 1. putting x = 0.99 (0.99)2 −1 f(0.99) = 0.99−1 −0.0199 = −0.01 = 1.99 2 putting x = 1.01 (1.01)2 −1 f(1.01) = 1.01−1 = 2.01  2 𝑥 2 −1 So, 𝑙𝑖𝑚 = 2 𝑥→1 𝑥−1 2.1 Investigation of continuity in different sets of numbers We know, some set of numbers are natural numbers, whole numbers integers, fractions, rational numbers, irrational numbers and real numbers. Here we shall discuss about continuity and discontinuity in their order in real number line. In real number line, numbers are extended in a line from – ∞ to ∞. a. Observe the number line with Natural number below. It is obvious that we can find natural numbers between 2 and 6. These are 3, 4, 5. Similarly, we can find natural numbers between 10 and 20. Obviously there are 11, 12, 13, 14, 15, 16, 17, 18, 19. But in number line there is no natural number between 4 and 5. 4.5 is also the number in number line but not natural number. Same property exists in case of integers also. There is no continuity in the set of integers. The set of integers does not hold a property of continuity. b. The set of whole numbers also holds the same property of discontinuity as that of the set of natural numbers. The set of whole numbers is discontinuous as can be seen below: 93 c. Consider the set of rational numbers Q, which have no common factors (or in lowest p form). 𝑄 = {𝑥|𝑥 = q where p and q are integers but q ≠ 0} All the numbers in number line are not only rational numbers. There are other numbers which are irrationals. Rational numbers and irrational numbers form the set of real numbers. Graphically: Therefore, the set of rational numbers Q holds the property of discontinuity in increasing or decreasing order. d. Continuity in set of real numbers The set of real numbers is represented by the points in the real number line. There is one-one correspondence between the real numbers and the points on the number line. Between any two real numbers on the line there correspond real numbers represented by the points between given two points. This establishes the fact that the set of real numbers is dense. Hence, the real number line is continuous. Exercise 2.1 1. Write the following sets in set builder or tabular form. (a) natural number (b) whole number (c) integers (d) rational numbers (e) irrational numbers (f) real numbers 2. (a) Show the first five natural number in a number line. (b) Write the natural numbers from 20 to 30. (c) Write the rational numbers from -5 to 5. (d) Form a set of whole numbers with first ten elements. Present them in a number line. 3. Study the following number line and examine the continuity or discontinuity in the set of numbers. (a) 94  (i) Set of natural numbers (ii) set of integers (iii) set of real numbers (b) (i) set of integers (ii) set of rational numbers (iii) set of whole number (iv) set of irrational numbers 4. Observe the following and find the continuity or discontinuity in their situation: (a) The movement of frog from one place to other (b) The movement of crocodile from one place to other (c) The height of a plant from first Sunday to second Sunday of the same month (d) The weight of a person (e) The flow of water in a river (f) The number of presence of students in your class for a week 5. Collect any three examples of continuity and three examples of discontinuity which can be shown in scale. Present your findings in classroom 2.2 Investigation of continuity and discontinuity in graphs Let us observe the following graphs for inequalities: (a) (b) What is the inequality? What is the inequality? (c) (d) What is the inequality? What is the inequality? 