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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 $\gamma=c$?



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 $\gamma=c$?

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Shiyu, the idea of a Q&A site is that the questions are (single) questions and answers are (single) answers. So please do not include answers or new questions in your original question. If you think you have found an answer to your own question, you can simply post it as an answer below and accept it. Then you can start a new thread with the new question. – TMM Apr 22 '12 at 16:00

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.

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Do you mean it is possible $\alpha=a,\beta=b,\gamma\ne c$? But I think if $\alpha=a,\beta=b$, then we have $\gamma=c$. $$ \left[ \begin{array}{ccc} a & b & c \\ a & b & c \\ a & b & \gamma \\ \end{array} \right] \left[ \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ \end{array} \right]=0 $$ I think from the above equation, it is clear $\gamma=c$ since $x_3\ne 0$. – Shiyu Apr 22 '12 at 14:34

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, $\beta=b,\ \gamma=c$.

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There's a mistake in your conclusion from (5). In (5) you have $(\beta - b) x_2 + (c - \gamma) x_3 = 0$ while in (7) you have the term $(b - \beta) x_2 + (c - \gamma) x_3 = -(\beta - b) x_2 + (c - \gamma) x_3$. Note the minus sign. – TMM Apr 22 '12 at 14:38
Thanks TMM, I didn't notice that. Thanks anyway, Kunjan. – Shiyu Apr 22 '12 at 15:14

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