JEE Main 2023 — Mathematics Algebra

For infinite solution $\Delta ={\Delta }_{x}={\Delta }_{y}={\Delta }_{z}=0$Δ=Δx​=Δy​=Δz​=0

$$\Delta =|\begin{matrix}\alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{matrix}|=0$$Δ=∣α11​1α1​11α​∣=0

$$\alpha ({\alpha }^{2}-1)-1(\alpha -1)+1(1-\alpha )=0$$α(α2−1)−1(α−1)+1(1−α)=0

$$\Rightarrow {\alpha }^{3}-3\alpha +2=0$$⇒α3−3α+2=0

$$\Rightarrow {\alpha }^{2}(\alpha -1)+\alpha (\alpha -1)-2(\alpha -1)=0$$⇒α2(α−1)+α(α−1)−2(α−1)=0

$$\Rightarrow (\alpha -1)({\alpha }^{2}+\alpha -2)=0$$⇒(α−1)(α2+α−2)=0

$$\Rightarrow \alpha =1,\alpha =-2,1$$⇒α=1,α=−2,1

If $\beta =1$β=1, then all planes are overlapping

$\therefore$∴ option (d) is correct.

Put $\alpha =2,\beta =1$α=2,β=1 the system equations as,

$$2x+y+z=1$$2x+y+z=1

$$x+2y+z=1$$x+2y+z=1

$$x+y+2z=1$$x+y+2z=1

Adding all three above equations we get,

$$x+y+z=\frac{3}{4}$$x+y+z=43​

$\therefore$∴ option (c) is correct.

Now, if $\alpha =-2,\beta =1$α=−2,β=1 then,

$\Delta =0,{\Delta }_{x}\neq 0$Δ=0,Δx​=0 (No solution)

$\therefore$∴ option (b) is correct.

Now, if $\alpha =2$α=2

$\Delta \neq 0$Δ=0 (unique solution exists)

$\therefore$∴ option (a) is incorrect