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Mathematics Question on Sets

Let αβγ=45\alpha \beta \gamma = 45; α,β,γ∈R\alpha, \beta, \gamma \in \mathbb{R}. If x(α,1,2)+y(1,β,2)+z(2,3,γ)=(0,0,0)x(\alpha, 1, 2) + y(1, \beta, 2) + z(2, 3, \gamma) = (0, 0, 0) for some x,y,z∈Rx, y, z \in \mathbb{R}, xyz≠0xyz \neq 0, then 6α+4β+γ6\alpha + 4\beta + \gamma is equal to ______.

Answer

Given αβγ=45\alpha \beta \gamma = 45, α,β,γ∈R\alpha, \beta, \gamma \in \mathbb{R},

x(α,1,2)+y(1,β,2)+z(2,3,γ)=(0,0,0).x(\alpha, 1, 2) + y(1, \beta, 2) + z(2, 3, \gamma) = (0, 0, 0).

Expanding, we get:

xα+y+2z=0,x \alpha + y + 2z = 0, x+yβ+3z=0,x + y \beta + 3z = 0, 2x+2y+zγ=0.2x + 2y + z \gamma = 0.

Since xyz≠0xyz \neq 0, the determinant of the coefficient matrix must be zero for a non-trivial solution:

∣α12 1β2 23γ∣=0.\begin{vmatrix} \alpha & 1 & 2 \\\ 1 & \beta & 2 \\\ 2 & 3 & \gamma \end{vmatrix} = 0.

Expanding the determinant:

α∣β2 3γ∣−1∣12 2γ∣+2∣1β 23∣=0.\alpha \begin{vmatrix} \beta & 2 \\\ 3 & \gamma \end{vmatrix} - 1 \begin{vmatrix} 1 & 2 \\\ 2 & \gamma \end{vmatrix} + 2 \begin{vmatrix} 1 & \beta \\\ 2 & 3 \end{vmatrix} = 0.

Calculating each minor:

∣β2 3γ∣=βγ−6,\begin{vmatrix} \beta & 2 \\\ 3 & \gamma \end{vmatrix} = \beta \gamma - 6, ∣12 2γ∣=γ−6,\begin{vmatrix} 1 & 2 \\\ 2 & \gamma \end{vmatrix} = \gamma - 6, ∣1β 23∣=3−2β.\begin{vmatrix} 1 & \beta \\\ 2 & 3 \end{vmatrix} = 3 - 2\beta.

Substituting:

α(βγ−6)−(γ−6)+2(3−2β)=0.\alpha (\beta \gamma - 6) - (\gamma - 6) + 2(3 - 2\beta) = 0.

Simplify:

αβγ−6α−γ+6+6−4β=0.\alpha \beta \gamma - 6\alpha - \gamma + 6 + 6 - 4\beta = 0.

Since αβγ=45\alpha \beta \gamma = 45:

45−6α−γ+12−4β=0.45 - 6\alpha - \gamma + 12 - 4\beta = 0.

Rearranging:

6α+4β+γ=55.6\alpha + 4\beta + \gamma = 55.