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THE THEORY OF MATRICES AND DETERMINANTS 509 60.5 The inverse or reciprocal of a 2 by 2 matrix TheinverseofmatrixA is A 1 such that AðA 1 D I, the unit matrix. Let matrix A be 12 34 and let the inverse matrix, A 1 be ab cd . Then, since A ðA 1 D I, 12 34 ð ab cd D 10 01 Multiplying the matrices on the left hand side, gives a C 2cbC 2d 3a C 4c 3b C 4d D 10 01 Equating corresponding elements gives: b C2d D 0, i.e. b D2d and 3a C4c D 0, i.e. a D 4 3 c Substituting for a and b gives: 4 3 c C 2c 2d C2d 3 4 3 c C 4c 32d C4d D 10 01 i.e. 2 3 c 0 0 2d D 10 01 showing that 2 3 c D 1, i.e. c D 3 2 and 2d D 1, i.e. d D 1 2 . Since b D2d, b D 1 and since a D 4 3 c, a D2. Thus the inverse of matrix 12 34 is ab cd that is, 21 3 2 1 2 . There is, however, a quicker method of obtaining the inverse of a 2 by 2 matrix. For any matrix pq rs the inverse may be obtained by: (i) interchanging the positions of p and s, (ii) changing the signs of q and r,and (iii) multiplying this new matrix by the reciprocal of the determinant of pq rs . Thus the inverse of matrix 12 34 is 1 4 6 4 2 31 D 21 3 2 1 2 as obtained previously. Problem 13. Determine the inverse of 3 2 74 The inverse of matrix pq rs is obtained by inter- changing the positions of p and s, changing the signs of q and r and multiplying by the reciprocal of the determinant pq rs . Thus, the inverse of 3 2 74 D 1 3 ð4 2 ð7 42 73 D 1 26 42 73 D 2 13 1 13 −7 26 3 26 Now try the following exercise Exercise 203 Further problems on the inverse of 2 by 2 matrices 1. Determine the inverse of 3 1 47 7 17 1 17 4 17 3 17 2. Determine the inverse of 1 2 2 3 1 3 3 5 7 5 17 8 4 7 4 2 7 6 3 7 510 ENGINEERING MATHEMATICS 3. Determine the inverse of 1.37.4 2.5 3.9 0.290 0.551 0.186 0.097 correct to 3 dec. places 60.6 The determinant of a 3 by 3 matrix (i) The minor of an element of a 3 by 3 matrix is the value of the 2 by 2 determinant obtained by covering up the row and column containing that element. Thus for the matrix 123 456 789 the minor of element 4 is obtained by covering the row (4 5 6) and the column 1 4 7 , leaving the 2 by determinant 23 89 , i.e. the minor of element 4is2 ð9 3 ð8 D6. (ii) The sign of a minor depends on its position within the matrix, the sign pattern being CC C CC . Thus the signed-minor of element 4 in the matrix 123 456 789 is 23 89 D6 D 6. The signed-minor of an element is called the cofactor of the element. (iii) The value of a 3 by 3 determinant is the sum of the products of the elements and their cofactors of any row or any column of the corresponding 3 by 3 matrix. There are thus six different ways of evaluating a 3 ð 3 determinant — and all should give the same value. Problem 14. Find the value of 341 207 1 3 2 The value of this determinant is the sum of the products of the elements and their cofactors, of any row or of any column. If the