Taking inverse of a complex matrix

Question

I checked previous posts and it says that I should just follow the same process as in real case. However, when I try to find the inverse of a complex matrix with coefficient method, should I take conjugate transpose of the coefficient matrix. If it is so, I think they don't give the same result. What do I do wrong ?

Answer 1

The simplest test:

Is $\begin{bmatrix} i\end{bmatrix}$ inverted as $\begin{bmatrix} i^{-1}\end{bmatrix}$ or $\begin{bmatrix}i^{*-1}\end{bmatrix}$ ?

Check by performing the product.

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Alternatively:

Consider the system of equations

$$\begin{cases}ax+by=u,\\cx+dy=v\end{cases}$$ where all variables are complex.

Now combining the two equations,

$$d(ax+by)-b(cx+dy)=(ad-bc)y=du-bv$$

so that

$$y=\frac{\begin{vmatrix}u&b\\ v&d\end{vmatrix}}{\begin{vmatrix}a&b\\ c&d\end{vmatrix}}$$ where the determinants are evaluated as usual.

As you see, there is no "room" for conjugation. Of course this extends to matrix inversion, and to any order.

Answer 2

It's exactly the same as it is for a real matrix. Of course you need to check that the matrix is non-singular, i.e that the determinant is non-zero.

Answer 3

It is the same method: $\left[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right]^{-1}=\frac1{\det A}\left[\begin{smallmatrix}d&-b\\-c&a\end{smallmatrix}\right]$. So, if$$A=\begin{bmatrix}2+i&1\\1&-2-i\end{bmatrix},$$then $\det A=-4-4i$. Therefore,\begin{align}A^{-1}&=\frac1{-4-4i}\begin{bmatrix}-2-i&-1\\-1&2+i\end{bmatrix}\\&=\begin{bmatrix}\frac38-\frac i8&\frac18-\frac i8\\ \frac18-\frac i8&-\frac38+\frac i8\end{bmatrix}.\end{align}