Use the fact that the determinant is multilinear. $\det\left(\begin{array}{c} r_1\\\vdots\\r_{i-1}\\ r_i+\alpha s_i\\r_{i+1}\\\vdots\\r_n\end{array}\right) = \det\left(\begin{array}{c}r_1\\\vdots\\r_{i-1}\\r_i\\r_{i+1}\\\vdots\\r_n\end{array}\right) + \alpha \det\left(\begin{array}{c} r_1\\\vdots\\r_{i-1}\\s_i\\r_{i+1}\\\vdots\\r_n\end{array}\right)$ where $r_1,\ldots,r_n$ are rows, $s_i$ is a row, $\alpha$ is a scalar, and $i$ is arbitrary.
Also use the fact that a determinant of a matrix with two identical rows is equal to $0$.
(Since you don't show your Gaussian elimination, I can't tell whether you made a mistake or performed an operation that would change the value of the determinant.)
Here's my computation of this, using multilinearity; when we exchange two rows, it multiplies the determinant by $-1$: $\begin{align*} \det\left(\begin{array}{c}4v_1+2v_4\\v_2\\v_3\\5v_1+2v_4\end{array}\right) &= \det\left(\begin{array}{c}4v_1\\v_2\\v_3\\5v_1+2v_4\end{array}\right) + \det\left(\begin{array}{c}2v_4\\v_2\\v_3\\5v_1+2v_4\end{array}\right)\\ &= \det\left(\begin{array}{c} 4v_1\\v_2\\v_3\\5v_1\end{array}\right) + \det\left(\begin{array}{c}4v_1\\v_2\\v_3\\2v_4\end{array}\right) + \det\left(\begin{array}{c}2v_4\\v_2\\v_3\\5v_1\end{array}\right) + \det\left(\begin{array}{c}2v_4\\v_2\\v_3\\2v_4\end{array}\right)\\ &= 4\cdot5\cdot\det\left(\begin{array}{c}v_1\\v_2\\v_3\\v_1\end{array}\right) + 4\cdot2\cdot\det\left(\begin{array}{c}v_1\\v_2\\v_3\\v_4\end{array}\right) \\ &\qquad\mathop{+} 2\cdot 5\cdot\det\left(\begin{array}{c}v_4\\v_2\\v_3\\v_1\end{array}\right) + 2\cdot2\cdot\det\left(\begin{array}{c}v_4\\v_2\\v_3\\v_4\end{array}\right)\\ &=20(0) + 8\det(A) +10(-1)\det\left(\begin{array}{c}v_1\\v_2\\v_3\\v_4\end{array}\right) + 4(0)\\ &= 8\det(A) - 10\det(A)\\ &= -2\det(A)\\ &= -2(8)\\ &= -16. \end{align*}$