The system `x+4y-2z=3,3x+y+5z=7` and `2x+3y+z=5` has

Options

A. infinite number of solutions
B. unique solution
C. trivial solution
D. no solution

Answer

To determine the nature of the solutions for the given system of equations, we will analyze the system using determinants. The equations are:

1. \( x + 4y - 2z = 3 \)
2. \( 3x + y + 5z = 7 \)
3. \( 2x + 3y + z = 5 \)

### Step 1: Form the Coefficient Matrix and Calculate the Determinant (Δ)

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

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

Now we calculate the determinant \( \Delta \) of matrix \( A \):

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

Using the determinant formula for a 3x3 matrix:

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

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 1 & 5 \\ 3 & 1 \end{vmatrix} = 1 \cdot 1 - 5 \cdot 3 = 1 - 15 = -14 \)

2. \( \begin{vmatrix} 3 & 5 \\ 2 & 1 \end{vmatrix} = 3 \cdot 1 - 5 \cdot 2 = 3 - 10 = -7 \)

3. \( \begin{vmatrix} 3 & 1 \\ 2 & 3 \end{vmatrix} = 3 \cdot 3 - 1 \cdot 2 = 9 - 2 = 7 \)

Now substituting back into the determinant calculation:

\[ \Delta = 1(-14) - 4(-7) - 2(7) \]
\[ = -14 + 28 - 14 = 0 \]

### Step 2: Analyze the Determinant

Since \( \Delta = 0 \), the system of equations may either have no solution or infinitely many solutions. We need to check the consistency of the equations.

### Step 3: Calculate the Determinants for Consistency (Δ1, Δ2, Δ3)

Next, we will calculate the determinants \( \Delta_1, \Delta_2, \Delta_3 \) to check for consistency.

1. **For \( \Delta_1 \)** (replace the first column with the constants):

\[ \Delta_1 = \begin{vmatrix} 3 & 4 & -2 \\ 7 & 1 & 5 \\ 5 & 3 & 1 \end{vmatrix} \]

Calculating \( \Delta_1 \):

\[ = 3 \begin{vmatrix} 1 & 5 \\ 3 & 1 \end{vmatrix} - 4 \begin{vmatrix} 7 & 5 \\ 5 & 1 \end{vmatrix} - 2 \begin{vmatrix} 7 & 1 \\ 5 & 3 \end{vmatrix} \]

Calculating the 2x2 determinants:

- \( \begin{vmatrix} 1 & 5 \\ 3 & 1 \end{vmatrix} = -14 \)

- \( \begin{vmatrix} 7 & 5 \\ 5 & 1 \end{vmatrix} = 7 \cdot 1 - 5 \cdot 5 = 7 - 25 = -18 \)

- \( \begin{vmatrix} 7 & 1 \\ 5 & 3 \end{vmatrix} = 7 \cdot 3 - 1 \cdot 5 = 21 - 5 = 16 \)

Substituting back:

\[ \Delta_1 = 3(-14) - 4(-18) - 2(16) \]
\[ = -42 + 72 - 32 = -2 \quad (\Delta_1 \neq 0) \]

### Conclusion

Since \( \Delta = 0 \) and \( \Delta_1 \neq 0 \), the system of equations has **no solution**.

### Final Answer

The system of equations has **no solution**.

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