#1 – Proof: variation of a determinant

By definition:

${\rm det}A=\frac{1}{n!}\epsilon^{\alpha_{1}\dots\alpha_{n}}\epsilon^{\beta_{1}\beta_{2}\dots\beta_{n}}\ A_{\alpha_{1}\beta_{1}}A_{\alpha_{2}\beta_{2}}\dots A_{\alpha_{n}\beta_{n}}=\epsilon^{\alpha_{1}\dots\alpha_{n}}\ A_{\alpha_{1}1}A_{\alpha_{2}2}\dots A_{\alpha_{n}n}$

so

$\frac{\partial\ {\rm det}A}{\partial A_{\mu\nu}}=\epsilon^{\alpha_{1}\alpha_{2}\dots\alpha_{n}}\left[\delta^{\mu}_{\alpha_{1}}\delta^{\nu}_{1}A_{\alpha_{2}2}\cdots A_{\alpha_{n}n}+A_{\alpha_{1}1}\delta^{\mu}_{\alpha_{2}}\delta^{\nu}_{2}A_{\alpha_{3}3}\cdots A_{\alpha_{n}n}+\dots+\right.$

$\left.+A_{\alpha_{1}1}A_{\alpha_{2}2}\cdots A_{\alpha_{n-1}n-1}\delta^{\mu}_{\alpha_{n}}\delta^{\nu}_{n}\right].$

Thus:

$\frac{\partial\ {\rm det}A}{\partial A_{\mu\nu}}A_{\mu\nu}=\epsilon^{\alpha_{1}\alpha_{2}\dots\alpha_{n}}\left[\underbrace{A_{\alpha_{1}1}A_{\alpha_{2}2}\cdots A_{\alpha_{n}n}+A_{\alpha_{1}1}A_{\alpha_{2}2}\cdots A_{\alpha_{n}n}+\cdots }_{\rm n\ times}\right]= n\cdot\epsilon^{\alpha_{1}\alpha_{2}\dots\alpha_{n}}\left[A_{\alpha_{1}1}A_{\alpha_{2}2}\cdots A_{\alpha_{n}n}\right]=n\cdot {\rm det}A.$

If $A$ is a metric tensor, $\frac{A_{\mu\nu}A^{\mu\nu}}{n}=\frac{{\delta^{\mu}}_{\mu}}{n}=\frac{{\rm tr}\ (1_{n\times n})}{n}=1,$ so we have:

$\frac{\partial\ {\rm det}A}{\partial A_{\mu\nu}}A_{\mu\nu}=\frac{A_{\mu\nu}A^{\mu\nu}}{n}\cdot n\cdot {\rm det}A.$

Equating terms proportional to $A_{\mu\nu}$:

$\frac{\partial\ {\rm det}A}{\partial A_{\mu\nu}}={\rm det}A\cdot A^{\mu\nu}.$

Finally:

$\delta\ {\rm det}A=\frac{\partial\ {\rm det}A}{\partial A_{\mu\nu}}\delta A_{\mu\nu}={\rm det}A\cdot A^{\mu\nu}\delta A_{\mu\nu}.$