Classify the following system of equations as consistent or inconsistent. If consistent, then solve them :
`2x+5y-z=9`, `3x-3y+2z=7`, `2x-4y+3z=1`

Answer

To classify the given system of equations and solve it, we will follow these steps:

### Given Equations:
1. \( 2x + 5y - z = 9 \) (Equation 1)
2. \( 3x - 3y + 2z = 7 \) (Equation 2)
3. \( 2x - 4y + 3z = 1 \) (Equation 3)

### Step 1: Form the Coefficient Matrix and Constant Matrix
The coefficient matrix \( A \) and the constant matrix \( B \) can be represented as follows:

\[ A = \begin{bmatrix} 2 & 5 & -1 \\ 3 & -3 & 2 \\ 2 & -4 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} 9 \\ 7 \\ 1 \end{bmatrix} \]

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

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

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} -3 & 2 \\ -4 & 3 \end{vmatrix} = (-3)(3) - (2)(-4) = -9 + 8 = -1 \)

2. \( \begin{vmatrix} 3 & 2 \\ 2 & 3 \end{vmatrix} = (3)(3) - (2)(2) = 9 - 4 = 5 \)

3. \( \begin{vmatrix} 3 & -3 \\ 2 & -4 \end{vmatrix} = (3)(-4) - (-3)(2) = -12 + 6 = -6 \)

Now substituting back into the determinant calculation:

\[ \text{det}(A) = 2(-1) - 5(5) - 1(-6) = -2 - 25 + 6 = -21 \]

### Step 3: Classify the System
Since \( \text{det}(A) \neq 0 \), the system of equations is **consistent** and has a **unique solution**.

### Step 4: Find the Solution
To find the solution, we can use the inverse of matrix \( A \):

\[ X = A^{-1}B \]

#### Step 4.1: Find the Adjoint of Matrix \( A \)
To find the inverse, we first need the adjoint of \( A \). The adjoint is calculated using the cofactors.

\[ \text{adj}(A) = \begin{bmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix} \]

Where \( C_{ij} \) are the cofactors of \( A \).

Calculating the cofactors:

1. \( C_{11} = -1 \)
2. \( C_{12} = -5 \)
3. \( C_{13} = -6 \)
4. \( C_{21} = -11 \)
5. \( C_{22} = 18 \)
6. \( C_{23} = -7 \)
7. \( C_{31} = 7 \)
8. \( C_{32} = -7 \)
9. \( C_{33} = -21 \)

Thus, the adjoint matrix is:

\[ \text{adj}(A) = \begin{bmatrix} -1 & -5 & -6 \\ -11 & 18 & -7 \\ 7 & -7 & -21 \end{bmatrix} \]

#### Step 4.2: Calculate the Inverse of Matrix \( A \)
The inverse of \( A \) is given by:

\[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) = \frac{1}{-21} \begin{bmatrix} -1 & -5 & -6 \\ -11 & 18 & -7 \\ 7 & -7 & -21 \end{bmatrix} \]

#### Step 4.3: Multiply \( A^{-1} \) by \( B \)
Now we calculate \( X = A^{-1}B \):

\[ X = \frac{1}{-21} \begin{bmatrix} -1 & -5 & -6 \\ -11 & 18 & -7 \\ 7 & -7 & -21 \end{bmatrix} \begin{bmatrix} 9 \\ 7 \\ 1 \end{bmatrix} \]

Calculating the product:

1. First row: \( -1(9) + -5(7) + -6(1) = -9 - 35 - 6 = -50 \)
2. Second row: \( -11(9) + 18(7) + -7(1) = -99 + 126 - 7 = 20 \)
3. Third row: \( 7(9) + -7(7) + -21(1) = 63 - 49 - 21 = -7 \)

Thus,

\[ X = \frac{1}{-21} \begin{bmatrix} -50 \\ 20 \\ -7 \end{bmatrix} = \begin{bmatrix} \frac{50}{21} \\ -\frac{20}{21} \\ \frac{7}{21} \end{bmatrix} \]

### Final Solution
The solution to the system of equations is:

\[ x = \frac{50}{21}, \quad y = -\frac{20}{21}, \quad z = \frac{1}{3} \]