Exercise 18
Let \(A\) be an \(n \times n\). Prove that \(\det(A^*) = \overline{A}\).


Proof:

Let \(A\) be a matrix of size \(n \times n\). We first will prove that \(\det(\overline{A}) = \overline{\det{A}}\) by induction on \(n\).

Base Case: \(n = 1\). \(\det(\overline{A}) = \overline{a_{11}} = \overline{\det{(A)}}\).

Inductive Case: Assume \(\det(\overline{A}) = \overline{\det{A}}\) for \(n-1\). We will prove this for \(n\). Computing the determinant of \(\overline{A}\) by cofactor expansion along the first row and applying the inductive hypothesis

$$ \begin{align*} \det(\overline{A}) &= (-1)^{1+1}\overline{a_{11}}\det{\widetilde{\overline{A_{11}}}} + (-1)^{1+2}\overline{a_{12}}\det{\widetilde{\overline{A_{12}}}} + ... + \\ &+ (-1)^{1+n}\overline{a_{1n}}\det{\widetilde{\overline{a_{1n}}}} \\ &= (-1)^{1+1}\overline{a_{11}}\overline{\det{\widetilde{A_{11}}}} + (-1)^{1+2}\overline{a_{12}}\overline{\det{\widetilde{A_{11}}}} + ... + \\ &+ (-1)^{1+n}\overline{a_{1n}}\overline{\det{\widetilde{A_{11}}}} \\ &= \overline{\det{A}} \end{align*} $$

So now we can apply this result to show that

$$ \begin{align*} \det(A^*) &= \det(\overline{A^t}) \\ &= \overline{\det(A^t)} \\ &= \overline{\det{A}} \quad \text{(we prove previously that $\det{A}=\det{A^t}$)} \end{align*} $$




References