If the system of equations :
$x+y-z=0$;
$3x-\alpha y-3z=0$;
$x-3y+z=0$
has non-zero solution, then $\alpha=$

Q: If the system of equations : $x+y-z=0$; $3x-\alpha y-3z=0$; $x-3y+z=0$ has non-zero solution, then $\alpha=$

To determine the value of $\alpha$ for which the given system of equations has a non-zero solution, we need to analyze the system: 1. $x + y - z = 0$ 2. $3x - \alpha y - 3z = 0$ 3. $x - 3y + z = 0$ For the system to have a non-zero solution, the determinant of the coefficient matrix must be zero. Let's write the coefficient matrix: \[ A = \begin{bmatrix} 1 & 1 & -1 \\ 3 & -\alpha & -3 \\ 1 & -3 & 1 \end{bmatrix} \] The determinant of matrix $A$ is calculated as follows: \[ \text{det}(A) = 1 \left( (-\alpha)(1) - (-3)(-3) \right) - 1 \left( 3(1) - (-3)(1) \right) - 1 \left( 3(-3) - (-\alpha)(1) \right) \] Simplifying each term: - First term: $1 \times (-\alpha - 9) = -\alpha - 9$ - Second term: $-1 \times (3 + 3) = -6$ - Third term: $-1 \times (-9 + \alpha) = 9 - \alpha$ Putting it all together: \[ \text{det}(A) = -\alpha - 9 - 6 + 9 - \alpha = -2\alpha - 6 \] For the system to have a non-zero solution, $\text{det}(A)$ must be zero: \[ -2\alpha - 6 = 0 \] Solving for $\alpha$: \[ -2\alpha = 6 \\ \alpha = -3 \] Thus, the correct value of $\alpha$ is $-3$, which corresponds to Option 4. **Explanation for Other Options:** - **Option 1 (-1):** If $\alpha = -1$, substituting into the determinant calculation gives $-2(-1) - 6 = 2 - 6 = -4$, which is not zero. Therefore, the system does not have a non-zero solution for $\alpha = -1$. - **Option 2 (0):** If $\alpha = 0$, substituting into the determinant calculation gives $-2(0) - 6 = -6$, which is not zero. Therefore, the system does not have a non-zero solution for $\alpha = 0$. - **Option 3 (1):** If $\alpha = 1$, substituting into the determinant calculation gives $-2(1) - 6 = -2 - 6 = -8$, which is not zero. Therefore, the system does not have a non-zero solution for $\alpha = 1$. In conclusion, only Option 4 ($\alpha = -3$) results in a determinant of zero, allowing for a non-zero solution to the system of equations.

Question

If the system of equations :
 $x+y-z=0$; 
$3x-\alpha y-3z=0$; 
$x-3y+z=0$ 
has non-zero solution, then $\alpha=$

Accepted Answer

To determine the value of $\alpha$ for which the given system of equations has a non-zero solution, we need to analyze the system:

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

For the system to have a non-zero solution, the determinant of the coefficient matrix must be zero. Let's write the coefficient matrix:

$$
A = \begin{bmatrix}
1 & 1 & -1 \
3 & -\alpha & -3 \
1 & -3 & 1
\end{bmatrix}
$$

The determinant of matrix $A$ is calculated as follows:

$$
\text{det}(A) = 1 \left( (-\alpha)(1) - (-3)(-3) \right) - 1 \left( 3(1) - (-3)(1) \right) - 1 \left( 3(-3) - (-\alpha)(1) \right)
$$

Simplifying each term:

- First term: $1 \times (-\alpha - 9) = -\alpha - 9$
- Second term: $-1 \times (3 + 3) = -6$
- Third term: $-1 \times (-9 + \alpha) = 9 - \alpha$

Putting it all together:

$$
\text{det}(A) = -\alpha - 9 - 6 + 9 - \alpha = -2\alpha - 6
$$

For the system to have a non-zero solution, $\text{det}(A)$ must be zero:

$$
-2\alpha - 6 = 0
$$

Solving for $\alpha$:

$$
-2\alpha = 6 \
\alpha = -3
$$

Thus, the correct value of $\alpha$ is $-3$, which corresponds to Option 4.

**Explanation for Other Options:**

- **Option 1 (-1):** If $\alpha = -1$, substituting into the determinant calculation gives $-2(-1) - 6 = 2 - 6 = -4$, which is not zero. Therefore, the system does not have a non-zero solution for $\alpha = -1$.

- **Option 2 (0):** If $\alpha = 0$, substituting into the determinant calculation gives $-2(0) - 6 = -6$, which is not zero. Therefore, the system does not have a non-zero solution for $\alpha = 0$.

- **Option 3 (1):** If $\alpha = 1$, substituting into the determinant calculation gives $-2(1) - 6 = -2 - 6 = -8$, which is not zero. Therefore, the system does not have a non-zero solution for $\alpha = 1$.

In conclusion, only Option 4 ($\alpha = -3$) results in a determinant of zero, allowing for a non-zero solution to the system of equations.