The values of `alpha` for which the system of equations `alphax-3y+z=0, x+alphay+3z=1, 3x+y+5z=2`, does not have unique solution are

Options

A. `-1, (11)/(5)`
B. `-1, (-11)/(5)`
C. `1, (-11)/(5)`
D. `1, (11)/(5)`

Answer

To find the values of \( \alpha \) for which the system of equations does not have a unique solution, we need to analyze the determinant of the coefficient matrix. The system of equations is given as:

1. \( \alpha x - 3y + z = 0 \)
2. \( x + \alpha y + 3z = 1 \)
3. \( 3x + y + 5z = 2 \)

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

\[ A = \begin{bmatrix} \alpha & -3 & 1 \\ 1 & \alpha & 3 \\ 3 & 1 & 5 \end{bmatrix} \]

### Step 2: Calculate the Determinant of the Coefficient Matrix
To find the values of \( \alpha \) for which the system does not have a unique solution, we need to set the determinant of matrix \( A \) to zero:

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

### Step 3: Expand the Determinant
Using the cofactor expansion along the first row:

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

Calculating the 2x2 determinants:

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

2. \( \begin{vmatrix} 1 & 3 \\ 3 & 5 \end{vmatrix} = 1 \cdot 5 - 3 \cdot 3 = 5 - 9 = -4 \)

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

Putting it all together:

\[ \text{det}(A) = \alpha (5\alpha - 3) + 3(-4) + (1 - 3\alpha) \]

### Step 4: Simplify the Determinant
Now simplify the expression:

\[ \text{det}(A) = 5\alpha^2 - 3\alpha - 12 + 1 - 3\alpha \]
\[ = 5\alpha^2 - 6\alpha - 11 \]

### Step 5: Set the Determinant to Zero
To find the values of \( \alpha \) for which the determinant is zero:

\[ 5\alpha^2 - 6\alpha - 11 = 0 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 5, b = -6, c = -11 \):

\[ \alpha = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 5 \cdot (-11)}}{2 \cdot 5} \]
\[ = \frac{6 \pm \sqrt{36 + 220}}{10} \]
\[ = \frac{6 \pm \sqrt{256}}{10} \]
\[ = \frac{6 \pm 16}{10} \]

Calculating the two possible values:

1. \( \alpha = \frac{22}{10} = \frac{11}{5} \)
2. \( \alpha = \frac{-10}{10} = -1 \)

### Final Answer
The values of \( \alpha \) for which the system of equations does not have a unique solution are:

\[ \alpha = -1 \quad \text{and} \quad \alpha = \frac{11}{5} \]