Let λ be a real number for which the system of linear equations x+y+z=6, 4x+λ y-λ z=λ -2 and 3x+2y-4z=-5 has infinitely many solutions. Then λ is a…

The system of 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 a infinitely many solution, 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.

Thus, for infinitely many solutions, we have $D=0$D=0

$$\Rightarrow |\begin{matrix}1 & 1 & 1 \\ 4 & \lambda & -\lambda \\ 3 & 2 & -4\end{matrix}|=0$$⇒∣143​1λ2​1−λ−4​∣=0

$$\Rightarrow 1(-4\lambda +2\lambda )-1(-16+3\lambda )+1(8-3\lambda )=0$$⇒1(−4λ+2λ)−1(−16+3λ)+1(8−3λ)=0

$$\Rightarrow -2\lambda +16-3\lambda +8-3\lambda =0$$⇒−2λ+16−3λ+8−3λ=0

$$\Rightarrow 24-3\lambda =0$$⇒24−3λ=0

$$\Rightarrow \lambda =3$$⇒λ=3

And, $\lambda =3$λ=3 satisfies the equation ${\lambda }^{2}-\lambda -6=0.$λ2−λ−6=0.