Find coefficient in linear equation system for non-trivial solution (easy question)

Question

Find the real value of $\alpha$ so that the system admits solution different from (0, 0, 0).

$\begin{cases} \alpha x + y = 0 \\ \alpha y + z = 0 \\ 8x + \alpha z = 0 \end{cases}$

a) 8 b) 2 c) 1 d) -2 e) -4

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I constructed the determinant and made sure it was non-zero.

$\delta = \begin{vmatrix} \alpha & 1 & 0 \\ 0 & \alpha & 1 \\ 8 & 0 & \alpha \\ \end{vmatrix} \neq 0 \iff \alpha^3 + 8 \neq 0 \iff \alpha \neq -2$

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The correct answer is (d). But, either I'm completely mistaken or that is the only value $\alpha$ could not have. So, I deduce that the question wording is incorrect. Maybe it should be "find the value (...) so that the system admits solution only equal to (0, 0, 0)"?

Answer 1

The equations always have the trivial solution $(0,0,0)$, the idea of the question is to find those values of $\alpha$ s.t. there are non-zero solutions.

If you write your system of equations in the form

$A \begin{bmatrix} x\\ y\\z \end{bmatrix} = \begin{bmatrix} \alpha & 1 & 0\\ 0 & \alpha & 1 \\ 8 & 0 & \alpha \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 0\\ 0\\0 \end{bmatrix}$

Then the matrix $A$ has an inverse unless $\det A =0$. If the matrix $A$ has an inverse (and $\det A \ne 0$) then the solution is $ \begin{bmatrix} x\\ y\\z \end{bmatrix} = A^{-1} \begin{bmatrix} 0\\ 0\\0 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\0 \end{bmatrix}$

So for non trivial solutions you need that $A$ isn't invertible.

Answer 2

Oh, I see.

The system is homogeneous -- all the independent coefficients are zero. So, surely exists the trivial solution: (0, 0, 0).

By Cramer, $\delta \neq 0 \iff \text{there's one and only one determined solution}$. In this case, it would be the trivial one.

For there to be other solutions beyond the trivial, we need $\delta = 0$! :)

Answer 3

If you have any other value of $\alpha \neq -2$, you system cannot be reduced, following 3 equations for 3 variables.

If, however, $\alpha=-2$, you no longer have 3 linear independent equations. This extends the set of solutions from $$\mathcal{S}_{\alpha \neq -2} = \{(x,y,z) \in \mathbb{R}:(x,y,z)=(0,0,0)\}$$ to $$\mathcal{S}_{\alpha \neq -2}=\{(x,y,z) \in \mathbb{R}: (x,y,z) = (0,0,0)+(1,2,4)t: t\in\mathbb{R}\}.$$

One can easily see that $(0,0,0)\in \mathcal{S}_{\alpha \neq -2}$.