proof of block determinants

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}.

\\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.