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# Integer Answer Type Questions: Matrices and Determinants | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

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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 z2r  + z4s = – 1 and z2s ((–z)r  + zr) = 0

⇒ z2r + zs + 1 = 0 and (–z)r + zr = 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.

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## Maths 35 Years JEE Main & Advanced Past year Papers

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## Maths 35 Years JEE Main & Advanced Past year Papers

132 docs|70 tests

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