Determinant of Block Diagonal Matrix

Theorem
Let $\mathbf A$ be a block diagonal matrix of order $n$.

Let $\mathbf A_1,\ldots,\mathbf{A}_k$ be the square matrices on the diagonal, i.e.:
 * $\displaystyle \mathbf A = \begin{bmatrix}

\mathbf A_1 & 0 & \cdots & 0 \\ 0 & \mathbf A_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mathbf A_k \end{bmatrix}$

Then the determinant $\det \left({\mathbf A}\right)$ of $\mathbf A$ satisfies:
 * $\displaystyle \det \left({\mathbf A}\right) = \prod_{i \mathop = 1}^k \det \left({\mathbf A_i}\right)$

Also see

 * Determinant of Diagonal Matrix, a special case of this theorem.