Determinant of 5x5 Matrix with 6 Adjacent Zeros

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Problem

Prove $ \ \ \begin{vmatrix} a_{1} & a_{2} & a_{3} & a_{4} & a_{5} \\ b_{1} & b_{2} & b_{3} & b_{4} & b_{5} \\ c_{1} & c_{2} & c_{3} & c_{4} & c_{5} \\ d_{1} & d_{2} & 0 & 0 & 0 \\ e_{1} & e_{2} & 0 & 0 & 0 \\ \end{vmatrix} = 0. $

Restrictions: All variables are nonzero.

Attempt

My first attempt was to reduce the bottom two rows to zero, and another attempt was to transform the matrix into a lower triangular matrix. Neither worked.

Notes

If anyone can provide me detailed steps on how to achieve the RHS, I'd be very appreciative! Thanks!

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Hint By Row Operations you can reduce the bottom two rows

$$\begin{vmatrix} a_{1} & a_{2} & a_{3} & a_{4} & a_{5} \\ b_{1} & b_{2} & b_{3} & b_{4} & b_{5} \\ c_{1} & c_{2} & c_{3} & c_{4} & c_{5} \\ d_{1} & d_{2} & 0 & 0 & 0 \\ e_{1} & e_{2} & 0 & 0 & 0 \\ \end{vmatrix} = \begin{vmatrix} a_{1} & a_{2} & a_{3} & a_{4} & a_{5} \\ b_{1} & b_{2} & b_{3} & b_{4} & b_{5} \\ c_{1} & c_{2} & c_{3} & c_{4} & c_{5} \\ 1 & 0 & 0 & 0 & 0 \\ 0 & d_1e_{2}-d_2e_2 & 0 & 0 & 0 \\ \end{vmatrix} $$ Then, by expnasion you get $$\begin{vmatrix} a_{1} & a_{2} & a_{3} & a_{4} & a_{5} \\ b_{1} & b_{2} & b_{3} & b_{4} & b_{5} \\ c_{1} & c_{2} & c_{3} & c_{4} & c_{5} \\ d_{1} & d_{2} & 0 & 0 & 0 \\ e_{1} & e_{2} & 0 & 0 & 0 \\ \end{vmatrix} = \left(d_1e_{2}-d_2e_2\right)\begin{vmatrix} a_{3} & a_{4} & a_{5} \\ b_{3} & b_{4} & b_{5} \\ c_{3} & c_{4} & c_{5} \\ \end{vmatrix} $$

which can be calculated explicitely and is not identically 0.