Mathematics-1.Matrices and Determinants with Solution

1. The value of is : (A) (  5) (B) 5 (  5) (C) 5 ( + 5) (D) 5 (  5) Solution: L.H.S.= + =5 +5 = 5 (  5) + 5  0 = 5 (  5) Hence (B) is the correct answer. 2. is equal to : (A) constant other than zero (B) zero (C) 100 (D) –1997 Solution: We know that in a determinant, this element in all the rows (columns) are in AP with same or different common difference, the value of the determinant is zero Hence (B) is the correct answer. 3. If a + b +c > 0 and a  b  c  0, then  = then : (A)  > 0 (B)  < 0 (C)  = 0 (D) Data insufficient Solution: The element of the row are in circular arrangement = (a + b + c) { (a  b)2 + (b  c)2 + (c  a)2 } < 0 Hence (B) is the correct answer. 4. The determinant is equal to zero if : (A) a, b, c are in A.P. (B) a, b, c are in G.P. (C) a, b, c are in H.P. (D) none of these Solution: R3  R3  (R1  + R2) = This determinant is zero if (a 2 + b  + c) (a c  b2) = 0  b2 = a c  a, b, c are in G.P. Hence (B) is the correct answer. 5. The determinant is divisible by : (A) 1 + x (B) (1 + x)2 (C) x2 (D) none of these Solution: The given determinant is symmetric in nature  = a2 b2 c2 (1 + x)3 + 2 a2 b2 c2  3 a2 b2 c2 (1 + x) = a2 b2 c2 [ (1 + x)3 + 2  3 (1 + x)] = a2 b2 c2 [x3 + 3 x2], divisible by x2 Hence (C) is the correct answer. 6. is equal to : (A) a positive number (B) a negative number (C) Zero (D) none of these Solution:  Applying R1  R3 = + Applying C2  C3 =  = 0 Hence (C) is the correct answer. 7. If f(x) = , then dx is equal to : (A) 1/4 (B) –1/3 (C) 1/2 (D) 1 Solution: Expanding the given determinant, f (x) = 4 cos3 x  3 cos x = cos 3 x f (x) dx = cos 3 x dx = =  Hence (B) is the correct answer. 8. For positive numbers x, y, z, the numerical value of the determinant is : (A) 0 (B) 1 (C) 2 (D) none of these Solution: = = = 0 Hence (A) is the correct answer. 9.  = is equal to : (A) p + q (B) a + b + c (C) x + y + z (D) 0 Solution: =  = 0 Hence (D) is the correct answer. 10. is equal to : (A) abc (B) a2b2c2 (C) ab + bc + ca (D) 0 Solution:  = = a b c = Applying R2  R2  R1 R3  R3  R1 = a b c = a b2 c2 (a  b) (a  c) = 0, R2 and R3 are identical Hence (D) is the correct answer. 11. is equal to : (A) 0 (B) 1 (C) –1 (D) none of these Solution: L.H.S. = =  = 0 Hence (A) is the correct answer. 12. If f (x) = then f (100) is equal to : (A) 0 (B) 1 (C) 100 (D) –100 Solution: f (x) = C3  C3  (C1 + C2) = = 0 f (100) = 0 Hence (A) is the correct answer. 13. is equal to : (A) 1 + åa2 (B) åa2 (C) (åa)2 (D) åa Solution:  C1  C1 + b C2 + c C3 = = C2  C2  b C1, C3  C3  c C1 = = 1 + a2 + b2 + c2 Hence (A) is the correct answer. 14. If A, B, C are the angles of a triangle and , then D is equal to: (A) constant other than zero (B) not a constant (C) 0 (D) none of these Solution: R2  R2  R1 R3  R3  R1 = = 0 Hence (C) is the correct answer. 15. If [.] denotes the greatest integer less than or equal to the real number under consideration and –1  x < 0; 0  y < 1 and 1  z < 2, then the value of the determinant is : (A) [z] (B) [y] (C) [x] (D) none of these Solution:  1  x < 1  [x] =  1 0  y < 1  [y] = 0 1  z < 2  [z] = 1 Given determinant equal to = 1 = [z] Hence (A) is the correct answer. 16. If D = = 0, then the non zero root of the equation is : (A) a+b +c (B) abc (C) – a – b –c (D) None of these Solution: C1  C1 + C2 + C3 = 0 (x +  a) = 0  x =  a  b  c Hence (C) is the correct answer. 17. If f (x) = then the value of f(x) dx is equal to : (A) 0 (B) 1 (C) 2 (D) none of these Solution: f (x) = =  f (x) f (x) is an odd function f (x) dx = 0 Hence (A) is the correct answer. 18. If A = , then A2 is equal to (A) unit matrix (B) null matrix (C) A (D) –A Solution: A2 = = I3 19. If A is the transpose of a square matrix A, then (A) |A|  |A| (B) |A| = |A| (C) |A| + |A| = 0 (D) |A| = |A| only when A is symmetric Solution: Value of a determinant remains unattached by interchanging the rows and columns  |A| = |AT| 20. If I = , J = and B = , then B equals (A) I cos + J sin (B) I sin + J cos (C) I cos – J sin (D) – I cos + J sin Solution: I cos  + J sin  = + = = B 21. If In is the identity matrix of order n, then (In)–1 (A) does not exist (B) equal to In (C) equals to O (D) nIn Solution: In. In = In  In1 = In 22. If for a matrix A, A2 + I = O where I is the identity matrix, then A equals (A) (B) (C) (D) Solution: A2 + I = 0  A2 =  I For A = =  I 23. If A = , then A40 equals (A) (B) (C) (D) none of these Solution: A = diag (1, 0, 1)  A40 = diag (140, 040, 140) = diag (1, 0, 1) 24 If A [aij] is a square matrix of order n ´ n such that aii = k for all i, then trace of A is equal to (A) kn (B) (C) nk (D) none of these Solution: Trace of a matrix = sum of principal diagonal elements = k + k + k + … n times = n k 25. If A = and I = , then the value of k so that A2 = 8A + kI is (A) 7 (B) - 7 (C) 0 (D) 5 Solution: A2 = 8 A + k I  Therefore 8 + k = 1  k =  7 26. If A = , then the value of |adj A| is (A) a27 (B) a9 (C) a6 (D) a2 Solution: |adj A| = |A|n1 = (a3)31 = a6 27. If A and B are symmetric matrices of order n (A  B), then (A) A + B is skew symmetric (B) A + B is symmetric (C) A + B is a diagonal matrix (D) A + B is a zero matrix Solution: A = AT amd B = BT (A + B)T = AT + BT = A + B  A + B is symmetric 28. If A = and B = then AB = (A) A3 (B) B2 (C) O (D) I Solution: A B = = O 29. If A = the A is (A) Idempotent (B) nilpotent (C)symmetric (D) none of these Solution: A2 = A3 = = O  A is a nilpotent matrix 30. If A = , then 19A–1 is equal to (A) A (B) 2A (C) (D) A Solution: A1 = A  19 A1 = A

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