Let n be the number obtained on rolling a fair die. If the probability that the system has a unique solution is k 6 , then the sum of k and all possible values of n is: [2026]

Question

Let n be the number obtained on rolling a fair die. If the probability that the system

$\begin{matrix} x - n y + z = 6 \\ x + (n - 2) y + (n + 1) z = 8 \\ (n - 1) y + z = 1 \end{matrix}$

Has a unique solution is $\frac{k}{6}$ , then the sum of k and all possible values of n is: [2026]

Smart Achievers · Accepted Answer

(4)

$x - n y + z = 6$

$x + (n - 2) y + (n + 1) z = 8$

$(n - 1) y + z = 1$

$| \begin{matrix} 1 & - n & 1 \\ 1 & (n - 2) & n + 1 \\ 0 & n - 1 & 1 \end{matrix} |$ $\neq 0$

$\Rightarrow n^{2} - 3 n + 2 \neq 0$

$n \neq 1, 2$

$For unique solution n = 3, 4, 5, 6$

$Now P (probability when system of equations has unique solution) = \frac{4}{6}$

$So k = 4$

$Now required sum = 4 + (3 + 4 + 5 + 6) = 22$

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