95 Here, (a) The inequality for solution region is -5 < x < 5. -5 and 5 both do not belong to the solution region. It is written as (-5, 5) = {x: -5 < x < 5} (b) The inequality for solution region is -5 < x < 5. The end point -5 belongs to the solution region but 5 does not belong to the region. (c) The inequality for solution region is -5 < x < 5. The end point -5 does not belong to the region but 5 belongs to the region. (d) The inequality for solution region is -5 and 5 both belong to the region. Again observe the following graphs from -5 to +5. (a) (b) (c) (d) 96 If we discuss about continuity and discontinuity of the above function at x = 0, we get the following results: (a) The function is defined from -5 to +5 in X-axis. At x = 0 there is no break, jump, gap or hole. So the function is continuous at x = 0. (b) The function is defined from -5 to +5 in X-axis At x = 0, the curve has no break or no gap, so the function representing by the curve is continuous at x = 0. (c) The function representing by the curve is defined from x = -5 to x = 5. At x = 0 the graph has a break so the function representing the curve is discontinuous at x = 0. (d) The function representing by the straight line is defined from x = –5 to x = 5. The graph has break/gap at x = 0, so it is discontinuous at x = 0. Let y = f(x) be a function defined from x = a to x = b. The function f(x) is said to be continuous at x = c if its graph has no 'break', 'jump', 'gap' or 'hole' at x = c otherwise, a function y = f(x) is said to be discontinuous at x = c. Example 1 Draw the graph of y = sinx when -180° < x < 360° and discuss about its continuity at x = -90° and x = 180°. Solution To draw the sine graph as in chapter one, list the value of sinx corresponding to x as following in the difference of 90°. Let 10 small division along horizontal axis represents 90° and along vertical axis 10 small divisions represent 1 unit. We get (-180° and along vertical axis small 10 small divisions represent unit. We get (-180°, 0) (-90°, -1), (0°, 0) (90°, 1), (180°, 0), (270°, -1) …., (360°, 0) coordinates to plot on the graphs paper. Join these coordinates by free hand. We get the graphs as below. 97  At x = -90° and x = 180°, the graph has no break and no jump, so the function is continuous at x = -90° and x = 180° Example 2 Examine the continuity or discontinuity of the graph defined from x = -3 to x = 8. Solution The function is defined from x = -3 to x = 8 in the graph. The graph has different steps (i.e. 5 steps.) The graphs is discontinuous at x = -1, x = 0, x = 1 and x = 4, But it is piece wise continuous for -3 < x < -1, -1 < x < 0, o < x < 1, 1 < x < 4 and 4 < x < 8 Exercise 2.2 1. From the following graphs, find: (a) domain of the function (b) Point of discontinuity (c) one point of continuity for the following graphs of function. (a) (b) 98  (c) (d) (e) (f) 2. Draw graph of the following functions and discuss about continuity at different points at most 3 points. (a) y = x + 2(-4 < x < 5) (b) y = x2(-6 < x < 6) (c) y = x3 (-10 < x < 10) (d) y = cosx(-180° < x < 360°) 3. Collect the different daily life examples of continuity and discontinuity and make a short report. 99 2.3 Symbolic representation of continuity 𝑥 2 −1 Let us suppose a function 𝑓(𝑥) = 𝑥−1 defined for - < x < . Also take different situations at x = 2 and x = 1 22 −1 At x = 2, f(2) = 2−1 =3 (1.99)2 −1 3.9601−1 for x < 2, let, x = 1.99, f(1.99) = 1.99−1 = 0.99 = 2.99 = 3 (nearly) for x > 2, let x = 2.01(say), f(2.01) = (2.01)2 −1 4.0401−1 = = 3.01 = 3 (App.) 2.01−1 1.01 At x = 2, lim– 𝑓(𝑥) = 2.99 = 3 (App.) 𝑥→2 lim 𝑓(𝑥) = 3.01 = 3 (App.) 𝑥→2+ f(2) = 3 So, the function has no jump, hole, break at x = 2. the function is continuous at x = 2 (1)2 −1 0 At x = 1, f(1) = = (does not exist) 1−1 0 There is gap at x = 1 But for x < 1 and x > 1 let us take x = 0.99 and x = 1.01, we get f(0.99) = 1.99 = 2 (nearly) and f(1.01) = 2.01 = 2(nearly) lim 𝑓(𝑥) = lim+ 𝑓(𝑥) = 2 𝑥→2– 𝑥→2 The function is discontinuous at x = 1 Let y = f(x), be a function defined at x = a, f(x) is said to be continuous at x = a if lim −𝑓(𝑥) 𝑥→2 = lim+ 𝑓(𝑥) = f(a). lim− 𝑓(𝑥) is read as when x is nearly a from left to the value of f(x) is 𝑥→𝑎 𝑥→𝑎 nearly approaches to f(a). lim+ +𝑓(𝑥) is read as when x is nearly approaches a from right, 𝑥→𝑎 f(x) is nearly approaches f(a). 