Advance Engineering Mathematics (H.K. Dass) - Unit 1 PDF

4 4 Determinants and Matrices DETERMINANTS AND MATRICES 4.1 INTRODUCTION In Engineering Mathematics, solution of simultaneous equations i...

4 4 Determinants and Matrices DETERMINANTS AND MATRICES 4.1 INTRODUCTION In Engineering Mathematics, solution of simultaneous equations is very important. In this chapter we shall study the system of linear equations with emphasis on their solution by means of determinants. 4.2 DETERMINANT The notation of determinants arises from the process of elimination of the unknowns of simultaneous linear equations. Consider the two linear equations in x, a1 x + b1 = 0... (1) a2 x + b2 = 0... (2) b1 From (1) x a1 Substituting the value of x in (2); we get the eliminant  b  a2   1   b2  0  a1  or a1b2 – a2b1 = 0... (3) From (1) and (2) by suppressing x, the eliminant is written as a1 b1 0... (4) a2 b2 when the two rows of a1, b1 and a2, b2 are enclosed by two vertical bars then it is called a determinant of second order. a1 b1 and a2 b2 Column 1 Column 2 Row 1 a1 b1.... Row 2 a2 b2.... Each quantity a1, b1, a2, b2 is called an element or a constituent of the determinant. From (3) and (4), we know that both expressions are eliminant, so we equate them. a1 b1 a1 b1  a1 b2  a2 b1 or = a1b2 – a2b1 a2 b2 a2 b2 223 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  224 Determinants and Matrices a1 b1 a1b2 – a2b1 is called the expansion of the determinant of. a2 b2 3 2 Example 1. Expand the determinant. 6 7 + – 3 2 Solution. = (3) × (7) – (2) × (6) = 21 – 12 = 9. Ans. 6 7 EXERCISE 4.1 Expand the following determinants : 4 6 3 7 1. 2 5 Ans. 8 2. Ans. – 26 2 4 8 5 5 2 3. 3 1 Ans. – 7 4. Ans. 23 4 3 4.3. DETERMINANT AS ELIMINANT Consider the following three equations having three unknowns, x, y and z. a1 x + b1 y + c1 z = 0...(1) a2 x + b2 y + c2 z = 0...(2) a3 x + b3 y + c3 z = 0...(3) From (2) and (3) by cross-multiplication, we get x y z    k (say) b2 c3  b3 c2 a3 c2  a2 c3 a2 b3  a3 b2 x = (b2 c3 – b3 c2) k y = (a3 c2 – a2 c3) k and z = (a2 b3 – a3 b2) k Substituting the values of x, y and z in (1), we get the eliminant a1 (b2c3 – b3c2) k + b1 (a3c2 – a2c3) k + c1 (a2b3 – a3b2) k = 0 or a1 (b2c3 – b3c2) – b1 (a2c3 – a3c2) + c1 (a2b3 – a3b2) = 0...(4) From (1), (2) and (3) by suppressing x, y, z the remaining can be written in the determinant as a1 b1 c1 a2 b2 c2  0...(5) a3 b3 c3 This is determinant of third order. As (4) and (5) both are the eliminant of the same equations. a1 b1 c1 a2 b2 c2  a1 (b2c3  b3c2 )  b1 (a2c3  a3c2 )  c1 (a2b3  a3b2 )  0  a3 b3 c3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  Determinants and Matrices 225 a1 b1 c1 b2 c2 a2 c2 a2 b2 or a2 b2 c2  a1  b1  c1 b3 c3 a3 c3 a3 b3 a3 b3 c3 4.4. MINOR The minor of an element is defined as a determinant obtained by deleting the row and column containing the element. Thus the minors a1, b1 and c1 are respectively. b2 c2 a2 c2 a2 b2 , and b3 c3 a3 c3 a3 b3 Thus a1 b1 c1 a2 b2 c2 = a1 (minor of a1) – b1 (minor of b1) + c1 (minor of c1). a3 b3 c3 4.5. COFACTOR Cofactor = (– 1)r+c Minor where r is the number of rows of the element and c is the number of columns of the element. The cofactor of any element of jth row and ith column is (– 1)i+j minor 1+1 Thus the cofactor of a1 = (– 1) (b2c3 – b3c2) = + (b2c3 – b3c2) The cofactor of b1 = (– 1)1+2 (a2c3 – a3c2) = – (a2c3 – a3c2) The cofactor of c1 = (– 1)1+3 (a2b3 – a3b2) = + (a2b3 – a3b2) The determinant = a1 (cofactor of a1) + a2 (cofactor of a2) + a3 (cofactor of a3). Example 2. Write down the minors and cofactors of each element and also evaluate the determinant. 