Consider the system $$ \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}' = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}. $$ where $a,b,c,d$ are real constants and $ad-bc=0.$
Question. Discuss all possible behaviors of the solutions and sketch the corresponding phase portraits
This is exactly how the question being asked. First and foremost is the determinant being zero is a "typo" ? Because if the determinant is zero, and the origin being not an isolated critical point, there will be infinitely many solutions.
If it's a typo and determinant is not zero, what's the most efficient and shorter way of classifying the zero solution. I can just go ahead and calculate the eigen values and argue for (stable/unstable)nodes, saddle-points, spiral points and centre. It'll be too long I presume. Any suggestions are appreciated.
For a linear system with two states, $x=\pmatrix{x_1&x_2}^T$ in general form: $$\dot{x}=Ax$$ the general form of the solution is: $$x(t)=\Phi(t)x(0)$$ where $\Phi(t)=e^{At}$ can be evaluated by converting $A$ to its Jordan canonical form. $$A=M^{-1}J M\implies e^{At}=M^{-1}e^{Jt}M$$ This means the general solution of the system is a linear combination of the elements of $e^{Jt}$. Now in your case where $\det A=0$, we can be sure that one of the eigenvalues are zero. Thus $$\begin{align} J&=\pmatrix{\lambda_1&0\\0&0}\text{ or}&\pmatrix{0&0\\0&0}&\text{ or }\pmatrix{0&1\\0&0}\\ e^{Jt}&=\pmatrix{e^{\lambda_1 t}&0\\0&1}\text{or}&\pmatrix{1&0\\0&1}&\text{ or }\pmatrix{1&t\\0&1} \end{align}$$ If $e^{Jt}$ is not a function of $t$ (meaning $J=0$), then the system is static. If $\lambda_1\neq 0$ you can analyze the system's behavior based on the real part of $\lambda_1$, i.e. one of the system's modes is: