Computing the determinant of a matrix which splits up into unequal blocks

Question

Suppose I have a matrix of the following form:

$M = \begin{bmatrix} A & \mathbb{v} \\ \mathbb{v}^T & a\\ \end{bmatrix}$

where $A$ is an $n \times n$ matrix, $\mathbb{v}$ is an $n \times 1$ column vector, and $a$ is a scalar.

Is there a way to compute the determinant of $M$ in terms of the determinant of $A$ and the quantities $\mathbb{v}$ and $a$?

More generally, can we do this if the row vector is different from the column vector?

$M = \begin{bmatrix} A & \mathbb{v} \\ \mathbb{w} & a\\ \end{bmatrix}$

Answer 1

Using Laplace's formula (where I am denoting the minor matrix with a hat) and expanding along the bottom row: $$\det M=\sum_{j=1}^{n+1} (-1)^{n+1+j}M_{n+1,j}\det \hat M_{n+1,j}$$ $$\det M=a\det A+\sum_{j=1}^{n} (-1)^{n+1+j}w_j \det \hat A_j$$ Here $\hat A_j$ is the matrix $A$ with the $j$-th column removed and $v$ appended to it. And $$\det \hat A_j=(-1)^{n-j}\det \bar A_j$$ where $\bar A_j$ is the matrix $A$ with the $j$-t column replaced by $v$. So: $$\det M=a\det A+\sum_{j=1}^{n} (-1)^{n+1+j}w_j (-1)^{n-j}\det \bar A_j=a\det A-\sum_{j=1}^{n}w_j\det \bar A_j$$ By Cramer's rule, if $\det A$ is non-zero, then $$\det \bar A_j=x_j\det A$$ where the vector $x$ is the solution to $Ax=v$. In other words $x=A^{-1}v$. Then the determinant can be factored out of the sum, and the remaining terms of the sum can be seen to be the scalar product between $w$ and $A^{-1}v$.

So, if $A$ is invertible: $$\det M=(a-w\cdot A^{-1}v)\det A$$

More general formulas can be found on Wikipedia as pointed out by another poster.

Answer 2

Yes, checkout the block matrix formulas on Wikipedia