If the system of equations `(.^(n)C_(3))x+(.^(n)C_(4))y+35z=0, (.^(n)C_(4))x+35y+(.^(n)C_(3))z=0 and 35x+(.^(n)C_(3))y+(.^(n)C_(4))z=0` has a non - trivial solution, then the value of n is equal to `(AA n in N, n ge 4)`

Options

A. 6
B. 7
C. 8
D. 9

Answer

To solve the problem, we need to determine the value of \( n \) such that the given system of equations has a non-trivial solution. This occurs when the determinant of the coefficients of the equations is equal to zero.

### Step 1: Write the system of equations in matrix form
The given equations are:
1. \( \binom{n}{3} x + \binom{n}{4} y + 35z = 0 \)
2. \( \binom{n}{4} x + 35y + \binom{n}{3} z = 0 \)
3. \( 35x + \binom{n}{3} y + \binom{n}{4} z = 0 \)

We can express this in matrix form as:
\[ \begin{bmatrix} \binom{n}{3} & \binom{n}{4} & 35 \\ \binom{n}{4} & 35 & \binom{n}{3} \\ 35 & \binom{n}{3} & \binom{n}{4} \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Set up the determinant
For the system to have a non-trivial solution, the determinant of the coefficient matrix must be zero:
\[ D = \begin{vmatrix} \binom{n}{3} & \binom{n}{4} & 35 \\ \binom{n}{4} & 35 & \binom{n}{3} \\ 35 & \binom{n}{3} & \binom{n}{4} \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
We can calculate the determinant using cofactor expansion. For simplicity, we will expand along the first row:
\[ D = \binom{n}{3} \begin{vmatrix} 35 & \binom{n}{3} \\ \binom{n}{3} & \binom{n}{4} \end{vmatrix} - \binom{n}{4} \begin{vmatrix} \binom{n}{4} & \binom{n}{3} \\ 35 & \binom{n}{4} \end{vmatrix} + 35 \begin{vmatrix} \binom{n}{4} & 35 \\ 35 & \binom{n}{3} \end{vmatrix} \]

Calculating these 2x2 determinants:
1. \( \begin{vmatrix} 35 & \binom{n}{3} \\ \binom{n}{3} & \binom{n}{4} \end{vmatrix} = 35 \cdot \binom{n}{4} - \binom{n}{3}^2 \)
2. \( \begin{vmatrix} \binom{n}{4} & \binom{n}{3} \\ 35 & \binom{n}{4} \end{vmatrix} = \binom{n}{4}^2 - 35 \cdot \binom{n}{3} \)
3. \( \begin{vmatrix} \binom{n}{4} & 35 \\ 35 & \binom{n}{3} \end{vmatrix} = \binom{n}{4} \cdot \binom{n}{3} - 35^2 \)

Substituting these back into the determinant expression gives:
\[ D = \binom{n}{3} (35 \cdot \binom{n}{4} - \binom{n}{3}^2) - \binom{n}{4} (\binom{n}{4}^2 - 35 \cdot \binom{n}{3}) + 35 (\binom{n}{4} \cdot \binom{n}{3} - 35^2) \]

### Step 4: Set the determinant to zero
Setting \( D = 0 \) leads to the equation:
\[ \binom{n}{3}^2 + \binom{n}{4}^2 + 35^2 - 3 \cdot 35 \cdot \binom{n}{3} \cdot \binom{n}{4} = 0 \]

### Step 5: Solve for \( n \)
This expression can be analyzed using the identity \( a^2 + b^2 + c^2 - 3abc = 0 \). This gives us two conditions:
1. \( \binom{n}{3} + \binom{n}{4} + 35 = 0 \) (not possible since they are positive)
2. \( \binom{n}{3} = \binom{n}{4} = 35 \)

From \( \binom{n}{3} = \binom{n}{4} \), we have:
- Either \( n = 3 + 4 = 7 \) or \( n = 3 \) or \( n = 4 \) (not valid since \( n \) must be greater than or equal to 4).

Thus, the only valid solution is:
\[ n = 7 \]

### Conclusion
The value of \( n \) is \( 7 \).