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Say $A$ is a $(n-1)\times (n-1)$ matrix and we augment it by a $n^{th}$ row and a column and get a $n \times n$ matrix $B$. Is there a nice way to relate $det(B)$ and $det(A)$ and the added row and column?

A close by thing I am reminded of is this, http://en.wikipedia.org/wiki/Matrix_determinant_lemma

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  • $\begingroup$ Well, $\text{det}(B)$ is a sum of $n$ terms and only one of them is a multiple of $\text{det}(A)$. So you can't expect much here without severely constraining that additional row and column, can you? $\endgroup$
    – Vidit Nanda
    Commented Jan 7, 2015 at 8:59

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yes, there is the Sherman-Morrison formula $$\det B=(\det A)(b-yA^{-1}x),$$ where $b, x$ and $y$ are blocks: $$B=\begin{pmatrix} A & x \\ y & b \end{pmatrix}.$$

Edit. After Hachino's comment, one can also write $$\det B=b\det A-y\hat Ax,$$ where $\hat A$ is the transpose of the cofactor matrix.

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  • $\begingroup$ This only works for $\det(A) \neq 0$... $\endgroup$
    – Michael Stoll
    Commented Jan 7, 2015 at 10:22
  • $\begingroup$ @Michael: of course $\endgroup$
    – Denis Serre
    Commented Jan 7, 2015 at 10:53
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    $\begingroup$ Not only, if one recalls that $det(A) A^{-1}$ is the adjunct matrix of $A$ (nothing to do with adjoints or symmetric matrices here), which is defined for any matrix. $\endgroup$
    – Hachino
    Commented Jan 7, 2015 at 12:18
  • $\begingroup$ @Hachino. Right ! $\endgroup$
    – Denis Serre
    Commented Jan 7, 2015 at 12:55
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    $\begingroup$ The formula (with the cofactor matrix) for $\det B$ is equation $(19)$ in Cauchy's 1815 memoir on determinants. $\endgroup$
    – Dan Fox
    Commented Jan 7, 2015 at 17:23

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