Chapter 11.1 Determinants PDF

30 0 60 300 Determinants 8.1.1 Definition. a2 x  b 2 y  0.....(ii) E3 (1) Consider two equations, a1 x  b1 y  0.....(i) and Multiplying (i) by b 2 and (ii) by b 1 and subtracting, dividing by x, we get, a1 b 2  a 2 b1  0 The result a1 b 2  a 2 b1 is represented by a1 a2 b1 b2 ID Which is know...

30 0 60 300 Determinants 8.1.1 Definition. a2 x  b 2 y  0.....(ii) E3 (1) Consider two equations, a1 x  b1 y  0.....(i) and Multiplying (i) by b 2 and (ii) by b 1 and subtracting, dividing by x, we get, a1 b 2  a 2 b1  0 The result a1 b 2  a 2 b1 is represented by a1 a2 b1 b2 ID Which is known as determinant of order two and a1 b 2  a 2 b1 is the expansion of this determinant. The horizontal lines are called rows and vertical lines are called columns. Now let us consider three homogeneous linear equations U a1 x  b1 y  c1 z  0 , a 2 x  b 2 y  c 2 z  0 and a 3 x  b 3 y  c 3 z  0 Eliminated x, y, z from above three equations we obtain D YG a1 (b 2 c 3  b 3 c 2 )  b1 (a 2 c 3  a 3 c 2 )  c1 (a 2 b 3  a 3 b 2 )  0 a1 The L.H.S. of (iii) is represented by a 2 a3 b1 b2 b3.....(iii) c1 c2 c3 Its contains three rows and three columns, it is called a determinant of third order. Note :  The number of elements in a second order is 2 2  4 and the number of elements in a third order determinant is 3 2  9. U (2) Rows and columns of a determinant : In a determinant horizontal lines counting from top 1 , 2nd, 3rd,….. respectively known as rows and denoted by R1 , R 2 , R 3 ,...... and vertical lines st ST counting left to right, 1st, 2nd, 3rd,….. respectively known as columns and denoted by C1 , C 2 , C 3 ,..... (3) Shape and constituents of a determinant : Shape of every determinant is square. If a determinant of n order then it contains n rows and n columns. i.e., Number of constituents in determinants = n2 (4) Sign system for expansion        ,    , are given by        of determinant : Sign system for order 2, order 3, order 4,…..       ,.....       8.1.2 Expansion of Determinants. Determinants 301 Unlike a matrix, determinant is not just a table of numerical data but (quite differently) a short hand way of writing algebraic expression, whose value can be computed when the values of terms or elements are known. (1) The 4 numbers a1 , b1 , a 2 , b 2 arranged as a1 a2 b1 b2 is a determinant of second order. These 60 numbers are called elements of the determinant. The value of the determinant is defined as a1 b1  a1 b 2  a 2 b 1. a2 b 2 E3 The expanded form of determinant has 2! terms. (2) The 9 numbers a r , b r , c r (r  1, 2, 3) arranged as a1 a2 a3 b1 b2 b3 c1 c2 c3 is a determinant of third ID order. Take any row (or column); the value of the determinant is the sum of products of the elements of the row (or column) and the corresponding determinant obtained by omitting the D YG U row and the column of the element with a proper sign, given by the rule (1)i j , where i and j are the number of rows and the number of columns respectively of the element of the row (or the a1 b 1 c 1 b c2 a c2 a b2 column) chosen. Thus a 2 b 2 c 2 = a1 2  b1 2  c1 2 b3 c3 a3 c 3 a3 b 3 a3 b 3 c 3 The diagonal through the left-hand top corner which contains the element a1 , b 2 , c 3 is called the leading diagonal or principal diagonal and the terms are called the leading terms. The expanded form of determinant has 3! terms. ST U Short cut method or Sarrus diagram method : To find the value of third order determinant, following method is also useful b1 c1 b2 c2 a1 b1 a1 a2 a3 a2 b2 = b3 b3 c3 a3 Taking product of R.H.S. diagonal elements positive and L.H.S. diagonal elements negative and adding them. We get the value of determinant as  a1 b 2 c 3  b1 c 2 a 3  c1 a 2 b 3  c1 b 2 a 3  a1 c 2 b 3  b1 a 2 c 3 Note : This method does not work for determinants of order greater than three. 8.1.3 Evaluation of Determinants. 