JEE Main 2022 — Mathematics Algebra

Given,

$$x+y+az=2$$x+y+az=2

$$3x+y+z=4$$3x+y+z=4

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

Now finding,

$$\Delta =|\begin{matrix}1 & 1 & \alpha \\ 3 & 1 & 1 \\ 1 & 0 & 2\end{matrix}|=-(\alpha +3)$$Δ=∣131​110​α12​∣=−(α+3)

$${\Delta }_{1}=|\begin{matrix}2 & 1 & \alpha \\ 4 & 1 & 1 \\ 1 & 0 & 2\end{matrix}|=-(3+\alpha )$$Δ1​=∣241​110​α12​∣=−(3+α)

$${\Delta }_{2}=|\begin{matrix}1 & 2 & \alpha \\ 3 & 4 & 1 \\ 1 & 1 & 2\end{matrix}|=-(\alpha +3)$$Δ2​=∣131​241​α12​∣=−(α+3)

$${\Delta }_{3}=|\begin{matrix}1 & 1 & 2 \\ 3 & 1 & 4 \\ 1 & 0 & 1\end{matrix}|=0$$Δ3​=∣131​110​241​∣=0

$$\alpha \neq -3,x=\frac{{\triangle }_{1}}{\triangle }=1,y=\frac{{\triangle }_{2}}{\triangle }=1,z=\frac{{\triangle }_{3}}{\triangle }=0,$$α=−3,x=△△1​​=1,y=△△2​​=1,z=△△3​​=0,

Now points (\alpha ,1),(1,\alpha )&(1,-1) are collinear

$$|\begin{matrix}\alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & -1 & 1\end{matrix}|=0$$∣α11​1α−1​111​∣=0

$$\Rightarrow \alpha (\alpha +1)-1(1-1)+1(-1-\alpha )=0$$⇒α(α+1)−1(1−1)+1(−1−α)=0

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

$$\alpha =\pm 1$$α=±1

So, $|1|+|-1|=2$∣1∣+∣−1∣=2