# determinant of a bigger matrix

In wiki (rule 6) it mentions that det({{A,B},{C,D}}) = det(AD-BC). I want to know if A,B,C,D are all square matrices of the same size, then does the above (or under other conditions) imply det({{A,B},{C,D}}) = det(A)det(D)-det(B)det(C)? How to prove them more vigorously?

$DET\begin{pmatrix} \mathbf A & 0 \\ \mathbf C & D \\ \end{pmatrix} , \mathbf A ,\mathbf 0 ,\mathbf C , \mathbf D$ are block matrices. The determinant needs a square matrix so the block matrices do not necessarily have to be square matrices but as a whole the orders of each block matrix have to be such that the determinant matrix is square. However, If you look at the wiki page you posted you will see that
$DET\begin{pmatrix} \mathbf A & \mathbf B \\ \mathbf C & \mathbf D \\ \end{pmatrix}= DET(\mathbf A\mathbf D - \mathbf B\mathbf C)$. The respective sizes of each matrix here has to be such that multiplication works but also the size of $\mathbf A \mathbf D$ and $\mathbf B \mathbf C$ has to be equal so that subtraction is possible, furthermore the size of $\mathbf A\mathbf D - \mathbf B\mathbf C$ should also be square so $DET$ can be computed hence $\mathbf A ,\mathbf B ,\mathbf C , \mathbf D$ must be square.