1 3 –2 4 –5 6 3 5 2   1 3 2   4 5 6 Solution. M11 = Minor of element (1) 3 5 2 5 6   ( 5)  2  6  5   10  30   40 5 2 Cofactor of element (1) = A11 = (– 1)1 + 1 M11 = (– 1)2 (– 40) = – 40 M12 = Minor of element (3) 1  3 2 4 6  4  5 6   4  2  3  6  8  18  10 3 2 3 5 2  Cofactor of element (–2) = A12 = (– 1)1 + 2 (– 10) = 10 M13 = Minor of element (– 2) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  226 Determinants and Matrices 1 3  2  4 5 4 5 6   4  5  (5)  3  20  15  35 =  3 5 3 5 2  Cofactor of element (– 2) = A13 = (– 1)1+3 M13 = (–1)4 35 = 35 M21 = Minor of element (4) 1 3 2  3 2 = 4 5 6   3  2  (2)  5  6  10  16  5 2 3 5 2  Cofactor of element (4) = A21 = (– 1)2+1 M21 = (– 1)2+1 (16) = – 16 M22 = Minor of element (– 5) 1 3 2  1 2 = 45 6   1 2  (2)  3  2  6  8  3 2 3 5 2  Cofactor of element (– 5) = A22 = (– 1)2+2 M22 = (– 1)2+2 (8) = 8 M23 = Minor of element (6) 1 3 2  1 3 = 4  5 6   1 5  3  3  5  9   4  3 5 3 5 2  Cofactor of element (6) = A23 = (–1)2+3 M23 = (– 1)2+3 (– 4) = 4 M31 = Minor of element (3) 1 3 2  3 2 4 5 6   3  6  (2)  (5)  18  10  8 =  5 6 3 5  2   Cofactor of element (3) = A31 = (– 1)3+1 M31 = (– 1)3+1 8 = 8 M32 = Minor of element (5) 1 3 2  1 2 = 4 5 6   1 6  (2)  4  6  8  14  4 6 35  2  Cofactor of element (5) = A32 = (– 1)3+2 M32 = (– 1)3+2 14 = – 14   M33 = Minor of element (2) 1 3 2  1 3 = 4 5 6   1 (5)  4  3  5  12  17  4 5 3  5 2 Cofactor of element (2) = A33 = (– 1)3+3 M33 = (– 1)3+3 (– 17) = – 17. 1 3 2 4 5 6 = 1 × (cofactor of 1) + 3 × (cofactor of 3) + (– 2) × [cofactor of (– 2)]. 3 5 2 = 1 × (– 40) + 3 × (10) + (– 2) × (35) = – 40 + 30 – 70 = – 80 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  Determinants and Matrices 227 Example 3. Find : (i) Minors (ii) Cofactors of the elements of the first row of the determinant 2 3 5 4 1 0 6 2 7 Solution. (i) The minor of the element (2) is 235  1 0 4 1 0   (1)  (7)  (0)  (2)  7  0  7  2 7 6 2 7 The minor of the element (3) is 23 5  4 0 4 1 0   (4)  (7)  (0)  (6)  28  0  28  6 7 6 2 7 The minor of the element (5) is 23 5  4 1 4 1 0   (4)  (2)  (1)  (6)  8  6  2  6 2 6 2 7 (ii) The cofactor of (2) = (– 1)1+1 (7) = + 7 The cofactor of (3) = (– 1)1+2 (28) = – 28 Ans. The cofactor of (5) = (– 1)1+3 (2) = + 2. 6 2 3 Example 4. Expand the determinant 2 3 5 6 2 3 4 2 1 Solution. 2 3 5 = 6 (cofactor of 6) + 2 (cofactor of 2) + 3 (cofactor of 3). 4 2 1 = 6 (3 × 1 – 5 × 2) – 2 (2 × 1 – 4 × 5) + 3 (2 × 2 – 3 × 4) = 6 (3 – 10) – 2 (2 – 20) + 3 (4 – 12) = 6 (– 7) – 2 (– 18) + 3 (– 8) = – 42 + 36 – 24 = – 30. Ans. 1 0 4 3 5 –1. Example 5. Evaluate the determinant 0 1 2 (i) With the help of second row, (ii) with the help of third column. Solution. 1 0 4 (i) 3 5 1 = 3 × (cofactor of 3) + 5 × (cofactor of 5) + (– 1) (cofactor of – 1). 