CHAPTER 4: DIFFERENTIATION AND INTEGRATION Derive the following functions a) = 2 − 5 b) = 3 + 2 2 − 6 + 7 c) = (5 + 3)( + 2) d) () = (5 + 3) 2 e) () = 1 − (2 − 1) 2 f) = 3 − 3 Exercise 4
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 95 Find if for a) = (2 − 1) 3 b) = 2 ( 2−1) 3 Answer: a) Let = 3 where = 2 − 1 = 3( 3−1 ) = 3 2 = 2 = × = 3 2 (2) = 6(2 − 1) 2 b) = 2 ( 2−1) 3 = 2( 2 − 1) −3 Let = 2 −3 where = 2 − 1 = −3(2 −3−1 ) = −6 −4 = 2 = × = −6 −4 (2) = −12( 2 − 1) −4 4.1.3 Rule of Differentiation c) Chain Rule Given = = × Example 6
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 96 Find if a) = ( 2 + 2 − 3) 4 b) = √4 − 2 Answer: a) = ( 2 + 2 − 3) 4 = 4 ( 2 + 2 − 3) 4 −1 × ( 2 + 2 − 3) = 4 ( 2 + 2 − 3) 4 −1 × (2 + 2) = (8 + 8) ( 2 + 2 − 3) 3 b) = √4 − 2 = (4 − 2) 1 2 = 1 2 (4 − 2) 1 2 −1 × (4 − 2) = 1 2 (4 − 2) − 1 2 × −2 = − 1 √4−2 d) Extended Power Rule (Alternative of Chain Rule) Given = = ( −1 ) × Example 7
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 97 1. Differentiate = (2 − ) 3 2. Differentiate = (3 4 − 5) 5 3. Differentiate = ( 3 − 2 2 − ) 2 4. Differentiate = 1 ( 3 +7) 4 Exercise 5
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 98 5. Differentiate = ( 2 ( 4−2) ) 4 6. Differentiate () = 2 3(1− 2) 5 7. Find ′() for () = √2 3 + 5
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 99 Find if = 2 ( − 1) 2 Answer: Let = 2 = ( − 1) 2 = 2 = 2( − 1) 2−1 = 2( − 1) = + = 2 [2( − 1)] + ( − 1) 2 2 = 2( − 1) [ + ( − 1)] = 2( − 1)(2 − 1) Find if = (5 + 3) 4 ( + 7) 3 Answer: Let = (5 + 3) 4 = ( + 7) 3 = 4(5 + 3) 4−1 ∙ 5 = 20(5 + 3) 3 = 3( + 7) 3−1 ∙ 1 = 3( + 7) 2 = + = (5 + 3) 4 [ 3( + 7) 2 ] + ( + 7) 3 [20(5 + 3) 3 ] = (5 + 3) 4 ( + 7) 2 [3(5 + 3) + 20( + 7)] = (5 + 3) 4 ( + 7) 2 [15 + 9 + 20 + 140] = (5 + 3) 4 ( + 7) 2 [35 + 149] e) Product Rule Example 8 Given = × = + Example 9
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 100 1. Differentiate = 4 (2 − ) 3 2. Differentiate = ( − 1)( 2 + 1) 5 3. Differentiate = ( 2 − 6) 5 ( 2 + 1) 6 Exercise 6
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 101 4. Find for = 8(3 − 2)( + 7) 4 5. Find for () = 3√2 + 5
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 102 Find ′() if () = 2 4 3−1 Answer: Let = 2 4 and = 3 − 1 = 4(2 4−1 ) = 8 3 = 3 ′ () = = − 2 = (3−1) (8 3) −(2 4) (3) (3−1) 2 = 24 4−8 3−6 4 (3−1) 2 = 18 4−8 3 (3−1) 2 f) Quotient Rule Given = = − 2 Example 10
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 103 Find if = 2 (1−) 2 Answer: Let = 2 and = (1 − ) 2 = 2( 2−1 ) = 2 = 2(1 − ) 2−1 ∙ −1 = −2(1 − ) = − 2 = (1−) 2(2) − 2[−2(1−)] [(1−) 2] 2 = 2(1−) 2 + 2 2(1−) (1−) 4 = (2)(1−)[(1−) +] (1−) 4 = 2 (1−) 3 Alternative solution: Solve by using Product Rule = 2 (1 − ) −2 Let = 2 and = (1 − ) −2 = 2( 2−1 ) = 2 = −2(1 − ) −2−1 ∙ −1 = 2(1 − ) −3 = + = 2 ∙ 2(1 − ) −3 + (1 − ) −2 (2) = 2 2 (1−) 3 + 2 (1−) 2 = 2 2 (1−) 3 + 2(1−) (1−) 2(1−) = 2 2 (1−) 3 + 2−2 2 (1−) 3 = 2 (1−) 3 Example 11
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 104 1. Differentiate = 10 3−5 2. Differentiate = 4 (−3 2) 2 Exercise 7
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 105 3. Differentiate () = (−2) 2 +1 4. Find for = √ 2+2 √3 4
