proof of cofactor expansion

Let $M\in ma{t}_N(K)$ be a $n\times n$-matrix with entries from a commutativefield $K$. Let $e_1,\mathrm{\dots },e_n$ denote the vectors of the canonical basis of$K^n$. For the proof we need the following

Lemma: Let $M_{ij}^{*}$ be the matrix generated by replacing the $i$-throw of $M$ by $e_j$. Then

$$det{M}_{ij}^{*}={(-1)}^{i+j}det{M}_{ij}$$

where $M_{ij}$ is the $(n-1)\times (n-1)$-matrix obtained from $M$ by removingits $i$-th row and $j$-th column.

Proof.

By adding appropriate of the $i$-th row of $M_{ij}^{*}$to its remaining rows we obtain a matrix with 1 at position $(i,j)$ and 0 atpositions $(k,j)$ ($k\ne i$). Now we apply the permutation

$$(12)\circ (23)\circ \mathrm{\dots }\circ ((i-1)i)$$

to rows and

$$(12)\circ (23)\circ \mathrm{\dots }\circ ((j-1)j)$$

to columns of the matrix. The matrix now looks like this:

• Row/column 1 is the vector $e_1$;

• under row 1 and right of column 1 is the matrix $M_{ij}$.

Since the determinant has changed its sign $i+j-2$ times, we have

$$det{M}_{ij}^{*}={(-1)}^{i+j}det{M}_{ij}.$$

Note also that only those permutations$\pi \in S_n$ are for the computation of the determinant of $M_{ij}^{*}$ where $\pi (i)=j$.∎

Now we start out with

$detM$$={\displaystyle \sum _{\pi \in S_n}}\mathrm{sgn}\pi \left({\displaystyle \prod _{j=1}^n}{m}_{j\pi (j)}\right)$

$={\displaystyle \sum _{k=1}^n}{m}_{ik}\left({\displaystyle \sum _{\pi \in S_n\mid \pi (i)=k}}\mathrm{sgn}\pi \left({\displaystyle \prod _{1\le j\le i}}{m}_{j\pi (j)}\right)\cdot 1\cdot \left({\displaystyle \prod _{i\le j\le n}}{m}_{j-\pi (j)}\right)\right).$

From the previous lemma, it follows that the associated with $M_{ik}$is the determinant of $M_{ij}^{*}$. So we have

$$detM=\sum _{k=1}^n{M}_{ik}\left({(-1)}^{i+k}det{M}_{ik}\right).$$