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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 $d=a=a+d$.

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  • $\begingroup$ The left hand side is a real number and the right hand side is a matrix. Do you mean the determinant of the right side, too? $\endgroup$ – Thomas Andrews Sep 20 '14 at 22:33
  • $\begingroup$ Also, the formula you've given for $\det(A)$ is actually the formula for $A^{-1}$. $\endgroup$ – Thomas Andrews Sep 20 '14 at 22:34
  • $\begingroup$ oh yes, the determinant is just ad-bc. $det(A)=ad-bc$ $\endgroup$ – mickey4691 Sep 20 '14 at 22:36
  • $\begingroup$ Then edit your question to indicate what you really mean. It is still impossible for $\det(A)$ to be equal to a $2\times 2$ matrix. $\endgroup$ – Thomas Andrews Sep 20 '14 at 22:54
  • $\begingroup$ Hint: trace is independent of basis, and depends only on the eigenvalues. $\endgroup$ – Adam Hughes Sep 20 '14 at 23:16
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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:

$$\frac{1}{2}det\begin{pmatrix} tr(A)&1\\ tr(A^2)&tr(A) \end{pmatrix}=\frac{1}{2}det\begin{pmatrix} a+d&1\\ a^2+2bc+d^2&a+d \end{pmatrix}=$$

$$=\frac{1}{2}[(a+d)^2-(a^2+2bc+d^2)]=\frac{1}{2}[a^2+d^2+2ad-a^2-2bc-d^2]=$$

$$=\frac{1}{2}[2ad-2bc]=ad-bc=det(A)$$

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