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 106 A curve has an equation of () = 3 − 7. Find the gradient of the tangent line that passed through the curve at point (−1, 3). Answer: Step 1: Differentiate f x () = 3 − 7 ′() = 3 2 Step 2: Substitute = −1 into ′() ′(−1) = 3(−1) 2 = 3 The gradient of tangent line, 1 is 3 A curve has an equation of = ( 2 − 7)( − 1). Find the gradient of the tangent line that passed through the curve at point ( 1 2 , −1). Answer: Step 1: Differentiate () = ( 2 − 7)( − 1) = 3 − 2 − 7 + 7 Step 2: Substitute = 1 2 into ′() = ( 1 2 ) 3 − ( 1 2 ) 2 − 7 ( 1 2 ) + 7 = − 29 4 The gradient of tangent, 1 is − 29 4 4.1.4 Calculate gradients of tangent to a curve by applying the differentiation techniques A tangent is a straight line that touches a curve at a given point. Meanwhile, normal is a straight line that is perpendicular to tangent at the same point. The gradient of the tangent, m1 at the given point on the curve is given by the derivative of the curve function Example 12 Example 13
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 107 Find the gradient of the tangent line that passed through the curve = 1 (1− 3) 3 at = −1. Answer: Step 1: Differentiate = (1 − 3 ) −3 = −3(1 − 3 ) −3−1 ∙ −3 3−1 = 9 2 (1− 3) 4 Step 2: Substitute = −1 into = 9(−1) 2 (1−(−1) 3) 4 = 9 16 The gradient of tangent, 1 is 9 16 4.1.5 Compute the tangent and normal equations Equation of tangent The standard equation of tangent is = + , where is the gradient of tangent Steps to construct the equation of a tangent to a curve: Step 1: Calculate the gradient of the tangent, Step 2: Substitute into the formula: − 1 = ( − 1 ) Where x1 and y1 are both x and y axis values. Step 3: Make y as the subject of the formula Example 14
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 108 A curve has an equation of () = 3 − 4 + 1. Find the equation of the tangent that passed through the curve at point (1, 0). Answer: Step 1: Calculate the gradient of the tangent () = 3 − 4 + 1 ′() = 3 2 − 4 = ′ (1) = 3(1) 2 − 4 = −1 Step 2: Substitute , and into the formula − 1 = ( − 1 ): = −, = 1, = 0 − 0 = −1 ( − 1) = −1 + 1 Equation of normal The standard equation of tangent is = + , where is the gradient of normal Steps to construct the equation of a tangent to a curve: Step 1: Calculate the gradient of the normal, = − 1 Step 2: Substitute into the formula: − 1 = ( − 1 ) Where x1 and y1 are both x and y axis values. Step 3: Make y as the subject of the formula Example 15
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 109 Find the equation of the normal lines that passed through the curve = 2 − 3 at point (−2, 1). Answer: Step 1: Calculate the gradient of the tangent = 2 − 3 ′ = 2 − 3 2 Gradient of tangent, = ′ = 2 − 3(−2) 2 = −10 Gradient of normal, = 1 −10 Step 2: Substitute , and into the formula − 1 = ( − 1 ): = − 1 −10 = 1 10 = −2, = 1 − 1 = 1 10 ( − (−2)) = 1 10 + 1 5 +1 = 1 10 + 6 5 1. Determine the equations of the tangent line at the curve of () = 2 3 − 2 + 4 at point = 4, = −7. Example 16 Exercise 8
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 110 2. Find the equation of the tangent and normal to the curve = ( − 1) 2 at the point = 3 with given the gradient of its normal is –1. 3. A tangent line with an equation of + = 1 passed the curve of() = 3 − 6 at the point = . Find the value of a.