302 Determinants If A is a square matrix of order 2, then its determinant can be easily found. But to evaluate determinants of square matrices of higher orders, we should always try to introduce zeros at maximum number of places in a particular row (column) by using the properties and then we should expand the determinant along that row (column). (1) R i to denote i th row. (2) R i  R j to denote the interchange of i th and j th rows. 60 We shall be using the following notations to evaluate a determinant : (3) R i R i  R j to denote the addition of  times the elements of j th row to the E3 corresponding elements of i th row. (4) R i ( ) to denote the multiplication of all element of i th row by . Similar notations are used to denote column operations if R is replaced by C. b1 b2 b3 a1 c1 c 2 and D'  b1 c1 c3 a2 b2 c2 a3 b 3. Then D'  D, D and D' are transpose of each other. c3  Since the determinant remains unchanged when rows and columns are D YG Note : value of determinant remains unchanged, if the rows and the columns are U P-1 : The interchanged. a1 If D  a 2 a3 ID 8.1.4 Properties of Determinants. U interchanged, it is obvious that any theorem which is true for ‘rows’ must also be true for ‘columns’. P-2 : If any two rows (or columns) of a determinant be interchanged, the determinant is unaltered in numerical value but is changed in sign only. a2 b 2 c 2 a1 b 1 c 1 D  a 2 b 2 c 2 and D'  a1 b1 c1. Then D'  D Let a3 b 3 c 3 a3 b 3 c 3 ST P-3 : If a determinant has two rows (or columns) identical, then its value is zero. a1 b 1 c 1 D  a1 b1 c1. Then, D = 0 Let a2 b 2 c 2 P-4 : If all the elements of any row (or column) be multiplied by the same number, then the value of determinant is multiplied by that number. ka1 kb1 kc1 a1 b 1 c 1 Let D  a 2 b 2 c 2 and D   a 2 b 2 c 2. Then D'  kD a3 b 3 c 3 a3 b 3 c3 P-5 : If each element of any row (or column) can be expressed as a sum of two terms, then the determinant can be expressed as the sum of the determinants. Determinants 303 a1  x a2 a3 e.g., b1  y c1  z a1 = a2 b2 c2 b3 c3 a3 b1 b2 b3 c1 x c 2 + a2 c3 a3 y b2 b3 z c2 c3  It should be noted that while applying P-6 at least one row (or column) must E3 Note : 60 P-6 : The value of a determinant is not altered by adding to the elements of any row (or column) the same multiples of the corresponding elements of any other row (or column) a1  ma 2 b1  mb 2 c1  mc 2 a1 b 1 c 1 e.g., D  a 2 b 2 c 2 and D' . Then D'  D a2 b2 c2 a 3  na1 b 3  nb1 c 3  nc1 a3 b 3 c 3 U D becomes zero on putting x   , then we say that (x   ) is factor 5 9 16 2 4. At x  2, D  0 (because C 1 and C 2 are identical at x  2 ) 8 D YG P-8 : If a determinant of determinant. x D  x2 e.g., if x3 ID remain unchanged. P-7 : If all elements below leading diagonal or above leading diagonal or except leading diagonal elements are zero then the value of the determinant equal to multiplied of all leading diagonal elements. a1 0 0 a1 b 1 c 1 a1 0 0 0 b 2 c 2  a 2 b 2 0  0 b 2 0  a1 b 2 c 3 e.g., a3 b 3 c 3 0 0 c3 0 0 c3 Hence (x  2) is a factor of D. Note : U  It should be noted that while applying operations on determinants then at least one row (or column) must remain unchanged.or, Maximum number of operations = order or determinant –1  It should be noted that if the row (or column) which is changed by multiplied a non zero number, then the determinant will be divided by that number. Solution: (a) n  2n 1 has the value [Pb. CET 1991; Rajasthan PET 2001; AIEEE 2003] (a) 0 (b)  Applying C1 C1  C 2  C3 , we get 1   n   2n   1   n   2n 1   n   2n Example: 2 1  2n n If n  3k and 1,  ,  2 are the cube roots of unity, then    2n ST Example: 1 n 1 n 1  2n n  2n 1 0 n  0 1 0  2n (c)  2  2n n  0 ( 1   n   2 n  0 if n is not multiple of 3) 1 1 x 1 1 1 1y 1  1 1 1z  1 1 1 (a) xyz 1     x y z  (d) 1 [Rajasthan PET 1992; Kerala (Engg.) 2002] (b) xyz (c) 1  1 1 1   x y z (d) 1 1 1   x y z 304 Determinants   xyz 1 y 1 z 1 x 1 1 x 1 y 1 y 1 z 1  1 1 1  xyz 1     x y z   1 1 1  xyz 1     x y z  Trick : Put x  1, y  1 1 1  xyz 1     x y z  1 z 1 0 0 1 (by C 2 C 2 1 0 y 1 0 1 z 1 0  1 1 1  xyz 1     0 1 x y z  2 1  2 and z  3 , then 1 3 1 1 1 1 y 1 z 1 1 1 z 1 1  2(11)  1(3)  1(1  3)  17. 