Matrices and Determinants
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- Ευτέρπη Κυπραίος
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1 Matrices and Determinants SUBJECTIVE PROBLEMS: Q 1. For what value of k do the following system of equations possess a non-trivial (i.e., not all zero) solution over the set of rationals Q? x + ky + 3z = 0 3x + ky z = 0 x + 3y 4z = 0 For that value of k, find all the solutions for the system. (IIT JEE Marks) Q. Let a, b, c be positive and not all equal. Show that the value of determinant a b c b c a c a b is negative. (IIT JEE Marks) Q 3. Without expanding a determinant at any stage, show that A and B are determinants of order 3 not involving x. 1 x x x x x 3x 1 3x 3x x x x x = xa +B where (IIT JEE Marks) Q 4. Show that the system of equations 3x y + 4z = 3 x + y 3z = - 6x + 5y + λ z = -3 Has at least one solution for any real number λ -5. Find the set of solutions if λ = -5 (IIT JEE Marks)
2 Q 5. Show that C C C C C C x x x x x1 x r r 1 r r r 1 r y y y y y1 y Cr Cr 1 Cr Cr Cr 1 Cr 1 Cr Cr 1 Cr Cr Cr 1 Cr (IIT JEE 1985 Marks) Q 6. Consider the system of linear equations in x, y, z: (Sin 3θ) x y + z = 0 (Cos θ x + y + z = x + 7y + 7z = 0 Find the values of θ for which this system has nontrivial solutions. (IIT JEE Marks) Q 7. Let a = a 1 n 6 ( a 1) n 4n 3 ( 1) a n n n Show that = c, a constant (IIT JEE Marks) Q 8. Let the three digit numbers A 8, 3 B9, and 6 C, where A, B, and C are integers between 0 and 9, be A 3 6 divisible by a fixed. Integer k. Show that the determinant 8 9 C is divisible by k. B (IIT JEE Marks) Q 9. If a p, b q, c r and p b c a q c a b r = 0. (IIT JEE Marks) Then find the value of p/p a + q/q b + r/r c
3 Q 10. For a fixed positive integer n, if n! ( n 1)! ( n )! D ( n 1)! ( n )! ( n 3)! ( n )! ( n 3)! ( n 4)! Then show that [D/ (n!) 3 4] is divisible by n. (IIT JEE Marks) Q 11. Let λ and α be real. Find the set of all values of λ for which the system of linear equations λ x + sin α y + cos α z =, x + cos α y + sin α z =, -x + sin α y cos α z = has a non trivial solution : for λ =, find all values of α. (IIT JEE Marks) Q 1. For all values of A, B, C and P, Q, R show that (IIT JEE Marks) cos( A P) cos( A Q) cos(a R) cos( B P) cos( B Q) cos(b R) 0 cos( C P) cos( C Q) cos(c R) Q 13. Let a > 0, d > 0. Find the value of the determinant (IIT JEE Marks) a a( a d ) ( a d )( a d ) ( a d) ( a d)( a d) ( a d)(a 3d) ( a d) (a d)(a 3d) ( a 3 d)( a 4 d) Q 14. bc ca ab Find the value of the determinant p q r (IIT JEE 1997C Marks) Where a, b and c are respectively the p th, q th and r th of a harmonic progression.