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 111 4.2.1 Indefinite integral as the inverse of differentiation In previous in section, derivative of a function () with respect to x is denoted as ′ (). Then, () is said to be an anti-derivative of ′ (). The inverse process of differentiation or finding anti-derivative of ′ () is called integration. Anti-derivative of ′ () is called indefinite integral or in notation, The symbol of integration ‘ ’ means sum as integration use to find total sum of an infinite number of a small areas. Integral constant, C applied to indefinite integral. Recall: () = 3 4 − 1 … … … (1) = 3 4 + 5 … … … (3) ′() = 12 3 … … … (2) = 12 3… … … (4) Integral constant, C is applied to indefinite integral because some equations have same derivative. It can be seen that derivative of functions (1) and (3) give same answer; (2) and (4) and constants in both equations (1 and 5) are disappeared during differentiation process. Because of that, the constant C is applied to the integral. 4.2 Describe integration Derivative Anti-derivative () ′()
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 112 1. Integrate ∫ 10 Answer: ∫ 10 = 10 + 2. Integrate ∫ 2 Answer: ∫ 2 = 2 + 3. Integrate ∫(−3) Answer: ∫(−3) = −3 + 4.2.2 Indefinite Integral of Algebraic Functions As integration is an inverse to differentiation process, the steps are opposed from differentiation process. The answers for indefinite integral of algebraic function are in algebraic term with addition of integral constant C. Integral of a Constant: Example 17
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 113 1. Integrate ∫ 4 Answer: ∫ 4 = 4 1 + 1 1 + 1 + = 2 2 + 2. Integrate ∫ 4 Answer: ∫ 4 = 4 + 1 4 +1 + = 5 5 + 3. Integrate ∫ √ 3 Answer: ∫ √ 3 = ∫ 1 3 = 1 3 +1 1 3 +1 + = 3 4 3 4 + 4. Integrate ∫ 1 4 Answer: ∫ 1 4 = ∫ − 4 = − 4 + 1 −4 +1 + = 1 3 3 + Integral of Power x: Example 18
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 114 1. Integrate ∫(2 + 7 ) 3 Answer: ∫(2 + 7 ) 3 = 3 (2+7 ) 3 + 1 2 (3+1) + = 3 (2+7 ) 4 8 + 2. Integrate ∫ 4(1 − 5 ) 2 Answer: ∫ 4(1 − 5 ) 2 = 4 (1−5 ) 2 + 1 −5 (2+1) + = − 4 (1−5 ) 3 15 + 3. Integrate ∫ 3 (3 + 2) 2 Answer: ∫ 6 (3 + 2) 2 = 6 ∫(3 + 2) −2 = 6 (3 + 2) − 2 + 1 3(− 2 +1) + = 6 (3 + 2) − 1 −3 + = 2 3 + 2 + Integral of (ax + b) n function: Example 19
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 115 1. Integrate ∫(2 + 1) Answer: ∫(2 + 7) = ∫ 2 + ∫ 1 = 2 1 + 1 1 + 1 + + = 2 + + 2. Integrate ∫(7 − 5 3 ) Answer: ∫ 7 − 5 3 = ∫ 7 − ∫ 5 3 = 7 − 5 3 + 1 3 + 1 + C = 7 + 5 4 4 + C 3. Integrate ∫(2 12 − 8 + 10 ) Answer: ∫(2 12 − 8 + 10 ) = ∫ 2 12 − ∫ 8 + ∫ 10 = 2 12 + 1 12 + 1 8 + 1 8 + 1 +10 + = 2 13 13 9 9 +10 + Sum and Difference Rule: Example 20
