Inverse and determinant of complex matrix

Question

Is the determinant calculated the same way as a real matrix?
Also when does $A^{-1}$ exist? Should the determinant be different from zero? a real number? or any complex number?

Answer 1

Simply just use the formula for the determinant of a matrix you already know, ensuring you consider the arithmetic of complex numbers where necessary (for example, division by $a+ib$ occurring somewhere would require you to multiply through by conjugates etc..

The inverse would not exist is if the determinant of the matrix with complex entries is zero. If it is non-zero, you can calculate the inverse.

Answer 2

Yes it is ; working in $\mathbb{R}$ or $\mathbb{C}$ does not change anything when dealing with determinant and inverses of matrices, though of course, the determinant of a complex matrice is a priori complex.

As a reminder, I recall one of the definitions of the determinant (there are others, which are equivalent):

$$\det(A)=\sum_{\sigma\in\Sigma_n}\varepsilon(\sigma)\prod_{i=1}^n a_{i,\sigma(i)}$$

(where $\varepsilon$ is a signature of the permutation $\sigma$, and $\Sigma_n$ the set of permutations of $\{ 1,\ldots,n\}$).

More generally, this is valid for matrices whose coefficients belong to a field. It is more involved if you work with matrices whose coefficients belong to a ring (say $\mathbb{Z}$). In general, if $R$ is a ring and $A\in M_n(R)$ then $\det(A)\in R$, and $$A\textrm{ is inversible (has in inverse in }M_n(R)) \Leftrightarrow \det(A)\textrm{ is inversible in }R,$$ which gives, for example if $R=\mathbb{Z}$: $$A\textrm{ has in inverse in }M_n(\mathbb{Z}) \Leftrightarrow \det(A)\in\{1,-1\}.$$