4 Q 15. Prove that for all values of θ, (IIT JEE Marks) sin cos sin 4 sin cos sin = sin cos sin Q 16. If matrix A = a 3 + b 3 + c 3. Q 17. a b c b c a c a b where a, b, c are real positive numbers, abc = 1 and A T A = I, then the value of (IIT JEE Marks) If M is 3 x 3 matrix, where det M = 1 and MM T = I, here I is a idetit atri, proe that det (M 1) = 0. (IIT JEE 004 Marks) Q 18. If a 1 0 a 1 1 f a x A 1 b d, B 0 d c, U g, V 0, X y 1 b c f g h h 0 z And AX = U has infinitely many solutions, prove that BX = V has no unique solution. Also show that if afd 0, then BX = V has no solution. (IIT JEE Marks)
5 Matrices and Determinants - Solutions Sol 1. We should have, 1 k 3 3 k k + 6) k (-1 + 4) + 3 (9 k) = 0 -k + 33 = 0 k = / Substituting k = / and putting x = b, where Q, we get the system as 33y + 6z = -b (1) 33y 4z = -6b () 3y 4z = -b (3) (1) () z = b z = / b (1) y = -b 1b/5 = -b/ y = -b/15 The solution is x = b, y = -b/15, z = b/5 Sol. The given determinant, on expanding along R 1, we get a b c b c a c a b = a (bc a ) b (b ac) + c (ab c ) = 3abc a 3 b 3 c 3 = - (a 3 + b 3 + c 3 3abc) = - (a + b + c) [a + b + c ab bc ca] = -1/ (a + b + c) [a + b + c ab bc ca] = -1/ (a + b + c) [(a b) + (b c) + (c a) ] As a, b, c > 0 a + b + c > 0 Also a b c (a b) + (b c) + (c a) > 0 Hence the given determinant is ve.
6 Sol 3. 1 x x x x x 3x 1 3x 3x x x x x = xa + B L. H. S. = 1 x x x x x 3x 1 3x 3x x x x x Operation R R R 1 and R 3 R 3 R 1 1 x x x x x 1 x x 1 x 3 x x x x x x x x 0 x x 1 x1 x x1 0 x x 1 x 3 x x 1 x x 1 x 0 x 1 x x 1 x 3 x x 1 Operating R 3 R 3 R and R R R 1 x x 1 x x x x x = xa + B = R. H. S. Hence Proved. Sol 4. The given system of equations is 3x y + 4z = 3 x + y 3z = -
7 6x + 5y + λz = -3 Then D = = 3 (λ + + λ + + 1) = 7 λ + = λ + For unique sol., D λ + or λ -5 System has unique SolFor λ - 5 For λ = - 5, D = 0 Then D = 3 ( ) + 1 (10 9) + 4 ( ) = = D = 3 (10 9) 3 ( ) + 4 (-3 + 1) = = 0 D = 3 ( ) + 1 (-3 + 1) + 3 (5 1) = = 0 D 1 = D = D 3 = 0 Infinite many solutions. Let z = k then equations becomes 3x y = 3 4k x + y = 3k
8 on solving we get x = 4 5k/7, y = 13k 9/7, z = k This satisfies the third equation. Sol 5. On L. H. S. = D, applying operations C C + C 1 and C 3 C 3 + C and using n C r + n C r + 1 = n + 1 C r + 1, we get C C C x x1 x1 r r 1 r y y1 y1 r r 1 r 1 1 Cr Cr 1 Cr D C C C Operating C 3 + C and using the same result, we get C C C x x1 x r r 1 r y y1 y r r 1 r 1 Cr Cr 1 Cr D C C C = R H S Hence Proved Sol 6. The system will have a non trivial solution if sin cos Expanding along C 1, we get 1) sin 3θ (-7 cos θ ) = 0 sin θ + cos θ 14 = 0 sin θ + cos θ = 0 sin 4 sin 3 θ + sin θ = 0 4 sin 3 θ + sin θ sin θ = sin θ sin θ sin θ + = Sin θ = or sin θ = / sin θ = -3/ not possible) θ = n π or θ = n π + -1) n π/, n Z.
