Pro-ble-ma-trix

Geometry Level 3

A = [ 0 1 sec ⁡ θ tan ⁡ θ − sec ⁡ θ tan ⁡ θ 1 0 1 ] \large A = \begin{bmatrix} 0 & 1 & \sec\theta\\ \tan\theta & -\sec\theta & \tan\theta\\ 1 & 0 & 1\end{bmatrix}

Let matrix A A be defined as above, where 0 < θ < π 2 0< \theta < \large \frac{\pi}{2} . If det ⁡ ( adj ( A ) ) = 81 \det(\text{adj}(A)) = 81 , then what is tan ⁡ 2 θ \tan^{2}\theta

Notation : adj ( A ) \text{adj}(A) denotes the adjoint of matrix A A .


The answer is 8.

Watsapol Kerdlarppol by Watsapol Kerdlarppol , 22, Thailand

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1 solution

Brian Moehring
Apr 4, 2017

By a direct computation (for instance, by cofactor expansion), ∣ A ∣ = sec ⁡ 2 θ . |A| = \sec^2\theta.

Further, since A adj ( A ) = ∣ A ∣ I 3 , A \text{ adj}(A) = |A|\, I_3, by taking the determinant of both sides, we have ∣ A ∣ ⋅ 81 = ∣ A ∣ 3 , |A| \cdot 81 = |A|^3, so sec ⁡ 2 θ = ∣ A ∣ ∈ { 0 , ± 9 } ⟹ sec ⁡ 2 θ = 9. \sec^2\theta = |A| \in \{0, \pm 9\} \implies \sec^2\theta = 9.

Finally, by the Pythagorean identity, tan ⁡ 2 θ = sec ⁡ 2 θ − 1 = 9 − 1 = 8 \tan^2\theta = \sec^2\theta - 1 = 9-1 = \boxed{8}

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