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Consider an $N \times N$ matrix $A$ with inverse $B$. In the case of $N=3$ we can directly check that $b_{ij} b_{kk} - b_{ik}b_{kj}$ is equal to $(-1)^{i+j} a_{ij}/\det(A)$. What is the extension of this result to $N>3$?

Another way to pose this question is as follows: in https://math.stackexchange.com/a/4245146/165163 it is shown that $b_{ik}b_{ki} = b_{ii}b_{kk} -\frac{\det A_{[ii,kk]}}{\det A}$. What is the extension of this relationship to the case with $i \neq j$?

Andres
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1 Answers1

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There is a result apparently known as Jacobi's formula for the minors of an inverse matrix. We need to define the following notation:

  • For subsets $I,J \subset \{1,\dots,N\}$, $A_{I,J}$ denotes the submatrix of $A$ using only the $i$th rows and $j$th columns of $A$ for $i \in I$ and $j \in J$
  • $I^c$ denotes the complement of $I$ (i.e. $I^c =\{i : 1 \leq i \leq n, i \notin I\}$)
  • $\sigma(I)$ denotes the sum of all elements of $I$

With that: if $I$ and $J$ are sets of $2$ elements, we have $$ \det(B_{I,J}) = (-1)^{\sigma(I) + \sigma(J)}\frac{\det A_{J^c,I^c}}{\det(A)}. $$ For $N = 3$, this gives you the formula you found. Taking $I = \{i,k\}$ and $J = \{j,k\}$ yields $$ \det(B_{I,J}) = b_{ij}b_{kk} - b_{ik}b_{kj},\\ (-1)^{\sigma(I) + \sigma(J)}\frac{\det A_{J^c,I^c}}{\det(A)} = (-1)^{i + j + 2k} \frac{a_{ij}}{\det(A)} = (-1)^{i + j} \frac{a_{ij}}{\det(A)}. $$ For $N = 4$, here is an example. Taking $I = \{1,2\}$ and $J = \{3,4\}$ yields $$ \det(B_{I,J}) = b_{ip}b_{jq} - b_{iq}b_{jp},\\ (-1)^{\sigma(I) + \sigma(J)} \frac{\det A_{J^c,I^c}}{\det(A)} = \frac{(-1)^{1+2+3+4}}{\det(A)}(\det A_{J^c,I^c}) = \frac 1{\det(A)} (a_{13}a_{24} - a_{14}a_{23}). $$

Ben Grossmann
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  • In the line for $\det(B_{I,J})$ in the $N=3$ case I think it should read $\det(B_{I,J}) = b_{jk} b_{kk} - b_{jk} b_{kj}$. – Andres Sep 16 '21 at 15:20
  • In the line of the main result, I think it should be $\det A_{I^c,J^c}$ rather than $\det A_{J^c,I^c}$ – Andres Sep 16 '21 at 15:24
  • @Andres I've corrected the expression for $B_{I,J}$. Regarding your second comment, the result is written correctly. The rows of $B$ that are selected determine the columns of $A$ that are selected and vice versa. – Ben Grossmann Sep 16 '21 at 15:31