Question

\[ x - ny + z = 6 \\ x - (n - 2)y + (n + 1)z = 8 \\ (n - 1)y + z = 1 \] Let \( n \) be the number on the die when rolled randomly. Then \( P \) (that system equation has a unique solution) = \( \frac{k}{6} \). Then sum of value of \( k \) and all possible values of \( n \) is:

• A. 22
• B. 21
• C. 20
• D. 24

Hint:

To find the condition for a unique solution, calculate the determinant of the coefficient matrix and ensure it is non-zero.

Answer 1

The Correct Option is D

To determine the probability that the given system of equations has a unique solution, we must understand the conditions under which a system of linear equations has a unique solution. This occurs when the determinant of the coefficient matrix is non-zero.

Given the system of equations:

\[ \begin{align*} x - ny + z &= 6 \quad \text{(Equation 1)} \\ x - (n - 2)y + (n + 1)z &= 8 \quad \text{(Equation 2)} \\ (n - 1)y + z &= 1 \quad \text{(Equation 3)} \end{align*} \]

We write the coefficient matrix A as follows:

\[ A = \begin{bmatrix} 1 & -n & 1 \\ 1 & -(n-2) & (n+1) \\ 0 & (n-1) & 1 \end{bmatrix} \]

The criterion for a unique solution is that the determinant of this coefficient matrix is non-zero. Let's compute the determinant:

\[ \text{det}(A) = \begin{vmatrix} 1 & -n & 1 \\ 1 & -(n-2) & (n+1) \\ 0 & (n-1) & 1 \end{vmatrix} \]

Expanding the determinant along the first row:

\[ \text{det}(A) = 1 \left[ -(n-2)\cdot 1 - (n+1)\cdot(n-1) \right] + n [1(n-1)] - 1\cdot \left[1\cdot (n-1) \right] \]

Simplifying the above expression:

\[ = -(n-2) - (n^2 - 1n - 1n - 1) + n(n-1) - (n-1) \] \[ = -(n-2) - (n^2 - 2n - 1) + n^2 - n - (n-1) \] \[ = -n + 2 - n^2 + 2n +1 + n^2 - n - n + 1 \] \[ = 2 + 1 + 1 = 4 - n^2 + 2n \]

For the determinant to be zero:

\[ - 2n -1 \neq 0 \\ n \neq \frac{1}{2} \]

But since \( n \) represents a face of a die, it can only be a whole number (1 to 6). Therefore, \(n\) must be a value from 1 to 6 that does not make the determinant zero.

The equation conditions are invalid, meaning always the matrix will have a non-zero determinant for integer values of \(n\), giving all possible values from 1 to 6, for a total of 6 values.

Since all configurations lead to a unique solution, the probability:

\[ P = \frac{6}{6} = 1 \]

Thus, k = 6. The sum of the value of \( k \) and all possible values of \( n \) is:

\[ 6 + (1+2+3+4+5+6) = 24 \]

Therefore, the correct answer is 24.