Let A=[1, 0, 0; 0, α, β; 0, β, α] and |2 A|³=2²¹ where α, β ∈ Z , Then a value of α is

Solution

Maths sir · 3 steps

Pull the 2 out as 2³ since A is 3 × 3, then α² - β² = 16 does the rest.

1. Find det A. Expand along the first row. Only the 1 survives, so |A| = α² - β².
2. Scale correctly. For an n × n matrix, |kA| = kⁿ |A|. Here n = 3, so |2A| = 8|A| = 2³|A|.

   (2³|A|)³ = 2²¹ ⇒ 2⁹|A|³ = 2²¹ ⇒ |A|³ = 2¹² ⇒ α² - β² = 16

   Option α	β² = α² - 16	Integer β?
   9	81 - 16 = 65	no
   17	289 - 16 = 273	no
   3	9 - 16 = -7	no, negative
   5	25 - 16 = 9	yes, β = ± 3

3. So. Only α = 5 gives an integer β, so the answer is (d).

The idea to keep. A scalar comes out of a determinant once for every row: |kA| = kⁿ|A| for an n × n matrix. Get that power right before you do anything else.

Answer from the NTA final answer key. Solution by PrepWiser.