second row or second column is selected, the element 0 will make the product of the element and its cofactor zero and reduce the amount of arithmetic to be done to a minimum. Supposing a second row expansion is selected. The minor of 2 is the value of the determinant remaining when the row and column containing the 2 (i.e. the second row and the first column), is covered up. Thus the cofactor of element 2 is 4 1 3 2 i.e. 11. The sign of element 2 is minus, (see (ii) above), hence the cofactor of element 2, (the signed-minor) is C11. Similarly the minor of ele- ment 7 is 34 1 3 i.e. 13, and its cofactor is C13. Hence the value of the sum of the products of the elements and their cofactors is 2 ð11C7 ð13, i.e., 341 207 1 3 2 D 211 C 0 C713 D 113 The same result will be obtained whichever row or column is selected. For example, the third column expansion is 1 20 1 3 7 34 1 3 C 2 34 20 D 6 C91 C16 D 113, as obtained previously. Problem 15. Evaluate 143 526 1 42 Using the first row: 143 526 1 42 D 1 26 42 4 56 12 C 3 52 1 4 D 4 C24 410 C 6 320 C2 D 28 C16 66 D −22 Using the second column: 143 526 1 42 D4 56 12 C 2 1 3 12 4 1 3 56 D410 C6 C22 3 C46 15 D 16 2 36 D −22 THE THEORY OF MATRICES AND DETERMINANTS 511 Problem 16. Determine the value of j2 1 Cj 3 1 j 1 j 0 j45 Using the first column, the value of the determi- nant is: j2 1 j j45 1 j 1 Cj 3 j45 C 0 1 Cj 3 1 j D j25 j 2 4 1 j5 Cj5 j12 C0 D j29 1 j5 j7 D j18 [5 j7 j5 Cj 2 7] D j18 [2 j12] D j18 C2 Cj12 D 2 Y j 30 or 30.07 66 86.19 ° Now try the following exercise Exercise 204 Further problems on 3 by 3 determinants 1. Find the matrix of minors of 4 76 240 574 16 8 34 14 46 63 24 12 2 2. Find the matrix of cofactors of 4 76 240 574 16 8 34 14 46 63 24 12 2 3. Calculate the determinant of 4 76 240 574 [212] 4. Evaluate 8 2 10 2 3 2 63 8 [328] 5. Calculate the determinant of 3.12.46.4 1.63.8 1.9 5.33.4 4.8 [242.83] 6. Evaluate j22j 1 Cj 1 3 5 j40 [2 j] 7. Evaluate 3 6 60 ° j21 0 1 Cj 2 6 30 ° 02j5 26.94 6 139.52 ° or 20.49 j17.49 60.7 The inverse or reciprocal of a 3 by 3 matrix The adjoint of a matrix A is obtained by: (i) forming a matrix B of the cofactors of A,and (ii) transposing matrix B to give B T ,whereB T is the matrix obtained by writing the rows of B as the columns of B T .Thenadj A = B T . The inverse of matrix A, A 1 is given by A −1 = adj A jAj where adj A is the adjoint of matrix A and jAj is the determinant of matrix A. Problem 17. Determine the inverse of the matrix 341 207 1 3 2 TheinverseofmatrixA, A 1 D adj A jAj . The adjoint of A is found by: (i) obtaining the matrix of the cofactors of the elements, and (ii) transposing this matrix. The cofactor of element 3 is C 07 3 2 D 21. The cofactor of element 4 is 27 1 2 D 11, and so on. 512 ENGINEERING MATHEMATICS The matrix of cofactors is 21 11 6 11 513 28 23 8 . The transpose of the matrix of cofactors, i.e. the adjoint of the