An ordered pair `(alpha, beta)` for which the system of linear equations `(1+alpha)x+betay+z=2`, `alphax+(1+beta)y+z=3` and `alphax+betay+2z=2` has unique solution is: (a) (2,4) (b) (-3,1) (c) (-4,2) (d) (1,-3)

Options

A. (1, -3)
B. (-3, 1)
C. (2, 4)
D. (-4, 2)

Answer

To find the ordered pair \((\alpha, \beta)\) for which the system of linear equations has a unique solution, we need to analyze the given equations:

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

### Step 1: Write the system in matrix form
We can express the system of equations in matrix form \(A\mathbf{x} = \mathbf{b}\), where:
\[ A = \begin{bmatrix} 1 + \alpha & \beta & 1 \\ \alpha & 1 + \beta & 1 \\ \alpha & \beta & 2 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 2 \\ 3 \\ 2 \end{bmatrix} \]

### Step 2: Find the determinant of the coefficient matrix
For the system to have a unique solution, the determinant of matrix \(A\) must be non-zero. We calculate the determinant \(D\) of matrix \(A\):
\[ D = \begin{vmatrix} 1 + \alpha & \beta & 1 \\ \alpha & 1 + \beta & 1 \\ \alpha & \beta & 2 \end{vmatrix} \]

### Step 3: Calculate the determinant
Using the determinant formula for a \(3 \times 3\) matrix:
\[ D = (1 + \alpha) \begin{vmatrix} 1 + \beta & 1 \\ \beta & 2 \end{vmatrix} - \beta \begin{vmatrix} \alpha & 1 \\ \alpha & 2 \end{vmatrix} + 1 \begin{vmatrix} \alpha & 1 + \beta \\ \alpha & \beta \end{vmatrix} \]

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

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

3. \(\begin{vmatrix} \alpha & 1 + \beta \\ \alpha & \beta \end{vmatrix} = \alpha \cdot \beta - \alpha \cdot (1 + \beta) = \alpha\beta - \alpha - \alpha\beta = -\alpha\)

Putting it all together:
\[ D = (1 + \alpha)(2 + \beta) - \beta \alpha - \alpha \]
Expanding this:
\[ D = 2 + \beta + 2\alpha + \alpha\beta - \beta\alpha - \alpha = 2 + \beta + \alpha \]

### Step 4: Set the determinant not equal to zero
For a unique solution, we need:
\[ D \neq 0 \implies 2 + \alpha + \beta \neq 0 \implies \alpha + \beta \neq -2 \]

### Step 5: Check the options
Now, we will check each option to see which one satisfies \(\alpha + \beta \neq -2\):

1. **Option (a)**: \((2, 4)\)
\(\alpha + \beta = 2 + 4 = 6 \neq -2\) (Valid)

2. **Option (b)**: \((-3, 1)\)
\(\alpha + \beta = -3 + 1 = -2\) (Invalid)

3. **Option (c)**: \((-4, 2)\)
\(\alpha + \beta = -4 + 2 = -2\) (Invalid)

4. **Option (d)**: \((1, -3)\)
\(\alpha + \beta = 1 - 3 = -2\) (Invalid)

### Conclusion
The only ordered pair \((\alpha, \beta)\) that results in a unique solution is:
**(2, 4)**