NCERT Solutions for Class 12 Maths Chapter 4 Determinants — Exercise 4.20

Question 1:Prove that the determinant sin cos –sin – 1 cos 1 x x x q q q q is independent of q.

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Question 2:Without expanding the determinant, prove that 2 2 3 2 2 3 2 2 3 1 1 1 = a a bc a a b b ca b b c c ab c c .

Question 3:Evaluate cos cos cos sin – sin – sin cos 0 sin cos sin sin cos a b a b a b b a b a b a .

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Question 4:If a, b and c are real numbers, and D = b c c a a b c a a b b c a b b c c a + + + + + + + + + = 0, Show that either a + b + c = 0 or a = b = c.

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Question 5:Solve the equation 0 x a x x x x a x x x x a + + = + , a ¹ 0

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Question 6:Prove that 2 2 2 2 2 2 + + + a bc ac c a ab b ac ab b bc c = 4a2b2c2

Question 7:If A–1 = ( ) 1 3 1 1 1 2 2 15 6 5 and B 1 3 0 , find AB 5 2 2 0 2 1 – – – – – – – –

       =      

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Question 8:Let A = 1 2 1 2 3 1 1 1 5   é ë êêê ù û úúú . Verify that (i) [adj A]–1 = adj (A–1) (ii) (A–1)–1 = A

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Question 9:Evaluate x y x y y x y x x y x y + + +

Question 10:Evaluate 1 1 1 x y x y y x x+ y + Using properties of determinants in Exercises 11 to 15, prove that:

Question 11:2 2 2 a a b + g b b g + a g g a + b = (b – g) (g – a) (a – b) (a + b + g)

Exercise Miscellaneous class 12 Math ncert solutions Chapter 4

Question 12:2 3 2 3 2 3 1 1 1 + + + x x px y y py z z pz = (1 + pxyz) (x – y) (y – z) (z – x), where π is any scalar.

1. Prove that the determinant sin cos sin 1

Question 13:3 3 a 3c a – a+b – a+c –b a b – b c – c – c+b + + + = 3(a + b + c) (ab + bc + ca)

Question 14:1 1 1 2 3 2 4 3 2 3 6 3 10 6 3 π p q π p q π p q + + + + + + + + + = 1

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Question 15:( ) ( ) ( ) sin cos cos sin cos cos 0 sin cos cos a a a + d b b b + d = g g g + d

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Question 16:Solve the system of equations 2 3 10 + + = 4 x y z 4 6 5 – + =1 x y z 6 9 20 + – = 2 x y z Choose the correct answer in Exercise 17 to 19

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Question 17:If a, b, c, are in A.P, then the determinant 2 3 2 3 4 2 4 5 2 x x x a x x x b x x x c + + + + + + + + + is (A) 0 (B) 1 (C) x (D) 2x

Question 18:If x, y, z are nonzero real numbers, then the inverse of matrix 0 0 A 0 0 0 0 x y z
   =     is (A) 1 1 1 0 0 0 0 0 0 x y z − − −
         (B) 1 1 1 0 0 0 0 0 0 x xyz y z − − −
         (C) 0 0 1 0 0 0 0 x y xyz z
       (D) 1 0 0 1 0 1 0 0 0 1 xyz
      

CBSE NCERT Solutions For Class 12 Determinants

Question 19:Let A = 1 sin 1 sin 1 sin 1 sin 1
q    − q q  − − q  , where 0 ≤ q ≤ 2p. Then (A) Det (A) = 0 (B) Det (A) Î (2, ∞) (C) Det (A) Î (2, 4) (D) Det (A) Î [2, 4]