For which values of $a,b,c$ is the matrix $A$ invertible? $A=\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}$
$$\Rightarrow\det(A)=\begin{vmatrix}b&c\\b^2&c^2\end{vmatrix}-\begin{vmatrix}a&c\\a^2&c^2\end{vmatrix}+\begin{vmatrix}a&b\\a^2&b^2\end{vmatrix}\\=ab^2-a^2b-ac^2+a^2c+bc^2-b^2c\\=a^2(c-b)+b^2(a-c)+c^2(b-a).$$

Clearly, $$\left\{\det(A)\neq0\left|\begin{matrix}c\neq b\\a\neq c\\b\neq a\\a,b,c\neq 0\end{matrix}\right.\right\}\\$$
Is it sufficient to say that the matrix is invertible provided that all 4 constraints are met? Would Cramer's rule yield more explicit results for $a,b,c$ such that $\det(A)\neq0$?
 A: The matrix is the special case of so-called Vandermonde Matrix (See Induction for Vandermonde Matrix). 
det($A$)$=(a-b)(a-c)(b-c)$. So if $a\neq b,a\neq c, b\neq c, \space$ det$(A)\neq 0$.
A: suppose the matrix is singular. then so is its transpose, which must annihilate some non-zero vector $(h,g,f)$
this gives three equations:
$$
fa^2 + ga+ h =0 \\
fb^2 + gb+ h =0 \\
fc^2 + gc+ h =0 \\
$$
we know from the algebra of fields that a quadratic equation can have at most two roots, so the three values $a,b,c$ cannot all be different
A: If $A$ is a square matrix, $\det(A)=\det(A^T)=\det\begin{pmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{pmatrix}$
If $B$ is obtained from $A$ by adding a multiple of a row of $A$ to another row in $A$, $\det(B)=\det(A)\Rightarrow\text{let }B=\begin{pmatrix}1&a&a^2\\0&b-a&b^2-a^2\\0&c-a&c^2-a^2\end{pmatrix}\\\Rightarrow\det(B)=\det(A)=\begin{vmatrix}b-a&b^2-a^2\\c-a&c^2-a^2\end{vmatrix}=(c^2-a^2)(b-a)-(b^2-a^2)(c-a)\\\\=(c-a)(b-a)((c-a)-(b-a))=(c-a)(b-a)(c-b)\\\therefore\det(A)\neq0\iff\begin{cases}a\neq c\\a\neq b\\b\neq c\end{cases}$
A: This is a transpose of the Vandermonde matrix: http://en.wikipedia.org/wiki/Vandermonde_matrix
According to the wikipedia page, the determinant of the matrix will be non-zero precisely when none of the elements $a, b$, or $c$ are the same.
A: You can rewrite the determinant as $(b-a)(c-a)(c-b)$, so $\det(A)$ is nonzero precisely when $a,b,c$ are all distinct (note that one of $a,b,c$ can be zero).
Your matrix is the $3\times 3$ case of the Vandermonde matrix, whose determinant is generally a similar product of differences.
A: Yes, this is enough. Basically, if you are trying to solve:
$$\begin{pmatrix}d_1&d_2&d_3\end{pmatrix}A = \begin{pmatrix}y_1&y_2&y_3\end{pmatrix}$$ you are looking for a polynomial $f(x)=d_1+d_2x+d_3x^2$ such that $f(a)=y_1$, $f(d_2)=y_2$, and $f(d_3)=y_3$. 
This we can find by simple interpolation of polynomials - given any $n$ distinct $a_i$ and $n$ distinct $b_i$ we can find a polynomial $p(x)$ of degree at most $n$ such that $p(a_i)=b_i$ for all $i$.
A: $$\det(A)= \det\begin{pmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{pmatrix}= \det\begin{pmatrix}1&a&a^2\\0&b-a&b^2-a^2\\0&c-a&c^2-a^2\end{pmatrix} \\
=\det\begin{pmatrix}b-a&(b-a)(b+a)\\c-a&(c-a)(c+a)\end{pmatrix} =(b-a)(c-a)\det\begin{pmatrix}1&b+a\\1&c+a\end{pmatrix}$$