4 C5 C7 12 C9 (a) 6 10 C4 C6 12 C8 Cm Cm 2 13 C m 4 is equal to zero, where m is (b) 4 10 C5 C7 12 C9 11 Cm Cm 2 13 C m 4 (c) 5 D YG Solution: (c) 11 12 ID 11 C4 C6 The value of   12 C8  11 11 12 1 z U Example: 3 10 11 1 (by R1 R1  R 2  R3 )  C1 and C3 C3  C1 ) 1 1 1  option (a) gives 1  2  3  1      17 1 2 3  10 1 1 y 1 y 60 Solution: (a) 1 x E3 1 (d) None of these =0 Applying C 2 C1  C 2 C 4  10 C 5 C 6  11 C 7 12 C 8  12 C 9 10 10 11 11 C4  C6 12 C8 11 Cm Cm 2  0    13 C m 4 12 10 C4 C6 12 C8 11 C5 C7 13 C9 12 (a) –2 Solution: (d) Cm Cm 2 = 0 13 C m 4 If a1 , a2 , a3 ,......., an ,...... are in G.P. then the value of the determinant ST Example: 4 11 12 [ C2 , C3 will be identical ] U Clearly m  5 satisfies the above result 11 log a n log a n  3 log a n  6 log a n  2 log a n  5 log a n  8 is [AIEEE 2004; IIT 1993] (b) 1 (c) 2 (d) 0  a1 , a2 , a3 ,......., an ,...... are in G.P.  a n21  a n.a n  2  2 log an1  log an  log an2 a n2 4  a n  3.a n  5  2 log an4  log an3  log an5 a n27  a n  6.a n  8  2 log an7  log an6  log an8 Putting these values in the second column of the given determinant, we get log a n 1  log a n  3 2 log a n  6 log a n 1 log a n  4 log a n 7 log a n  log a n  2 log a n  3  log a n  5 log a n  6  log a n  8 log a n  2 log a n  5 log a n  8 Determinants 305 [C2 is the sum of two elements, first identical with C1 and second with C3 ] 1 1 The value of (2 x  2  x )2 (3 x  3  x )2 (2 x  2  x )2 (3 x  3  x )2 (c) 30  x (b) 30 x (a) 0 Solution: (a) 1 (5 x  5  x )2 (5 x  5  x )2 Applying R 2 R 2  R 3 1   2.2 x.2.2  x (2 x  2  x ) 2 1 2.3 x.2.3  x (3 x  3  x ) 2 1 2.5 x.2.5  x (5 x  5  x ) 2 4 1 1 x (2  2  x ) 2 Trick : Putting x  0 , we get option (a) is correct 1 1  0 [ R1 and R2 are identical ] x x 2 (5  5 ) If x, y, z are integers in A.P. lying between 1 and 9 and x51, y41 and z31 are three digit numbers then 5 4 3 the value of x 51 y 41 z 31 x y z (a) x  y  z Solution: (c) 1 1 x (3  3  x ) 2 is (b) x  y  z  x 51  100 x  50  1 , z 31  100 z  30  1 (c) 0 (d) None of these U y 41  100 y  40  1 ID Example: 6 (d) None of these E3 Example: 5 1 (0 )  0 2 60  D YG 5 4 3    100 x  50  1 100 y  40  1 100 z  30  1 x y z Applying R 2 R 2  100 R 3  10 R1 5  1 x 4 1 y 3 1  x  2y  z z  x, y, z are in A.P. ,  x  2 y  z  0 ,    0 U Example: 7 0 If a  b  c , the value of x which satisfies the equation x  a x b ST (a) x  0 Solution: (a) x a 0 x c x b x  c  0 is 0 [EAMCET 1988; Karnataka CET 1991; MNR 1980; MP PET 1988, 99, 2001; DCE 2001] (c) x  b (b) x  a (d) x  c Expanding determinant, we get,   (x  a)[(x  b)(x  c)]  (x  b)[(x  a)(x  c)]  0  2 x 3  (2 ab)x  0  Either x  0 or x 2  ab. Since x  0 satisfies the given equation. 0 a b Trick : On putting x  0 , we observe that the determinant becomes  x 0  a 0  c  0 b c 0  x  0 is a root of the given equation. Example: 8 sin x The number of distinct real roots of cos x cos x (a) 0 (b) 2 cos x sin x cos x cos x   cos x  0 in the interval   x  is 4 4 sin x (c) 1 (d) 3 306 Determinants Solution: (c) 1 cos x (2 cos x  sin x ) 1 sin x 1 cos x cos x cos x  0 sin x Applying, R 2 R 2  R1 and R3 R3  R1 cos x sin x  cos x 0 cos x 0  0  (2 cos x  sin x )(sin x  cos x )2  0 sin x  cos x      tan x  2, 1 But tan x  2 in  , . Hence tan x  1  x  4 4 4   E3 (a) 16 (b) 18 (c) 17 (d) 19 Since it is an identity in  so satisfied by every value of . Now put   0 in the given equation, we t Example: 10 0 1 3 1 2  4  12  30  18 3 4 0 1 x (x  1) If f (x )  , then f (100 ) is equal to 2x x (x  1) (x  1)x 3 x (x  1) x (x  1)(x  2) (x  1)x (x  1) (b) 1 D YG (a) 0 Solution: (a) [IIT 1981] ID Solution: (b) have U Example: 9 2  3   1   3 2     4 , then value of t is If p4  q 3  r 2  s  t    1   3   4 3 60 1 (2 cos x  sin x ) 0 0 (c) 100 [IIT 1999, MP PET 2000] (d) –100 1 x (x  1) f (x )  2x x (x  1) (x  1)x 3 x (x  1) x (x  1)(x  2) (x  1)x (x  1) Applying C 3 C 3  C 2 , we get f (x )  1 x 1 2x x (x  1) 2x  0. Hence f (100 )  0 3 x (x  1) x (x  1)(x  2) 3 x (x  1) U 1 0 0 2 2 0 The value of 4 4 3 5 5 5 6 6 6 ST Example: 11 (a) 6! 0 0 0 4 6 0 0 0 0 5 is (b) 5! (c) 1. 2 2. 3. 4 3. 5 4. 6 4 (d) None of these Solution: (b) The elements in the leading diagonal are 1, 2, 3, 4, 5. On one side of the leading diagonal all the elements are zero.  The value of the determinant = The product of the elements in the leading diagonal = 1.2.3.4.5 = 5! Example: 12 The determinant a b a  b b c b   c  0 if a  b b   c 0 (a) a, b, c are in A.P. (b) a, b, c are in G.P. or ( x   ) is a factor of ax 2  2bx  c  0 (c) a, b, c are in H.P. [DCE 2000, 2001] Determinants 307 (d)  is a root of the equation Solution: (b) Applying R 3 R 3  R1  R 2 , we get a b b c 0 0 a  b b  c 0 2  a  b   b   c   (a 2  2b   c)(ac  b 2 )  0  a 2  2b   c  0 or b 2  ac 60  x   is a root of ax 2  2bx  c  0 or a, b, c are in G.P.  ( x   ) is a factor of ax 2  2bx  c  0 or a, b, c are in G.P. 8.1.5 Minors and Cofactors. where then determinant of minors ID M 12 M 22 M 32 determinant a13 a 23 , a 33 M 13 M 23 , M 33 M 11  minor of a11  a 22 a32 a a 23 , M 12  minor of a12  21 a31 a33 D YG M 11 M  M 21 M 31 the a12 a 22 a 32 U Consider a11   a 21 a 31 E3 (1) Minor of an element : If we take the element of the determinant and delete (remove) the row and column containing that element, the determinant left is called the minor of that element. It is denoted by M ij M 13  minor of a13  a 21 a31 a 23 a33 a 22 a32 Similarly, we can find the minors of other elements. Using this concept the value of determinant can be   a11 M 11  a12 M 12  a13 M 13 or,   a 21 M 21  a 22 M 22  a 23 M 23 or,   a 31 M 31  a 32 M 32  a 33 M 33. U (2) Cofactor of an element : The cofactor of an element aij (i.e. the element in the ith row and j th column) is defined as (1)i j times the minor of that element. It is denoted by C ij or A ij or Fij. ST Cij  (1)i j M ij a11   a 21 a 31 If a12 a 22 a 32 C 11 a13 a 23 , then determinant of cofactors is C  C 21 C 31 a 33 C 12 C 22 C 32 C 13 C 23 , where C 33 C11  (1)11 M 11   M 11 , C12  (1)1 2 M 12   M 12 and C13  (1)1 3 M 13   M 13 Similarly, we can find the cofactors of other elements. Note : The sum of products of the element of any row with their corresponding cofactor is equal to the value of determinant i.e.  a11 C11  a 21 C 21  a 31 C 31   a11 C 11  a12 C12  a13 C13 308 Determinants where the capital letters C 11 , C 12 , C 13 etc. denote the cofactors of a11 , a12 , a13 etc.  In general, it should be noted ai1 C j1  ai2 C j 2  ai3 C j3  0, if i  j or 1 2 The cofactor of the element 4 in the determinant 8 0 (a) 4 The cofactor of element 4, a1 If   a 2 a3 in the 2nd row (d) –4 and 3rd column is 1 3 1 (1) 2 3 8 0 1 0 2 1 = c1 c2 c3 and A1 , B1 , C1 denote the cofactors of a1 , b1 , c1 respectively, then the value of the determinant A1 A2 A3 B1 B2 B3 C1 C2 C3 D YG is U b1 b2 b3 (a)  Solution: (b) [MP PET 1987] (c) –10 {1(2)  3(8  0)  1.16 }  10 Example: 14 5 1 4 2 is 1 1 1 1 ID Solution: (b) (b) 10 3 3 0 2 60 Example: 13 If ' is the determinant formed by replacing the elements of a determinant , '  n 1 , where n  by their corresponding cofactors, then if   0 , then C  0 is the order of the determinant. E3  a1i C1 j  a 2i C 2 j  a 3 i C 3 j  0, if i  j (b) 2 a1 We know that .'  a 2 a3 b1 b2 b3 A1 c1 c 2. A2 A3 c3 B1 B2 B3 C1 C2 C3 (c) 3 a1 A1  0 0 (d) 0 0 a2 A2 0 0  0 0 0  0  0  3 a3 A3 0 0   '  2 U Trick : According to property of cofactors '  n 1  2 Example: 15 If the value of a third order determinant is 11, then the value of the square of the determinant formed by the cofactors will be (a) 11 n 1 ST Solution: (d) ( Hence n  3) '   3 1  (b) 121 (c) 1331 (d) 14641    (11)  121. But we have to find the value of the square of the determinant, so 2 2 required value is (121 ) 2  14641. 8.1.6 Product of two Determinants. Let the two determinants of third order be, a1 b1 c1 D1  a2 a3 b2 b3 c2 c3 1 and D2   2 3 1  1  2  2. Let D be their product. 3  3 Determinants 309 (1) Method of multiplying (Row by row) : Take the first row of D1 and the first row of D2 i.e. and  1 ,  1 ,  1 a1 , b1 , c1 multiplying the corresponding elements and add. The result is a1 1  b1  1  c1 1 is the first element of first row of D. Now similar product first row of D1 and second row of D2 gives a1 2  b1  2  c1 2 is the 60 second element of first row of D, and the product of first row D1 and third row of D2 gives a1 3  b1  3  c1 3 is the third element of first row of D. The second row and third row of D is obtained by multiplying second row and third row of D1 with 1st , 2nd , 3rd row of D2 , in the E3 above manner. Hence, Note : b1 b2 b3  a1 1  b1  1  c1 1 a1 2  b1  2  c1 2 1  1  2  2  a 2 1  b 2  1  c 2 1 a 2 2  b 2  2  c 2 2 a 3 1  b 3  1  c 3  1 a 3 2  b 3  2  c 3  2 3  3 1 c1 c2   2 c3 3 We can also multiply rows by columns or columns by rows or columns by U columns. For all values of A, B, C and P, Q, R the value of D YG Example: 16 (a) 0 Solution: (a) a1 3  b1  3  c1 3 a 2 3  b 2  3  c 2 3 a 3 3  b 3  3  c 