matrix, is obtained by writing the rows as columns, and is 21 11 28 11 5 23 613 8 . From Problem 14, the determinant of 341 207 1 3 2 is 113. Hence the inverse of 341 207 1 3 2 is 21 11 28 11 5 23 613 8 113 or 1 113 21 11 28 11 −5 −23 −613 −8 Problem 18. Find the inverse of 152 3 14 367 Inverse D adjoint determinant The matrix of cofactors is 17 9 15 23 13 21 18 10 16 . The transpose of the matrix of cofactors (i.e. the adjoint) is 17 23 18 9 13 10 15 21 16 . The determinant of 152 3 14 367 D 17 24 521 C 12 218 3 D17 C45 30 D2 Hence the inverse of 152 3 14 367 D 17 23 18 9 13 10 15 21 16 2 D 8.5 −11.5 −9 −4.56.55 −7.510.58 Now try the following exercise Exercise 205 Further problems on the inverse of a 3 by 3 matrix 1. Write down the transpose of 4 76 240 574 4 25 747 604 2. Write down the transpose of 36 1 2 5 2 3 7 10 3 5 351 6 2 3 0 1 2 7 3 5 3. Determine the adjoint of 4 76 240 574 16 14 24 8 46 12 34 63 2 4. Determine the adjoint of 36 1 2 5 2 3 7 10 3 5 2 5 3 3 5 42 1 3 10 2 3 10 18 1 2 2 3 6 32 THE THEORY OF MATRICES AND DETERMINANTS 513 5. Find the inverse of 4 76 240 574 1 212 16 14 24 8 46 12 34 63 2 6. Find the inverse of 36 1 2 5 2 3 7 10 3 5 15 923 2 5 3 3 5 42 1 3 10 2 3 10 18 1 2 2 3 6 32 61 The solution of simultaneous equations by matrices and determinants 61.1 Solution of simultaneous equations by matrices (a) The procedure for solving linear simultaneous equations in two unknowns using matrices is: (i) write the equations in the form a 1 x C b 1 y D c 1 a 2 x C b 2 y D c 2 (ii) write the matrix equation corresponding to these equations, i.e. a 1 b 1 a 2 b 2 ð x y D c 1 c 2 (iii) determine the inverse matrix of a 1 b 1 a 2 b 2 , i.e. 1 a 1 b 2 b 1 a 2 b 2 b 1 a 2 a 1 , (from Chapter 60) (iv) multiply each side of (ii) by the inverse matrix, and (v) solve for x and y by equating corre- sponding elements. Problem 1. Use matrices to solve the simultaneous equations: 3x C 5y 7 D 0 1 4x 3y 19 D 0 2 (i) Writing the equations in the a 1 x C b 1 y D c form gives: 3x C 5y D 7 4x 3y D 19 (ii) The matrix equation is 35 4 3 ð x y D 7 19 (iii) The inverse of matrix 35 4 3 is 1 3 ð3 5 ð4 3 5 43 i.e. 3 29 5 29 4 29 3 29 (iv) Multiplying each side of (ii) by (iii) and re- membering that A ðA 1 D I, the unit matrix, gives: 10 01 x y D 3 29 5 29 4 29 3 29 ð 7 19 Thus x y D 21 29 C 95 29 28 29 57 29 i.e. x y D 4 1 (v) By comparing corresponding elements: x = 4 and y = −1 Checking: equation (1), 3 ð4 C5 ð1 7 D 0 D RHS equation (2), 4 ð4 3 ð1 19 D 0 D RHS THE SOLUTION OF SIMULTANEOUS EQUATIONS BY MATRICES AND DETERMINANTS 515 (b) The procedure for solving linear simultaneous equations in three unknowns using matri- ces is: (i) write the equations in the form a 1 x C b 1 y Cc 1 z D d 1 a 2 x C b 2 y Cc 2 z D d 2 a 3 x C b 3 y Cc 3 z D d 3 (ii) write the matrix equation corresponding to these equations, i.e. a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 