The system of lienar equations
`x-y+z=1`
`x+y-z=3`
`x-4y+4z=alpha` has

Options

A. a unique solution when `alpha=2`
B. a unique number when `alpha!=-2`
C. an infinite number of solutions when `alpha=2`
D. an infinite number of solutions, when `alpha=-2`

Answer

To solve the system of linear equations given by:

1. \( x - y + z = 1 \) (Equation 1)
2. \( x + y - z = 3 \) (Equation 2)
3. \( x - 4y + 4z = \alpha \) (Equation 3)

we will analyze the determinant of the coefficient matrix and find the conditions under which the system has a unique solution, infinite solutions, or no solution.

### Step 1: Form the Coefficient Matrix
The coefficient matrix \( A \) for the system of equations is:

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

### Step 2: Calculate the Determinant of Matrix \( A \)
To determine the nature of the solutions, we need to calculate the determinant of matrix \( A \):

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

Expanding the determinant along the first row:

\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 1 & -1 \\ -4 & 4 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 1 & -1 \\ 1 & 4 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 1 \\ 1 & -4 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 1 & -1 \\ -4 & 4 \end{vmatrix} = (1)(4) - (-1)(-4) = 4 - 4 = 0 \)

2. \( \begin{vmatrix} 1 & -1 \\ 1 & 4 \end{vmatrix} = (1)(4) - (-1)(1) = 4 + 1 = 5 \)

3. \( \begin{vmatrix} 1 & 1 \\ 1 & -4 \end{vmatrix} = (1)(-4) - (1)(1) = -4 - 1 = -5 \)

Substituting back into the determinant calculation:

\[ \text{det}(A) = 1 \cdot 0 + 1 \cdot 5 + 1 \cdot (-5) = 0 + 5 - 5 = 0 \]

### Step 3: Analyze the Determinant
Since \( \text{det}(A) = 0 \), the system does not have a unique solution. We need to check for conditions of infinite solutions or no solutions.

### Step 4: Formulate the Augmented Matrix
The augmented matrix for the system is:

\[ \begin{bmatrix} 1 & -1 & 1 & | & 1 \\ 1 & 1 & -1 & | & 3 \\ 1 & -4 & 4 & | & \alpha \end{bmatrix} \]

### Step 5: Calculate \( \Delta_1 \) for Infinite Solutions
To find the condition for infinite solutions, we calculate \( \Delta_1 \):

\[ \Delta_1 = \begin{vmatrix} 1 & -1 & 1 \\ 1 & 1 & -1 \\ 1 & -4 & \alpha \end{vmatrix} \]

Expanding this determinant:

\[ \Delta_1 = 1 \cdot \begin{vmatrix} 1 & -1 \\ -4 & \alpha \end{vmatrix} - (-1) \cdot \begin{vmatrix} 1 & -1 \\ 1 & \alpha \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 1 \\ 1 & -4 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 1 & -1 \\ -4 & \alpha \end{vmatrix} = (1)(\alpha) - (-1)(-4) = \alpha - 4 \)

2. \( \begin{vmatrix} 1 & -1 \\ 1 & \alpha \end{vmatrix} = (1)(\alpha) - (-1)(1) = \alpha + 1 \)

3. \( \begin{vmatrix} 1 & 1 \\ 1 & -4 \end{vmatrix} = (1)(-4) - (1)(1) = -4 - 1 = -5 \)

Substituting back into the determinant calculation:

\[ \Delta_1 = 1 \cdot (\alpha - 4) + 1 \cdot (\alpha + 1) + 1 \cdot (-5) = \alpha - 4 + \alpha + 1 - 5 = 2\alpha - 8 \]

Setting \( \Delta_1 = 0 \) for infinite solutions:

\[ 2\alpha - 8 = 0 \implies \alpha = 4 \]

### Step 6: Calculate \( \Delta_2 \) for No Solutions
Now we will calculate \( \Delta_2 \):

\[ \Delta_2 = \begin{vmatrix} 1 & -1 & 1 \\ 1 & 1 & 3 \\ 1 & -4 & 4 \end{vmatrix} \]

Expanding this determinant:

\[ \Delta_2 = 1 \cdot \begin{vmatrix} 1 & 3 \\ -4 & 4 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 1 \\ 1 & -4 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 1 & 3 \\ -4 & 4 \end{vmatrix} = (1)(4) - (3)(-4) = 4 + 12 = 16 \)

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

3. \( \begin{vmatrix} 1 & 1 \\ 1 & -4 \end{vmatrix} = -5 \) (as calculated earlier)

Substituting back into the determinant calculation:

\[ \Delta_2 = 1 \cdot 16 + 1 \cdot 1 + 1 \cdot (-5) = 16 + 1 - 5 = 12 \]

### Step 7: Conclusion
Since \( \text{det}(A) = 0 \) and \( \Delta_1 = 0 \) leads to \( \alpha = 4 \) for infinite solutions, and \( \Delta_2 \neq 0 \) indicates no solutions for other values of \( \alpha \), we conclude:

- For \( \alpha = 4 \), the system has infinite solutions.
- For \( \alpha \neq 4 \), the system has no solutions.

Thus, the conditions for the system of equations are:

- **Unique solution**: Not possible (determinant is zero).
- **Infinite solutions**: When \( \alpha = 4 \).
- **No solutions**: For any \( \alpha \) other than 4.