JEE MAIN - Mathematics (2021 - 22th July Evening Shift - No. 6)
The values of $$\lambda$$ and $$\mu$$ such that the system of equations $$x + y + z = 6$$, $$3x + 5y + 5z = 26$$, $$x + 2y + \lambda z = \mu $$ has no solution, are :
$$\lambda$$ = 3, $$\mu$$ = 5
$$\lambda$$ = 3, $$\mu$$ $$\ne$$ 10
$$\lambda$$ $$\ne$$ 2, $$\mu$$ = 10
$$\lambda$$ = 2, $$\mu$$ $$\ne$$ 10
Explanation
$$x + y + z = 6$$ ..... (i)
$$3x + 5y + 5z = 26$$ .... (ii)
$$x + 2y + \lambda z = \mu $$ ..... (iii)
$$5 \times (i) - (ii) \Rightarrow 2x = 4 \Rightarrow x = 2$$
$$\therefore$$ from (i) and (iii)
$$y + z = 4$$ ..... (iv)
$$2y + \lambda z = \mu - 2$$ .....(v)
$$(v) - 2 \times (iv)$$
$$ \Rightarrow (\lambda - 2)z = \mu - 10$$
$$ \Rightarrow z = {{\mu - 10} \over {\lambda - 2}}$$ & $$y = 4 - {{\mu - 10} \over {\lambda - 2}}$$
$$\therefore$$ For no solution $$\lambda$$ = 2 and $$\mu$$ $$\ne$$ 10.
$$3x + 5y + 5z = 26$$ .... (ii)
$$x + 2y + \lambda z = \mu $$ ..... (iii)
$$5 \times (i) - (ii) \Rightarrow 2x = 4 \Rightarrow x = 2$$
$$\therefore$$ from (i) and (iii)
$$y + z = 4$$ ..... (iv)
$$2y + \lambda z = \mu - 2$$ .....(v)
$$(v) - 2 \times (iv)$$
$$ \Rightarrow (\lambda - 2)z = \mu - 10$$
$$ \Rightarrow z = {{\mu - 10} \over {\lambda - 2}}$$ & $$y = 4 - {{\mu - 10} \over {\lambda - 2}}$$
$$\therefore$$ For no solution $$\lambda$$ = 2 and $$\mu$$ $$\ne$$ 10.
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