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JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Determinants

Let

A = 100 0 α β 0 β α

and

∣2 A ∣^{3} = 2^{21}

where

α, β \in Z

,

Then a value of

α

is

(A)3
(B)5✓ Correct
(C)17
(D)9

Answer: (B)

Solution

A = 100 0 α β 0 β α

,

det A = α^{2} - β^{2}

.

∣2 A ∣ = 2^{3} ∣ A ∣ = 8 (α^{2} - β^{2})

, and

∣2 A ∣^{3} = 2^{21} ⟹ ∣2 A ∣ = 2^{7} = 128

.

8 (α^{2} - β^{2}) = 128 ⟹ α^{2} - β^{2} = 16 ⟹ (α - β) (α + β) = 16

For integer

α, β

, try

α = 5

:

β^{2} = 25 - 16 = 9 ⟹ β = \pm 3

, valid. (Checking the other listed options, 3, 17, and 9 give no integer

β

.)

Key confirmed: B (

α = 5

).

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Single correct option

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