JEE Main 2021 — Mathematics Algebra

Given system of linear equations are $x+y-z=2,x+2y+\alpha z=1,2x-y+z=\beta$x+y−z=2,x+2y+αz=1,2x−y+z=β

Using Cramer Rule, we can say for infinite solutions

$$\Delta ={\Delta }_{1}={\Delta }_{2}={\Delta }_{3}=0$$Δ=Δ1​=Δ2​=Δ3​=0

Here, $\Delta =|\begin{matrix}1 & 1 & -1 \\ 1 & 2 & \alpha \\ 2 & -1 & 1\end{matrix}|=0$Δ=∣112​12−1​−1α1​∣=0

Apply ${R}_{1}\leftrightarrow {R}_{1}+{R}_{3}$R1​↔R1​+R3​

$$\Rightarrow |\begin{matrix}3 & 0 & 0 \\ 1 & 2 & \alpha \\ 2 & -1 & 1\end{matrix}|=0$$⇒∣312​02−1​0α1​∣=0

$$\Rightarrow 3(2+\alpha )=0$$⇒3(2+α)=0

$$\Rightarrow \alpha =-2$$⇒α=−2

And ${\Delta }_{2}=|\begin{matrix}1 & 2 & -1 \\ 1 & 1 & -2 \\ 2 & \beta & 1\end{matrix}|=0$Δ2​=∣112​21β​−1−21​∣=0

$$\Rightarrow 1(1+2\beta )-2(1+4)-(\beta -2)=0$$⇒1(1+2β)−2(1+4)−(β−2)=0

$$\Rightarrow \beta -7=0$$⇒β−7=0

$$\Rightarrow \beta =7$$⇒β=7

$$\therefore \alpha +\beta =5$$∴α+β=5