Can I claim $\alpha=a$, $\beta=b$ and $\gamma=c$ based on this equation?

Question

Let $a, b, c$ be positive constants. $x=[x_1,x_2,x_3]^T\in\mathbb{R}^3$ is unknown but I know $x_i\ne 0$ for all $i=1,2,3$. Now I have the following equation $$ \left[ \begin{array}{ccc} \alpha & b & c \\ a & \beta & c \\ a & b & \gamma \\ \end{array} \right] \left[ \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ \end{array} \right]=Ax=0 $$ where $\alpha, \beta, \gamma$ are unknown constants to be determined. Considering $x\ne 0$, can I claim $\alpha=a$, $\beta=b$ and $\gamma=c$?

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EDIT:

At the first glance, it is a very simple question. The following is a wrong solution I got in the first place.

Wrong answer: We can claim $\alpha=a$, $\beta=b$ and $\gamma=c$, because if there exists $x\ne0$ solving $Ax=0$, $A$ must be rank deficient. So the rows of $A$ are linearly dependent. Consider the first two rows, since the third entry is $c$, the first two rows are linearly dependent iff $\alpha=a$ and $\beta=b$. Then it is easy to show $\gamma=c$.

Correct answer: If there exists $x\ne0$ solving $Ax=0$, $A$ is definitely rank deficient. However, it is possible that any two rows of $A$ are linearly independent but three rows are linearly dependent. The following is an example: Let $a=2,b=2,c=3$ and $\alpha=1\ne a,\beta=1\ne b,\gamma=4\ne c$. Then $$ A= \left[ \begin{array}{ccc} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 2 & 2 & 4 \\ \end{array} \right] $$ There exists $x=[1,1,-1]^T$ solving $Ax=0$. Note any two rows of $A$ are linearly independent but $A$ is rank deficient.

Note the structure of $A$ is still very special. Just for future discussion, a new question is: In addition to the equation $Ax=0$, under what other conditions can we claim $\alpha=a$, $\beta=b$ and $\gamma=c$?

Answer 1

No, you cannot claim from that information that ALL of those equalities hold. If two of these equalities hold, then there will be an $x$ satisfying $x \neq 0$ and $Ax=0$, where A is your matrix.

Answer 2

We have $$ \alpha x_{1}+bx_{2}+cx_{3}=0\tag{1}$$ $$ax_{1}+\beta x_{2}+cx_{3}=0\tag{2}$$ and $$ ax_{1}+bx_{2}+\gamma x_{3}=0\tag{3}$$ Now $(1)-(2)$ gives $$(\alpha-a)x_{1}+(b-\beta)x_{2}=0\tag{4}$$ $(2)-(3)$ gives $$(\beta-b)x_{2}+(c-\gamma)x_{3}=0\tag{5}$$ $(1)-(3)$ gives $$ (\alpha-a)x_{1}+(c-\gamma)x_{3}=0\tag{6}$$ Now,$(4)+(6)$ gives $$ 2(\alpha-a)x_{1}+(b-\beta)x_{2}+(c-\gamma)x_{3}=0$$ $$\Rightarrow 2(\alpha-a)x_{1}=0(\because (5))$$ $$\Rightarrow \alpha=a (\because x\neq0)$$ Similarly you can get, $\beta=b,\ \gamma=c$.