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Mathematics Question on Determinants

The system of equations α x+y+z=α−1\alpha \, x + y + z = \alpha - 1 x+αy+z=α−1x + \alpha y + z = \alpha - 1 x+y+αz=α−1x + y + \alpha z = \alpha - 1 has infinite solutions, if α\alpha is

A

-2

B

either - 2 or 1

C

not - 2

D

1

Answer

-2

Explanation

Solution

α x+y+z=α−1\alpha \ x + y + z = \alpha - 1 x+αy+z=α−1x + \alpha y + z = \alpha - 1 x+y+αz=α−1x + y + \alpha z = \alpha - 1 Δ=∣α11 1α1 11α∣\Delta = \begin{vmatrix}\alpha&1&1\\\ 1 &\alpha&1\\\ 1&1&\alpha\end{vmatrix} =α(α2−1)−1(α−1)+1(1−α)= \alpha\left(\alpha ^{2} - 1\right)-1\left(\alpha -1\right)+1\left(1-\alpha \right) =α(α−1)(α+1)−1(α−1)−1(α−1)= \alpha \left(\alpha -1\right)\left(\alpha +1\right)-1\left(\alpha -1\right)-1\left(\alpha -1\right) For infinite solutions ,Δ=0, \Delta = 0 ⇒(α−1)[α2+α−1−1]=0 \Rightarrow \left(\alpha -1\right)\left[\alpha ^{2}+\alpha -1-1\right]=0 ⇒(α−1)[α2+α−2]=0\Rightarrow \left(\alpha -1\right)\left[\alpha ^{2} +\alpha -2\right] = 0 ⇒(α−1)[α2+2α−α−2]=0\Rightarrow \left(\alpha -1\right)\left[\alpha ^{2}+2\alpha -\alpha -2\right]=0 ⇒(α−1)[α(α+2)−1(α+2)]=0 \Rightarrow\left(\alpha -1\right)\left[\alpha \left(\alpha +2\right)-1\left(\alpha +2\right)\right]= 0 (α−1)=0,α+2=0\left(\alpha -1\right) = 0, \alpha +2=0 ⇒α=−2,1;\Rightarrow\alpha =-2,1; But α≠1, \alpha \ne 1 , ∴α=−2\therefore \alpha = -2