Define the determinant $\operatorname{det}(A)$ of an $n \times n$ complex matrix A. Let $A_{1}, \ldots, A_{n}$ be the columns of $A$, let $\sigma$ be a permutation of $\{1, \ldots, n\}$ and let $A^{\sigma}$ be the matrix whose columns are $A_{\sigma(1)}, \ldots, A_{\sigma(n)}$. Prove from your definition of determinant that $\operatorname{det}\left(A^{\sigma}\right)=\epsilon(\sigma) \operatorname{det}(A)$, where $\epsilon(\sigma)$ is the sign of the permutation $\sigma$. Prove also that $\operatorname{det}(A)=\operatorname{det}\left(A^{t}\right) .$

Define the adjugate matrix $\operatorname{adj}(A)$ and prove from your $\operatorname{definitions}$ that $A \operatorname{adj}(A)=$ $\operatorname{adj}(A) A=\operatorname{det}(A) I$, where $I$ is the identity matrix. Hence or otherwise, prove that if $\operatorname{det}(A) \neq 0$, then $A$ is invertible.

Let $C$ and $D$ be real $n \times n$ matrices such that the complex matrix $C+i D$ is invertible. By considering $\operatorname{det}(C+\lambda D)$ as a function of $\lambda$ or otherwise, prove that there exists a real number $\lambda$ such that $C+\lambda D$ is invertible. [You may assume that if a matrix $A$ is invertible, then $\operatorname{det}(A) \neq 0$.]

Deduce that if two real matrices $A$ and $B$ are such that there exists an invertible complex matrix $P$ with $P^{-1} A P=B$, then there exists an invertible real matrix $Q$ such that $Q^{-1} A Q=B$.