Integer Answer Type Questions: Matrices and Determinants | JEE Advanced | 35 Years Chapter wise Previous Year Solved Papers for JEE PDF Download

Q. 1. Let w be the complex number   Then the number of distinct complex numbers z satisfying            

Ans. 0

Solution. 

∴ z = 0 is the only solution.

Q. 2. Let k be a positive r eal n umber and let 

If det (adj A) + det (adj B) = 10⁶. then [k] is equal to 

[Note  : adj M denotes the adjoint of square matrix M and [k] denotes the largest integer less than or equal k.

Ans. 4

Solution.  

= (1 + 2k) (8k – 4k + 4k²+ 1) = (2k+ 1)3

Also B = 0 as B is skew symmetric of odd order..

∴ |Adj A| + |Adj B| = |A|² + |B|² = 106

⇒ (2k + 1)⁶ = 10⁶ ⇒ 2k + 1 = 10 ⇒ k= 4.5

∴ [k] = 4

Q. 3. Let M be a 3 × 3 matrix satisfying  Then the sum of the diagonal entries of M is

Ans. 9

Solution.

∴ Sum of diagonal elements = a₁ + b₂ + c₃ = 0 + 2 + 7 = 9

Q. 4. The total number of distinc  for which 

Ans. 2

Solution. 

Operating C₂ – C₁, C₃ – C₁ for both the determinants, we get

⇒ x³ (–4 + 6) + x⁶ (48 – 36) = 10
⇒ 2x³ + 12x⁶ = 10 ⇒ 6x⁶ + x³ – 5 = 0

Q. 5.   and r, s ∈ {1, 2, 3}. Let  and I be the identity matrix of order 2. Then the total number of ordered pairs (r, s) for which P² = –I is

Ans. 1

Solution.

For P² = – I we should have z^2r  + z^4s = – 1 and z^2s ((–z)^r  + z^r) = 0

⇒ z^2r + zs + 1 = 0 and (–z)^r + z^r = 0

⇒ r is odd and s = r but not a multiple of 3.
Which is possible when s = r = 1
∴  only one pair is there.