Question

If |2A|³ = 2²¹ and A = \(\begin{bmatrix} 1 & 0 & 0 \\[0.3em] 0 & \alpha & \beta \\[0.3em] 0 & \beta & \alpha \end{bmatrix}\) then α is (if α, β ∈ I) 

(1) 5 

(2) 3 

(3) 9 

(4) 17

Answer 1

The correct option is (1) 5.

|2A| = 2⁷ 

8|A| = 2⁷ 

|A| = 2⁴ 

Now |A| = α² – β² = 2⁴ 

α² = 16 + β² 

α² – β² = 16 

(α – β) (α + β) = 16 

⇒ α + β = 8 and α – β = 2 

⇒ α = 5, and β = 3