Let n be the number obtained on rolling a fair die.

Question

Let n be the number obtained on rolling a fair die. If the probability that the system
$$ \begin{aligned} & x-\mathrm{n} y+z=6 \ & x+(\mathrm{n}-2) y+(\mathrm{n}+1) z=8 \ & \quad(\mathrm{n}-1) y+z=1 \end{aligned} $$
has a unique solution ...

Accepted Answer

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

$\begin{vmatrix} 1 \quad  -n \quad  1 \ 1 \quad  (n-2) \quad  n+1 \ 0 \quad  n-1 \quad  1 \end{vmatrix}=0\Rightarrow n=1,2n=-1$

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 $=4+(3+4+5+6)=22$

Solutions · 1