Expressing the determinant in terms of the trace of a matrix and the trace of its square How can I prove that
$$\det(A) = \frac{ 1 }{ 2 }  \begin{vmatrix}\operatorname{tr}(A) & 1 \\ \operatorname{tr}(A^{2}) & \operatorname{tr}(A)\end{vmatrix}$$
where vertical bars mean the determinant? 
This is what I have so far.
Let $A = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]$.
Then  $\det(A) = \frac{ 1 }{ ad-bc } \left[\begin{matrix} d & -b \\ -c & a\end{matrix}\right]$.  As such, $\operatorname{tr}(A) = a+d$, $ad+bc=2$ and $d=a=a+d$.

If I understand what you want to prove, for a 2x2 matrix we have:
$$A=\begin{pmatrix}
a&b\\
c&d
\end{pmatrix},\text{ } 
det(A)=ad-bc\text{ , } Tr(A)=a+d$$
$$A^2=AA=\begin{pmatrix}
a&b\\
c&d
\end{pmatrix}
\begin{pmatrix}
a&b\\
c&d
\end{pmatrix}=\begin{pmatrix}
a^2+bc&ab+bd\\
ac+cd&bc+d^2
\end{pmatrix}$$
$$tr(A^2)=a^2+2bc+d^2$$
Now we have that:
$$\frac{1}{2}det\begin{pmatrix}
tr(A)&1\\
tr(A^2)&tr(A)
\end{pmatrix}=\frac{1}{2}det\begin{pmatrix}
a+d&1\\
a^2+2bc+d^2&a+d
\end{pmatrix}=$$
$$=\frac{1}{2}[(a+d)^2-(a^2+2bc+d^2)]=\frac{1}{2}[a^2+d^2+2ad-a^2-2bc-d^2]=$$
$$=\frac{1}{2}[2ad-2bc]=ad-bc=det(A)$$
