JEE Main 2021 — Mathematics Algebra

Given,

$$2x+y-z=3$$2x+y−z=3

$$x-y-z=\alpha$$x−y−z=α

$$3x+3y+\beta z=3$$3x+3y+βz=3

Let A=[\begin{matrix}2 & 1 & -1 \\ 1 & -1 & -1 \\ 3 & 3 & \beta \end{matrix}]&B=[\begin{matrix}3 \\ \alpha \\ 3\end{matrix}]

For Equation to have infinitely many solutions |A|=0&(Adj.A).B=0

$$|A|=|\begin{matrix}2 & 1 & -1 \\ 1 & -1 & -1 \\ 3 & 3 & \beta \end{matrix}|=0$$∣A∣=∣213​1−13​−1−1β​∣=0

$$\Rightarrow -3\beta -3=0$$⇒−3β−3=0

$$\Rightarrow \beta =-1$$⇒β=−1

Now,

Minor of $A=[\begin{matrix}2\beta +3 & \beta +3 & 6 \\ \beta +3 & 2\beta +3 & 3 \\ -2 & -1 & -3\end{matrix}]$A=[2β+3β+3−2​β+32β+3−1​63−3​]

Cofactor of $A=[\begin{matrix}-(2\beta +3) & -(\beta +3) & -6 \\ -(\beta +3) & -(2\beta +3) & -3 \\ 2 & 1 & 3\end{matrix}]$A=[−(2β+3)−(β+3)2​−(β+3)−(2β+3)1​−6−33​]

Adjoint of $A=[\begin{matrix}-(2\beta +3) & -(\beta +3) & 2 \\ -(\beta +3) & -(2\beta +3) & 1 \\ -6 & -3 & 3\end{matrix}]$A=[−(2β+3)−(β+3)−6​−(β+3)−(2β+3)−3​213​]

Now, $(adj.A).B=0$(adj.A).B=0

$$\Rightarrow [\begin{matrix}-(2\beta +3) & -(\beta +3) & 2 \\ -(\beta +3) & -(2\beta +3) & 1 \\ -6 & -3 & 3\end{matrix}][\begin{matrix}3 \\ \alpha \\ 3\end{matrix}]=0$$⇒[−(2β+3)−(β+3)−6​−(β+3)−(2β+3)−3​213​][3α3​]=0

$$[\begin{matrix}-6\beta -9-\alpha \beta -3\alpha +6 \\ -3\beta -9-2\alpha \beta -3\alpha +3 \\ -18-3\alpha +9\end{matrix}]=0$$[−6β−9−αβ−3α+6−3β−9−2αβ−3α+3−18−3α+9​]=0

$$\Rightarrow -18-3\alpha +9=0$$⇒−18−3α+9=0

$$\Rightarrow \alpha =-3$$⇒α=−3

Therefore,$|\alpha +\beta -\alpha \beta |=|-3-1-3|=|-7|=7$∣α+β−αβ∣=∣−3−1−3∣=∣−7∣=7