9 Sol 7. We have a = a1 n 6 ( a 1) n 4n a n n n 3 ( 1) NOTE THIS STEP Then n a 1 a (1 1) n 6 ( 1) n 6 (1 1) n 4n ( 1) n 4n (1 1) ( 1) n n n n n n (n1) n 6 (n 1) n 4n 3 3 (n 1) n n n = (n1) n (n 1) n 4n (n1) n n1 n n ( 1)(n 1 n 4 n 6 nn ( 1) n 6 n n n n n n n ( n1) ( n 1) n ( n 1) 1 3 n( n 1) 3n 3 n( n 1) (Taking n(n 1)/1 common from C 1 and n from C ) = 0 (as C 1 and C 3 are identical) Thus, n a = a 1 a 1 n a = c (a constant) Where c = 0
10 Sol 8. Given that A, B, C are integers between 0 and 9 and the three digit numbers A 8, 3 B9 and 6C are divisible by a fixed integer k. Now, D = A C B On operating R R + 10 R R 1, we get A 3 6 A 3 6 A8 3B9 6C kn kn kn 1 3 B B [As per question A8, 3 B9 and 6 C are divisible by k, therefore, A 8 = kn 1 3 B9 = kn 6 C = kn 3] A 3 6 k n n n = k x some integral value. 1 3 B D is divisible by k. Sol 9. p b c Consider a q c 0 a b r Operating R 1 R 1 R and R R R 3 we get p a ( q b) c 0 q b c r 0 a b r Taking (p q), (q b) and (r c) common from (C 1, C and C 3 resp. we get
11 p a) (q b) (r c) a b r p a q b r c Expanding along R 1 p a) (q b) (r c) [1 (r/r c + b/q b) + a/p a]=0 As given that p a, q b, r c r/r c + b/q b + a/p a = 0 r/r c + q (q b)/q b + p (p a)/p a = 0 r/r c + q/ q b 1 + p/p a 1 = 0 p/p a + q/q b + r/r c = Sol 10. n! ( n 1)! ( n )! D ( n 1)! ( n )! ( n 3)! ( n )! ( n 3)! ( n 4)! = n! (n + 1)! (n + )! 1 n 1 ( n )( n 1) 1 n ( n 3)( n ) 1 n 3 ( n 4)( n 3) Operating R R R 1 and R 3 R 3 R, we get D = (n!) 3 (n + 1) (n + ) 1 n 1 n 3n 0 1 n n 6 Operating R 3 R 3 - R D = (n!) 3 (n + 1) (n + ) 1 n 1 n 3n 0 1 n = (n!) 3 (n + a) (n + ) 1 [] D/n! 3 = (n + 1) (n + )
12 D/n! 3 4 = (n 3 + 4n + 5n + ) 4 = (n 3 + 4n + 5n) = n (n + 4n + 5) D/n! 3 4 is divisible by n. Sol 11. Given that λ, α R and system of linear equations λ x + sin α y + cos α z = x + cos α y + sin α z = -x sin α y cos α z = Has a non trivial solution, therefore D = 0 sin cos 1 cos sin 0 1 sin cos λ - cos α sin α) sin α - cos α + sin α + cos α sin α + cos α = -λ + sin α cos α sin α + sin α cos α + cos α = λ = cos α sin α + sin α cos α λ = cos α + sin α For λ =, Cos α + sin α = / cos α + / sin α = / cos α cos π/ + sin α sin π/ = / cps α π/ = cos π/ α π/ = nπ ± π/ α = πr + π/ + π/; nπ π/ + π/ α = n π + π/ or n π Sol 1. L. H. S cos Acos P sin Asin P cos(a Q) cos( A R) cos Bcos P sin Bsin P cos( B Q) cos( B R) cos C cos P sin C sin P cos( C Q) cos( C R)
13 cos A cos( AQ) cos( AR) = cos P cos B cos(b Q) cos( B R) cosc cos( C Q) cos( C R) sin A cos( AQ) cos( AR) sin P sin B cos(b Q) cos( B R) sin C cos( C Q) cos( C R) Operating; C C C 1 (cos Q); C 3 C 3 - C 1 (cos R) on first determinant and C C 3 (sin Q) C 1 and C 3 C 3 (sin R) C 1 on second determinant, we get cos A sin Asin Q sin Asin R cos P cos B sin Bsin Q sin Bsin R cos C sin C sin Q sin C sin R sin A cos Acos Q cos Acos R cos P sin B cos Bcos Q cos Bcos R sin C cos C cos Q cos C cos R = cos P sin Q sin R cos A sin A sin A cos B