# Does this Vandermonde-like matrix always have nonvanishing determinant?

$$\newcommand{\l}{\lambda}$$ $$\newcommand{\bb}{\mathbb}$$ $$\newcommand{\a}{\alpha}$$ Let $$\{\lambda_1, \lambda_2, \dots, \lambda_5\} \subset \bb R$$ be distinct nonzero real numbers. Let $$a_1, b_1, c_1, d_1, e_1, a_2, b_2, c_2, d_2, e_2$$ be $$10$$ fixed parameters such that the matrix \begin{align*}A(\l_1, \dots, \l_5) := \begin{pmatrix} a_1 \l_1^2 & b_1 \l_2^2 & c_1 \l_3^2 & d_1 \l_4^2 & e_1\l_5^2 \\ a_1 \l_1 & b_1 \l_2 & c_1 \l_3 & d_1 \l_4 & e_1\l_5 \\ a_2 \l_1^3 & b_2 \l_2^3 & c_2 \l_3^3 & d_2 \l_4^3 & e_2\l_5^3 \\ a_2 \l_1^2 & b_2 \l_2^2 & c_2 \l_3^2 & d_2 \l_4^2 & e_2\l_5^2\\ a_2 \l_1 & b_2 \l_2 & c_2 \l_3 & d_2 \l_4 & e_2\l_5 \end{pmatrix} \end{align*} has nonvanishing determinant. Will it possible that for any $$5$$ disctinct nonzero numbers $$\{\a_1, \dots, \a_5\}$$, the matrix $$A(\a_1, \dots, \a_5)$$ would have nonvanishing determinant?

I tried to reduce the matrix to have $$2$$ alphabets with subscripts $$1$$ and $$3$$ alphabets with subscripts $$2$$ to be nonzero, then the determinant would be calculated in a manner very similar to the Vandermonde matrix. But apparently, without making the matrix singular, it seems like only restricted row operations are allowed and this takes me to nowhere.