Find the values of $a$ which make this matrix not invertible
\begin{pmatrix} 2 & a-3 & a^2 \\ 2 & 1 & 4 \\ 1 & a & 0 \\ \end{pmatrix}
Now I know that if the determinant is $0$ then it is not invertible, however when I attempt to solve for the determinant it yields $2a^3-a^2-4a-12$ and the roots of this equation are highly complex.
Using the rule of Sarrus it is easy to see that the determinant is zero if and only if $a$ is a real root of $$ 2a^3 - a^2 - 4a - 12=0. $$ This gives $a=2.38875419925$. Perhaps somewhere there is a typo in this exercise. Indeed, if the upper right corner is $a$, and not $a^2$, then, using semiclassical methods, $a$ is a root of $$ (2a + 3)(a - 4)=0.$$