$\begin{array}{l}A = \left[\begin{array}{cccc}1&-2&m&1\\3&-2&2&4\\-1&-2&2&3\\2&m&2&1\end{array}\right]\\ \text{Khai triển theo dòng 1, ta có:}\\ A_{11} = (-1)^{1 + 1}\left|\begin{array}{ccc}-2&2&4\\-2&2&3\\m&2&1\end{array}\right| \xrightarrow{h_2 \to h_2 - h_1} \left|\begin{array}{ccc}-2&2&4\\0&0&-1\\m&2&1\end{array}\right|=-2m -4\\ A_{12} = (-1)^{1 + 2}\left|\begin{array}{ccc}3&2&4\\-1&2&3\\2&2&1\end{array}\right| \xrightarrow{\begin{array}{l}h_2 \to h_2 - h_1\\h_3 \to h_3 - h_1\end{array}} \left|\begin{array}{ccc}3&2&4\\-4&0&-1\\-1&0&-3\end{array}\right| \xrightarrow{h_1 \to h_1 + 4h_2}\left|\begin{array}{ccc}-13&2&0\\-4&0&-1\\-1&0&-3\end{array}\right|=22\\ A_{13} = (-1)^{1 + 3}\left|\begin{array}{ccc}3&-2&4\\-1&-2&3\\2&m&1\end{array}\right| \xrightarrow{h_2 \to h_2 -h_1}\left|\begin{array}{ccc}3&-2&4\\-4&0&-1\\2&m&1\end{array}\right| =-13m - 4\\ A_{14} = (-1)^{1 + 4}\left|\begin{array}{ccc}3&-2&2\\-1&-2&2\\2&m&2\end{array}\right| \xrightarrow{h_2 \to h_2 - h_1}\left|\begin{array}{ccc}3&-2&2\\-4&0&0\\2&m&2\end{array}\right|=8m + 16\\ Ta\,\,được:\\ \det(A) = 1.(-2m - 4) + (-2).22 + m(-13m - 4) + 1.(8m + 16) = -13m^2 + 2m - 32\\ Ta\,\,có:\\ 2A = \left[\begin{array}{cccc}2&-4&2m&2\\6&-4&4&8\\-2&-4&4&6\\4&2m&4&2\end{array}\right]\\ \Rightarrow \det(2A) = 8\det(A)\\ \text{Theo đề ta được:}\\ \det(2A) = -164\\ \Leftrightarrow 8(-13m^2 + 2m - 32) = -164\\ \Leftrightarrow 104m^2 -16m - 92 = 0\\ \Leftrightarrow m = \dfrac{2 \pm 3i\sqrt{66}}{26} \end{array}$
