QJEEM20Maeb80: JEE Main 2020 Mathematics - Class 12 JEEM20Maeb80 Solution: Let S be the set of all R for which the system of linear…

For no solution :

$\Delta$ = 0 and $\Delta$ ₁/ $\Delta$ ₂/ $\Delta$ ₃ $\ne$ 0

$\Delta$ = $$\left|\begin{array}{ccc}2& -1& 2\\ 1& -2& \lambda \\ 1& \lambda & 1\end{array}\right|$$ = 0

$\Rightarrow$ 2(–2 – $\lambda$ ²) + 1 (1 – $\lambda$ ) + 2( $\lambda$ + 2) = 0

$\Rightarrow$ –2 $\lambda$ ² + $\lambda$ + 1 = 0

$\Rightarrow$ $\lambda$ = 1, $- {1 \over 2}$

When $\lambda$ = 1

2x – y + 2z = 2 ...(1)

x – 2y + z = –4 ...(2)

x + y + z = 4 ...(3)

Adding (2) and (3), we get

2x – y + 2z = 0 (contradiction) hence no solution.

$\therefore$ $\lambda$ = 1 belongs to set S.

When $\lambda$ = $- {1 \over 2}$

2x – y + 2z = 2 ...(1)

x – 2y $- {1 \over 2}$ z = –4 ...(2)

x $- {1 \over 2}$ y + z = 4 ...(3)

(1) and (3) contradict each other, hence no solution.

$\therefore$ $\lambda$ = $- {1 \over 2}$ belongs to set S.