3  3 ID a1 D  a2 a3 (b) cos A cos B cos C The determinant cos( A  P ) cos( A  Q) cos( A  R) cos( B  P ) cos( B  Q) cos( B  R) is cos(C  P ) cos(C  Q) cos(C  R) (c) sin A sin B sin C can be (d) cos P cos Q cos R expanded as cos A cos P  sin A sin P cos A cos Q  sin A sin Q cos A cos R  sin A sin R cos A sin A 0 cos P sin P 0 cos B cos P  sin B sin P cos B cos Q  sin B sin Q cos B cos R  sin B sin R  cos B sin B 0  cos Q sin Q 0  0 cos C cos P  sin C sin P cos C cos Q  sin C sin Q cos C cos R  sin C sin R cos C sin C 0 cos R sin R 0 log 3 512 log 3 8 log 4 3 log 2 3 log 8 3   log 4 9 log 3 4 log 3 4 U Example: 17 (a) 7 log 3 512 log 3 8 (c) 13 (d) 17 log 4 3 log 2 3 log 8 3  log 512 log 9 log 3 log 8   log 3 log 4 log 3 log 4             log 4 9 log 3 4 log 3 4 log 4 log 4 log 3   log 2 log 3 log 8 log 3   log 3 ST Solution: (b) (b) 10 [Tamilnadu (Engg.) 2002]  log 2 9 log 3 2 log 2 3     2 log 2 2  log 3 log 2   log 2 2 log 2 2    log 2  log 2 3    92 3   2      2    10   2 2  3  8.1.7 Summation of Determinants. f (r) a l Let  r  g(r) b m , where a, b, c, l, m and n are constants, independent of r. h(r) c n 310 Determinants n n Then,  r  r 1  f (r) a  g(r) b m.  h(r) c l r 1 n Here function of r can be the elements of only one row or r 1 n n 60 r 1 one column. 15 8 35 9 , then D1  D2  D3  D4  D5  25 10 (a) 0 Solution: (d) (b) 25 (c) 625 1 15 8 2 15 8 3  D1  1 35 9 , D2  4 35 9 , D 3  9 27 1 25 10 8 25 10  D1  D2  D3  D4  D5  E3 p If D p  p 2 p3 15 55 225 75 175 125 4 15 8 35 9 , D 4  16 64 25 10 (d) None of these 5 15 8 35 9 , D5  25 125 25 10 15 8 35 9 25 10 ID Example: 18 40 45 50 N Example: 19 The value of  n U n , if U n  n 2 n3 (a) 0 N  Solution: (a) n 1 (b) 1 N ( N  1) 2 N ( N  1)(2 N  1) Un  6 2  N ( N  1)    2   1 3N 2 U ST N ( N  1) 12 (c) –1 5 2N  1 2N  1  Applying C 3 C 3  C 2  identical] 1 5 2 N  1 2 N  1 is 3N 2 3N D YG n 1 U  15 (3125 )  75 (7375 )  40 (32500 )  46875  553125  1300000  700000 N ( N  1) 12 (d) None of these 6 1 5 4N  2 2N  1 2N  1 3 N ( N  1) 3 N 2 3N 3N 6 1 6 4N  2 2N  1 4N  2 =0 3 N ( N  1) 3 N 2 3 N ( N  1) [ C1 and C3 are 8.1.8 Differentiation and Integration of Determinants. (1) Differentiation of a determinant : (i) Let (x ) be a determinant of order two. If we write (x ) | C1 C2 | , where C1 and C 2 denote the 1st and 2nd columns, then ' (x )  C'1 C2  C1 C2 where C' i denotes the column which contains the derivative of all the functions in the i th column C i. In a similar fashion, if we write (x )  R1 R' R , then   (x )  1  1 R2 R2 R 2 (ii) Let (x ) be a determinant of order three. If we write (x )  C1 C2 C3 , then Determinants 311 ' (x )  C'1 C3  C1 C2 C3  C1 C '2 C2 C '3 R '1 R1 R1 R1 and similarly if we consider ( x )  R 2 , then ' (x )  R2  R '2  R2 R3 R3 R3 R '3 60 (or column) consists functions of x and other rows (or columns) are f3 (x ) b3 , c3 E3 (iii) If only one row constant, viz. f1 (x ) f2 (x ) Let (x )  b1 b2 c1 c2 n n n f1 (x ) f2 (x ) f3 (x ) f '1 (x ) f '2 (x ) f '3 (x ) n b2 b3 then ' (x )  b1 and in general  (x )  b1 b2 b3 c1 c2 c3 c1 c2 c3 x If 1  a a b x a b b x and  2   Example: 21 b x a d ( 1 )  3(x 2  ab)  3  2 dx y If y  sin mx , then the value of the determinant y 3 y6 (a) m 9 U (b) m 2 y y3 y6 y1 y4 y7 sin mx y2 y 5   m 3 cos mx y8  m 6 sin mx ST Solution: (d) (d) 1  332 / 2 b x b d  x 2  ab ( 1 )  3(x 2  ab) and  2  b  x 3  3 abx  a x dx x D YG x 1  a a [MNR 1986; Kurukshetra CEE 1998; UPSEAT 2000] d ( 1 )  2( 2 ) 2 (c) dx d ( 1 )  3  2 (b) dx (a)  1  3( 2 ) 2 Solution: (b) x b are the given determinants, then a x U Example: 20 ID where n is any positive integer and f n (x ) denotes the n th derivative of f (x ). m cos mx m 4 sin mx  m 7 cos mx y1 y4 y7 (c) m 3 y2 d ny y 5 , where y n  is dx n y8 (d) None of these  m sin mx m 5 cos mx m 8 sin mx 2 Taking  m 6 common from R 3 , R1 and R 3 becomes identical. Hence the value of determinant is zero. f (x ) g(x ) h(x ) b c , where a, b, c, l, m and n are (2) Integration of a determinant : Let (x )  a l m n constants.   b a (x )dx  b b b a a a  f (x )dx  g(x )dx  h(x )dx a l b m c n 312 Determinants Note : If the elements of more than one column or rows are functions of x then the integration can be done only after evaluation/expansion of the determinant. cos x 1  cos x cos x 1  sin x  cos x , then sin x 1 (a) 1/4 Solution: (d)  /2  (x ) dx is equal to 0 (b) 1/2 (c) 0 (d) –1/2 60 1 If (x )  1  sin x sin x Applying C 3 C 3  C 2  C1 1 (x )  1  sin x sin x   /2  cos x cos x sin x (x )dx   0 0 0  cos x  cos x (1  sin x )   sin x cos x 1  /2 1  cos 2 x  1  /2 sin 2 xdx    2 2  0 2 0   E3 Example: 22 1 1 (cos   cos 0)   4 2 ID 8.1.9 Application of Determinants in solving a system of Linear Equations. Consider a system of simultaneous linear equations is given by a1 x  b 1 y  c 1 z  d 1   a2 x  b 2 y  c 2 z  d 2  a 3 x  b 3 y  c 3 z  d 3  D YG U.....(i) A set of values of the variables x, y, z which simultaneously satisfy these three equations is called a solution. A system of linear equations may have a unique solution or many solutions, or no solution at all, if it has a solution (whether unique or not) the system is said to be consistent. If it has no solution, it is called an inconsistent system. If d 1  d 2  d 3  0 in (i) then the system of equations is said to be a homogeneous system. Otherwise it is called a non-homogeneous system of equations. Theorem 1 : (Cramer’s rule) The solution of the system of simultaneous linear equations a2 x  b 2 y  c 2 a1 x  b1 y  c1.....(i) and.....(ii) a D D1 , y  2 , where D  1 a2 D D U is given by x  c b1 , D1  1 c2 b2 b1 b2 and D2  a1 a2 c1 , c2 provided that D  0  a b1  is the determinant of the coefficient matrix  1 .  a2 b 2  The determinant D1 is obtained by replacing first column in D by the column of the Here D  ST Note :  a1 a2 b1 b2 right hand side of the given equations. The determinant D 2 is obtained by replacing the second column in D by the right most column in the given system of equations. (1) Solution of system of linear equations in three variables by Cramer’s rule : Theorem 2 : (Cramer’s Rule) The solution of the system of linear equations a1 x  b1 y  c1 z  d 1.....(i) a2 x  b 2 y  c 2 z  d 2.....(ii) a3 x  b 3 y  c 3 z  d 3.....(iii) Determinants 313 d1 D1  d 2 d3 b1 b2 b3 Note : a1 D3 D2 and z  , where D  a 2 y D D a3 a1 c1 c 2 , D2  a 2 a3 c3 Here D obtained d1 d2 d3 a1 c1 c 2 , and D3  a 2 a3 c3 b1 b2 b3 b1 b2 b3 c1 c2 , c3 d1 d 2 , Provided that D  0 d3 60 D is given by x  1 , D is the determinant of the coefficient matrix. The determinant D1 is by replacing the elements in first column of D by d 1 , d 2 , d 3. D 2 is E3 obtained by replacing the element in the second column of D by d 1 , d 2 , d 3 and to obtain D 3 , replace elements in the third column of D by d 1 , d 2 , d 3. Theorem 3 : (Cramer’s Rule) Let there be a system of n simultaneous linear equation n unknown given by ID a11 x 1  a12 x 2 ....  a1n x n  b1 a 21 x 1  a 22 x 2 ....  a 2 n x n  b 2 the j th a12 a 22  an 2  a1n  a 2n and let D j , be the determinant obtained from D after replacing D YG a11 a Let D  21  a n1 U    a n1 x 1  a n 2 x 2 ....  a nn x n  b n  anm b1 b D D D column by 2. Then, x 1  1 , x 2  2 ,....., x n  n , Provided that D  0  D D D bn U (2) Conditions for consistency Case 1 : For a system of 2 simultaneous linear equations with 2 unknowns ST (i) If D  0 , then the given system of equations is consistent and has a unique solution D D given by x  1 , y  2. D D (ii) If D  0 and D1  D2  0 , then the system is consistent and has infinitely many solutions. (iii) If D  0 and one of D1 and D 2 is non-zero, then the system is inconsistent. Case 2 : For a system of 3 simultaneous linear equations in three unknowns (i) If D  0 , then the given system of equations is consistent and has a unique solution D D D given by x  1 , y  2 and z  3 D D D (ii) If D  0 and D1  D 2  D3  0 , then the given system of equations is consistent with infinitely many solutions. 