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Determinant
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=== Block matrices === The formula for the determinant of a <math>2 \times 2</math> matrix above continues to hold, under appropriate further assumptions, for a [[block matrix]], i.e., a matrix composed of four submatrices <math>A, B, C, D</math> of dimension <math>m \times m</math>, <math>m \times n</math>, <math>n \times m</math> and <math>n \times n</math>, respectively. The easiest such formula, which can be proven using either the Leibniz formula or a factorization involving the [[Schur complement]], is :<math>\det\begin{pmatrix}A& 0\\ C& D\end{pmatrix} = \det(A) \det(D) = \det\begin{pmatrix}A& B\\ 0& D\end{pmatrix}.</math> If <math>A</math> is [[Invertible matrix|invertible]], then it follows with results from the section on multiplicativity that :<math>\begin{align} \det\begin{pmatrix}A& B\\ C& D\end{pmatrix} & = \det(A)\det\begin{pmatrix}A& B\\ C& D\end{pmatrix} \underbrace{\det\begin{pmatrix}A^{-1}& -A^{-1} B\\ 0& I_n\end{pmatrix}}_{=\,\det(A^{-1})\,=\,(\det A)^{-1}}\\ & = \det(A) \det\begin{pmatrix}I_m& 0\\ C A^{-1}& D-C A^{-1} B\end{pmatrix}\\ & = \det(A) \det(D - C A^{-1} B), \end{align}</math> which simplifies to <math>\det (A) (D - C A^{-1} B)</math> when <math>D</math> is a <math>1 \times 1</math> matrix. A similar result holds when <math>D</math> is invertible, namely :<math>\begin{align} \det\begin{pmatrix}A& B\\ C& D\end{pmatrix} & = \det(D)\det\begin{pmatrix}A& B\\ C& D\end{pmatrix} \underbrace{\det\begin{pmatrix}I_m& 0\\ -D^{-1} C& D^{-1}\end{pmatrix}}_{=\,\det(D^{-1})\,=\,(\det D)^{-1}}\\ & = \det(D) \det\begin{pmatrix}A - B D^{-1} C& B D^{-1}\\ 0& I_n\end{pmatrix}\\ & = \det(D) \det(A - B D^{-1} C). \end{align}</math> Both results can be combined to derive [[Sylvester's determinant theorem]], which is also stated below. If the blocks are square matrices of the ''same'' size further formulas hold. For example, if <math>C</math> and <math>D</math> [[commutativity|commute]] (i.e., <math>CD=DC</math>), then<ref>{{Cite journal|first=J. R.|last= Silvester|title= Determinants of Block Matrices|journal= Math. Gaz.|volume=84 |issue= 501|year=2000 | pages= 460β467| jstor=3620776|url= https://hal.archives-ouvertes.fr/hal-01509379/document|doi= 10.2307/3620776|s2cid= 41879675}}</ref> :<math>\det\begin{pmatrix}A& B\\ C& D\end{pmatrix} = \det(AD - BC).</math> This formula has been generalized to matrices composed of more than <math>2 \times 2</math> blocks, again under appropriate commutativity conditions among the individual blocks.<ref>{{cite journal|last1=Sothanaphan|first1=Nat|title=Determinants of block matrices with noncommuting blocks|journal=Linear Algebra and Its Applications|date=January 2017|volume=512| pages=202β218| doi=10.1016/j.laa.2016.10.004|arxiv=1805.06027|s2cid=119272194}}</ref> For <math>A = D</math> and <math>B = C</math>, the following formula holds (even if <math>A</math> and <math>B</math> do not commute). :<math>\det\begin{pmatrix}A & B\\ B & A\end{pmatrix} = \det\begin{pmatrix}A+B & B\\ B+A & A\end{pmatrix} = \det\begin{pmatrix}A+B & B\\ 0 & A-B\end{pmatrix} = \det(A+B) \det(A-B).</math>
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