0 1 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  228 Determinants and Matrices 0 4 1 4 1 0 = 3 × (– 1)2+1 1 2 + 5 × (– 1)2+2 0 2 + (– 1) × (– 1)2+3 0 1 = – 3 × (0 – 4) + 5 (2 – 0) + (1 – 0) = 12 + 10 + 1 = 23 Ans. 1 0 4 (ii) 3 5 1 = 4 × (cofactor of 4) + (– 1) (cofactor of (– 1)) + 2 × (cofactor of 2) 0 1 2 3 5 1 0 1 0 = 4 × (– 1)1+3 + (– 1) (– 1)2+3 + 2 × (– 1)3+3 0 1 0 1 3 5 = 4 × (3 – 0) + (1 – 0) + 2 (5 – 0) = 12 + 1 + 10 = 23 Ans. 0 1 2 3 1 0 2 0 Example 6. Expand the fourth order determinant 2 0 1 3 1 2 1 0 0 2 0 1 2 0 1 2 0 1 3  1 (–1) 2 1 3 Solution. Given determinant = (0) (–1)1 + 1 2 1 0 1 1 0 1 0 0 1 0 2 2 0 3  3 (–1)1  4 2 0 1 + 2 (–1)1+ 3 1 2 0 1 2 1 1 2 0 1 0 0 1 0 2 = 0 2 1 3 2 2 0 3 3 2 0 1 1 1 0 1 2 0 1 2 1 1 2 0 1(1 0  3  1)  2 (2  0  3  1)  0 (2  1  1  1) 2 1 3 Now   3 6 03 1 1 0 1 0 0 1 (0  0  3  2)  0 (2  0  3  1)  0(2  2  0  1) 2 0 3  6 1 2 0 1 0 2  1 ( 0 1  1  2)  0(2  1 1  1)  2 (2  2  0 1) 2 0 1  2– 0  8  6 1 2 1 0 1 2 3 1 0 2 0   3  2(  6 )  3(6) Now 2 0 1 3   3 – 12  18   33 Ans. 1 2 1 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  Determinants and Matrices 229 EXERCISE 4.2 Write the minors and co factors of each element of the following determinants and also evalutate the determinant in each case : 2 3 M11 = – 9, M12 = 4, M21 = 3, M22 = – 2 1. 4 9 A11 = – 9, A12 = – 4, A21 = – 3, A22 = – 2 |A|=6 Ans. cos   sin  M11 = cos , M12 = sin  M21 = – sin , M22 = cos  2. sin  cos  A11 = cos , A12 = – sin , A21 = sin , A22 = cos  | A | = 1 Ans. 42 1 6 M11 = 2, M12 = 0, M13 = – 14, M21 = – 16, M22 = 0 M23 = 112, M31 = – 38, M32 = 0, M33 = 266 3. 28 7 4 A11 = 2, A12 = 0, A13 = – 14, A21 = 16, A22 = 0 14 3 2 A23 = – 112, A31 = – 38, A32 = 0, A33 = 266, | A | = 0 Ans. M11 = (ab2 – ac2), M12 = (ab – ac), M13 = (c – b), M21 = a2b – bc2 1 a bc M22 = (ab – bc), M23 = (c – a), M31 = (ca2 – cb2), M32 = ca – bc, M33 = (b – a), 4. 1 b ca A11 = (ab2 – ac2), A12 = (ac – ab), A13 = (c – b), A21 = bc2 – a2b 1 c ab A22 = (ab – bc), A23 = (a – c), A31 = (ca2 – cb2), A32 = (bc – ca), A33 = (b – a) | A | = (a – b) (b – c) (c – a). Ans. Expand the following determinants : 2 3 4 5 0 7 a h g 5. 5 1 6 6. 8 6 4 7. h b f 7 8 9 2 3 9 g f c Ans. | A | = 5 Ans. | A | = 42 Ans. | A | = abc + 2fgh – af 2 – bg2 – ch2 Expand the following determinants by two methods : (i) along the-third row. (ii) along the-third column. 1 3 2 3 2 4 2 3 2 8. 4 1 2 9. 1 2 1 10. 1 2 3 3 5 2 0 1 1 2 1 3 Ans. | A | = 40 Ans. | A | = – 7 Ans. | A | = – 37 log 3 512 log 4 3 15 11. Ans. | A | = log 3 8 log 4 9 2 12. If a, b, c are all positive and are the pth, qth, rth terms of a G.P. respectively; then prove that 3 2 5 7 log a p 1 1 4 3 0 log b q 1  0 13. 6 4 2 1 Ans. 96 log c r 1 2 1 0 3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  230 Determinants and Matrices 4.6 RULESOF SARRUS (For third order determinants only). After writing the determinant, repeat the first two columns as below = (a1b2c3 + b1c2a3 + c1a2b3) + (– c1b2a3 – a1c2b3 – b1a2c3) Example 7. Expand the determinant 2 3 4   1 5 3 by Rule of Sarrus. 3 0 5 Solution. =(2) × (5) × (5) + (3) × (3) × (3) + (4) × (1) × (0) – (4) × (5) × (3) – (2) × (3) × (0) – (3) × (1) × (5) = 50 + 27 + 0 – 60 – 0 – 15 = 2 Ans. EXERCISE 4.3 Expand the following determinants by Rule of Sarrus. 