ð x y z D d 1 d 2 d 3 (iii) determine the inverse matrix of a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 (see Chapter 60) (iv) multiply each side of (ii) by the inverse matrix, and (v) solve for x, y and z by equating the corresponding elements. Problem 2. Use matrices to solve the simultaneous equations: x C y C z 4 D 0 1 2x 3y C 4z 33 D 0 2 3x 2y 2z 2 D 0 3 (i) Writing the equations in the a 1 x Cb 1 y Cc 1 z D d 1 form gives: x C y C z D 4 2x 3y C 4z D 33 3x 2y 2z D 2 (ii) The matrix equation is 111 2 34 3 2 2 ð x y z D 4 33 2 (iii) The inverse matrix of A D 111 2 34 3 2 2 is given by A 1 D adj A jAj The adjoint of A is the transpose of the matrix of the cofactors of the elements (see Chapter 60). The matrix of cofactors is 14 16 5 0 55 7 2 5 and the transpose of this matrix gives adj A D 14 0 7 16 5 2 555 The determinant of A, i.e. the sum of the products of elements and their cofactors, using a first row expansion is 1 34 2 2 1 24 3 2 C 1 2 3 3 2 D 1 ð14 1 ð16 C1 ð5 D 35 Hence the inverse of A, A 1 D 1 35 1407 16 5 2 555 (iv) Multiplying each side of (ii) by (iii), and remembering that AðA 1 D I, the unit matrix, gives 100 010 001 ð x y z D 1 35 1407 16 5 2 555 ð 4 33 2 x y z D 1 35 14 ð4 C0 ð33 C 7 ð2 16 ð4 C5 ð 33 C2 ð2 5 ð4 C5 ð33 C 5 ð2 D 1 35 70 105 175 D 2 3 5 (v) By comparing corresponding elements, x = 2, y = −3, z = 5, which can be checked in the original equations. 516 ENGINEERING MATHEMATICS Now try the following exercise Exercise 206 Further problems on solving simultaneous equations using matrices In Problems 1 to 5 use matrices to solve the simultaneous equations given. 1. 3x C4y D 0 2x C 5y C 7 D 0[x D 4,y D3] 2. 2p C 5q C 14.6 D 0 3.1p C1.7q C2.06 D 0 [p D 1.2, q D3.4] 3. x C2y C 3z D 5 2x 3y z D 3 3x C4y C 5z D 3 [x D 1, y D1, z D 2] 4. 3a C 4b 3c D 2 2a C 2b C2c D 15 7a 5b C4c D 26 [a D 2.5, b D 3.5, c D 6.5] 5. p C 2q C 3r C7.8 D 0 2p C5q r 1.4 D 0 5p q C7r 3.5 D 0 [p D 4.1, q D1.9, r D2.7] 6. In two closed loops of an electrical cir- cuit, the currents flowing are given by the simultaneous equations: I 1 C 2I 2 C 4 D 0 5I 1 C 3I 2 1 D 0 Use matrices to solve for I 1 and I 2 . [I 1 D 2, I 2 D3] 7. The relationship between the displace- ment, s, velocity, v, and acceleration, a, of a piston is given by the equations: s C2 v C2a D 4 3s v C4a D 25 3s C 2 v a D4 Use matrices to determine the values of s, v and a. [s D 2, v D3, a D 4] 8. In a mechanical system, acceleration x, velocity v and distance x are related by the simultaneous equations: 3.4Rx C 7.0Px 13.2x D11.39 6.0Rx C 4.0Px C3.5x D 4.98 2.7Rx C 6.0Px C7.1x D 15.91 Use matrices to find the values of Rx, Px and x. [Rx D 0.5, Px D 0.77, x D 1.4] 61.2 Solution of simultaneous equations by determinants (a) When solving linear simultaneous equations in two unknowns using determinants: (i) write the equations in the form a 1 x C b 1 y Cc 1 D 0 a 2 x C b 2 