There are three $2 \times 2$ matrices $A$, $B$ and $C4 they satisfy the following relation
\begin{equation} A_{ij}=B_{ij}+[CB+(CB)^T]_{ij}x+(CBC^T)_{ij}x^2, \end{equation}
where $x$ is an arbitrary variable. I can not obtain the relation
\begin{equation} \det(A) = \det(B) \left( 1 + \mbox{Tr}(C)x + \det(C)x^2 \right)^2. \end{equation}
Who can give some advice to calculate the determinant, or tell me whether the relations is right. I would thank very much for any answers.
Are you sure the parenhteses are right?.
If $C=I$, then $A=B+(B+B^T)x+Bx^2$
And if $B=\begin{pmatrix}0&1\\0&0\end{pmatrix}$
Then $A=\begin{pmatrix}0&1+x+x^2\\x&0\end{pmatrix}$
whose determinant is $-x(1+x+x^2)\neq 0=\det B(1+2x+x^2)$, a part from three values of $x$.
Thus the claimed inequality cannot be true.
On the other hand, consider that for 2x2 matrices we have
$1+Tr(C)x+\det(C)x^2=\det(I+xC)$
Thus if $A=(I+Cx)B(I+C^tx)=B+BC^tx+CBx+CBC^tx^2=B+(CB+BC^T)x+CBC^Tx^2$ then $\det(A)=\det(B)\det(I+Cx)\det((I+Cx)^T)=\det B(\det(I+Cx))^2=\det B(1+Tr(C) x+\det(C)x^2)$
as desired.