3 2 4 1 4 2 6 3 7 9 25 6 1. 5 1 1 2. 2 5 3 3. 32 13 37 4. 7 13 5 2 6 7 3 6 4 10 4 11 9 23 6 Ans. – 155 Ans. 0 Ans. 10 Ans. 6 ax c b 5. If a + b + c = 0, solve the equation c bx a 0 b a cx 2 2 2 Ans. x   (a  b  c  ab  bc  ca) , x = 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  Determinants and Matrices 231 4.7 PROPERTIES OF DETERMINANTS Property (i) The value of a determinant remains unaltered, if the rows are interchanged into columns (or the columns into rows). Consider the determinant. a1 b1 c1   a2 b2 c2 a3 b3 c3 = a1 (b2c3 – b3c2) – b1 (a2c3 – a3c2) + c1 (a2b3 – a3b2) = a1b2c3 – a1b3c2 – a2b1c3 + a3b1c2 + a2b3c1 – a3b2c1 = (a1b2c3 – a1b3c2) – (a2b1c3 – a2b3c1) + (a3b1c2 – a3b2c1) = a1 (b2c3 – b3c2) – a2 (b1c3 – b3c1) + a3 (b1c2 – b2c1) a1 a2 a3  b1 b2 b3 Proved. c1 c2 c3 Property (ii) If two rows (or two columns) of a determinant are interchanged, the sign of the value of the determinant changes. Interchanging the first two rows of , we get a2 b2 c2  '  a1 b1 c1 a3 b3 c3 = a2 (b1c3 – b3c1) – b2 (a1c3 – a3c1) + c2 (a1b3 – a3b1) = a2b1c3 – a2b3c1 – a1b2c3 + a3b2c1 + a1b3c2 – a3b1c2 = – [(a1b2c3 – a1b3c2) – (a2b1c3 – a3b1c2) + (a2b3c1 – a3b2c1)] = – [(a1 (b2c3 – b3c2) – b1 (a2c3 – a3c2) + c1 (a2b3 – a3b2)] a1 b1 c1 =  a2 b2 c2    Proved. a3 b3 c3 Property (iii) If two rows (or columns) of a determinant are identical, the value of the determinant is zero. a1 b1 c1 Let   a1 b1 c1 , so that the first two rows are identical. a3 b3 c3 By interchanging the first two rows, we get the same determinant . By property (ii), on interchanging the rows, the sign of the determinant changes. or  =– or 2=0 or  = 0 Proved. Property (iv) If the elements of any row (or column) of a determinant be each multiplied by the same number, the determinant is multiplied by that number. ka1 kb1 kc1   a2 b2 c2 a3 b3 c3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  232 Determinants and Matrices = ka1 (b2c3 – b3c2) – kb1 (a2c3 – a3c2) + kc1 (a2b3 – a3b2) = k [a1 (b2c3 – b3c2) – b1 (a2c3 – a3c2) + c1 (a2b3 – a3b2)] a1 b1 c1  k a2 b2 c2  k . a3 b3 c3 Example 8. Prove that a2 a bc 1 1 1 2 2 2 2 b b ca   a b c c2 c ab a3 b3 c3 a2 a bc 2 b b ca Solution. c2 c ab By multiplying R1, R2, R3 by a, b and c respectively we get a3 a2 abc a3 a2 1  1 b b abc  abc b 3 2 3 2 b 1 abc abc c3 c2 abc c3 c 2 1 a3 a2 1 1 a2 a3 3 2  b b 1   1 b2 b3 c3 c2 1 1 c2 c3 1 1 1 2 2  a b c2 By changing rows into columns a3 b3 c3 Proved Example 9. Without expanding and or evaluating, show that 2 3 2 a a 1 bcd a a a 1 2 3 2 b b 1 cda b b b 1  2 3 2 c c 1 dab c c c 1 2 3 2 d d 1 abc d d d 1 Solution 2 a a 1 bcd a3 a2 a abcd 2 3 2 R1 aR1 b b 1 cda 1 b b b abcd  R2 bR2 c 2 c 1 dab abcd c3 c2 c abcd R3 cR3 d2 d 1 abc d3 d2 d abcd R4 dR4 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  Determinants and Matrices 233 a3 a2 a 1 a3 a2 a 1 3 2 3 2 abcd b b b 1 b b b 1  1  abcd c3 c2 c 1 C4 C4 c3 c2 c 1 Proved abcd 3 2 3 2 d d d 1 d d d 1 1 a a2 1 a bc 2 Example 10. Prove that 1 b b  1 b ca (Try yourself) 1 c c 2 1 c ab Property (v) The value of the determinant remains unaltered if to the elements of one row (or column) be added any constant multiple of the corresponding elements of any other