y Cc 2 D 0 and then (ii) the solution is given by x D x D y D y D 1 D where D x D b 1 c 1 b 2 c 2 i.e. the determinant of the coefficients left when the x -column is covered up, D y D a 1 c 1 a 2 c 2 i.e. the determinant of the coefficients left when the y-column is covered up, and D D a 1 b 1 a 2 b 2 i.e. the determinant of the coefficients left when the constants-column is covered up. Problem 3. Solve the following simultaneous equations using determinants: 3x 4y D 12 7x C5y D 6.5 THE SOLUTION OF SIMULTANEOUS EQUATIONS BY MATRICES AND DETERMINANTS 517 Following the above procedure: (i) 3x 4y 12 D 0 7x C 5y 6.5 D 0 (ii) x 4 12 5 6.5 D y 3 12 7 6.5 D 1 3 4 75 i.e. x 46.5 125 D y 36.5 127 D 1 35 47 i.e. x 26 C60 D y 19.5 C84 D 1 15 C28 i.e. x 86 D y 64.5 D 1 43 Since x 86 D 1 43 then x D 86 43 D 2 and since y 64.5 D 1 43 then y D 64.5 43 D −1.5 Problem 4. The velocity of a car, accelerating at uniform acceleration a between two points, is given by v D u Cat, where u is its velocity when passing the first point and t is the time taken to pass between the two points. If v D 21 m/s when t D 3.5s and v D 33 m/s when t D 6.1s, use determinants to find the values of u and a, each correct to 4 significant figures. Substituting the given values in v D u Cat gives: 21 D u C3.5a1 33 D u C6.1a2 (i) The equations are written in the form a 1 x C b 1 y Cc 1 D 0, i.e. u C3.5a 21 D 0 and u C6.1a 33 D 0 (ii) The solution is given by u D u D a D a D 1 D , where D u is the determinant of coefficients left when the u column is covered up, i.e. D u D 3.5 21 6.1 33 D 3.533 216.1 D 12.6 Similarly, D a D 1 21 1 33 D 133 211 D12 and D D 13.5 16.1 D 16.1 3.51 D 2.6 Thus u 12.6 D a 12 D 1 2.6 i.e. u D 12.6 2.6 D 4 .846 m=s and a D 12 2.6 D 4 .615 m=s 2 , each correct to 4 significant figures Problem 5. Applying Kirchhoff’s laws to an electric circuit results in the following equations: 9 Cj12I 1 6 Cj8I 2 D 5 6 Cj8I 1 C 8 Cj3I 2 D 2 Cj4 Solve the equations for I 1 and I 2 . Following the procedure: (i) 9 Cj12I 1 6 Cj8I 2 5 D 0 6 Cj8I 1 C 8 Cj3I 2 2 Cj4 D 0 (ii) I 1 6 Cj8 5 8 Cj3 2 Cj4 D I 2 9 Cj12 5 6 Cj8 2 Cj4 D 1 9 Cj12 6 Cj8 6 Cj88 Cj3 518 ENGINEERING MATHEMATICS I 1 20 Cj40 C40 Cj15 D I 2 30 j60 30 Cj40 D 1 36Cj12328Cj96 I 1 20 Cj55 D I 2 j100 D 1 64 Cj27 Hence I 1 D 20 Cj55 64 Cj27 D 58.52 6 70.02 ° 69.46 6 22.87 ° D 0.84 66 47.15 ° A and I 2 D 100 6 90 ° 69.46 6 22.87 ° D 1.44 66 67.13 ° A (b) When solving simultaneous equations in three unknowns using determinants: (i) Write the equations in the form a 1 x C b 1 y Cc 1 z C d 1 D 0 a 2 x C b 2 y Cc 2 z C d 2 D 0 a 3 x C b 3 y Cc 3 z C d 3 D 0 and then (ii) the solution is given by x D x D y D y D z D z D 1 D where D x is b 1 c 1 d 1 b 2 c 2 d 2 b 3 c 3 d 3 i.e. the determinant of the coefficients obtained by covering up the x column. D y is a 1 c 1 d 1 a 2 c 2 d 2 a 3 c 3 d 3 i.e., the determinant of the coefficients