100 Example 1 1. A function f(x) is defined as follows: f(x) = 3x + 5 for 0 < x < 1 9x – 1 for x > 1 Examine the continuity at x = 1 Solution: for x = 1, f(x) = 9x – 1 Now, f(1) = 9 x 1 – 1 = 9 – 1 = 8 for x > 1, f(x) = 9x – 1 let us take x = 1.01 (nearly x = 1) f(1.01) = 9 x (1.01) – 1 = 9.09 – 1 = 8.09 = 8 (nearly) since, lim— 𝑓(𝑥) = 𝑓(1) = lim+ 𝑓(𝑥), so the function is continuous at x = 1 𝑥→1 𝑥→1 Exercise 2.2 1. a. Explain lim— 𝑓(𝑥), lim+ 𝑓(𝑥) and f(2) in words. 𝑥→2 𝑥→2 b. If f(x) = x + 2, what is f(2)? c. If f(x) = 2x – 1, what is lim— 𝑓(𝑥)? (take x = 0.99) 𝑥→2 d. If f(x) = 3x + 1, What is lim+ 𝑓(𝑥) (take x = 1.01) 𝑥→2 2. If f(x) = 3x + 2, a. find f(2.001), f(2.0001), f(1.999), f(1.9999) b. find lim— 𝑓(𝑥), lim+ 𝑓(𝑥) 𝑎𝑛𝑑 𝑓(2) 𝑥→2 𝑥→2 c. Is it continuous at x = 2? 𝑥 2 −9 3. If f(x) = , find: 𝑥−3 a. f(2.999) and f(3.001) b. f(2.999) and f(3.001) equal after approximation c. Discuss about continuity at x = 3 101 4. Examine the continuity at points mentioned below: a. f(x) = 2 - x2 , for x < 2, at x = 2 x – 4, for x > 2 x, for x < 0 b. f(x) = 0, for x = 0, at x = 0 x2 , for x > 0 2 c. f(x) = for x < 3, at x = 3 5−𝑥 5 – x for x > 3 5. Take a quadratic function and test its continuity at particular point. 102  Unit 3 Matrix 3.0 Review Observe the following table: Articles/Shops A B Pen Rs. 40 Rs. 50 Copy Rs. 35 Rs. 30 Bag Rs. 400 Rs. 450 If we interchange the position of shops and arties, what will change in matrix? Discuss. The above table shows the price of three articles in two shops. The profit from each unit of pen, copy and bag are Rs 4. Rs. 6 and Rs. 50 respectively. a) Write the information in the above table as a 3x2 matrix A. b) Write a row matrix B that represents the profit, per units of each type of product. c) Find the product of B and A. d) State what the elements of BA represent? 40 50 The product of matrix can be written as (4 6 50) ( 35 30 ) 400 450 Answer the above questions. 7.1 Determinant of a Matrix. a11 a12 pen book a b 1 2 A = [a a22 ] , B = [copy eraser] , C = [c d] , D = [3 8] 21 i) What kinds of matrix are these? ii) What is the order of each matrix? iii) Corresponding to each matrix, is there a number? Determinant is a function which associates each square matrix with a number. The determinants of above matrices are denoted by det (A), det (B), det (C), det (D) or |A|, |B|, |C|, |D| ‘’ is the single notation for determinant of any square matrix. 103  i.e. If A be a square matrix, its determinant is denoted by det (A) or |A| and is a number associated to that matrix A. A matrix whose determinant is Zero is said to be singular. Let A = [a11] be a square matrix of order 1x1 then det (A) is a11, If a11 = -5 then det (A) = -5. If a11 = 5, det (A) = 5. [The determinant of 1x1 matrix is simply the element of the matrix.] a11 a12 if, A = [a ] 21 a 22 a11 a12 The determinant of A is denoted by |A| = |a | = a11a22 – a21a12 21 a 22 Product of elements leading diagonal – product of elements of secondary diagonal. 2 1 1 2 Let us take two matrices A = [ ] and B = [ ] 0 2 2 4 2 1 Now, determinant of A = | | =2x2–1x0=4–0=4 0 2 1 2 and determinant of B = | | =1x4–2x2=4–4=0 2 4 A is non-singular matrix but B is singular matrix. Example 1 Evaluate 2 3 −2 −√5 𝑥 4 a) | | b) | | c) | | 4 5 −√5 3 𝑥 𝑥2 Solution 2 3 a) | | = 2 × 5 − 3 × 4 = 10 – 12 = − 2 4 5 −2 −√5 b) | | = (−2) × 3— (−√5)(−√5) = −6 − 5 = − 11 −√5 3 𝑥 4 c) | | = 𝑥 2. 