Question

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

• A. 21
• B. 24
• C. 20
• D. 22

Hint:

A system of linear equations has a unique solution if and only if the determinant of its coefficient matrix is non-zero.

Answer 1

The Correct Option is A

Step 1: Write the coefficient matrix of the system.
\[ A = \begin{pmatrix} 1 & -n & 1 \\ 1 & n-2 & n+1 \\ 0 & n-1 & 1 \end{pmatrix} \]
Step 2: Find the determinant of the coefficient matrix.
\[ |A| = \begin{vmatrix} 1 & -n & 1\\ 1 & n-2 & n+1 \\ 0 & n-1 & 1 \end{vmatrix} \] \[ |A| = 1[(n-2)(1)-(n+1)(n-1)] + n[1-(n+1)0] + 1[(n-1)] \] \[ |A| = -n^2 + 2n - 1 \]
Step 3: Condition for unique solution.
For a unique solution, \[ |A| \neq 0 \Rightarrow -n^2 + 2n - 1 \neq 0 \] \[ (n-1)^2 \neq 0 \Rightarrow n \neq 1 \]
Step 4: Find probability and required sum.
Possible values of \(n\) on a die are \(1,2,3,4,5,6\). Unique solution exists for all values except \(n=1\).
\[ \text{Favourable outcomes} = 5 \Rightarrow \text{Probability} = \frac{5}{6} \] Thus \(k=5\) and sum of all possible values of \(n\) giving unique solution: \[ 2+3+4+5+6 = 20 \] \[ k + \text{sum} = 5 + 16 = 21 \]
Final Answer: \[ \boxed{21} \]