JEE Main 2019 — Mathematics Algebra

The system of linear equations ${a}_{1}x+{b}_{1}y+{c}_{1}z+{d}_{1}=0,{a}_{2}x+{b}_{2}y+{c}_{2}z+{d}_{2}=0$a1​x+b1​y+c1​z+d1​=0,a2​x+b2​y+c2​z+d2​=0 and ${a}_{3}x+{b}_{3}y+{c}_{3}z+{d}_{3}=0$a3​x+b3​y+c3​z+d3​=0 have infinitely many solutions, if

$$D=|\begin{matrix}{a}_{1} & {b}_{1} & {c}_{1} \\ {a}_{2} & {b}_{2} & {c}_{2} \\ {a}_{3} & {b}_{3} & {c}_{3}\end{matrix}|=0,{D}_{1}=|\begin{matrix}{d}_{1} & {b}_{1} & {c}_{1} \\ {d}_{2} & {b}_{2} & {c}_{2} \\ {d}_{3} & {b}_{3} & {c}_{3}\end{matrix}|=0,{D}_{2}=|\begin{matrix}{a}_{1} & {d}_{1} & {c}_{1} \\ {a}_{2} & {d}_{2} & {c}_{2} \\ {a}_{3} & {d}_{3} & {c}_{3}\end{matrix}|=0$$D=∣a1​a2​a3​​b1​b2​b3​​c1​c2​c3​​∣=0,D1​=∣d1​d2​d3​​b1​b2​b3​​c1​c2​c3​​∣=0,D2​=∣a1​a2​a3​​d1​d2​d3​​c1​c2​c3​​∣=0

and ${D}_{3}=|\begin{matrix}{a}_{1} & {b}_{1} & {d}_{1} \\ {a}_{2} & {b}_{2} & {d}_{2} \\ {a}_{3} & {b}_{3} & {d}_{3}\end{matrix}|=0.$D3​=∣a1​a2​a3​​b1​b2​b3​​d1​d2​d3​​∣=0.

Hence, for the system of equations $x+y+z=5,x+2y+3z=9$x+y+z=5,x+2y+3z=9 and $x+3y+\alpha z=\beta$x+3y+αz=β have infinitely many solution, if

$$D=|\begin{matrix}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \alpha \end{matrix}|=0$$D=∣111​123​13α​∣=0

$$\Rightarrow (2\alpha -9)-(\alpha -3)+(3-2)$$⇒(2α−9)−(α−3)+(3−2)

$$\Rightarrow 2\alpha -9-\alpha +3+1=0$$⇒2α−9−α+3+1=0

$$\Rightarrow \alpha =5.$$⇒α=5.

And, ${D}_{1}=|\begin{matrix}5 & 1 & 1 \\ 9 & 2 & 3 \\ \beta & 3 & \alpha \end{matrix}|=0$D1​=∣59β​123​13α​∣=0

$$\Rightarrow |\begin{matrix}5 & 1 & 1 \\ 9 & 2 & 3 \\ \beta & 3 & 5\end{matrix}|=0$$⇒∣59β​123​135​∣=0

$$\Rightarrow 5(10-9)-(45-3\beta )+(27-2\beta )=0$$⇒5(10−9)−(45−3β)+(27−2β)=0

$$\Rightarrow 5-45+3\beta +27-2\beta =0$$⇒5−45+3β+27−2β=0

$$\Rightarrow \beta =13.$$⇒β=13.

Hence, $\beta -\alpha =13-5=8.$β−α=13−5=8.