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 116 Derive the following functions. a) = 2 b) = 1 4 c) = 4 √ 3 d) = 1 6 5 e) () = −3 3 f) () = −4 −6 g) = 4 3 3 Exercise 9
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 117 1. If ′ () = 3 2 − 2 is an anti-derivative of () and given that (−1) = 2, then find the equation of () Answer: () = ∫ ′() = ∫(3 2 − 2) = 3 2 + 1 2 + 1 − 2 + () = 3 − 2 + Use the given condition, (−1) = 2 to find the value of C () = 3 − 2 + 2 = (−1) 3 − 2(−1) + 2 = −1 + 2 + = 1 ∴ () = 3 − 2 + 1 4.2.3 Indefinite integral of an algebraic function with initial conditions given Initial condition or initial value is a point or number used to find the exact value of integral constant, C. Example 21
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 118 1. Integrate a) dx x 3 4 b) t t t dt (16 3) 7 2 c) (10w 9w 4w)dw 4 2 d) 3x 1x 1dx 2 Exercise 10
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 119 e) x dx 2 1 f) dx x x x x 7 5 2 3 2 4.2.4 Definite integrals Definite integral is an integral which has lower and upper limits. In notation: From the above notation, and represent the lower and upper limits of the integral. Definite integral used to calculate the area under the curve from the boundary of upper to lower limit.
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 120 Find the value of t dt 5 2 1 from limit a = 1 to b = 2. Answer: C t t dx t 2 5 1 1 2 1 5 1 5 C t t 2 6 1 6 Substitute a = 1 to b = 2 into integral function F C C 3 1 6 1 1 2 1 1 6 F C C 3 29 6 2 2 2 1 2 6 Upper limit minus lower limit F F C C 3 1 3 29 2 1 10C *The value of t dt 5 2 1 from limit a = 1 to b = 2 or written as t dt 2 1 5 2 1 is 10C . *Observe that C are cancelled during the process. Example 22
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 121 Evaluate 0 1 2 2x 4x 2 dx Answer: 0 1 0 2 1 1 1 1 2 2 1 1 4 2 1 2 4 2 2 x x x x x dx 0 1 2 3 2 2 3 2 x x x 2 1 2 1 3 2 1 2 0 2 0 3 2 0 2 3 2 3 3 10 Evaluate 2 1 4 9 dx x x x Answer: 2 1 2 1 3 2 1 4 4 9 9 9 dx x dx x x x x dx x x x 2 1 4 2 1 3 1 9 4 9 3 1 x x x x 1 4 1 2 4 2 4 4 4 3 6 4 27 4.2.5 Solve the problems related to definite integrals Example 23 Example 24
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 122 Solve x dx 1 2 4 3 Answer: 1 2 1 4 1 2 4 2 4 1 2 3 2 3 x x dx 1 2 4 8 2 3 x 8 2 2 3 8 2 1 3 4 4 4096 2401 4096 625 256 111 1. Given ( ) 2, 4 2 f x dx find the value of k if ( ) 6. 4 2 f x k dx Example 25 Exercise 11
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 123 2. Find the value of: a) dr r 3 1 3 ) 4 ( b) x x dx 3 1 2 (5 4 3 ) c) x x dx 6 4 ( 5)(7 8)
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 124 3. Solve the definite integral of 3 t 3 from x = 0 to x = 1. 4. Evaluate the integral of 3 2 1 x from x = 1 to x = 3.
