`4x + 3y + 2z =1`
`x-y + 3z = 4`
`2x + 5y - 4z = 6` has n solution than n =

Explanation

To determine the number of solutions \( n \) for the system of equations given by:

1. \( 4x + 3y + 2z = 1 \)
2. \( x - y + 3z = 4 \)
3. \( 2x + 5y - 4z = 6 \)

we will follow these steps:

### Step 1: Form the Coefficient Matrix and Calculate the Determinant

The coefficient matrix \( A \) is formed from the coefficients of the variables \( x, y, z \) in the equations:

\[ A = \begin{bmatrix} 4 & 3 & 2 \\ 1 & -1 & 3 \\ 2 & 5 & -4 \end{bmatrix} \]

Now, we will calculate the determinant of matrix \( A \).

\[ \text{det}(A) = 4 \begin{vmatrix} -1 & 3 \\ 5 & -4 \end{vmatrix} - 3 \begin{vmatrix} 1 & 3 \\ 2 & -4 \end{vmatrix} + 2 \begin{vmatrix} 1 & -1 \\ 2 & 5 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} -1 & 3 \\ 5 & -4 \end{vmatrix} = (-1)(-4) - (3)(5) = 4 - 15 = -11 \)

2. \( \begin{vmatrix} 1 & 3 \\ 2 & -4 \end{vmatrix} = (1)(-4) - (3)(2) = -4 - 6 = -10 \)

3. \( \begin{vmatrix} 1 & -1 \\ 2 & 5 \end{vmatrix} = (1)(5) - (-1)(2) = 5 + 2 = 7 \)

Now substituting these values back into the determinant calculation:

\[ \text{det}(A) = 4(-11) - 3(-10) + 2(7) \]
\[ = -44 + 30 + 14 = -44 + 44 = 0 \]

### Step 2: Analyze the Determinant

Since \( \text{det}(A) = 0 \), we have two cases to consider:

1. **No solution**: If the system is inconsistent.
2. **Infinitely many solutions**: If the system is dependent.

### Step 3: Calculate \( \Delta_1 \)

To determine whether we have no solution or infinitely many solutions, we will calculate \( \Delta_1 \) by replacing the first column of \( A \) with the constants from the right-hand side of the equations:

\[ \Delta_1 = \begin{vmatrix} 1 & 3 & 2 \\ 4 & -1 & 3 \\ 6 & 5 & -4 \end{vmatrix} \]

Calculating \( \Delta_1 \):

\[ \Delta_1 = 1 \begin{vmatrix} -1 & 3 \\ 5 & -4 \end{vmatrix} - 3 \begin{vmatrix} 4 & 3 \\ 6 & -4 \end{vmatrix} + 2 \begin{vmatrix} 4 & -1 \\ 6 & 5 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} -1 & 3 \\ 5 & -4 \end{vmatrix} = -11 \) (as calculated before)

2. \( \begin{vmatrix} 4 & 3 \\ 6 & -4 \end{vmatrix} = (4)(-4) - (3)(6) = -16 - 18 = -34 \)

3. \( \begin{vmatrix} 4 & -1 \\ 6 & 5 \end{vmatrix} = (4)(5) - (-1)(6) = 20 + 6 = 26 \)

Now substituting these values back into the determinant calculation:

\[ \Delta_1 = 1(-11) - 3(-34) + 2(26) \]
\[ = -11 + 102 + 52 = 143 \]

### Step 4: Conclusion

Since \( \Delta_1 \neq 0 \), we conclude that the system of equations has no solution.

Thus, the number of solutions \( n = 0 \).