𝑥 – 4𝑥 = 𝑥 3 – 4𝑥 𝑥 𝑥2 Example 2 If A = ( 2 1 ) then find |A| −3 −4 Solution Here, A = ( 2 1 ) −3 −4 Now, |A| = | 2 1 | = 2 × (−4) – 1 × (−3) = −8 + 3 = − 5 −3 −4 104 Example 3 Solve for x 𝑥−1 𝑥 | |=0 𝑥2 + 1 𝑥2 + 𝑥 + 1 Solution 𝑥−1 𝑥 Here, | 2 |=0 𝑥 +1 𝑥2 + 𝑥 + 1 or, (x – 1) (x2 + x + 1) – x(x2 + 1) = 0 or, x3 – 1 – (x3 + x) = 0 or, - 1 – x = 0 or, -1 = x x=-1 Example 4 2 4 If I is the identity matrix of order 2 x 2 and 𝐴 = ( ), find the determinant of 4A + 3I 3 4 Solution: Here, 2 4 1 0 A=( ) and I = ( ) 3 4 0 1 2 4 1 0 Now, 4A + 3I = 4( ) + 3( ) 3 4 0 1 8 16 3 0 =( )+( ) 12 16 0 3 8 + 3 16 + 0 11 16 =( )=( ) 12 + 0 16 + 3 12 19 |4A + 3I| = |11 16| 12 19 = 11 × 19 − 16 × 12 = 209 − 192 = 17 Exercise 3.1 1. a) Define determinant of a square matrix. b) What do you mean by singular matrix? 2. a) Evaluate 2 3 5 3 −√2 𝑦2 −2 i) | | ii) | | iii) |√5 | iv) | | 4 5 −1 −4 √2 √5 𝑦 −3 3 4 −𝑎 −𝑏 b) If 𝐴 = ( ), then, find |A| c) If 𝐵 = ( ), then find |B| 5 2 𝑏 𝑎 105  𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 d) If 𝐴 = ( ), then find |A| −𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 𝑐𝑜𝑠50° 𝑠𝑖𝑛10° e) If 𝐴 = ( ), then find |A| 𝑠𝑖𝑛50° 𝑐𝑜𝑠10° 3. Solve for x 𝑥 5 4𝑥 2 a) | | = 24 b) | | = 12 −4 𝑥 6 3 𝑥+3 4 𝑥−1 𝑥−2 c) | |=7 d) | |=0 𝑥−3 5 𝑥 𝑥−3 4. If I is the identity Matrix of order 2 x 2, find the following: 2 1 a) If A = ( ) , then |2A + 3I| 1 2 2 0 b) If A = ( ) , then|A2 + 2A + 4I| −1 3 4 2 c) B= ( ) , then |B 2 − 2B + 3I| 1 −2 m + 2 −3 5. If A = ( ) and |A| = 3m – 1 then find A2 + 3A m + 5 −4 6. Construct a matrix A of order 2 x 2 whose element aij = 3i + 2j and find |3A2+2A| 2 3 −1 2 7. If 𝐴 = ( ) and 𝐵 = ( ) verify that |AB| = |A| |B| 4 5 2 1 3.2 Inverse of a matrix 1 1. What is the product of a and 𝑎? 4 −3 −2 3 2. If 𝐴 = ( ) 𝑎𝑛𝑑 𝐵 = ( ), Find AB and BA. What conclusion do you get 3 −2 −3 4 from AB and BA? Discuss in the class. [For an n x n matrix A, if there is a matrix B for which AB = I = BA, then B is the inverse of A. B is written as A-1 and A is written as B–1 ab e f Let, A = [ ] and A−1 = [ ] c d g h b e f a ae + bg af + bh AA−1 = [ ][ ] =[ ] c d g h ce + dg cf + dh 1 0 We know, AA−1 = [ ] 0 1 106 Now, ae + bg = 1 …….. (i) ce + dg = 0 ……………(ii) ace + bcg = c [Equation (i) x C – equation (ii) x a] ace + adg = 0 (-) (-) (-) bcg – adg = c or, g(bc – ad) = c c or, g = bc−ad −c or, g = ad−bc [Equation (i) × d – equation (ii) × b] Again, d[ae + bg = 1] = ade + bdg = d ………… (i) b[ce + dg = 0] = bce + bdg = 0 ………….. (ii) 𝑜𝑟, (𝑎𝑑 − 𝑏𝑐)𝑒 = 𝑑 𝑑 𝑜𝑟, 𝑒 = 𝑎𝑑 − 𝑏𝑐 Again, af + bh = 0 ……….. (iii) cf + dh = 1 ………...(iv) solving (iii) and (iv) we get, c[ab + bh =0] = acf + bch = 0 [Equation (iii) × c – equation (iv) × b] a[cf + dh =1] = acf + adh = a (𝑏𝑐 − 𝑎𝑑)ℎ = −𝑎 (𝑎𝑑 − 𝑏𝑐)ℎ = 𝑎 𝑎 𝑜𝑟, ℎ = 𝑎𝑑 − 𝑏𝑐 𝑑[𝑎𝑓 + 𝑏ℎ = 0] → 𝑎𝑑𝑓 + 𝑏𝑑ℎ = 0 [Equation (iii) × d – equation (iv) × b] 𝑏[𝑐𝑓 + 𝑑ℎ = 1] → 𝑏𝑐𝑓 + 𝑏𝑑ℎ = 𝑏 (𝑎𝑑 − 𝑏𝑐)𝑓 = −𝑏 𝑏 𝑜𝑟, 𝑓 = − 𝑎𝑑 − 𝑏𝑐 𝑑 𝑏 𝑒 𝑓 1 𝑑 −𝑏 𝑁𝑜𝑤, 𝐴−1 =[ ] = [𝑎𝑑−𝑐 − 𝑏𝑐 𝑎𝑑 − 𝑏𝑐 ] = 𝑎 [ ] 𝑔 ℎ 𝑎𝑑 − 𝑏𝑐 −𝑐 𝑎 𝑎𝑑 − 𝑏𝑐 𝑎𝑑 − 𝑏𝑐 1 𝑑 −𝑏 𝑆𝑜, 𝐴−1 = [ ] |𝐴| −𝑐 𝑎 Note: 1. We can find the inverse of 2 x 2 matrix by interchanging the elements of leading diagonal, changing the sign of elements of secondary diagonal and dividing new matrix by determinant of given matrix. 2. At exists of and only if |A|  0, ie A is non-singular. 107 Let A and B are two non-singular matrices, then i) A. A-1 = A-1. A = I ii) (AB)-1 = B–1. A-1 iii) (A-1) = A iv) (AT)–1 = (A–1)T Example 1 2 3 If A = ( ) , does A−1 exist? 4 5 Solution: Here, 2 3 𝐴=( ) 4 5 2 3 |A|=| | = 2 x 5 – 3 x 4 = 10 – 12 = - 2 4 5 Since |A|≠0, so, A-1 exists Example 2 𝑦−3 𝑦−1 For what value of y, the matrix ( ) does not have its inverse? 𝑦 2(𝑦 − 1) solution: Here, y−3 y−1 Let A = ( ) y 2(y − 1) y−3 y−1 |A| = | | = 2(y - 1) (y - 3) –y (y - 1) y 2(y − 1) = 2 (y2 – y -3y + 3) – (y2 – y) = 2y2 – 8y + 6 – y2 + y = y2 + 7y + 6 -1 if A does not exist, then |A| = 0 or, y2 – 7y + 6 = 0 or, y2 – 6y – y +6 = 0 or, y(y – 6) -1 (y – 6) = 0 or, (y – 6) (y – 1) =0 either, y- 6 = 0 or y – 1 = 0 either, y = 6 or y = 1 𝑦−3 𝑦−1 for y = 6 or y = 1, the matrix ( ) does not have its inverse. 