$\Delta = \begin{vmatrix} a & a^2 & 0 \\ 1 & 2a+b & a+b \\ 0 & 1 & 2a+3b \end{vmatrix}$ is divisible by
Step-by-Step Solution
Key Concept: Expand the determinant along the first column to obtain $\Delta = a\begin{vmatrix} 2a+b & a+b \\ 1 & 2a+3b \end{vmatrix} - 1\begin{vmatrix} a^2 & 0 \\ 1 & 2a+3b \end{vmatrix}$, then factor the resulting expression to identify common polynomial divisors like $(a+b)$ and $(2a+3b-1)$.
For the determinant $\Delta = \begin{vmatrix} a & a^2 & 0 \\ 1 & 2a+b & a+b \\ 0 & 1 & 2a+2b \end{vmatrix}$, expand along the first column to get $\Delta = a\begin{vmatrix} 2a+b & a+b \\ 1 & 2a+2b \end{vmatrix} - 1\begin{vmatrix} a^2 & 0 \\ 1 & 2a+2b \end{vmatrix}$. Simplifying yields $\Delta = a(a+b)(2a+3b-1)$ after factoring.
Correct Answer: 1,3