314 Determinants (iii) If D  0 and at least one of the determinants D1 , D2 , D3 is non-zero, then given of equations is inconsistent. (3) Algorithm for solving a system of simultaneous linear equations by Cramer’s rule (Determinant method) Step 1 : Obtain D, D1 , D 2 and D 3 by x  60 Step 2 : Find the value of D. If D  0 , then the system of the equations is consistent has a unique solution. To find the solution, obtain the values of D1 , D 2 and D 3. The solutions is given D D D1 , y  2 and z  3. If D  0 go to step 3. D D D E3 Step 3 : Find the values of D1 , D 2 , D 3. If at least one of these determinants is non-zero, then the system is inconsistent. If D1  D 2  D3  0, then go to step 4 Note:  The system ID Step 4 : Take any two equations out of three given equations and shift one of the variables, say z on the right hand side to obtain two equations in x, y. Solve these two equations by Cramer’s rule to obtain x, y, in terms of z. of following homogeneous equations a1 x  b 1 y  c 1 z  0 , a 2 x  b 2 y  c 2 z  0 , a 3 x  b 3 y  c 3 z  0 is always consistent. c1 c 2  0 , then this system has the unique solution x  y  z  0 known c3 U b1 b2 b3 D YG a1 If   a 2 a3 as trivial solution. But if   0 , then this system has an infinite number of solutions. Hence for non-trivial solution   0. Example: 23 If the system of linear equations x  2ay  az  0, x  3 by  bz  0, x  4 cy  cz  0 has a non-zero solution, then a, b, c (a) Are in A.P. [AIEEE 2003] (c) Are in H.P. (d) Satisfy a  2b  3c  0 1 2a a System of linear equations has a non-zero solution, then 1 3 b b  0 1 4c c U Solution: (c) (b) Are in G.P. ST 1 0 a Applying C 2 C 2  2C 3 ; 1 b b  0 1 2c c Applying R3 R3  R2 and R2 R2  R1 1 0 0 Example: 24 0 a 2 1 1   ;  a, b, c are in H.P. b b  a  0  b(c  b)  (b  a)(2c  b)  0. On simplification b a c 2c  b c  b If the system of equations x  ay  0, az  y  0 and ax  z  0 has infinite solutions, then the value of a is [IIT Screening 2003] (a) –1 (b) 1 (c) 0 (d) No real values Determinants 315 Example: 25 1 a 0 0 1 a  0  1  a(a) 2  0  a 3  1  a  1 a 0 1 If the system of equations ax  y  z  0, x  by  z  0 and x  y  cz  0 , where a, b, c  1 has a nontrivial solution, then the value of (a) –1 Solution: (c) 1 1 1 is   1a 1b 1c (b) 0 (c) 1  a(b  1)(c  1)  1(1  a)(c  1)  1(1  a)(b  1)  0 1 1 a 1 1 1 1 1 1 1       0  1 0  1a 1a 1b 1c 1a 1b 1c 1b 1c ID  If the system of equations x  2y  3 z  1, (k  3)z  3, (2k  1)x  z  0 is inconsistent, then the value of k (a) –3 (c) 0 (d) 2 D YG 1 1 2 3 2 3 0 0 k  3  0  k  3 and D1  3 0 0  0 , Hence system is inconsistent for k  3. 2k  1 0 0 0 1 1 The equations x  y  z  6, x  2y  3 z  10, (a) m  3, n  R x  2y  mz  n give infinite number of values of the triplet (b) m  3, n  10 [] (c) m  3, n  10 (d) None of these Each of the first three options contains m  3. When m  3 , the last two equations become x  2 y  3 z  10 and x  2 y  3 z  n. U Solution: (c) 1 2 [Roorkee 2000] For the equations to be inconsistent D  0  D (x, y, z) if (b) U is Example: 27. E3 a 1 1 1  a b 1 0 0 1a 0 c 1 Solution: (a) (d) None of these a 1 1 As the system of the equations has a non-trivial solution 1 b 1  0 1 1 c Applying R 2 R 2  R1 and R3 R3  R1 Example: 26 60 Solution: (a) Obviously, when n  10 these equations become the same. So we are left with only two independent equations to find the values of the three unknowns. Consequently, there will be infinite solutions. The value of  for which the system of equations 2 x  y  z  12 , x  2y  z  4, x  y  z  4 has no ST Example: 28 solution is (a) 3 Solution: (d) [IIT Screening 2004] (b) –3 (c) 2 (d) –2 2 1 1 D  1  2 1  3   6. For no solution the necessary condition is 3  6  0    2. It can be see 1 1  that for   2 , there is no solution for the given system of equations. 8.1.10 Application of Determinants in Co-ordinate Geometry. (1) Area of triangle whose vertices are (x r , y r ); r  1, 2, 3 is 316 Determinants x1 1 1   [x 1 (y 2  y 3 )  x 2 (y 3  y 1 )  x 3 (y 1  y 2 )]  x 2 2 2 x3 y1 y2 y3 1 1 1 (2) If ar x  b r y  c r  0, (r  1, 2, 3) are the sides of a triangle, then the area of the triangle is 1 2 C1 C 2 C 3 b1 b2 b3 c1 c2 c3 2 , where C1  a 2 b 3  a 3 b 2 , C 2  a 3 b1  a1 b 3 , C 3  a1 b 2  a 2 b1 a1 b1 c1 a2 a3 b2 b3 c2. c3 are the E3  a1 a2 a3 60 given by cofactors of the elements c1 , c 2 , c 3 respectively in the determinant (3) The equation of a straight line passing through two points ( x 1 , y 1 ) and ( x 2 , y 2 ) is y y1 y2 1 1 0 1 ID x x1 x2 D YG U a1 (4) If three lines ar x  b r y  c r  0 ; (r  1, 2, 3) are concurrent if a 2 a3 b1 b2 b3 c1 c2  0 c3 (5) If ax 2  2hxy  by 2  2 gx  2 fy  c  0 represents a pair of straight lines then a h g abc  2 fgh  af  bg  ch  0  h b f g f c 2 2 2 ST U (6) The equation of circle through three non-collinear points A(x 1 , y 1 ), B(x 2 , y 