𝑦 2(𝑦 − 1) 108 Example 3 1 2 If 𝐴 = ( 3 ) then find A-1 if exist 2 6 1 2 Solution: Here, 𝐴 = ( 3 ) 2 6 1 1 |𝐴| = |3 2| = × 6 − 2 × 2 = 2 − 4 = −2 ≠ 0 3 2 6 Hence A-1 exists. 𝑎11 𝑎12 𝑎11 𝑎12 Let (𝑎 𝑎 ) be the inverse matrix of A. i.e. 𝐴−1 = (𝑎 𝑎22 ) 21 22 21 Now, AA-1 = I = A-1A 1 a11 a12 1 0 or, (3 2) (a )=( ) 21 a 22 0 1 2 6 1 1 a + 2a 21 a + 2a 22 1 0 or, (3 11 3 12 )=( ) 0 1 2a11 + 6a 21 2a12 + 6a 22 1 Now, 3 𝑎11 + 2𝑎21 = 1……………. (i) [equating the corresponding element] and 2𝑎11 + 6𝑎21 = 0 ……………… (ii) Multiplying equation (i) by 3 and subtracting equation (ii), we get 𝑎11 + 6𝑎21 = 3 2𝑎11 + 6𝑎21 = 0 – – – – 𝑎11 = 3 or, 𝑎11 = −3 1 Again, multiplying equation (ii) by and subtracting from a11 + 6a21 = 3 2 𝑎11 + 6𝑎21 = 3 𝑎11 + 3𝑎21 = 0 – – – 3a21 = 3 or, a21 = 1 Again, 1 3 𝑎12 + 2𝑎22 = 0………..(iii) 2𝑎12 + 6𝑎22 = 1……..(iv) 109  1 by 3 × equation (iii) – equation (iv) 2 𝑎12 + 6𝑎22 = 0 1 𝑎12 + 3𝑎22 = – – – 2 1 3𝑎22 = − 2 1 𝑜𝑟, 𝑎22 = − 6 by 3 × equation (iii) – equation (iv) we get, 𝑎12 + 6𝑎22 = 0 2𝑎12 + 6𝑎22 = 1 – – – –a12 = –1 a12 = 1 𝑎11 𝑎12 −3 1 Hence, A−1 = (𝑎 𝑎22 ) = ( 1 − ) 1 21 6 Alternatively: 1 |𝐴| = |3 2| = 1 × 6 − 2 × 2 = −2 3 2 6 A-1 exists 1 𝐴 = (3 2) 2 6 Interchanging the elements of leading diagonal and changing sign of secondary 6 −2 diagonal, we get. adj. of A = (−2 1 ) 3 1 Now, 𝐴−1 = |𝐴| (adj. of A) 1 6 −2 1 6 −2 𝐴−1 = ( 1 )= ( 1) |𝐴| −2 (−2) −2 3 3 −1 −1 ( )×6 ( ) × (−2) −3 1 =( 2 2 )=( −1) −1 −1 1 1 ( ) × (−2) ( )× 6 2 2 3 110 Example 4 2 3 3 1 If𝐴 = ( ) and B = ( ) find A-1 and B-1. Verify that (AB)–1 = B–1.A–1 4 5 4 5 Solution: Here, 2 3 |A| 2 3 A=( ), =| | = 10 − 12 = −2  0 4 5 4 5 So A-1 exists 3 1 B=( ) 4 5 3 1 |B| = | | = 15 − 4 = 11  0 4 5 So, B-1 exists Now, 1 5 −1 B −1 = ( ) 11 −4 3 1 5 −3 A−1 = ( ) −2 −4 2 1 5 −1 5 −3 B −1 A−1 = ( )( ) −22 −4 3 −4 2 1 5 × 5 + (−1) × (−4) 5 × (−3) + (−1) × 2 = ( ) −22 (−4) × 5 + 3 × (−4) (−4) × (−3) + 3 × 2 1 25 + 4 −15 − 2 = ( ) −22 −20 − 12 12 + 6 1 29 −17 =− ( ) 22 −32 18 2 3 3 1 AB = ( )( ) 4 5 4 5 2×3+3×4 2×1+3×5 =( ) 4×3+5×4 4×1+5×5 6 + 12 2 + 15 =( ) 12 + 20 4 + 25 18 17 =( ) 32 29 18 17 |AB| = | | 32 29 = 18 × 29 − 17 × 32 = 522 − 544 111  = −22 1 29 −17 (AB)−1 = ( ) |AB| −32 18 1 29 −17 = ( ) −22 −32 18 (AB)−1 = B −1 A−1 proved. Exercise 3.2 1. a) Define inverse of a matrix. 𝑎11 𝑎12 b) Under which condition does A-1 exist for 𝐴 = (𝑎 𝑎22 )? 21 -1 2. Does A exist? Give reason. a) A = (−8 6) b) 𝐴 = (tanA SecA ) −4 3 SecA tanA −a −b c) A = ( ) ; a ≠ 0, b ≠ 0 b −a 3. Find the inverse of each of the following matrices, if exist. a) A = [−2 3] b) B = [ 4 −7 ] c) C = [−2 −5] −3 4 −3 2 −3 −8 2 3 4. a) If P = (1 0) and 𝑄 = [ ] then find (PQ)-1 0 1 1 2 b) If M = [0 1] 𝑎𝑛𝑑 N = [2 3] find (NM)-1 1 0 1 2 2 3 5. If 𝐴 = [ ] then find 𝐴2 + 𝐴𝐴−1 + 2𝐼, where I is the 2x2 identify matrix. 4 7 2 6 2 1 6. (a) If A = ( ) and B = ( ) show that (AB)–1 = B–1A–1. 3 10 5 3 2 3 7 3 (b) If A = ( ) and B = ( ) verify that (AB)–1 = B–1A–1. 5 6 14 6 3.3 Solutions of system of linear equations by using matrix method Let us take a matrix equation: 2 −3 𝑥 1 ( ) (𝑦) = ( ) 1 1 2 When we multiply, 2x – 3y = 1 …………… (i) 112 x + y = 2 ………………. (ii) equation (i) and (ii) are simultaneous equation in x and y The above equations can be written as AX = B x Where A = (2 −3) , B = (1) and X = (y) 1 1 2 We have, AX = B A-1(AX) = A-1B [Multiplying by A-1 on both sides] or, (A-1A)X = A-1B [By associative law] -1 or, IX = A B [A-1A = I] or, X = A-1B [IX = X] Since, matrix multiplication is not commutative (AB ≠ BA) in general, case must be taken to multiply on the left by A-1 Example 1 Solve the given system of equations by matrix method: x = 2, x + y = 7 Solution Here, x + 0.y = 2……. (i) x + y = 7 ………….. (ii) Writing equation (i) and (ii) in matrix form, we have, 1 0 𝑥 2 ( )( ) = ( ) 1 1 𝑦 7 AX=B or, X = A–1B