$\newcommand{\l}{\lambda}$ $\newcommand{\bb}{\mathbb}$ $\newcommand{\a}{\alpha}$ 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.


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