If x+y-z=0, 3x-`alphay`-3z=0, x-3y+z=0 has non zero solution , then `alpha`=

Options

A. `-1`
B. 0
C. 1
D. `-3`

Answer

To solve the problem, we need to find the value of \( \alpha \) such that the system of equations has a non-zero solution. The given equations are:

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

### Step 1: Write the system of equations in matrix form
We can express the system of equations in matrix form \( A \mathbf{v} = 0 \), where \( A \) is the coefficient matrix and \( \mathbf{v} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \).

The coefficient matrix \( A \) is:
\[ A = \begin{bmatrix} 1 & 1 & -1 \\ 3 & -\alpha & -3 \\ 1 & -3 & 1 \end{bmatrix} \]

### Step 2: Set up the determinant of the coefficient matrix
For the system to have a non-zero solution, the determinant of the matrix \( A \) must be equal to zero:
\[ \text{det}(A) = 0 \]

### Step 3: Calculate the determinant
The determinant of matrix \( A \) can be calculated using the formula for the determinant of a 3x3 matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where:
- \( a = 1, b = 1, c = -1 \)
- \( d = 3, e = -\alpha, f = -3 \)
- \( g = 1, h = -3, i = 1 \)

Calculating the determinant:
\[ \text{det}(A) = 1((- \alpha)(1) - (-3)(-3)) - 1(3(1) - (-3)(1)) + (-1)(3(-3) - (-\alpha)(1)) \]

This simplifies to:
\[ = 1(-\alpha - 9) - 1(3 + 3) - (9 + \alpha) \]
\[ = -\alpha - 9 - 6 - 9 - \alpha \]
\[ = -2\alpha - 24 \]

### Step 4: Set the determinant to zero
Now, we set the determinant to zero:
\[ -2\alpha - 24 = 0 \]

### Step 5: Solve for \( \alpha \)
Solving for \( \alpha \):
\[ -2\alpha = 24 \]
\[ \alpha = -12 \]

Thus, the value of \( \alpha \) is:
\[ \alpha = -12 \]

### Final Answer
The value of \( \alpha \) is \( -12 \).

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