Now, A−1 = |A| ( 1 0) = 1 ( 1 0) = ( 1 0) 1 1 −1 1 −1 1 −1 1 So, X = 𝐴−1 B 1 0 2 1×2+0×7 2+0 2 =( )( ) = ( )=( )=( ) −1 1 7 (−1) × 2 + 1 × 7 −2 + 7 5 𝑥 2  (𝑦 ) = ( ) 5 Example 2 Solve the following equation by using matrix method 2x + 5 = 4 (y + 1) – 1; 3x + 4 = 5 (y + 1) – 3 Solution: Here, 2x + 5 = 4(y + 1) – 1 or, 2x + 5 = 4y + 4 – 1 or, 2x – 4y = -2 ………….. (i) Again, 3x + 4 = 5(y + 1) – 3 or, 3x + 4 = 5y +5 – 3 113 or, 3x – 5y = -2 ………… (ii) Writing equation (i) and (ii) in matrix form, we have 2 −4 𝑥 −2 ( )( ) = ( ) 3 −5 𝑦 −2 or, Ax = B 𝑥 Let, 𝐴 = (2 −4) , X = (𝑦) 𝐵 = (−2) 3 −5 −2 or, x = A-1B 2 −4 |𝐴| = | | = (-5) x 2 – (-4) x 3 = -10 + 12 = 2 (A-1 Exists) 3 −5 1 −5 4 so, 𝐴−1 = ( ) 2 −3 2 Now, x = A–1B 1 −5 4 −2 1 (5) × (−2) + 4 × (−2) 𝑥= ( )( ) = ( ) 2 −3 2 −2 2 (−3) × (−2) + 2 × (−2) 𝑥 1 10 − 8 1 2 (𝑦 ) = ( )= ( ) 2 6−4 2 2 1 𝑥 ×2 1 ( 𝑦 ) = (2 )=( ) 1 1 ×2 2 𝑥 1 (𝑦) = ( ) 1 Comparing corresponding components of equal matrices, we get x = 1 and y = 1 Example 3 2𝑥+4 40−3𝑥 Solve: =𝑦= 5 4 Solution: Here, 2𝑥+4 𝑦= 5 2 4 or, 𝑦 = 𝑥 + 5 5 2 4 or, 𝑥 − 𝑦 = − …….. (i) 5 5 40 − 3𝑥 𝑦= 4 40 3 or, 𝑦 = − 𝑥 4 4 3 or, 𝑥 + 𝑦 = 10 ……….(ii) 4 writing equation (i) and (ii) in matrix form we have, 2 −1 𝑥 −4 (5 ) (𝑦 ) = ( 5 ) 3 1 10 4 Let, AX = B 114 X = A-1B ……………. (iii) for A-1 2 −1 2 3 2 3 8 + 15 23 |𝐴| = |5 |= × (1) + = + = = 3 5 4 5 4 20 20 1 4 23 1 1 20 1 1 𝐴−1 = 1 ÷ ( 3 2) = ( 3 2) 20 − 23 − 4 5 4 5 so, from equation (iii) X = A-1B 20 1 1 − 4 𝑜𝑟, X = ( 3 2) ( 5) 23 − 4 5 10 4 20 1 × (− ) + 1 × 10 5 = ( −3 −4 2 ) 23 ( ) × ( ) + × 10 4 5 5 −4+50 46 20 46 20 20 5 × 8 = 5 ( 3+20 ) = ( ) 23 23 = (23 20 5 23) =( ) 23 × 4 5 5 23 5 𝑥 8 (𝑦) = ( ) 4  x = 8 and y = 4 Example 4 The sum of two numbers is 20 and their difference is 4. a) Write the equation in matrix form. b) Solve them by matrix method. Solution Let the numbers be x and y (x>y) by question: x + y = 20 ………. (i) x – y = 4 ……… (ii) Writing equation (i) and (ii) in matrix form 1 1 𝑥 20 ( ) (𝑦 ) = ( ) 1 −1 4 Let, AX = B or, X = A-1B ……… (iii) 1 1 For A-1 : |𝐴| = | | 1 −1 = 1 × (−1) − 1 × 1 = −1 − 1 = −2  0 (A-1 exists) Now, 1 −1 −1 𝐴−1 = ( ) −2 −1 1 115 Now, X = A−1 B 𝑥 1 −1 −1 20 1 (−1) × 20 + (−1) × 4 (𝑦 ) = ( )( ) = ( ) −2 −1 1 4 −2 (−1) × 20 + 1 × 4 1 −20 − 4 1 −24 = ( )= ( ) −2 −20 + 4 −2 −16 1 (− ) × (−24) =( 2 ) 1 (− ) × (−16) 2 𝑥 12 (𝑦 ) = ( ) 8  The required numbers are 12 and 8 Exercise 3.3 1. a) If AA-1 = A-1A = I and AX = B, Write X in terms of A-1 and B. b) Write the matrix form of 2x + y = 4 3x + 2y = 7 2. Solve the following system of linear equations using matrix method: (a) x + y = 5, x – y = 1 (b) 3x + 5y = 3, 4x + 3y = 4 1 𝑥 (d) ( 2 1 3𝑥+5𝑦 5𝑥−2𝑦 (c) = =3 )( ) = ( ) 8 3 −3 1 𝑦 2 8 5 𝑥 2 1 1 1 1 (e) ( )( ) = ( ) (f) + = 8, −𝑦+1=0 7 4 𝑦 1 𝑥 2𝑦 2𝑥 5 1 5 2 (g) = 𝑥 − 1, =𝑥−4 𝑦 𝑦 h) 4(x – 1) + 5(y + 2) = 10, 5(x – 1) –3 (y + 2) +6 = 0 (i) 5x + 7y = 31xy (j) 7x +5y = 29xy 3. Write equations and solve the following by matrix method: a) The cost of a pen and copy is 120. The cost of 2 copy is Rs. 100. b) The sum of present ages of a father and Son is 53 years. After 2 years, the age of father will be 47 years. 4. Write yourself two simultaneous equations in (x, y) related to your daily life and solve them by matrix method. 116 3.4 Cramer’s rule Let, two linear equations in x and y are, 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 = 0 ………… (i) 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 = 0 ………… (ii) Multiplying equation (i) by b2 and equation (ii) by b1 and subtracting we get a1b2x + b1b2y + b2c1 = 0 a2b1x + b1b2y + b1c2 = 0 – – – a1b2x – a2b1x + (b2c1 – b1c1) = 0 or, (a1b2 – a2b1)x = –(b2c1 – b-1c2) 𝑏1 𝑐2 − 𝑏2 𝑐1 𝑜𝑟, 𝑥 = 𝑎1 𝑏2 − 𝑎2 𝑏1 Again, multiplying equation (i) by a2 and equation (ii) by a1 and subtracting equation (ii) from (i). We get, a1a2x + a2b1y + a2c1 = 0 a1a2x + a1b2y + a1c2 = 0 – – – (a2b1 – a1b2)y + (a2c1 –a1c2) = 0 (a2b1 – a1b2)y = –(a2c1 – a1c2) or, (a2b1 – a1b2)y = a1c2 – a2c1 a c –a c or, y = a 1b2 – a2 b1 2 1 1 2 If a1x + b1y = c1 and a2x + b2y = c2 The above solution becomes b1 c1 b1 c2 − b2 c1 |b2 c2 | 𝑥= = a1 b2 − a2 b1 a b1 | 1 | a2 b2 a1 c1 a1 c2 −a2 c1 |a | 2 c2 and y = = a b1 a1 b2 −a2 b1 | 1 | a2 b2 D𝑥 Dy i.e. 