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(D) The system of equations has a unique solution if the determinant of the coefficient matrix $D 
eq 0$.$D = \begin{vmatrix} 1 & -n & 1 \ 1 & n-2 &

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(D) The system of equations has a unique solution if the determinant of the coefficient matrix $D 
eq 0$.$D = \begin{vmatrix} 1 & -n & 1 \ 1 & n-2 & n+1 \ 0 & n-1 & 1 \end{vmatrix}$Expanding along the first column:$D = 1((n-2)(1) - (n+1)(n-1)) - 1((-n)(1) - (1)(n-1)) + 0$$D = (n-2 - (n^2-1)) - (-n - n + 1)$$D = (n-2 - n^2 + 1) - (-2n + 1)$$D = -n^2 + n - 1 + 2n - 1 = -n^2 + 3n - 2$For a unique solution, $D 
eq 0$, so $-n^2 + 3n - 2 
eq 0$, which implies $n^2 - 3n + 2 
eq 0$.$(n-1)(n-2) 
eq 0$, so $n 
eq 1$ and $n 
eq 2$.Since $n$ is the outcome of a fair die, $n \in \{1, 2, 3, 4, 5, 6\}$.The values of $n$ for which the system has a unique solution are $n \in \{3, 4, 5, 6\}$.The number of such values is $4$, so the probability is $\frac{4}{6}$.Thus, $k = 4$.The sum of $k$ and all possible values of $n$ is $4 + (3 + 4 + 5 + 6) = 4 + 18 = 22$.

Let $n$ be the number obtained on rolling a fair die. If the probability that the system of equations
$x-ny+z=6$
$x+(n-2)y+(n+1)z=8$
$(n-1)y+z=1$
has a unique solution is $\frac{k}{6}$, then the sum of $k$ and all possible values of $n$ is:

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Let $n$ be the number obtained on rolling a fair die. If the probability that the system of equations$x-ny+z=6$$x+(n-2)y+(n+1)z=8$$(n-1)y+z=1$has a unique solution is $\frac{k}{6}$, then the sum of $k$ and all possible values of $n$ is: