If the system of linear equations given by
x+y+z=3
2x+y-z=3
x+y-z=1
is consistent and if `(alpha ,beta,gamma)` is a solution then `2alpha+2beta+gamma` = ______

Answer

To solve the given system of linear equations and find the value of \(2\alpha + 2\beta + \gamma\), we can follow these steps:

### Step 1: Write the system of equations in matrix form
The given equations are:
1. \(x + y + z = 3\)
2. \(2x + y - z = 3\)
3. \(x + y - z = 1\)

We can represent this system in matrix form as \(AX = B\), where:
\[ A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 1 & 1 & -1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix} \]

### Step 2: Calculate the determinant of matrix \(A\)
To find the inverse of matrix \(A\), we first need to calculate its determinant:
\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 1 & -1 \\ 1 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & -1 \\ 1 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} \]

Calculating the minors:
1. \(\begin{vmatrix} 1 & -1 \\ 1 & -1 \end{vmatrix} = 1 \cdot (-1) - (-1) \cdot 1 = -1 + 1 = 0\)
2. \(\begin{vmatrix} 2 & -1 \\ 1 & -1 \end{vmatrix} = 2 \cdot (-1) - (-1) \cdot 1 = -2 + 1 = -1\)
3. \(\begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = 2 \cdot 1 - 1 \cdot 1 = 2 - 1 = 1\)

Now substituting back:
\[ \text{det}(A) = 1 \cdot 0 - 1 \cdot (-1) + 1 \cdot 1 = 0 + 1 + 1 = 2 \]

### Step 3: Calculate the adjoint of matrix \(A\)
The adjoint of matrix \(A\) is calculated using the cofactors:
\[ \text{adj}(A) = \begin{pmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{pmatrix} \]

Calculating the cofactors:
1. \(C_{11} = \begin{vmatrix} 1 & -1 \\ 1 & -1 \end{vmatrix} = 0\)
2. \(C_{12} = -\begin{vmatrix} 2 & -1 \\ 1 & -1 \end{vmatrix} = -(-1) = 1\)
3. \(C_{13} = \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = 1\)
4. \(C_{21} = -\begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -(-2) = 2\)
5. \(C_{22} = \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -2\)
6. \(C_{23} = -\begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = -(-1) = 1\)
7. \(C_{31} = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = -1\)
8. \(C_{32} = -\begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = -(-3) = 3\)
9. \(C_{33} = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = -1\)

Thus, the adjoint matrix is:
\[ \text{adj}(A) = \begin{pmatrix} 0 & 1 & 1 \\ 2 & -2 & 1 \\ -1 & 3 & -1 \end{pmatrix} \]

### Step 4: Calculate the inverse of matrix \(A\)
The inverse of matrix \(A\) is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{2} \begin{pmatrix} 0 & 1 & 1 \\ 2 & -2 & 1 \\ -1 & 3 & -1 \end{pmatrix} \]

### Step 5: Solve for \(X\)
Now, we can find \(X\) using:
\[ X = A^{-1}B \]
Calculating:
\[ X = \frac{1}{2} \begin{pmatrix} 0 & 1 & 1 \\ 2 & -2 & 1 \\ -1 & 3 & -1 \end{pmatrix} \begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix} \]

Calculating the product:
1. First row: \(0 \cdot 3 + 1 \cdot 3 + 1 \cdot 1 = 3 + 1 = 4\)
2. Second row: \(2 \cdot 3 - 2 \cdot 3 + 1 \cdot 1 = 6 - 6 + 1 = 1\)
3. Third row: \(-1 \cdot 3 + 3 \cdot 3 - 1 \cdot 1 = -3 + 9 - 1 = 5\)

Thus,
\[ X = \frac{1}{2} \begin{pmatrix} 4 \\ 1 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 \\ 0.5 \\ 2.5 \end{pmatrix} \]

### Step 6: Find \(2\alpha + 2\beta + \gamma\)
From the solution, we have:
\(\alpha = 2\), \(\beta = 0.5\), \(\gamma = 2.5\).

Calculating:
\[ 2\alpha + 2\beta + \gamma = 2(2) + 2(0.5) + 2.5 = 4 + 1 + 2.5 = 7.5 \]

### Final Answer
Thus, the value of \(2\alpha + 2\beta + \gamma\) is \(7.5\).

---