obtained by covering up the y column. D z is a 1 b 1 d 1 a 2 b 2 d 2 a 3 b 2 d 3 i.e. the determinant of the coefficients obtained by covering up the z column. and D is a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 i.e. the determinant of the coefficients obtained by covering up the constants column. Problem 6. A d.c. circuit comprises three closed loops. Applying Kirchhoff’s laws to the closed loops gives the following equations for current flow in milliamperes: 2I 1 C 3I 2 4I 3 D 26 I 1 5I 2 3I 3 D87 7I 1 C 2I 2 C 6I 3 D 12 Use determinants to solve for I 1 , I 2 and I 3 . (i) Writing the equations in the a 1 x C b 1 y Cc 1 z C d 1 D 0formgives: 2I 1 C 3I 2 4I 3 26 D 0 I 1 5I 2 3I 3 C 87 D 0 7I 1 C 2I 2 C 6I 3 12 D 0 (ii) The solution is given by I 1 D I 1 D I 2 D I 2 D I 3 D I 3 D 1 D , where D I 1 is the determinant of coefficients obtained by covering up the I 1 column, i.e., D I 1 D 3 4 26 5 387 2612 D 3 387 6 12 4 587 2 12 C 26 5 3 26 D 3486 C 4114 2624 D −1290 D I 2 D 2 4 26 1 387 7612 D 236 522 412 C609 C 266 21 D972 C2388 C390 D 1806 [...]... 2 j4 I1 2I3 156 0° 3y C 4z D 33 3x j4 I2 j4 C 2 I2 2x b1 b2 b3 i.e the z-column has been replaced by the R.H.S b column (This is the same as Problem 2 and a comparison of methods may be made) Following the above method: 1 1 1 3 4 DD 2 3 2 2 D1 6 8 1 4 12 C1 4 Dx D 33 2 D4 6 4 1 3 2 8 C1 66 1 4 Dy D 2 33 3 2 D1 66 9 D 14 C 16 C 5 D 35 1 4 2 1 6 1 4 2 8 4 66 8 D 56 C 74 4 60 D 70 12 C 1 4 99 74 C 64 95... 37 (c) 42 (b) 47 (a) 52 (c) 57 (d) 3 (a) 8 (c) 13 (c) 18 (d) 23 (a) 28 (b) 33 (a) 38 (d) 43 (d) 48 (c) 53 (b) 58 (c) 4 (a) 9 (b) 14 (d) 19 (a) 24 (b) 29 (c) 34 (d) 39 (c) 44 (c) 49 (b) 54 (a) 5 (c) 10 (d) 15 (a) 20 (a) 25 (d) 30 (d) 35 (d) 40 (b) 45 (b) 50 (a) 55 (a) Index Abscissa 231 Adjoint of matrix 511 Algebra 34, 44 Algebraic expression 57 Amplitude 187, 190 And-function 48 3 And-gate 49 5 Angle... (b) 33 (a) 38 (a) 43 (a) 48 (a) 53 (b) 4 (d) 9 (b) 14 (d) 19 (b) 24 (c) 29 (d) 34 (d) 39 (d) 44 (c) 49 (d) 54 (c) 5 (c) 10 (b) 15 (a) 20 (a) 25 (d) 30 (b) 35 (d) 40 (b) 45 (d) 50 (a) 55 (a) Multiple choice questions on chapters 44 –61 (page 522) 5 (b) 10 (a) 15 (d) 20 (d) 25 (a) 30 (a) 35 (b) 40 (d) 45 (b) 50 (c) 55 (b) 60 (d) 1 (b) 6 (a) 11 (b) 16 (b) 21 (d) 26 (a) 31 (c) 36 (a) 41 (d) 46 (d) 51 (c) 56... 17 (d) 22 (c) 27 (a) 32 (d) 37 (d) 42 (a) 47 (b) 52 (d) 57 (b) 3 (b) 8 (c) 13 (c) 18 (b) 23 (a) 28 (b) 33 (d) 38 (c) 43 (d) 48 (c) 53 (b) 58 (a) 4 (a) 9 (c) 14 (c) 19 (d) 24 (c) 29 (d) 34 (a) 39 (b) 44 (b) 49 (c) 54 (a) 59 (a) 1 (d) 6 (d) 11 (d) 16 (a) 21 (b) 26 (b) 31 (b) 36 (c) 41 (b) 46 (b) 51 (a) 56 (b) 2 (b) 7 (c) 12 (a) 17 (c) 22 (c) 27 (c) 32 (a) 37 (c) 42 (c) 47 (d) 52 (c) 3 (a) 8 (d) 13 (d)... (d) 33 (d) 38 (a) 43 (b) 48 (b) 53 (d) 58 (d) 4 (b) 9 (c) 14. (d) 19 (c) 24 (b) 29 (b) 34 (c) 39 (b) 44 (c) 49 (c) 54 (d) 59 (b) Multiple choice questions on chapters 27 43 (page 369) 5 (a) 