If $A^{-1}$ exists, then $$\begin{pmatrix} A & B \\ C & D \end{pmatrix} =\begin{pmatrix} I & O \\ CA^{-1} & I \end{pmatrix}\begin{pmatrix} A & B \\ O & D-CA^{-1}B \end{pmatrix}.$$ So $$\det \begin{pmatrix} A & B \\ C & D \end{pmatrix} =\det \begin{pmatrix} I & O \\ CA^{-1} & I \end{pmatrix}\det \begin{pmatrix} A & B \\ O & D-CA^{-1}B \end{pmatrix}. $$ Each of the first matrices in the decompositions are triangular. Hence their determinants equal $1$ . This means that the determinant of the original matrix equals the determinant of either of the second matrices in the decomposition. Therefore $$\det \begin{pmatrix} A & B \\ C & D \end{pmatrix} =\det(A)\det(D-CA^{-1}B).$$ The second formula follows by using a similar trick.