JEE MAIN - Mathematics (2023 - 24th January Morning Shift - No. 2)
Let N denote the number that turns up when a fair die is rolled. If the probability that the system of equations
$$x + y + z = 1$$
$$2x + \mathrm{N}y + 2z = 2$$
$$3x + 3y + \mathrm{N}z = 3$$
has unique solution is $${k \over 6}$$, then the sum of value of k and all possible values of N is :
18
21
20
19
Explanation
For unique solution $\Delta \neq 0$
i.e. $\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & N & 2 \\ 3 & 3 & N\end{array}\right| \neq 0$
$\Rightarrow\left(N^{2}-6\right)-(2 N-6)+(6-3 N) \neq 0$
$\Rightarrow N^{2}-5 N+6 \neq 0$
$\therefore N \neq 2$ and $N \neq 3$
$\therefore $ Probability of not getting 2 or 3 in a throw of dice $=\frac{2}{3}$
As given $\frac{2}{3}=\frac{k}{6} \Rightarrow k=4$
$\therefore$ Required value $=1+4+5+6+4=20$
i.e. $\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & N & 2 \\ 3 & 3 & N\end{array}\right| \neq 0$
$\Rightarrow\left(N^{2}-6\right)-(2 N-6)+(6-3 N) \neq 0$
$\Rightarrow N^{2}-5 N+6 \neq 0$
$\therefore N \neq 2$ and $N \neq 3$
$\therefore $ Probability of not getting 2 or 3 in a throw of dice $=\frac{2}{3}$
As given $\frac{2}{3}=\frac{k}{6} \Rightarrow k=4$
$\therefore$ Required value $=1+4+5+6+4=20$
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