Prove that determinant is zero iff $x= 2, 3$

Question

Prove that $$\begin{vmatrix} 1 & x & x^{2}\\ 1 & 2 & 2^{2}\\ 1 & 3 & 3^{2} \end{vmatrix}=0\Leftrightarrow x= 2, 3$$

I see that if determinant is zero, two lines must equal, in this problem, there are between line 1 and line 2, 3, but how can I prove there is no other solutions ?? And what for $x^{3}, x^{4}, etc$ instead ?? Thks

Answer 1

Why not to expand the determinant and find all the solutions? $$\begin{vmatrix} 1 & x & x^{2}\\ 1 & 2 & 2^{2}\\ 1 & 3 & 3^{2} \end{vmatrix} = \begin{vmatrix} 1 & x & x^{2}\\ 0 & 2-x & 2^{2}-x^2\\ 0 & 1 & 5 \end{vmatrix} = \begin{vmatrix} 2-x & 2^{2}-x^2\\ 1 & 5 \end{vmatrix}$$
$$=10-5x-(4-x^2) = x^2-5x+6=(x-2)(x-3)$$ Which equals to zero iff $x=2,3$

Answer 2

The Leibniz formula for determinants implies that the determinant in question is a quadratic polynomial in $x$. Since you have found two different roots, they are all the roots.

Answer 3

$$p(x) = \begin{vmatrix} 1 & x & x^{2}\\ 1 & 2 & 2^{2}\\ 1 & 3 & 3^{2} \end{vmatrix}= (x-2)(x-3)$$

Why? See that $p(x)$ has degree at most $2$, and it has two roots $2,3$ (by inspection). So it must be of the form $\alpha(x-2)(x-3)$ for some $\alpha \in \mathbb{R}$. $\alpha = 1$, but you don't really need this information to conclude that $2$ and $3$ are the only roots of $p(x)$, since we know that any polynomial $f(x)$ has roots $x_1,...,x_n$, iff $\lambda f(x)$ has roots $x_1,...,x_n$ for some scalar $\lambda\in\mathbb{R}$.

It should be clear now!

As mentioned in the comments, you can look at Vandermonde determinants for a generalization.