Test: Single Correct MCQs: Matrices and Determinants | JEE Advanced

For every ‘det, with value 1’ (∈ B) we can find a det. with value – 1 by changing the sign of one entry of ‘1’.
Hence there are equal number of elements in B and C.
∴ (b) is the correct option

Operating R₁→ R₁ – R₂ + R₃

Then the given system of equations becomes
X + Y – Z = 1
X – Y + Z= 1, – X + Y + Z = 1
This is the new system of equations For new system, we have

∴ New system of equations has unique solution.

If A and B are square matrices of same degree then matrices A and B can be added or subtracted or multiplied. By algebra of matrices the only correct option is A + B = B + A

Expanding along C₁, we get
Δ = (1 + a² – 2a cos dx) [sin (p + d) x cos px  – sin px cos (p + d)x]
⇒ Δ = (1 + a² – 2a cos dx) [sin {(p + d)x – px}]
⇒ Δ = (1 + a² – 2a cos dx) [sin dx]
which is independent of p.

= 0 [∵ C₁ and C₂ are identical] which is free of x, so the function is true for all values of x. Therefore, at x = 100, f (x) = 0, i.e., f (100) = 0

For the given homogeneous system to have non zero solution determinant of coefficient matrix should be zero; i.e.,

Given that 
Also 1 + ω + ω² = 0 and ω³ = 1
Now given det. is

[Using ω = –1 – ω² and ω³ = 1]

Operating C₁ → C₁ + C₂ + C3

   (as 1 + ω + ω2 = 0)

Expanding along C₁, we get
3 (ω2 – ω4) =3 (ω2 – ω) = 3ω (ω– 1)

For infinitely many solutions the two equations become identical

Given that 

and A² = B

∵ There is no common value
∴ There is no real value of a for which A² = B

The given system is, x + ay = 0
az + y = 0
ax + z = 0

It is system of homogeneous equations therefore, it will have infinite many solutions if determinant of coefficient matrix is zero. i.e.,

Since the system has no solution, Δ = 0 and any one amongst D_x, D_y, D_z is non-zero.

Now, | A | = α² – 4
⇒ (α² – 4)³ = 125 = 5³
⇒  α² – 4 = 5
⇒  α = + 3

∴ Characteristic eqn of above matrix A is given by

Also by Cayley Hamilton thm (every square matrix satisfies its characteristic equation) we obtain
A³ – 6A² + 11A – 6I = 0

Multiplying by A^–1, we get

Comparing it with given relation,

we get c = – 6 and d = 11

We observe that Q = P A PT
⇒ Q² = (P A P^T) (P A P^T)
= P A (P^T P) A P^T = PA (I A) P^T
∴ P A² P^T
Proceeding in the same way, we get Q²⁰⁰⁵ = P A²⁰⁰⁵ PT

and proceeding in the same way 

The given points are P (- sin(β- α), – cosβ) , Q(cos(β - a), sinβ)
R(cos(β- α+θ), sin(β- θ))

Operating C₃ - C₁ sin_θ- C₂ cosθ , we get

 = (1 - sinθ - cosθ)[cosβ cos(β -α) - sin β sin(β-α)]
⇒ Δ = [1 - (sinθ+ cosθ)] cos(2β-α)

∴ Δ≠ 0 ⇒ the three points are non collinear..

Then the given system is equivalent to

Which represents three distinct planes. But three planes can not intersect at two distinct points, therefore no such system exists.

For the given matrix to be non-singular

⇒ 1 – (a + c) ω + a cω² ≠ 0 ⇒ (1 – aω) (1 – cω) ≠ 0
⇒ a ≠ ω² and c ≠ ω² where ω is complex cube root of unity.
As a, b and c are complex cube roots of unity
∴ a and c can take only one value i.e. w while b can take 2 values i.e. ω and ω². ∴ Total number of distinct matrices = 1 × 1 × 2 = 2

We have

= 2¹² x |P| = 2¹² x 2 = 2¹³

We have P^T = 2P + I
⇒ P = 2P^T + I ⇒ P = 2(2P + I) + I
⇒ P = 4P + 3I ⇒ P + I = 0
⇒ PX + X = 0 ⇒ PX = – X

Now P⁵⁰ = (I + A)⁵⁰ = ⁵⁰C₀ I⁵⁰ + ⁵⁰C₁ I⁴⁹ A + ⁵⁰C₂ I⁴⁸ A² + O        
= I + 50A + 25 × 49 A².
∴ Q = P⁵⁰ – I = 50A + 25 × 49 A².
⇒ q₂₁ = 50 × 4 = 200
⇒ q₃₁ = 50 × 16 + 25 × 49 × 16 = 20400
⇒ q₃₂ = 50 × 4 = 200