row (or column) respectively. a1 b1 c1 Let   a2 b2 c2 a3 b3 c3 On multiplying the second column by l and the third column by m and adding to the first column we get a1  lb1  mc1 b1 c1  '  a2  lb2  mc2 b2 c2 a3  lb3  mc3 b3 c3 a1 b1 c1 b1 b1 c1 c1 b1 c1  a2 b2 c2  l b2 b2 c2  m c2 b2 c2 a3 b3 c3 b3 b3 c3 c3 b3 c3 = + 0 + 0 (Since columns are identical) = Proved 9 9 12 Example 11. Evaluate, using the properties of determinant 1 3 – 4 9 9 12 1 9 12 Solution. Let   1 3 4 1 9 12 Applying : R1 R1  3R 2 and R 3 R 3  3R 2 , we get 12 18 0 2 3 0  1 3 4  6  2 1 3 4 4 18 0 2 9 0 2 3 Expand by C3  = 6×2×4 2 9 = 48 (2 × 9 – 2 × 3) = 48 × 12 = 576. Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  234 Determinants and Matrices 265 240 219 Example 12. Without expanding evaluate the determinant   240 225 198 219 198 181 Solution. Applying C1 C1  C3 and C 2 C 2  C3 , we get 46 21 219   42 27 198 38 17 181 Applying C1 C1  2C2 and C3 C3  10 C2 , we get 4 21 9   12 27  72 4 17 11 Applying R1 R1  R3 and R2 R2  3R3 0 4 2 0 2 1   0 78  39  2(39) 0 2 1 [Taking 2 common from R1 and 39 common from R2] 4 17 11 4 17 11 = 78 × 0 = 0 (Since R1 and R2 are identical) Ans. bc ca ab Example 13. Show that   c  a a b bc =0 ab bc ca bc ca ab Solution. Let  ca ab bc ab bc ca Applying C1 C1  C2  C3 , we get 0 ca ab  0 ab bc 0 [C1 consists of all zeros.] 0 bc ca sin cos  sin( +  ) Example 14. Without expanding, evaluate the determinant sin  cos  sin(  +  ). sin  cos  sin(  + ) sin  cos  sin (  ) Solution. Let   sin  cos  sin (  ) sin  cos  sin (  ) sin  cos  sin  cos   cos  sin     sin  cos  sin  cos   cos  sin  sin  cos  sin  cos   cos  sin  [ sin (A + B) = sin A cos B + cos A sin B] Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  Determinants and Matrices 235 sin  cos  0    sin  cos  0 [Applying C3 C3  cos . C1  sin . C2 ] sin  cos  0   =0 [ C3 consists of all zeros] Ans. 2x – 1 x + 7 x+4 Example 15. Solve the determinantal equation x 6 2 0 x–1 x+1 3 2x  1 x  7 x4 Solution. Given equation x 6 2 0 x 1 x 1 3 0 0 x 1 x 6 2 0 By applying R1 R1 – (R2 + R3), we get x 1 x 1 3 On expanding by first row, we get (x – 1) (x2 + x – 6x + 6) = 0  (x – 1) (x – 2) (x – 3) = 0  x = 1, 2, 3 Ans. Example 16. Using the properties of determinants, show that x+ y x x 3 5x + 4y 4x 2x = x. 10x + 8y 8x 3x x y x x Solution. Let   5x  4 y 4x 2x 10 x  8 y 8x 3x Operate : R2 R2  2 R1 ; R3 R3  3R1 x y x x   3x  2 y 2x 0 3x  2 y 2x Expand by C3 x 7 x  5 y 5x 0 7 x  5 y 5x = x [5x (3x + 2y) – 2x (7x + 5y)] = x [15x2 + 10 xy – (14x2 + 10 xy)] = x3. Proved. Example 17. Using the properties of determinants, evaluate the following : 2 2 0 ab ac a 2b 0 bc 2 2 2 a c cb 0 2 2 0 ab ac Solution. Let   a 2b 0 bc 2 a 2c cb2 0 0 a a 2 2 2 Take a2, b2 and c2 common from C1, C2 and C3 respectively, a b c b 0 b c c 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  236 Determinants and Matrices 0 0 a 2 2 2 Operate : C2 C2  C3 ,   a b c b b b c c 0 2 2 2 b b 3 2 2 3 3 3 Expand by R1,   a b c.a  a b c (bc  bc)  2a b c. Ans. c c Example 18. Using properties of determinants, prove that x y z 2 2 2 x y z = x y z (x – y)(y – z) (z – x). 