125 5.1 Construct matrix 5.1.1 Define matrix 5.1.2 Identify characteristics of a matrix: element, order 5.1.3 Identify different types of matrices and their basic properties 5.1.4 Write identity matrix 5.1.5 Describe basic operations on matrices 5.1.6 Calculate the operations of addition, subtraction and multiplication involving matrices 5.1.7 Compute the determinants of square matrices of orders 2 and 3 and use the properties of determinants 5.1.8 Write the transposition of a matrix 5.1.9 Calculate the determinant of a matrix 5.1.10 Identify singular and non-singular matrices 5.1.11 Simplify the inverse of a matrix 5.2 Explain system of linear equations 5.2.1 Explain the procedure to solve linear equations 5.2.2 Solve systems of linear equations 5.2.3 Solve simultaneous equations with three variables using Cramer’s Rule CHAPTER 5: MATRICES AND LINEAR ALGEBRA The term matrix was introduced by the 19th-century English mathematician James Sylvester, but it was his friend the mathematician Arthur Cayley who developed the algebraic aspect of matrices in two papers in the 1850s. Cayley first applied them to the study of systems of linear equations, where they are still very useful. They are also important because, as Cayley recognized, certain sets of matrices form algebraic systems in which many of the ordinary laws of arithmetic (e.g., the associative and distributive laws) are valid but in which other laws (e.g., the commutative law) are not valid. Matrices have also come to have important applications in computer graphics, where they have been used to represent rotations and other transformations of images. (https://www.britannica.com)
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 126 State the order of matrix ( − ) Answer: ( 7 0 6 7 3 −8 ) Therefore order of matrix is 3 x 2 5.1.1 Define matrix A rectangular arrangement of data in rows and columns which consist of a set of numbers (elements). 5.1.2 Identify characteristics of a matrix: element, order Element in a matrix is defined by which i is the numbers of row and j is the numbers of column. = [ 11 12 13 21 22 23 31 32 33 ] The order (or size, or dimension) of a matrix is defined by m x n matrix where m = the number of rows, and n = the number of columns. 5.1 Construct Matrix Example 1 2 1 3 1 2 Row (i) Column (j)
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 127 Find the value of element Answer: = [ 59 8 7 43 57 11 4 63 22 ] From 31, number of row is 3 and number column is 1. Therefore, the element is 4. 1. Find the following elements in the matrix B = [ 2 1 6 5 0 4 5 9 −8 11 6 7 ] i. 24 ii. 13 iii. 32 2. State the order of matrix H and find element at ℎ21 = [ 0 −1 0.75 −1 −3 2 0 1 −0.5 ] Exercise 1 Example 2
