MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathbf{M}$ be an integer square $2 \times 2$ block matrix $$ \mathbf{M} = \left( \begin{array}{cc} \mathbf{A} & \mathbf{B} \\ \mathbf{C} & \mathbf{D} \end{array} \right) , $$ where $\mathbf{A}$ and $\mathbf{D}$ are square matrices (not necessarily of the same size). Is there a way of testing if $\mathbf{M}$ is regular ($\det (\mathbf{M}) = \pm 1$) in terms of $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ and $\mathbf{D}$?

For example, is it known some expression of $\det (\mathbf{M})$ in terms of $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ and $\mathbf{D}$ without any extra assumption (such as some regularity or certain commutativity relations) on the blocks? And if you assume that $\mathbf{A}$ is an $1 \times 1$ matrix (i.e. $\mathbf{A} \in \mathbb{Z}$)?

share|cite|improve this question

You can get conditions from the Jacobi identity or (especially in your special case where $\mathbf{A} \in \mathbb{Z}$) from the formulas here. (third equation, in particular)

share|cite|improve this answer

When $A$ is a $1 \times 1$ matrix:

Let $d = \det D$.

  1. if $d \neq 0$, then $\det M = d (A - B D^{-1} C) = dA - B \ adj(D) \ C$, where $adj(D)$ is the adjugate matrix of $D$.
  2. since both sides of this equality are continuous functions of the elements of $D$, this equality holds also when $d = 0$, i.e. when $D$ is singular.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.