3 3 3 x y z x y z 1 1 1 2 2 2 Solution. Let  x y z  xyz x y z 3 3 3 2 2 2 x y z x y z 0 0 1 Operate : C1 C1  C2 ; C2 C2  C3 ,   xyz x y yz z 2 2 2 2 2 x  y y z z x y yz 1 1 On expanding by R1,   xyz 2 2 2 2  xyz ( x  y) ( y  z) x  y y z x y yz = xyz (x – y) (y – z) (z – x). Proved. Example 19. Using the properties of determinants, show that a+x y z 2 x a+ y z = a (a + x + y + z). x y a+z ax y z Solution. Let  x a y z x y az a a 0 Operate : R1 R1  R2 ,  x a y z x y az a 0 0 Operate : C2 C2  C1 ,   x a y  x z x y x a z a yx z On expanding by R1 a = a [(a + y + x) (a + z) – (y + x) z] yx az = a [a2 + az + (y + x) a + (y + x) z – (y + x) z] = a2 (a + x + y + z). Proved. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  Determinants and Matrices 237 Example 20. If is the one of the imaginary cube roots of unity, find the value of the determinant           1  2 2 Solution. The given determinant    1 2  1  By R1 R1  R2  R3, we get 1    2 1    2 1    2 0 0 0 2 2    1    1 [1 + + 2 = 0] 2 2  1   1  =0 (Since each entry in R1 is zero) Ans. a b c Example 21. Without expanding the determinant, show that (a + b + c) is a factor of b c a. a b c c a b Solution. Let  b c a c a b abc b c 1 b c Operate : C1 C1  C2  C3 ,  abc c a  (a  b  c) 1 c a abc a b 1 a b  (a + b + c) is a factor of . Proved. Example 22. Using properties of determinants, prove that : x+4 x x x x+4 x = 16 (3x + 4) x x x+4 x4 x x   x x4 x Solution. Let x x x4 3x  4 x x Operate : C1 C1  C2  C3 ,   3x  4 x  4 x 3x  4 x x4 1 x x 1 x x  (3x  4) 1 x  4 x  (3x  4) 0 4 0 R2  R1 4 0  (3x  4) 1 x x4 0 0 4 R3  R1 0 4  16(3x  4) Proved. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  238 Determinants and Matrices 1 a b+c Example 23. Without expanding the determinant, prove that 1 b c + a = 0. 1 c a +b 1 a bc Solution. Let  1 b ca 1 c ab 1 a abc 1 a 1 Operate : C3 C3  C2 ,   1 b a  b  c  (a  b  c) 1 b 1 1 c abc 1 c 1 =0 ( C1 and C3 are identical). Proved. 1 a2 bc a 1 b 2 ca = 0 Example 24. Without expanding the determinant, prove that b 1 c2 ab 1 a2 c bc a  1 b 2 Solution. Let ca b 1 c2 ab c Multiply R1 by a, R2 by b and R3 by c. 1 a 3 abc 1 a3 1   1 1 b abc  1. abc 1 b 1  1  0  0. 3 3 abc abc 1 c 3 abc 1 c3 1 (Since C1 and C3 are identical) Proved. 2 1 a a Example 25. Evaluate 1 b b2 2 1 c c Solution. Let  be the given determinant. Applying R2 R2  R1 and R3 R3  R1, we get, 1 a a2 1 a a 2 2 2 [Taking out (b – a) common  0 ba b a  (b  a) (c  a) 0 1 b  a from R2 and (c – a) from R3] 0 c  a c2  a 2 0 1 ca 2 1 a a = (b – a) (c – a) 0 1 b  a [Applying R3 R3  R2 ] 0 0 cb Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  Determinants and Matrices 239 1 ba = (b – a) (c – a) [Expanding along C1] 0 cb = (b – a) (c – a) (c – b). Ans. Example 26. Using properties of determinants, prove that : 1 a a3 1 b b 3 = (a – b)(b – c)(c – a)(a + b + c) 1 c c3 1 a a3 3 Solution. Let = 1 b b 1 c c3 0 a  b a3  b3 3 3 3 3 ab a b Operate : R1 R1  R2 ; R2 R2  R3 ,   0 b  c b  c  1. 