Homogenous equations

Algebra Level 4

Given α \alpha , β \beta and γ \gamma are angles in an euclidean triangle, does the system of equations below have NON-trivial solution? { ( tan ⁡ α ) x + y + z = 0 x + ( tan ⁡ β ) y + z = 0 x + y + ( tan ⁡ γ ) z = 0 \large \begin{cases} (\tan{\alpha})x+y+z &=0 \\ x+ (\tan{\beta})y+z &=0 \\ x+y+(\tan{\gamma})z &= 0 \end{cases}

Yes, it has always has No, it always hasn't Uncertain, it depends on shapes of triangles.

Tommy Li by Tommy Li , 22, Hong Kong

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1 solution

Tommy Li
Sep 2, 2019

Note that tan ⁡ α + tan ⁡ β + tan ⁡ γ = tan ⁡ α tan ⁡ β tan ⁡ γ \tan{\alpha}+\tan{\beta}+\tan{\gamma} =\tan{\alpha}\tan{\beta}\tan{\gamma} \ where α + β + γ = π \alpha+\beta+\gamma=\pi

Determinant of coefficient matrix = ∣ tan ⁡ α 1 1 1 tan ⁡ β 1 1 1 tan ⁡ γ ∣ = − ( tan ⁡ α + tan ⁡ β + tan ⁡ γ ) + tan ⁡ α tan ⁡ β tan ⁡ γ + 2 = 2 ≠ 0 =\begin{vmatrix} \tan{\alpha} & 1 &1 \\ 1& \tan{\beta}&1 \\ 1&1&\tan{\gamma} \end{vmatrix} =-(\tan{\alpha}+\tan{\beta}+\tan{\gamma}) +\tan{\alpha}\tan{\beta}\tan{\gamma} + 2 =2 \neq 0

No, it always hasn't non-trivial solution because the the determinant always does not equal to 0.

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