𝑥 = ,y = D D b1 c1 a1 c1 a1 b1 Where D𝑥 = | | D = |a c2 | and D = |a 2 | b2 c2 𝑦 2 b2 117 The method of solving system of linear equations by using determinant is known a Cramer's rule. Note: (i) Grabriel Cramer is a Swiss mathematician who introduced technique to solve a system of linear equations using determinant. (ii) If D = 0, Dx ≠ 0, Dy ≠ 0, the system of equations has no solutions, because 0x ≠ 0, 0y ≠ 0. (iii) We always write the equation in the form 𝑎1 𝑥 + b1 y = c and 𝑎2 𝑥 + 𝑏2 𝑦 = 𝑐2 c1 i.e. for finding Dx(D1), substitute first column of D by (c ) and for finding Dy (D2), 2 c1 substitute the second column by (c ). 2 Example 1 Solve 2(x -1) = y and 3 (x – 1) = 4y using Cramer’s rule Solution Here, 2(x – 1) = y or, 2x – y = 2 …………. (i) 3(x – 1) = 4y or, 3x – 4y = 3 ………… (ii) Now, arranging the coefficients and constant term Coefficient of x Coefficient of y Constant term 2 -1 2 3 -4 3 2 −1 Now, 𝐷 = | | = 2 x (-4) – 3 x (-1) = - 8 + 3 = - 5 3 −4 2 −1 𝐷𝑥 = | | = −5 3 −4 2 2 𝐷𝑦 = | |=2×3−2×3=0 3 3 𝐷 −5 𝑥 = 𝐷𝑥 = −5 = 1 𝐷𝑦 0 𝑦 = 𝐷 = −5 = 0 (𝑥, 𝑦) = (1,0) 118 Example 2 Solve the following system of linear equations using Cramer’s rule. 2 1 𝑥+𝑦 =1 𝑥+𝑦 =1 3 2 Solution Coefficient of x Coefficient of y Constant term 2 1 1 3 1 1 1 2 2 2 1 𝐷 = |3 | 1 1 2 2 1 4−3 1 = ×1−1× = = (≠ 0) 3 2 6 6 1 1 𝐷𝑥 = |1 | 1 2 1 1 1 =1×1−1× =1− = 2 2 2 2 1 𝐷𝑦 = |3 | 1 1 2 2 2 1 1 1 1 2 − 3 −1 = × −1× = − = = 3 2 2 3 2 6 6 1 𝐷𝑋 2 6 𝑁𝑜𝑤, 𝑥 = = = =3 𝐷 1 2 6 −1 𝐷𝑦 𝑥= = 6 = −1 𝐷 1 6 (x, y) = (3, 1) 119 Exercise 3.4 1. In the equation 𝑎1 𝑥 + 𝑏1 𝑦 = 𝑐1 and 𝑎2 𝑥 + 𝑏2 𝑦 = 𝑐2 , according to Crammer’s rule, a) Write D in determinant form. b) Write Dx in determinant form. c) Write Dy in determinant form. d) What is necessary condition for obtaining unique solution? 2. Solve the following system of linear equations using Cramer’s rule: a) 2x – y = 3 y + 3x = 7 b) y = 2x 3 𝑥− 𝑦+1=0 2 4 5 c) 𝑥 + 𝑦 = 28 7 3 + = 67 𝑥 𝑦 d) 4(x – 1) + 5(y + 2) = 10 5(x – 1) – 3(y + 2) + 6 = 0 e) 3xy – 10y = 6x 5xy + 3x = 21y f) 3y + 4x = 2xy and 18y – 4x = 5xy 3. Ask the price of any five daily uses goods. Make two different system of equation a and b in terms of x and y. Solve these equations by Cramer’s rule and present your findings in the classroom. 120  Unit 4 Coordinate Geometry 4.0 Review In a class four friends are seated at 11 the points A, B, C and D as shown in 10 graph. Sarita and Bimal walk into the 9 class and after observing for a few minute Sarita asked following 8 questions to Bimal. D 7 Discuss about these answers in 6 group and write conclusions. 5 1. Find the equation of AB, BC, CD 4 and AD B C 3 2. Find the slope of each of the 2 lines AB, BC, CD and AD. 1 The slope of a line parallel to X-axis is A 0. The slope of Y-axis is not defined. The 1 2 3 4 5 6 7 8 9 10 11 slope is independent of the sense of the line segment (i.e. slope of AB = slope of BA). The equation of a straight line is the relation between x and y, which satisfies the co-ordinates of each and every point on the line and not by those of any other point. The equation of a line with slope m and passing through a point (x1, y1) is given by y – y1 = m(x – x1) and the equation of the line through the two given points (x1, y1) and 𝑦 −𝑦 (x2, y2) is given by y - y1 = 2 1 (x-x1). 