行列式の性質を利用すると

$ $$\begin{eqnarray*} \begin{vmatrix} 3 & 4 & 4 & 3 & 2 \\ 2 & 3 & 4 & 2 & 2 \\ 2 & 4 & 4 & 4 & 4 \\ 4 & 4 & 4 & 3 & 3 \\ 3 & 3 & 4 & 4 & 2 \end{vmatrix} & = & 4 \times \begin{vmatrix} 3 & 4 & 1 & 3 & 2 \\ 2 & 3 & 1 & 2 & 2 \\ 2 & 4 & 1 & 4 & 4 \\ 4 & 4 & 1 & 3 & 3 \\ 3 & 3 & 1 & 4 & 2 \end{vmatrix}\\[1em] & = & (-4) \times \begin{vmatrix} 1 & 4 & 3 & 3 & 2 \\ 1 & 3 & 2 & 2 & 2 \\ 1 & 4 & 2 & 4 & 4 \\ 1 & 4 & 4 & 3 & 3 \\ 1 & 3 & 3 & 4 & 2 \end{vmatrix} \\[1em] & = & (-4) \times \begin{vmatrix} 1 & 4 & 3 & 3 & 2 \\ 0 & -1 & -1 & -1 & 0 \\ 0 & 0 & -1 & 1 & 2 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & -1 & 0 & 1 & 0 \end{vmatrix}\\[1em] & = & (-4) \times \begin{vmatrix} -1 & -1 & -1 & 0 \\ 0 & -1 & 1 & 2 \\ 0 & 1 & 0 & 1 \\ -1 & 0 & 1 & 0 \end{vmatrix}\\[1em] & = & 4 \times \begin{vmatrix} 1 & 1 & 1 & 0 \\ 0 & -1 & 1 & 2 \\ 0 & 1 & 0 & 1 \\ -1 & 0 & 1 & 0 \end{vmatrix}\\[1em] & = & 4 \times \begin{vmatrix} 1 & 1 & 1 & 0 \\ 0 & -1 & 1 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 2 & 0 \end{vmatrix}\\[1em] & = & 4 \times \begin{vmatrix} -1 & 1 & 2 \\ 1 & 0 & 1 \\ 1 & 2 & 0 \end{vmatrix}\\[1em] & = & 4 \times \begin{vmatrix} -1 & 1 & 2 \\ 0 & 1 & 3 \\ 0 & 3 & 2 \end{vmatrix}\\[1em] & = & (-4) \times \begin{vmatrix} 1 & 3 \\ 3 & 2 \end{vmatrix}\\[1em]  & =& (-4) \times (2 - 9) = 28\end{eqnarray*}$$ $