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 128 5.1.3 Identify different types of matrices and their basic properties No. Matrices Types Example 1. Single matrix A matrix with a single column/ single row. (1 2) – Row matrix ( 7 5 6 ) – Column matrix 2. Zero matrix / Null matrix A matrix with all elements that has zero value. ( 0 0 0 0 ) or ( 0 0 0 0 0 0 ) or ( 0 0 0 0 0 0 0 0 0 ) or ( 0 0 0 0 0 0 ) 3. Square matrix The numbers of rows and columns are both equal ( 5 2 8 0 ) or ( 5 7 6 4 8 2 1 6 3 ) 4. Upper triangular matrix A square matrix which the lower triangular elements is 0 ( −6 4 −2 0 5 1 0 0 3 ) 5. Lower triangular matrix A square matrix which the upper triangular elements is 0 ( −6 0 0 −2 5 0 4 1 3 ) 6. Diagonal matrix A square matrix which has zero elements on the upper and lower triangular sides. ( −6 0 0 0 5 0 0 0 3 ) or ( 2 0 0 4 ) 7. Identity/unit matrix () Similar with diagonal matrix but the leading diagonal element is number 1. = ( 1 0 0 0 1 0 0 0 1 ) or = ( 1 0 0 1 ) 8. Scalar matrix Similar with diagonal and identity matrix but the leading element is consistent with same number besides 1. ( 5 0 0 0 5 0 0 0 5 ) or ( 3 0 0 0 3 0 0 0 3 ) or ( −6 0 0 0 −6 0 0 0 −6 )
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 129 State the characteristics of square matrix. Answer: The numbers of rows and columns are both equal. Write 2 types of identity matrix in terms of 3 x 3 matrix and 2 x 2 matrix. Answer: = ( 1 0 0 0 1 0 0 0 1 ) = ( 1 0 0 1 ) 1. Write 2 types of matrices with its examples. 2. Identify the types of matrices for the following matrix. i. ( 3 0 0 0 7 0 5 1 9 ) ii. ( 3 0 0 0 6 0 0 0 3 ) 5.1.4 Write identity matrix Identity matrix is the matrix that has 1's on the main diagonal and 0's elsewhere such as below: Example 3 Exercise 2 Example 4 = ( 1 0 0 0 1 0 0 0 1 ) or = ( 1 0 0 1 )
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 130 Solve the following matrices equation: i. ( 2 −4 0 −7 3 6 ) + ( 5 −1 0 6 3 3 ) 1. State the definition of identity matrix. 2. Match the following matrices with its examples. 5.1.5 Describe basic operations on matrices Addition (or subtraction) of matrices is performed by adding (or subtracting) elements which is similar in positions. Addition is only valid if the two matrices have the same order. Matrix multiplication is the result of multiplying a row matrix by a column matrix. This calculations also called a dot product. The number of columns in the row matrix must equal the number of rows in the column matrix. If not, the operation is undefined. The division operation can be explained with the help of the identity matrix and the inverse operation. 5.1.6 Calculate the operations of addition, subtraction and multiplication involving matrices Addition and Subtraction Exercise 3 Example 5
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 131 Find the solution for following matrices equation: i. ( 2 −4 0 −7 3 6 ) + 3 ( 5 −1 0 6 3 3 ) ii. ( 7 4 5 2 6 4 1 −3 1 ) − ( −5 7 6 4 1 2 1 6 −3 ) Answer: i. = ( 2 −4 0 −7 3 6 ) + 3 ( 5 −1 0 6 3 3 ) = ( 2 −4 0 −7 3 6 ) + ( 3 × 5 3 × (−1) 3 × 0 3 × 6 3 × 3 3 × 3 ) = ( 2 + 15 −4 + (−3) 0 + 0 −7 + 18 3 + 9 6 + 9 ) = ( 17 −7 0 11 12 15) ii. ( 7 4 5 2 6 4 1 −3 1 ) − ( −5 7 6 4 1 2 1 6 −3 ) = ( 7 − (−5) 4 − 7 5 − 6 2 − 4 6 − 1 4 − 2 1 − 1 −3 − 6 1 − (−3) ) = ( 12 −3 −1 −2 5 2 0 −9 4 ) ii. ( 0 9 −2 3 ) − ( 7 4 5 8 9 7 ) Answer: i. = ( 2 + 5 −4 + (−1) 0 + 0 −7 + 6 3 + 3 6 + 3 ) = ( 7 −5 0 −1 6 9 ) ii. Can’t be solved since the matrix orders are not same. Example 6