sin B sin B cosc sin C sin C + sin P cos Q cos R sin A cos A cos A sin B cos B cos B sinc cos C cos C = [Both determinants become zero as C C 3] = 0 = R. H. S. Sol 13. Let us denote the given determinant by. Taking 1/a (a + d) (a + d) as common from R 1, 1/(a + d) (a + d) (a + 3d) from R and 1/(a + d) (a + 3d) (a + 4d) from R 3 we get = 1/a (a + d) (a + d) 3 (a + 3d) a + d 1 Where 1 ( a d)(a d) a d a ( a d)(a 3d) a 3d a d ( a 3 d)(a 4d) a 4d a d Applying R 3 R 3 R and R R R 1, we get 1 ( a d)(a d) a d a ( a d)( d) d d ( a 3 d)( d) d d
14 Applying R 3 R 3 R, we get ( a d)( a d) a d a 1 ( a d)( d) d d d 0 0 Expanding along R 3, we get 1 = (d) a d a d d = (d) (d) (a + d a) = 4d 4 Thus, = d 4 /a (a + d) (a + d) 3 (a + 3d) (a + 4d). Sol14. Given that a, b, c are pth, qth and rth terms of an H. P. 1/a, 1/b, 1/c are pth, qth, rth terms of an A. P. = + = + = + } (1) Now given determinant is = = abc Substituting the values of 1/a, 1/b, 1/c from (1), we get = abc Operating R 1 R 1 (A D) R 3 D R, we get = abc = 0
15 Sol 15. R R + R 3, sin cos sin 4 = 0 sin cos cos cos sin cos sin cos sin = sin cos sin sin cos sin 4 sin cos sin = 0 Sol 16. Given that A T A = I Where A = [ ] A T = [ ] A T A = I [ ] [ ] = [ ] [ ] = [ ] a + b + c = (1) And ab + bc + ca = () abc = 1 (given) (3) Now a 3 + b 3 + c 3 3abc = (a + b + c) (a + b + c ab bc ca) a 3 + b 3 + c 3 = (a + b + c) [1 0] + 3 x 1 [Using (1), () and (3)]
16 a 3 + b 3 + c 3 = a + b + c + 3 Now (a + b + c) = a + b + c + (ab + bc + ca) a + b + c + = 1 a + b + c = [ a + b + c -1 as a, b, c all are + ve numbers] we get a 3 + b 3 + c 3 = 4 ALTERNATE SOLUTION Given that A T A = I A T A = A T A = A A = [ I = ] A = (1) From given matrix A = [ ] A = [ ] = a 3 + b 3 + c 3 3abc () (a 3 + b 3 + c 3 3abc = 1 or -1 But for a 3, b 3 c 3 using AM GM We get a 3 + b 3 + c 3 /3 We must have a 3 + b 3 + c 3 3abc = 1 a 3 + b 3 + c 3 = x 1 = 4 [Using abc = 1] Sol 17. We are given that MM T = I where M is a square matrix of order 3 and det M = 1. Consider det (M I) = det (M MM T ) [Given MM T = I] = det [M (I M T )] [ AB = A B ] = -(det M) (det (M T I)) = -1 [det (M T I] [ det M = ]
17 = - det (M - I) [ det M T I) = det [( M I) T = det (M I)] det (M I) = 0 det (M I) = 0 Hence Proved Sol18. Given that, A =[ ], X =[ ], U = [ ] And AX = U has infinite many solution. A = 0 = A 1 = A = A 3 Now, A = 0 [ ] = a (bc bd) 1 (c d) = 0 ab 1) (c d) = 0 ab = or c = d (1) And A 1 = = 0 f bc bd) - 1 (gc hd) = 0 f b c d) = gc hd () A = = 0 a gc hd) f (c d) = 0 a gc hd) = f (c d) A 3 = = 0 abh bg) 1 (h g) + f (b d) = 0 ab h g) (h g) = 0 ab = or h = g Taking c = d h = g and ab from, and Now taking BX = V
18 Where B =, V = Then B = = 0 [ In vew pf c = d and g = h, C and C and C 3 are indentical BX = V has no unique solution And B 1 = = c = d, g = h B = B 3 = a cf = a d f c = d a d f If adf then B = B 3 Hence no solution exist.
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