The equations `2x-3y+6z=4, 5x+7y-14z=1 3x+2y-4z=0,` have

Options

A. unique solution
B. no solution
C. infinitely many solutions
D. None of the above

Explanation

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

1. **Write the system of equations**:
\[ \begin{align*} 2x - 3y + 6z &= 4 \quad \text{(1)} \\ 5x + 7y - 14z &= 1 \quad \text{(2)} \\ 3x + 2y - 4z &= 0 \quad \text{(3)} \end{align*} \]

2. **Form the coefficient matrix \(A\)** and the constant matrix \(B\)**:
\[ A = \begin{bmatrix} 2 & -3 & 6 \\ 5 & 7 & -14 \\ 3 & 2 & -4 \end{bmatrix}, \quad B = \begin{bmatrix} 4 \\ 1 \\ 0 \end{bmatrix} \]

3. **Calculate the determinant of matrix \(A\)**:
\[ \text{det}(A) = 2 \begin{vmatrix} 7 & -14 \\ 2 & -4 \end{vmatrix} - (-3) \begin{vmatrix} 5 & -14 \\ 3 & -4 \end{vmatrix} + 6 \begin{vmatrix} 5 & 7 \\ 3 & 2 \end{vmatrix} \]

- Calculate the minors:
\[ \begin{vmatrix} 7 & -14 \\ 2 & -4 \end{vmatrix} = (7)(-4) - (-14)(2) = -28 + 28 = 0 \]
\[ \begin{vmatrix} 5 & -14 \\ 3 & -4 \end{vmatrix} = (5)(-4) - (-14)(3) = -20 + 42 = 22 \]
\[ \begin{vmatrix} 5 & 7 \\ 3 & 2 \end{vmatrix} = (5)(2) - (7)(3) = 10 - 21 = -11 \]

- Substitute back into the determinant formula:
\[ \text{det}(A) = 2(0) + 3(22) + 6(-11) = 0 + 66 - 66 = 0 \]

4. **Interpret the determinant**:
Since \(\text{det}(A) = 0\), the system of equations is either inconsistent or has infinitely many solutions.

5. **Check for consistency**:
We can use the augmented matrix \([A|B]\) and perform row operations to check for consistency:
\[ \begin{bmatrix} 2 & -3 & 6 & | & 4 \\ 5 & 7 & -14 & | & 1 \\ 3 & 2 & -4 & | & 0 \end{bmatrix} \]

- Perform row operations to reduce this matrix to row echelon form.

6. **Row operations**:
- Multiply Row 1 by \( \frac{1}{2} \):
\[ \begin{bmatrix} 1 & -\frac{3}{2} & 3 & | & 2 \\ 5 & 7 & -14 & | & 1 \\ 3 & 2 & -4 & | & 0 \end{bmatrix} \]

- Replace Row 2 with Row 2 - 5 * Row 1:
\[ \begin{bmatrix} 1 & -\frac{3}{2} & 3 & | & 2 \\ 0 & \frac{43}{2} & -29 & | & -9 \\ 3 & 2 & -4 & | & 0 \end{bmatrix} \]

- Replace Row 3 with Row 3 - 3 * Row 1:
\[ \begin{bmatrix} 1 & -\frac{3}{2} & 3 & | & 2 \\ 0 & \frac{43}{2} & -29 & | & -9 \\ 0 & \frac{11}{2} & -13 & | & -6 \end{bmatrix} \]

7. **Continue row reduction** to check for contradictions or free variables.

8. **Conclusion**:
If there are no contradictions, the system has infinitely many solutions. If a row leads to a contradiction (like \(0 = k\) where \(k \neq 0\)), then the system is inconsistent.