# setting coefficients in linear system for $n-1$ pivots

Question: Given a system of linear equations $$ax_1 + ax_2 + ax_3 = 2\\ x_1+ ax_2 + ax_3 = 0\\ 2x_1 + 3x_2 + ax_3 = 1$$ For what 2 values of $a$ will the system's augmented matrix have less than 3 pivots?

I'm not looking for an answer to the question, but I'm currently using trial and error to try and form a row $0\,0\,0\,0$, and was wondering if there's some conceptual understating I'm missing that would point to a more logical strategy for finding $a$?

$$\begin{bmatrix} a & a & a \\ 1 & a & a \\ 2 & 3 & a \\ \end{bmatrix}$$
I believe you get a cubic polynomial. Set the resulting expression equal to zero and solve for $a$. Check each of those solutions to see if the resulting original equations have zero solutions or infinitely many solutions.