Let $n$ be the number obtained on rolling a fair die. If the probability that the system
$x-ny+z=6$
$x+(n-2)y+(n+1)z=8$
$(n-1)y+z=1$
has a unique solution is $\dfrac{k}{6}$, then the sum of $k$ and all possible values of $n$ is:
Step-by-Step Solution
Key Concept: Unique solution $\Leftrightarrow\det(A)\neq0$. Compute $\det(A)=-(n-1)(n-2)$. This is zero when $n=1$ or $n=2$. For a fair die $n\in\{1,2,3,4,5,6\}$, unique solution for $n\in\{3,4,5,6\}$.
$k=4$, valid $n\in\{3,4,5,6\}$. Sum $=4+3+4+5+6=22$.
Correct Answer: 4