Classify the following system of equations as consistent or inconsistent. If consistent, then solve them :
`x-y+3z=6`, `x+3y-3z=-4`, `5x+3y+3z=14`

Answer

To classify the given system of equations as consistent or inconsistent, and to solve them if they are consistent, we will follow these steps:

### Given Equations:
1. \( x - y + 3z = 6 \) (Equation 1)
2. \( x + 3y - 3z = -4 \) (Equation 2)
3. \( 5x + 3y + 3z = 14 \) (Equation 3)

### Step 1: Write the system in matrix form
We can express the system of equations in the form \( Ax = b \), where \( A \) is the coefficient matrix, \( x \) is the variable matrix, and \( b \) is the constant matrix.

\[ A = \begin{bmatrix} 1 & -1 & 3 \\ 1 & 3 & -3 \\ 5 & 3 & 3 \end{bmatrix}, \quad x = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad b = \begin{bmatrix} 6 \\ -4 \\ 14 \end{bmatrix} \]

### Step 2: Calculate the determinant of matrix \( A \)
To determine if the system is consistent, we need to calculate the determinant of the coefficient matrix \( A \).

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

Using the formula for the determinant of a 3x3 matrix:

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

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 3 & -3 \\ 3 & 3 \end{vmatrix} = (3)(3) - (-3)(3) = 9 + 9 = 18 \)

2. \( \begin{vmatrix} 1 & -3 \\ 5 & 3 \end{vmatrix} = (1)(3) - (-3)(5) = 3 + 15 = 18 \)

3. \( \begin{vmatrix} 1 & 3 \\ 5 & 3 \end{vmatrix} = (1)(3) - (3)(5) = 3 - 15 = -12 \)

Now substituting back into the determinant calculation:

\[ \text{det}(A) = 1 \cdot 18 + 1 \cdot 18 + 3 \cdot (-12) = 18 + 18 - 36 = 0 \]

### Step 3: Analyze the determinant
Since \( \text{det}(A) = 0 \), the system may have either no solution or infinitely many solutions. We will proceed to check the consistency by using row operations.

### Step 4: Form the augmented matrix and perform row operations
The augmented matrix \([A|b]\) is:

\[ \begin{bmatrix} 1 & -1 & 3 & | & 6 \\ 1 & 3 & -3 & | & -4 \\ 5 & 3 & 3 & | & 14 \end{bmatrix} \]

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

1. Subtract Row 1 from Row 2:
\[ R_2 \leftarrow R_2 - R_1 \Rightarrow \begin{bmatrix} 1 & -1 & 3 & | & 6 \\ 0 & 4 & -6 & | & -10 \\ 5 & 3 & 3 & | & 14 \end{bmatrix} \]

2. Subtract 5 times Row 1 from Row 3:
\[ R_3 \leftarrow R_3 - 5R_1 \Rightarrow \begin{bmatrix} 1 & -1 & 3 & | & 6 \\ 0 & 4 & -6 & | & -10 \\ 0 & 8 & -12 & | & -16 \end{bmatrix} \]

3. Now, simplify Row 3 by subtracting 2 times Row 2:
\[ R_3 \leftarrow R_3 - 2R_2 \Rightarrow \begin{bmatrix} 1 & -1 & 3 & | & 6 \\ 0 & 4 & -6 & | & -10 \\ 0 & 0 & 0 & | & 4 \end{bmatrix} \]

### Step 5: Analyze the final row echelon form
The last row \( [0 \; 0 \; 0 | 4] \) indicates that we have a contradiction (0 cannot equal 4). Therefore, the system of equations is inconsistent.

### Conclusion
The given system of equations is inconsistent, meaning there is no solution.