𝑥2 −𝑥1 4.1 Angle between two straight lines Let AB and CD be two straight lines with inclination θ2 and θ1 respectively. Answer the following questions: i. What is the slope (m1) of CD in terms of θ1? ii. What is the slope (m2) of AB in terms of θ2? iii. What is the relation between θ, θ1, θ2? As we know θ1 and θ2 are angles made by the lines with positive direction, so 121 m1 = tan θ1 and m2 = tan θ2 Again, we have θ1 = θ + θ2 [exterior units] or, θ = θ1 – θ2 taking tangents on both sides, tan θ = tan (θ1 – θ2) tan θ1 –tan θ2 or, tan θ = 1+𝑡𝑎𝑛𝜃1. 𝑡𝑎𝑛𝜃2 1m −m2 or, tan θ = 1+m 1 m2 𝑚1 − 𝑚2 𝑜𝑟, tan(180° − 𝜃) = −𝑡𝑎𝑛𝜃 = − ( ) 1 + 𝑚1 𝑚2 𝑚1 − 𝑚2 ∴ 𝑡𝑎𝑛𝜃 = ± ( ) 1 + 𝑚1 𝑚2 Remember: (i) ‘By the angle between two lines’ we mean ‘the acute or obtuse angle between the lines’ 𝑚 −𝑚2 1 1 𝑚 −𝑚2 (ii) If 1+𝑚 is positive, then θ is acute angle. If 1+𝑚 is negative, θ is obtuse 1 𝑚2 1 𝑚2 angle. (iii) The complete angle formula is used when the angle is given. (iv) If the lines are parallel, θ = 0° or 180° then m1 = m2 (v) If m1 = m2, the lines are coincident to each other and if m1 m2 = -1, the lines are perpendicular to each other. Example 1 Find the angle between the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 Solution: Slope of the lines a1x + b1y + c1 = 0 is coefficient of x a m1 = - coefficient of y = - b1 (say) 1 coefficient of x a Slope of the line a2x + b2y + c2 = 0 is m2 = - = - b2 (say) coefficient of y 2 If θ be the angle between the lines then, 122  m1 − m2 tanθ = ± 1 + m1 m2 a1 a 2 a2 b1 − a1 b2 −+ a 2 b1 − a1 b2 b1 b2 b1 b2 b1 b2 = ±[ a1 a 2 ] = ± [ b1 b2 + a1 a 2 ] = ± ( )×( ) 1+. b1 b2 b1 b2 + a1 a 2 b1 b2 b1 b2 a2 b1 − a1 b2 = ±[ ] a1 a2 + b1 b2 a2 b1 − a1 b2 ∴ θ = tan−1 (± ( )) a1 a2 + b1 b2 Example 2 Find the acute angle between the lines.: √3𝑥– 𝑦 + 8 = 0 𝑎𝑛𝑑 𝑦 + 10 = 0 Solution Slope of the line √3𝑥– 𝑦 + 8 = 0 is 𝑚1 = √3 (Since, y = √3𝑥 + 8) Slope of the line y + 10 = 0 is m2 = 0. If the θ be the angle between the lines then, 𝑚1 − 𝑚2 𝑡𝑎𝑛𝜃 = ± ( ) 1 + 𝑚1 𝑚2 √3 − 0 𝑜𝑟, 𝑡𝑎𝑛𝜃 = ± ( ) 1 + √3 × 0 = ± (√3) Taking (+ve) sign: tanθ = √3 tanθ = tan60° or, θ = 60°  Required acute angle is 60° Example 3 Find the obtuse angle between the lines 2x – y + 3 = 0 and x – 3y + 2 = 0 Solution Slope of line 2x – y + 3 = 0 is coefficient of 𝑥 2 m1= – = −( )= 2 coefficient of y −1 123  Slope of line x – 3y + 2 = 0 coefficient of 𝑥 1 1 m2 = – = − (−3 ) = coefficient of y 3 If θ is the obtuse angle between the lines, then 𝑚1 − 𝑚2 𝑡𝑎𝑛𝜃 = ± ( ) 1 + 𝑚1 𝑚2 1 2− 𝑜𝑟, 𝑡𝑎𝑛𝜃 = ± ( 3 ) 1 1+2×3 6−1 𝑜𝑟, 𝑡𝑎𝑛𝜃 = ± ( 3 ) 3+2 3 5 3 𝑜𝑟, 𝑡𝑎𝑛𝜃 = ± ( × ) 3 5 𝑜𝑟, 𝑡𝑎𝑛𝜃 = ± (1) For obtuse angle take 𝑡𝑎𝑛𝜃 = −1 𝑜𝑟, 𝑡𝑎𝑛𝜃 = 𝑡𝑎𝑛135° 𝑜𝑟, 𝜃 = 135°,  Obtuse angle between the lines is 135° Example 4 If the lines ax + 5y – 16 = 0 and 6x + 10y – 9 = 0 are perpendicular to each-other, find the value of a. Solution Slope of the line ax+5y-16=0 is coefficient of 𝑥 a −a m1 = – = − (5 ) = coefficient of y 5 Slope of the line 6x + 10y – 9 = 0 coefficient of 𝑥 6 −3 m2 = – coefficient of y = − (10 ) = 5 Since the lines are perpendicular to each-other, so m1 m2 = −1 124  −𝑎 −3 𝑜𝑟, × = −1 5 5 3𝑎 𝑜𝑟, = −1 25 𝑜𝑟, 3𝑎 = −25 25  𝑎=− 3 Example 5 Find the equation of the line which passes through the point(3,4) and parallel to the line 3x + 4y – 12 = 0 Solution Slope of the line 3x + 4y – 12 = 0 is coefficient of 𝑥 3 m1 = - = − coefficient of y 4 Slope of the parallel line to the given line 3 𝑚=− [∵ 𝑚1 = 𝑚2 ] 4 The line passes through the point, (𝑥1 , 𝑦1 ) = (3, 4) We know, equation of the line: 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) 3 or, 𝑦 − 4 = − 4 (𝑥 − 3) or, 4(𝑦 − 4) = −3(𝑥 − 3) or, 4𝑦 − 16 = −3𝑥 + 9 or, 3𝑥 + 4𝑦 − 16 − 9 = 0 or, 3𝑥 + 4𝑦 − 25 = 0  The required equation of line is 3x + 4y – 25 = 0 Example 6 Find the equation of the lines which passes through (2 , 4) and make angle 60° with the line 𝑦 = −√3𝑥 + 2 125 Solution Let M(2, 4) be the point. MA and MC be the lines which make 60°. With the line y = −√3𝑥 + 2 Slope of the given line = −√3 = m2 (say) M (2,4) Slope of unknown lines be m1 = (m) say. If θ be the angle between the lines, then 𝑚1 − 𝑚2