Find all constants where a matrix is symmetric I have a matrix like below:
$$M = \begin{bmatrix} 2 && a-2b+c && 2a+b+c \\ 3 && 5 && -2 \\ 0 && a+c && 7\end{bmatrix}$$
In order for the matrix to be symmetric, the following constraints on $a$, $b$, and $c$ must hold:
$$a-2b+c = 3 \\ 2a+b+c = 0 \\ a+c = -2$$
How would I find the domain of $a$, $b$, and $c$? The set of equations above can form a matrix, and solving the system would give an individual value for each, but this gives a lone example of values that work, how would I find the range?
Am I reading too much into the question? Are the set of equations I've provided the ranges?
 A: I guess that you are overthinking it. You've found that system correctly. In fact: $$\begin{cases} a-2b+c = 3 \\ 2a+b+c = 0 \\ a+c = -2 \end{cases} \implies \begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix}  = \begin{bmatrix} 3 \\ 0 \\ -2\end{bmatrix}, $$
and since the determinant of that matrix is non zero, we have an unique solution $(a,b,c)$ for that system. In other words, we have only one choice of $a,b$ and $c$ to make the initial matrix symmetric.
A: Hint:
$$a-2b+c = 3 \\ 2a+b+c = 0 \\ a+c = -2$$ 
Rank of coefficient matrix is full i.e. $3$, Hence system has a unique solution.
Now eliminate $b$ from 1),2) then you will get a new equation in $a,c$ then solve equation with 3). you will get the value of $a,b,c$.
A: Just for variety, avoiding explicit linear algebra:
It is suggested to me to subtract $a+c$ from $a-2b+c$, because that simultaneously eliminates $a$ and $c$. By the symmetry, the result should be $3-(-2)$. So $$(a-2b+c)-(a+c)=5$$ $$-2b=5$$ $$b=-\frac{5}{2}$$
Now subtracting two expressions like $$(2a+b+c)-2(a+c)=0-2(-2)$$ $$b-c=4$$ so $c=b-4=-\frac{13}{2}$. And solving for $a$ is now trivial. 
