There is no solution for the equations
`x + 4y - 2z=3`
`3x + y + 5z = 7`
` 2x+3y +z=5 `. True or False.

Answer

To determine whether the given system of equations has no solution, 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: Formulate the Coefficient Matrix and the Constant Matrix

The coefficient matrix \( A \) and the constant matrix \( B \) can be represented as follows:

\[ A = \begin{bmatrix} 1 & 4 & -2 \\ 3 & 1 & 5 \\ 2 & 3 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 3 \\ 7 \\ 5 \end{bmatrix} \]

### Step 2: Calculate the Determinant of the Coefficient Matrix \( \Delta \)

To check for no solution, we calculate the determinant \( \Delta \) of the coefficient matrix \( A \):

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

Using the determinant formula for a 3x3 matrix, we expand along the first row:

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

Calculating the 2x2 determinants:

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

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

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

Now substituting back into the determinant calculation:

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

### Step 3: Calculate the Determinants \( \Delta_1, \Delta_2, \Delta_3 \)

Next, we need to calculate \( \Delta_1, \Delta_2, \Delta_3 \) to check for the conditions of no solution.

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

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

Expanding \( \Delta_1 \):

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

Calculating the 2x2 determinants:

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

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

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

Substituting back:

\[ \Delta_1 = 3(-14) - 4(-18) - 2(16) \]
\[ = -42 + 72 - 32 = -2 \quad (\text{non-zero}) \]

**For \( \Delta_2 \)** (replace the second column with constants):

\[ \Delta_2 = \begin{vmatrix} 1 & 3 & -2 \\ 3 & 7 & 5 \\ 2 & 5 & 1 \end{vmatrix} \]

Following similar steps as above, we will find that \( \Delta_2 \) is also non-zero.

**For \( \Delta_3 \)** (replace the third column with constants):

\[ \Delta_3 = \begin{vmatrix} 1 & 4 & 3 \\ 3 & 1 & 7 \\ 2 & 3 & 5 \end{vmatrix} \]

This will also yield a non-zero value.

### Step 4: Conclusion

Since \( \Delta = 0 \) and \( \Delta_1, \Delta_2, \Delta_3 \) are non-zero, the system of equations has no solution.

Thus, the statement "There is no solution for the equations" is **True**.