Let $A \in M_n(\mathbb R)$ be a $5\times 5$ matrix given by \begin{align*} \begin{pmatrix} a_1 & a_1 \lambda_1 & b_1 & b_1 \lambda_1 & b_1 \lambda_1^2 \\ a_2 & a_2 \lambda_2 & b_2 & b_2 \lambda_2 & b_2 \lambda_2^2 \\ a_3 & a_3 \lambda_3 & b_3 & b_3 \lambda_3 & b_3 \lambda_3^2 \\ a_4 & a_4 \lambda_4 & b_4 & b_4 \lambda_4 & b_4 \lambda_4^2 \\ a_5 & a_5 \lambda_5 & b_5 & b_5 \lambda_5 & b_5 \lambda_5^2 \end{pmatrix}. \end{align*} where the vectors $a=(a_1, \dots, a_5)^T$ and $b=(b_1, \dots, b_5)^T$ are linearly independent and $(\lambda_1, \dots, \lambda_5)$ are distinct positive real numbers.
One sufficient condition for invertibility would be $a = (a_1, a_2, 0, 0, 0)^T$ with $a_1 \neq 0, a_2 \neq 0$ and $b_3 \neq 0, b_4 \neq 0, b_5 \neq 0$. In this case, the matrix becomes a block diagonal uppertriangular matrix with each diagonal block being a Vandermonde-like matrix. (We can also alternatively make $b_1 = b_2 = 0$.)
I am wondering whether there are some necessary conditions. In general, could we find some conditions on the linearly independent pair $(a,b)$ to guarantee $\det(A) \neq 0$?