2 ), C(x 3 , y 3 ) is Example: 29 x2 x 12 x 22 x 32  y2  y 12  y 22  y 32 x x1 x2 x3 y y1 y2 y3 1 1 0 1 1 The three lines ax  by  c  0, bx  cy  a  0, cx  ay  b  0 are concurrent only when (a) a  b  c  0 (b) a 2  b 2  c 2  ab  bc  ca (c) a 3  b 3  c 3  ab  bc  ca (d) None of these a b c Solution: (a, b)Three lines are concurrent if b c a  0 or, 3 abc  a 3  b 3  c 3  0  a 3  b 3  c 3  3 abc c a b Also, a 3  b 3  c 3  3 abc  0  (a  b  c)(a 2  b 2  c 2  ab  bc  ca)  0  (a  b  c)  0 or a 2  b 2  c 2  ab  bc  ca. 8.1.11 Some Special Determinants. Determinants 317 (1) Symmetric determinant : a h element aij  a ji  i, j e.g., h b g f A determinant is called symmetric determinant if for its every g f c 60 (2) Skew-symmetric determinant : A determinant is called skew symmetric determinant if for 0 3 1 its every element aij   a ji  i, j e.g.,  3 0 5 1 5 0 E3 Note :  Every diagonal element of a skew symmetric determinant is always zero.  The value of a skew symmetric determinant of even order is always a perfect square and that of odd order is always zero. ID 0 c b For example (i)  c 0 a  c(0  ab)  b(ac  0)  abc  abc  0 b a 0 U 0 a  0  a 2  a 2 (Perfect square) (ii) a 0 0 ab e  f 0 lm  0 (iii) b  a f e m l 0 D YG (3) Cyclic order : If elements of the rows (or columns) are in cyclic order. i.e. (i) a b c 1 2 2 2 a b c  a2 bc ca ab a3 ST (iii) (iv) Note : 1 b2 b3 a bc abc a a2 b ca abc  b b 2 c ab abc c c2 U (ii) 1 a a2 1 b b 2  (a  b )(b  c)(c  a) 1 c c2 1 a a3 1 b b3 1 c 2  (a  b)(b  c)(c  a)(ab  bc  ca) c3 a3 b 3  abc(a  b)(b  c)(c  a) c3 1 c  (a  b)(b  c)(c  a)(a  b  c) c3 (v) a b c b c a  (a 3  b 3  c 3  3 abc) c a b  These results direct applicable in lengthy questions (As behavior of standard results) Example: 30 Solution: (d) 0 i  100   100  i 0 500  i i  1000 i  500 1000  i is equal to 0 (a) 100 (b) 500 (c) 1000 This determinant of skew symmetric of odd order, hence is equal to 0. (d) 0 318 Determinants [Pb. CET 1997; DCE 2002] (b) (a  b)(b  c)(c  a) (a) a 2  b 2  c 2 Solution: (c) (c) (a  b)(b  c)(c  a) 1 a a2   1 b b2 1 c c2 Applying R1 R1  R 2 Example: 32 If 1 a a3 1 b b3 0 0 1  (a  b)(b  c)(a  c) 0 1 b  c  (a  b)(b  c)(a  c)(1)  (a  b)(b  c)(c  a) 1 c c2 ID 0 0 (a  c)  (a  b)(b  c) 0 1 b  c 1 c c2 E3 Applying R1 R1  R 2 & R 2 R 2  R3 0 a  b a2  b 2 0 1 ab  0 b  c b 2  c 2  (a  b)(b  c) 0 1 b  c ; 1 c c2 1 c c2 (d) None of these 60 Example: 31 1 a a2 1 b b2  1 c c2 1 c  (a  b)(b  c)(c  a)(a  b  c) Where a, b, c are all different, then the determinant c3 Solution: (b) (b) x  1 (a  b  c) 3 D YG (a) a  b  c  0 U 1 1 1 (x  a) 2 (x  b) 2 ( x  c) 2 vanishes when (x  b )(x  c) (x  c)(x  a) (x  a)(x  b) 1  a a3 1 b b3 (c) x  1 (a  b  c) 2 1 c  (a  b)(b  c)(c  a)(a  b  c) c3 (d) x  a  b  c …… (i) 1 1 1 2 2 (x  a) (x  b) ( x  c) 2 0 Now, (x  b )(x  c) (x  c)(x  a) (x  a)(x  b) U 1  (x  a)(x  b )(x  c) (x  a) (x  b) ( x  c) 3 3 (x  a) (x  b) ( x  c) 3 0 (x  a)(x  b)(x  c) (x  a)(x  b)(x  c) (x  a)(x  b )(x  c) ST Applying C1 C1 (x  a), C 2 C 2 (x  b), C 3 C 3 (x  c) (x  a) (x  a)3 1 (x  b) (x  b)3 1 ( x  c) (x  c)3  0 1 [(x  a)  (x  b)][(x  b)  (x  c)][(x  c)  (x  a)](x  a  x  b  x  c)  0 (b  a)(c  b)(a  c)[3 x  (a  b  c)]  0 or x  Example: 33 1 (a  b  c) 3 [ a  b  c]  If  ,  and  are the roots of the equations x 3  px  q  0 then value of the determinant   [AMU 1990] (a) p Solution: (d) (c) p 2  2q (b) q Since ,  ,  are the roots of x  px  q  0 ,        0 3 (d) 0       is Determinants 319        Applying R1 R1  R 2  R 3 , We get, x 1 x2  2 x (x  1) If   x (x  1) x  1 x (x 2  2)  p 0 x 6  p 1 x 5  p 2 x 4  p 3 x 3  p 4 x 2  p 5 x  p 6 ,then (p 5 , p 6 )  2 x  2 x (x  1) x 1 (a) (–3, –9) Solution: (b) 0 0 0 =    0    (b) (–5, –9) Putting x  0 in both sides, we get, (c) (–3, –5) (d) (3, –9) E3 Example: 34                   60    1 2 0 0 1 0  p 6  p 6  9 by expansion. 2 0 1 p 5 is the coefficient of x or constant term in the differentiation of determinant. ID Differentiate both sides, 1 2x 2x  1 x 1 x 2  2 x (x  1) x 1 x 2  2 x (x  1) 2 2 x (x  1) x  1 x (x  2)  2 x  1 1 3 x  2  x (x  1) x  1 x (x 2  2) x 2  2 x (x  1) x 1 x 2  2 x (x  1) x  1 2x 2x  1 1 U  6 p 0 x 5  5 p1 x 4  4 p 2 x 3  3 p 3 x 2  2 p 4 x  p 5 D YG Putting x  0 both sides, we get p 5  5 ;  (p 5 , p 6 )  (5, 9). ST U ***