How to calculate determinant of a block matrix?

Question

How to calculate the following determinant?

$$\det \begin{pmatrix}A&v\\v^T& b\end{pmatrix}$$

Answer 1

Assuming that the matrix $\bf A$ is invertible,

$$ \begin{pmatrix} {\bf I} & {\bf 0} \\ - {\bf c}^\top {\bf A}^{-1} & 1\end{pmatrix} \begin{pmatrix} {\bf A} & {\bf c} \\ {\bf c}^\top & b\end{pmatrix} = \begin{pmatrix} {\bf A} & {\bf c} \\ {\bf 0}^\top & b - {\bf c}^\top {\bf A}^{-1} {\bf c}\end{pmatrix}$$

Hence,

$$ \underbrace{\det \begin{pmatrix} {\bf I} & {\bf 0} \\ - {\bf c}^\top {\bf A}^{-1} & 1\end{pmatrix}}_{= 1} \cdot \det \begin{pmatrix} {\bf A} & {\bf c} \\ {\bf c}^\top & b \end{pmatrix} = \underbrace{\det \begin{pmatrix} {\bf A} & {\bf c} \\ {\bf 0}^\top & b - {\bf c}^\top {\bf A}^{-1} {\bf c}\end{pmatrix}}_{= \det ({\bf A}) \left( b - {\bf c}^\top {\bf A}^{-1} {\bf c} \right)} $$

and, thus,

$$ \det \begin{pmatrix} {\bf A} & {\bf c} \\ {\bf c}^\top & b \end{pmatrix} = \color{blue}{\det ({\bf A}) \left( b - {\bf c}^\top {\bf A}^{-1} {\bf c} \right)} $$

Take a look at the Schur complement.

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_{Related: Determinant of a block lower triangular matrix}

Answer 2

1. The determinant of any sized square matrix is computed recursively, as follows:

$$ \begin{vmatrix} a & b & c & d & e \\ f&g&h&i&j\\k&l&m&n&o \\ p&q&r&s&t \\ u&v&w&x&y \end{vmatrix} = a \begin{vmatrix} g&h&i&j\\l&m&n&o \\ q&r&s&t \\ v&w&x&y \end{vmatrix} - b \begin{vmatrix} f&h&i&j\\k&m&n&o \\ p&r&s&t \\ u&w&x&y \end{vmatrix} + c \begin{vmatrix} f&g&i&j\\k&l&n&o \\ p&q&s&t \\ u&v&x&y \end{vmatrix} - d \begin{vmatrix} f&g&h&j\\k&l&m&o \\ p&q&r&t \\ u&v&w&y \end{vmatrix} + e \begin{vmatrix} f&g&h&i\\k&l&m&n \\ p&q&r&s \\ u&v&w&x \end{vmatrix}$$

2. The $+, -, +, -$ sign alternates per term, always starting with $+$.

3. The determinants of the smaller submatrices are computed the same recursive manner, until the limiting $2x2$ sized matrix is reached, and then the recursion exits with an explicit computation

$$ \begin {vmatrix} a&b \\ c&d \end{vmatrix} = ad - bc$$

4. There are likely some clever identities/shortcuts knowing the block structure of the full matrix.