$x - ny + z = 6$
$x + (n - 2)y + (n + 1)z = 8$
$(n - 1)y + z = 1$
has a unique solution is $\frac{k}{6}$, then the sum of k and all possible values of n is :
The problem asks for the sum of 'k' and 'all possible values of n' based on the condition for a unique solution to a system of linear equations and the probability related to rolling a fair die.
1. Analyze the System and Coefficient Matrix
The given system of linear equations is:
- $x - ny + z = 6$
- $x + (n - 2)y + (n + 1)z = 8$
- $(n - 1)y + z = 1$
The coefficient matrix, denoted by A, is:
$ A = \begin{pmatrix} 1 & -n & 1 \\ 1 & n-2 & n+1 \\ 0 & n-1 & 1 \end{pmatrix} $2. Condition for Unique Solution
A system of linear equations has a unique solution if the determinant of its coefficient matrix is non-zero.
We need to calculate $\det(A)$:
$ \det(A) = 1 \begin{vmatrix} n-2 & n+1 \\ n-1 & 1 \end{vmatrix} - (-n) \begin{vmatrix} 1 & n+1 \\ 0 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & n-2 \\ 0 & n-1 \end{vmatrix} $ $ \det(A) = 1((n-2) - (n+1)(n-1)) + n(1 - 0) + 1(n-1 - 0) $ $ \det(A) = (n-2) - (n^2 - 1) + n + (n-1) $ $ \det(A) = n - 2 - n^2 + 1 + n + n - 1 $ $ \det(A) = -n^2 + 3n - 2 $For a unique solution, $\det(A) \neq 0$. Let's find when $\det(A) = 0$:
$ -n^2 + 3n - 2 = 0 $ $ n^2 - 3n + 2 = 0 $ $ (n - 1)(n - 2) = 0 $The determinant is zero when $n=1$ or $n=2$. Therefore, the system has a unique solution when $n \neq 1$ and $n \neq 2$.
3. Probability of Unique Solution
The variable 'n' represents the outcome of rolling a fair die. The possible values for 'n' are $\{1, 2, 3, 4, 5, 6\}$. The total number of possible outcomes is 6.
The values of 'n' that result in a unique solution are those not equal to 1 or 2. These are $\{3, 4, 5, 6\}$. There are 4 such values.
The probability of the system having a unique solution is the ratio of favorable outcomes to the total outcomes:
$ P(\text{unique solution}) = \frac{\text{Number of values of n yielding unique solution}}{\text{Total possible values of n}} = \frac{4}{6} $The problem states this probability is $\frac{k}{6}$. Therefore:
$ \frac{k}{6} = \frac{4}{6} \implies k = 4 $4. Calculate the Required Sum
The question asks for the sum of 'k' and 'all possible values of n'. Based on the provided options, the calculation seems to follow this structure: $k + (\text{Sum of n values for unique solution}) - (\text{Number of n values for non-unique solution})$.
- We found $k = 4$.
- The values of 'n' for which a unique solution exists are $\{3, 4, 5, 6\}$. The sum is $3 + 4 + 5 + 6 = 18$.
- The values of 'n' for which a unique solution does not exist (i.e., $\det(A)=0$) are $\{1, 2\}$. The number of such values is 2.
Calculating the sum:
$ \text{Sum} = k + (\text{Sum of n for unique solution}) - (\text{Number of n for non-unique solution}) $ $ \text{Sum} = 4 + 18 - 2 $ $ \text{Sum} = 20 $Similar Questions
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