MCQs (One or More Correct Option): Matrices and Determinants | JEE Advanced

Operating C₃→ C₃ – C₁ α – C₂, we get

⇒ (ac – b²) (aα² + 2bα + c) = 0
⇒ either ac – b² = 0 or aα² + 2b α + c = 0
⇒ either a, b, c are in G.P. or (x – α) is a factor of ax² + 2bx + c
⇒ (b) and (e) are the correct answers.

[∴ C₂ and C₃ are identical]
⇒ x + iy = 0   ⇒ x = 0, y = 0

[As a skew symmetric matrix of order 3 cannot be non singular, therefore the data given in the question is inconsistent.]
We have  M²N² (M^T N)^–1 (MN^–1)T = M²N²N^–1 (M^T)^–1 (N^–1)^T
M^T
= M² N (M^T)^–1 (N^–1)^T M^T = –M²NM^–1 N^–1M
(∵ M^T = –M, N^T = –N and (N^–1)T = (N^T)^–1
= – M (NM) (NM)^–1 M             (∵ MN = NM)
= – MM = –M²

ANSWER :- a,d

Solution :-  |adj P| = |P|^2 as (adj(P)| = |P|^(n-1))

|adj P| = 1(3-7) -4(6-7) +4(2-1) = 4

Hence, |P| = 2 or -2

(a)(N' M N)' = (M N)'N = N'M 'N = N'M N or –N'M N According as M  is symm. or skew symm. ∴correct
(b) (MN – NM)' = (MN)' – (NM)' = N'M' – M'N' = NM – MN = –(MN – NM)
∴ It is skew symm. Statement B is also correct.
(c)(MN)' = N'M' = NM ¹ MN
∴ Statement C is incorrect
(d) (adj M) (adj N) = adj (MN) is incorrect.

It shows P² = 0 if n is a multiple of 3.
So for P² ≠ 0, n should not be a multiple of 3 i.e. n can take values 55, 58, 56

 where a, b, c are integers.

M is invertible if 

∴ (a) is not correct.
If [bc] = [ab] ⇒ b = a = c ⇒ ac = b²
∴ (b) is not correct.

∴ M is invertible.
(c) is correct
As ac ≠ (integer)2 ⇒ ac ≠ b²
∴ (d) is correct.

Given MN = NM, M ≠ N² and M² = N⁴.
Then M² = N⁴ ⇒ (M + N²) (M – N ²) = 0
⇒ (i) M + N² = 0 and M – N² ≠ 0
(ii) |M + N²| = 0 and |M – N²| = 0
In each case |M + N²| = 0
∴ |M² + MN²| = |M| |M + N²| = 0
∴ (a) is correct and (c) is not correct.
Also we know if |A| = 0, then there can be many matrices U, such that AU = 0
∴ (M² + MN²)U = 0 will be true for many values of U.
Hence (b) is correct.
Again if AX = 0 and |A| = 0, then X can be non-zero.
∴ (d) is not correct.

⇒ 2α²(–2α) = –324α  ⇒ α³ – 81α = 0 ⇒ α = 0, 9, –9

X ' = –X, Y ' = –Y, Z ' = Z
(Y³Z⁴ – Z⁴Y³)' = (Z⁴)'(Y³)' – (Y³)'(Z⁴)'
= (Z')4(Y')3 – (Y')3(Z')4
= –Z⁴Y³ + Y³Z⁴ = Y³Z⁴ – Z⁴Y³

∴ Symmetric matrix.
Similarly X⁴⁴ + Y⁴⁴ is symmetric matrix and X⁴Z³ – Z³X⁴ and X²³ + Y²³ are skew symmetric matrices.

Comparing the third elements of 2nd row on both sides, we get

ax + 2y = λ
3x – 2y = μ

For unique solution, 
∴ (b) is the correct option.
For infinite many solutions and a = – 3

∴ (c) is the correct option.

⇒ system has no solution.
⇒ (d) is the correct option.