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 132 Solve the matrix problem: = ( 2 −1 3 1 4 5 ) ( 4 −5 −1 −2 0 3 ) Answer: = ( 2(4) + (−1)(−1) + 3(0) 2(−5) + (−1)(−2) + 3(3) 1(4) + 4(4) + 5(0) 1(−5) + 4(−2) + 5(3) ) = ( 9 1 0 2 ) Multiplication The guidelines for multiplication of matrices can be represented as follows: Reminder: Check the order of matrices before start the calculation. ( ) × ( ) = ( + + ) 1. Given that = ( −1 2 0 4 5 3 ) = ( 7 1 −3 2 0 6 ) = ( 1 2 −4 9 ) = ( 11 5 0 −2 ) = ( 0 1 1 0 0 3 ) Find the following (if possible): i. + Exercise 4 Example 7
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 133 ii. − iii. − iv. 3 v. 2 + vi. 5 − 4 Answer: i) ( 6 3 −3 6 5 9 ) ii) ( 10 3 4 −11) iii) Can’t be solved iv) ( −3 6 0 12 15 9 ) v) ( 13 9 −8 16) vi) Can’t be solved
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 134 2. = ( −2 4 2 4 12 4 7 −20 −9 ) = ( 5 7 6 4 8 2 1 6 3 ) Find the multiplying matrices of: i. ii. iii. 2 Answer: i) ( 8 30 2 72 148 60 −54 −165 −25 ) ii) ( 60 −16 −16 38 72 22 43 16 −1 ) iii)( 59 127 62 54 104 46 32 73 27 )
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 135 Find the determinant of each of the following matrices: i. = [ 6 −3 2 −4 ] ii. = [ 6 1 9 4 2 5 8 3 7 ] Answer: (a) det() = | 6 −3 2 −4 | = (6)(−4) − (−3)(2) = −18 5.1.7 Compute the determinants of square matrices of orders 2 and 3 and use the properties of determinants The determinant of a square matrix is a real number corresponding to the matrix. The determinant of A, det (A) (or || ) is defined as follows: (a) Let = [ 11 12 21 22 ] Then, () = | 11 12 21 22 | = 1122 − 1221 (b) Let = [ 11 12 13 21 22 23 31 32 33 ] Then, () = | 11 12 13 21 22 23 31 32 33 | = 11 | 22 23 32 33 | − 12 | 21 23 31 33 | + 13 | 21 22 31 32 | = 11(2133 − 2331) − (12)(2133 − 2331) + 13(2132 − 2231) Example 8
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 136 (b) det( ) = | 6 1 9 4 2 5 8 3 7 | = (6) | 2 5 3 7 | − (1) | 4 5 8 7 | + (9) | 4 2 8 3 | = 6(2 ∙ 7 − 5 ∙ 3) − (1)(4 ∙ 7 − 5 ∙ 8) + 9(4 ∙ 3 − 2 ∙ 8) = 6(−1) − (−12) + 9(−4) = −30 1. Find the determinant of each matrix. i. [ 1 3 −2 5 ] Answer: i) 11 ii. [ 2 4 6 12] Answer: ii) 0 2. Find the determinant of the matrix. = [ 1 1 4 1 −1 −3 1 2 0 ] Answer: i) 15 3. If the determinant of matrix A is 62. Find the element value for z. = [ 1 3 −5 −2 4 6 0 9 ] Answer: = −7 Exercise 5
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 137 Find transpose for matrix = [ 4 0 3 5 1 7 ] ; = [ 4 7 3 3 2 3 2 1 2 ] Answer: = [ 4 3 1 0 5 7 ] = [ 4 3 2 7 2 1 3 3 2 ] Let = [ 1 5 2 2 2 1 ] and = [ 1 0 1 1 1 0 ]. Find and () Answer: = [ 1 5 2 2 2 1 ][ 1 0 1 1 1 0 ] = [ 1 5 2 2 2 1 ][ 1 1 0 1 1 0 ] = [ 3 6 3 4 ] () = ([ 1 5 2 2 2 1 ][ 1 0 1 1 1 0 ]) = ′ ℎ . 5.1.8 Write the transposition of a matrix Transposition of a matrix is a concept of forming a new matrix by switching its rows with its columns. Transposition of matrix A is denoted by . The superscript "T" means transpose. Example 9 Example 10