3 3 bc b c 1 c c3 (Expanding by C1) 2 2 1 a  ab  b  (a  b) (b  c) 2 2 1 b  bc  c 2 2 0 (a  c )  (ab  bc) Operate : R1 R1  R2 ,   (a  b) (b  c) 2 2 1 b  bc  c = (a – b). (b – c). (– 1) [(a2 – c2) + b (a – c)] = (a – b). (b – c) (c – a) (a + b + c). Proved. a–b–c 2a 2a Example 27. Evaluate 2b b–c–a 2b 2c 2c c–a–b abc abc abc Solution. By R1 R1 + R2 + R3, we get 2b bca 2b 2c 2c cab 1 1 1  (a  b  c) 2b b  c  a 2b 2c 2c cab 1 0 0  (a  b  c) 2b  (a  b  c) 0 C2  C1 2c 0  (a  b  c) C3  C1 On expanding by first row = (a + b + c) (a + b + c)2 = (a + b + c)3. Ans. 1 1 1 Example 28. Show, without expanding x y z = (x – y)(y – z)(z – x) 2 2 2 x y z Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  240 Determinants and Matrices 0 0 1 x y yz Solution. By C1 – C2, C2 – C3, we get  x y yz z  2 2 2 2 x  y y z 2 2 2 2 2 x  y y z z On expanding by first row, we get 1 1  ( x  y) ( y  z ) = (x – y) (y – z) (y + z – x – y) = (x – y) (y – z) (z – x). x y yz Proved.    Example 29. Prove that                            Solution. Let    2 . 2 2       2 2 2 Applying R3 R1  R3           2 2 = ( +  +  )    2 [Taking out (+  + ) common from R3] 1 1 1    2 2 2 2 2 Applying C2 C2  C1  (     )      C3 C3  C1 1 0 0  1 1 2  (    ) (  ) (  )    1 0 0 1 1  (    ) (  ) (  ).1 [Expanding along R3]   = (+ + ) (– ) (– ) (+ – – ) = (+ + ) (– ) (– ) (– ) Proved. 2 –a ab ac 2 2 2 2 Example 30. Prove that ba – b bc = 4 a b c ac bc – c 2  a2 ab ac 2 Solution. Let  ba  b bc ac bc  c 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  Determinants and Matrices 241 a a a Taking a, b, c common from R1, R2 and R3 respectively, we get   abc b b b 1 1 1 c c c 2 2 2 abc 1 1 1 [Taking a, b, c common from C1, C2 and C3 respectively] 1 1 1 1 0 0 2 2 2 abc 1 0 2 [Applying C2 C2  C1, C3 C3  C1 ] 1 2 0 0 2 = a2b2c2 (– 1) 2 0 [Expanding along R1] = a2b2c2 (– 1) (0 – 4) = 4a2 b2 c2 Proved. 3a –a + b –a + c Example 31. Show that –b + a 3b – b + c = 3(a + b + c) (ab + bc + ca) –c + a –c + b 3c 3a a  b a  c Solution. Let   b  a 3b b  c c  a c  b 3c a  b  c a  b a  c Applying C1 C1  C2  C3, we get  abc 3b b  c abc c  b 3c 1 a  b a  c  (a  b  c) 1 3b b  c [Taking (a + b + c) common from C1] 1 c  b 3c 1 a  b a  c = (a  b  c) 0 2b  a  b  a [Applying R2 R2  R1 , R3 R3  R1 ] 0  c  a 2c  a 2b  a b  a  (a  b  c) [Expanding along C1] c  a 2 c  a = (a + b + c) [(2b + a) (2c + a) – (– b + a) (– c + a)] = (a + b + c) {(4bc + 2ab + 2ca + a2 – (bc – ab – ac + a2)} = (a + b + c) (3bc + 3ab + 3ca) = 3 (a + b + c) (ab + bc + ca) Proved. Property (vi) If each element of a row (or column) of a determinant consists of the algebraic sum of n terms, the determinant can be expressed as the sum of n determinants, a1  p1  q1 b1 c1 Let   a 2  p2  q 2 b2 c2. a3  p3  q3 b3 c3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  242 Determinants and Matrices = (a1 + p1 + q1) (b2c3 – b3c2) – (a2 + p2 + q2) (b1c3 – b3c1) + (a3 + p3 + q3) (b1c2 – b2c1) = a1 (b2c3 – b3c2) – a2 (b1c3 – b3c1) + a3 (b1c2 – b2c1) + p1 (b2c3 – b3c2) – p2 (b1c3 – b3c1) + p3 (b1c2 – b2c1) + q1 (b2c3 – b3c2) – q2 (b1c3 – b3c1) + q3 (b1c2 – b2c1) a1 b1 c1 p1 b1 c1 q1 b1 c1  a2 b2 c2  p2 b2 c2  q2 b2 c2 Proved. a3 b3 c3 p3 b3 c3 q3 b3 c3 2 3 a a a –1 Example 32. If b b2 b3 – 1 = 0, prove that abc = 1. 2 3 c c c –1 a a2 a3  1 a a2 a3 a a2 1 2 3 2 3 2 Solution. b b b 1  0  b b b  b b 1  0 2 3 2 3 2 c c c 1 c c c c c 1 1 a a2 a a2 1 2 2 abc 1 b b  b b 1 0  2 2 1 c c c c 1 (Taking out common a, b, c from R1, R2 and R3 from 1st determinant) 1 a a2 a 1 a2 2 2 abc 1 b b  b 1 b 0  (Interchanging C2 and C3) 2 2 1 c c c 1 c 1 a a2 1 a a2 2 2 abc 1 b b  1 b b 0  (Interchanging C1 and C2 ) 2 2 1 c c 1 c c 2 1 a a 2 ( a b c  1) 1 b b  0  2 1 c c  abc  1  0 Proved.  