How does this transformation works?

Question

given

$$-\frac{N}{2}\ln|\Sigma|-\frac12 \operatorname{Tr}\left[\Sigma^{-1}\sum_{n=1}^N (t_n-y_n) (t_n-y_n)^T\right]$$

how did we get this $\Sigma$ if I was to set the derivative wrt to $\Sigma^{-1}$ for the above to 0.

$$\Sigma = \frac{1}{N}\sum_{n=1}^N (t_n-y_n) (t_n-y_n)^T$$

knowing that $\frac{∂}{∂A}ln|A|= (A^{-1})^T $

Answer 1

$$-\frac{N}{2}\ln|\Sigma|-\frac12 \operatorname{Tr}\left[\Sigma^{-1}\sum_{n=1}^N (t_n-y_n) (t_n-y_n)^T\right]=\frac{N}{2}\ln|\Sigma^{-1}|-\frac12 \operatorname{Tr}\left[\Sigma^{-1}\sum_{n=1}^N (t_n-y_n) (t_n-y_n)^T\right]$$

Differentiating with respect to $\Sigma^{-1}$ and equate it to $0$,

$$\frac{N}{2}(\Sigma^{-1})^{-T}-\frac12\sum_{n=1}^N(t_n-y_n)(t_n-y_n)^T=0$$

$$\Sigma^T=\frac1{N}\sum_{n=1}^N(t_n-y_n)(t_n-y_n)^T$$

Taking transpose and due to the right hand side is symmetric,

$$\Sigma=\frac1{N}\sum_{n=1}^N(t_n-y_n)(t_n-y_n)^T$$