Matrix/Adjugate matrix/Formula with determinant/Fact/Proof

Proof

Let ${}M={\left(a_{ij}\right)}_{ij}$. Let the coefficients of the adjugate matrix be denoted by

${}b_{ik}=(-1)^{i+k}\det M_{ki}\,.$

The coefficients of the product ${}(M^{\operatorname {adj} })\cdot M$ are

${}c_{ij}=\sum _{k=1}^{n}b_{ik}a_{kj}=\sum _{k=1}^{n}(-1)^{i+k}a_{kj}\det M_{ki}\,.$

In case ${}j=i$, this is ${}\det M$, as this sum is the expansion of the determinant with respect to the ${}j$-th column. So let ${}j\neq i$, and let ${}N$ denote the matrix that arises from ${}M$ by replacing in ${}M$ the ${}i$-th column by the ${}j$-th column. If we expand ${}N$ with respect to the ${}i$-th column, then we get

${}0=\det N=\sum _{k=1}^{n}(-1)^{i+k}a_{kj}\det M_{ki}=c_{ij}\,.$

Therefore, these coefficients are ${}0$, and the first equation holds.
The second equation is proved similarly, where we use now the expansion of the determinant with respect to the rows.

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