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 138 1. If = [ 2 5 7 7 4 9 3 0 6 ] . Transpose and identify element 23 2. Given = [ 23 34 76 54 42 57 63 75 25 ] and = [ 96 42 37 75 52 64 33 24 18 ]. Find: i. ( + ) ii. + Answer: i)[ 119 129 96 76 94 99 113 121 43 ] ii) [ 119 129 96 76 94 99 113 121 43 ] Exercise 6
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 139 3. Let = [ 1 5 2 2 2 1 ] and = [ 1 1 0 1 1 0 ]. Find Answer: [ 3 3 6 4 ] 5.1.9 Calculate the determinant of a matrix The determinant of a 2x2 matrix can also be calculated by using below formula: Meanwhile, the determinant of a 3x3 matrix can be calculated by using below formula: [ ] = | | = − [ ℎ ] = | ℎ | − + ℎ = ( − ℎ) − ( − ) + (ℎ − )
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 140 1. Find the determinant of each matrix. i. [ 3 2 −1 5 ] ii. [ 3 −5 9 1 2 3 5 8 2 ] Answer: i. [ 3 2 −1 5 ] = 3 × 5 − 2 × (−1) = 17 ii. [ 3 5 9 1 2 3 5 8 2 ] = 3 | 2 3 8 2 | − (−5) | 1 3 5 2 | + 9 | 1 2 5 8 | = 3(4 − 24) + 5(2 − 15) + 9(8 − 10) = - 143 5.1.10 Identify singular and non-singular matrices Singular Matrix Non-singular matrix A matrix is singular if its . Non-singular matrix is square matrix whose determinant is not equal to zero = [ 1 4 2 8 ] || = (1 × 8) − (4 × 2) = 0 Therefore, A is a singular matrix. = [ 3 4 1 2 ] || = (3 × 2) − (4 × 1) = 2 Therefore, A is a non-singular matrix. Example 11
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 141 Determine the singularity of the following matrices. i. = [ 3 6 2 4 ] ii. = [ 1 0 2 2 −1 0 1 2 1 ] Answer: i. || = (3 × 4) − (6 × 2) = 0 Therefore, B is a singular matrix. ii. det() = | 1 0 2 2 −1 0 1 2 1 | = (1) | −1 0 2 1 | − (0) | 2 0 1 1 | + (2) | 2 −1 1 2 | = 1(−1 − 0) − 0(2 − 0) + 2(4 + 1) = 9 Therefore, C is a non-singular matrix. 1. Find the determinant value and determine the type of matrix. i. = [ 7 5 5 2 ] Example 12 Exercise 7
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 142 ii. = [ 1 0 0 0 1 0 0 0 1 ] Answer: i) -11 ii) 1 5.1.11 Simplify the inverse of a matrix In order to get the inverse of matrix for order 2 x 2, check the determinant value first from the elements by referring below notation. Swap the diagonal elements and negate the elements before divide it by the determinant such as in the following formula. −1 = 1 det() [ − − ] = 1 − [ − − ] = [ ] and − ≠ 0
CHAPTER 5: MATRICES AND LINEAR ALGEBRA 143 Find the inverses of matrix = [ 5 10 8 5 ] Answer: −1 = 1 − [ − − ] = 1 (5)(5) − (10)(8) [ 5 −10 −8 5 ] = 1 −55 [ 5 −10 −8 5 ] = [ −5 55 10 55 8 55 5 55 ] To compute the inverse of a matrix 3 × 3 or larger order of matrix, there are some of necessary steps must be fulfilled. Example 13 Step 1: DETERMINANT (det) Step 2: MINOR (M) Step 3: COFACTOR (cof) Step 4: ADJOINT (adj) Step 5: Inverse matrix (3x3) formula − = () ∙ ()