Factorize \begin{vmatrix} a & -b & -a & b\\ b & a & -b & -a\\ c & -d & 2c & -2d\\ d & c & 2d & 2c \end{vmatrix} by $$\begin{vmatrix} A & -A\\ B & 2B \end{vmatrix} =3^n |A||B|,$$ where A and B are n power matrices.
Here is my thinking: $$\begin{vmatrix} a & -b & -a & b \\ b & a & -b & -a \\ c & -d & 2c & -2d \\ d & c & 2d & 2c \end{vmatrix} \xrightarrow[C_4 +C_2]{C_3 +C_1} \begin{vmatrix} a & -b & 0 & 0 \\ b & a & 0 & 0 \\ c & -d & 3c & -3d \\ d & c & 3d & 3c \end{vmatrix} \xrightarrow[R_1 \leftrightarrow R_2 ,\; R_2 \leftrightarrow R_3]{C_3 \leftrightarrow C_2} -\begin{vmatrix} b & 0 & a & 0 \\ c & 3c & -d & -3d \\ a & 0 & -b & 0 \\ d & 3d & c & 3c \end{vmatrix}$$
$$\begin{vmatrix} A & -A \\ B & 2B \end{vmatrix} = \begin{vmatrix} A & 0 \\ B & 3B \end{vmatrix} = 3^n |A||B|$$
$$-\begin{vmatrix} b & 0 & a & 0 \\ c & 3c & -d & -3d \\ a & 0 & -b & 0 \\ d & 3d & c & 3c \end{vmatrix} = -\left(\begin{vmatrix} b & 0 \\ c & 3c \end{vmatrix} \begin{vmatrix} a & 0 \\ -d & -3d \end{vmatrix} \begin{vmatrix} a & 0 \\ d & 3d \end{vmatrix} \begin{vmatrix} -b & 0 \\ c & 3c \end{vmatrix}\right)$$
$$= -( 3bc\times (-3ab) \times 3ab\times (-3bc)) = -81a^2 b^2 c^2 d^2$$
The result I got seems to be weird. It can never turn to expression of first degree.
Please let me know where is wrong. Thanks for your help.
Don't know what "$n-$ power matrices" means.
If $A,B$ are square matrices of the same size, we have
$$ \begin{pmatrix} A & -A\\ B & 2B \end{pmatrix} \begin{pmatrix} I & 0\\ \frac{-1}{2}I & I \end{pmatrix} = \begin{pmatrix} \frac{3}{2}A & -A\\ 0 & 2B \end{pmatrix} $$ which is block triangular, while the determinant is $\frac{3}{2} \det A \; \; 2 \det B = 3 \det A \det B $