Determinant of a block matrix including non-square matrices

We suppose the following matrix multiplication by the original matrix and own transposed form:

$$ \begin{aligned} MM^{\text{T}}=& \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} A^{\text{T}} & C^{\text{T}} \\ B^{\text{T}} & D^{\text{T}} \end{bmatrix} \\=& \begin{bmatrix} AA^{\text{T}}+BB^{\text{T}} & AC^{\text{T}}+BD^{\text{T}} \\ CA^{\text{T}}+DB^{\text{T}} & CC^{\text{T}}+DD^{\text{T}} \end{bmatrix} \end{aligned} $$

Then, each diagonal block matrix becomes a square form. Therefore we can apply the determinant formula of a block matrix:

$$ \begin{aligned} \det(MM^{\text{T}})=&\det(M)^2 \\=& \det\left| \begin{array}{cc} AA^{\text{T}}+BB^{\text{T}} & AC^{\text{T}}+BD^{\text{T}} \\ CA^{\text{T}}+DB^{\text{T}} & CC^{\text{T}}+DD^{\text{T}} \end{array} \right| \geq 0 \end{aligned} $$

Hence we obtain:

$$ \begin{aligned} &\det(M)= \\ \pm& \sqrt{ \det \Big( AA^{\text{T}}+BB^{\text{T}} \Big) \det \Big( (CC^{\text{T}}+DD^{\text{T}})- (CA^{\text{T}}+DB^{\text{T}}) (AA^{\text{T}}+BB^{\text{T}})^{-1} (AC^{\text{T}}+BD^{\text{T}}) \Big) } \end{aligned} $$