abc  1 b+c c+a a+b a b c q+r r+ p p+q =2 p q r Example 33. Show that y+z z+x x+ y x y z Solution. The above determinant can be expressed as the sum of 8 determinants as given below: bc ca ab b c a b a a b c b b a b qr r p pq  q r p  q p p  q r q  q p q y z z x x y y z x y x x y z y y x y Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  Determinants and Matrices 243 c c a c a a c c b c a b  r r p  r p p  r r q r p q + z z x z x x z z y z x y b c a c a b  q r p 000000 r p q y z x z x y a b c a b c a b c 2 2 = (1) p q r  (1) p q r 2 p q r x y z x y z x y z              0 Example 34. Prove that       Proved.        Solution. Given determinant    The above determinant can be expressed as the sum of 8 determinants. Each of the 8 determinants has either two identical columns or identical rows. Each of the resulting determinant is zero. Hence the result. Proved. x l m 1  x n 1 Example 35. Prove that  (x   ) (x   ) (x   )   x 1    1 x l m 1 x l m 1  x n 1 0 x n 1 Solution.  (C1 C1 – C4)   x 1 0  x 1    1 0   1 [On expanding by first column we get] x n 1 x n 1  (x )  x 1  (x  ) 0 x 1 (C1 C1 – C3)   1 0  1  ( x  ) ( x  ) ( x  ) [On expanding by first column] Proved. Example 36. Show that x = – (a + b + c) is one root of the equation: x+a b c b x+c a = 0 and solve the equation completely.. c a x+b xabc b c Solution. By C1 C1 + C2 + C3, we get xabc xc a 0 xabc a xb Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  244 Determinants and Matrices 1 b c ( x  a  b  c) 1 x  c a 0  1 a xb 1 b c  ( x  a  b  c) 0 x  b  c ac  0, R2 R2  R1; R3 R3  R1 0 ab xbc On expanding by first column, we get (x + a + b + c) [(x – b + c) (x + b – c) – (a – b) (a – c)] = 0  (x + a + b + c) [x2 – (b – c)2 – (a2 – ac – ab + bc)] = 0  (x + a + b + c) (x2 – b2 – c2 + 2bc – a2 + ac + ab – bc] = 0  (x + a + b + c) (x2 – a2 – b2 – c2 + ab + bc + ca) = 0 Either x + a + b + c = 0  x = – (a + b + c) or x2 – a2 – b2 – c2 + ab + bc + ca = 0 2 2 2  x=  a  b  c  ab  bc  ca Hence, x = – (a + b + c) is one root of the given equation. Proved. (b + c)2 a2 a2 2 2 2 Example 37. Find the value of b (c + a) b c2 c2 (a + b)2 (b  c)2  a 2 a2  a2 a2 2 2 2 2 2 b b (c  a)  b b Solution. By C1 – C3, C2 – C3, we get c 2  (a  b) 2 c 2  (a  b) 2 (a  b)2 (a  b  c) (b  c  a) 0 a2 2  0 (a  b  c) (c  a  b) b (a  b  c) (c  a  b) (a  b  c) (c  a  b) (a  b)2 On taking out (a + b + c) as common from 1st and 2nd column, we get bca 0 a2 2 2  (a  b  c ) 0 cab b cab c  a  b (a  b ) 2 a  b  c 0 a2 2 2  (a  b  c) 0 a  b  c b R3 R3  (R1  R2 )  2b  2a 2ab Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/  Determinants and Matrices 245 a  b  c 0 a2 2 2   2 (a  b  c) 0 abc b b a  ab On expanding by first row, we get = – 2 (a + b + c)2 [(– a + b + c) {– ab (a – b + c) – ab2} + a2 {0 – b (a – b + c)}] = – 2 (a + b + c)2 [(– a + b + c) (– a2b – abc) – a2b (a – b + c)] = – 2ab (a + b + c)2 [(– a + b + c) (– a – c) – a (a – b + c)] = – 2ab (a + b + c)2 (a2 + ac – ab – bc – ac – c2 – a2 + ab – ac] = – 2ab (a + b + c)2 (– bc – ac – c2) = 2abc (a + b + c)2 (b + a + c) = 2abc (a + b + c)3. Ans. a+x a – x a – x