The system of equation `102x-95y+88z=81,3x+10y+17z=24, 57x+50y+43z=36` has

Options

A. Many solutions
B. No solution
C. a unique solution
D. none of these

Answer

To determine the nature of the solutions for the given system of equations:

1. **Write the equations**:
\[ 102x - 95y + 88z = 81 \quad \text{(1)} \]
\[ 3x + 10y + 17z = 24 \quad \text{(2)} \]
\[ 57x + 50y + 43z = 36 \quad \text{(3)} \]

2. **Set up the determinant** \(D\) using the coefficients of \(x\), \(y\), and \(z\):
\[ D = \begin{vmatrix} 102 & -95 & 88 \\ 3 & 10 & 17 \\ 57 & 50 & 43 \end{vmatrix} \]

3. **Calculate the determinant** \(D\):
- Expand the determinant using the first row:
\[ D = 102 \begin{vmatrix} 10 & 17 \\ 50 & 43 \end{vmatrix} - (-95) \begin{vmatrix} 3 & 17 \\ 57 & 43 \end{vmatrix} + 88 \begin{vmatrix} 3 & 10 \\ 57 & 50 \end{vmatrix} \]

4. **Calculate the minors**:
- For the first minor:
\[ \begin{vmatrix} 10 & 17 \\ 50 & 43 \end{vmatrix} = (10 \cdot 43) - (17 \cdot 50) = 430 - 850 = -420 \]
- For the second minor:
\[ \begin{vmatrix} 3 & 17 \\ 57 & 43 \end{vmatrix} = (3 \cdot 43) - (17 \cdot 57) = 129 - 969 = -840 \]
- For the third minor:
\[ \begin{vmatrix} 3 & 10 \\ 57 & 50 \end{vmatrix} = (3 \cdot 50) - (10 \cdot 57) = 150 - 570 = -420 \]

5. **Substituting back into the determinant**:
\[ D = 102(-420) + 95(-840) + 88(-420) \]
\[ D = -42840 - 79800 - 36960 \]
\[ D = -159600 \]

6. **Determine the nature of the solutions**:
- Since \(D \neq 0\), the system has a **unique solution**.

### Final Answer:
The system of equations has a **unique solution**.