10 (c) 15 (a) 20 (d) 25 (a) 30 (d) 35 (a) 40 (d) 45 (d) 50 (a) 55 (d) 60 (c) Multiple choice questions on chapters 17–26 (page 225) 1 (d) 6 (a) 11 (b) 16 (d) 21 (b) 26 (b) 31 (b) 36 (a) 41 (d) 46 (c) 51 (c) 56 (d)... distribution 3 34 series 1 14, 115 theorem 115 practical problems 120 Blunder 24 BODMAS 2, 40 Boolean algebra 48 3 laws and rules of 48 8 Boyle’s law 42 Brackets 38 Calculations 24, 26 Calculator 26, 95, 100, 178 Calculus 375 Cancelling 1 Cartesian axes 231 co-ordinates 1 94 complex numbers 291 Centre of gravity 46 6 Centroids 46 6 Chain rule 389 Change of limits, integration 41 6 Charles’ law 42 Chord 139 Circle... D 1 74 3 5 2 35 4 2 3 6 21 35 48 C 45 C 132 D 129 I3 D 6 I3 1 D 1161 129 1290 = 10 mA, 129 1806 I2 D = 14 mA 129 and 5 4 3 6 30 C 6 Thus I2 I1 D D 1290 1806 giving I1 D 26 2 1791 C 858 D −1161 228 C 4 12 C 609 C D and 26 87 12 7 8 In problems 1 to 5 use determinants to solve the simultaneous equations given 5y D 17.6 2x 22 D 0 [x D 1.2, y D 2.8] 2 2.3m 4. 4n D 6. 84 8.5n 6.7m D 1.23 [m D 6 .4, n D 4. 9]... of area 47 5 volumes 46 1 Integration using algebraic substitutions 41 4 partial fractions 42 6 Â t D tan substitution 43 0 2 trigonometric substitutions 41 8 Intercept 232 Interpolation 237, 352 Interval estimate 360 Inverse functions 275 proportion 3, 42 trigonometric functions 276 Inverse matrix 509, 511 Invert-gate 49 5 Karnaugh maps 49 1 Lagging angles 188, 190, 287 Lamina 46 6 Laws of algebra 34 growth... Multiple-choice questions 127, 2 24, 369, 522 Nand-gate 49 5 Napierian logarithms 89, 100 Natural logarithms 89, 100 Newton–Raphson method 123 Non-terminating decimal 5 Nor-gate 49 6 Normal curve 340 distribution 340 equations 351 probability paper 344 standard variate 340 Normals 40 3 Nose-to-tail method 282 Not-function 48 4 Not-gate 49 5 Number sequences 106 Numerator 1 Numerical integration 43 9 Octagon 131 Octal... law 42 Horizontal bar chart 308 Hyperbola 267 rectangular 268 Hyperbolic logarithms 89, 100 Identity 57 trigonometric 208 Imaginary number 291 Improper fraction 1 INDEX Independent event 326 variable 42 Indices 9, 36 laws of 9 Indicial equations 92 Integral calculus 40 7 Integrals, standard 40 8 Integration 40 7 areas under and between curves 44 8 by parts 43 4 Centroids 46 6 mean values 45 7 r.m.s values 45 9 . 32. (d) 33. (d) 34. (a) 35. (b) 36. (a) 37. (d) 38. (c) 39. (b) 40 . (d) 41 . (d) 42 . (a) 43 . (d) 44 . (b) 45 . (b) 46 . (c) 47 . (b) 48 . (c) 49 . (c) 50. (c) 51. (c) 52. (d) 53. (b) 54. (a) 55. (b) 56 32. (a) 33. (a) 34. (d) 35. (d) 36. (c) 37. (c) 38. (a) 39. (d) 40 . (b) 41 . (b) 42 . (c) 43 . (a) 44 . (c) 45 . (d) 46 . (b) 47 . (d) 48 . (a) 49 . (d) 50. (a) 51. (a) 52. (c) 53. (b) 54. (c) 55. (a) 56 exercise Exercise 2 04 Further problems on 3 by 3 determinants 1. Find the matrix of minors of 4 76 240 57 4 16 8 34 14 46 63 24 12 2 2. Find the matrix of cofactors of 4 76 240 57 4 16