How to prove the Cofactor Expansion Theorem for Determinant of a Matrix?

Question

I understand that the definition of a determinant of a matrix implies that you can expand over the first row over the matrix, but how does that itself imply that you can expand over any row the matrix. Is there a proper proof to this, perhaps using induction on the size of matrix or something? Thanks in advance.

Answer 1

Below is a proof I found here. The idea is to do induction: since the minors are smaller matrices, one can calculate them via the desired row/column.

One first checks by hand that the determinant can be calculated along any row when $n=1$ and $n=2$. $\newcommand\detname{\,\det}$ $\newcommand\matrixentry[2]{#1_{#2}}$ $\newcommand\submatrix[3]{#1(#2|#3)}$

For the induction, we use the notation $\submatrix{A}{i_1,i_2}{j_1,j_2}$ to denote the $(n-2)\times (n-2)$ matrix obtained from $A$ by removing the rows $i_1$ and $i_2$, and the columns $j_1$ and $j_2$. We assume as inductive hypothesis that for square matrices with $n-1$ rows or less, the determinant can be calculated along any row/column.

For the calculation of the minors there will be a column missing when we express the minor in terms of the entries of the original matrix, so one needs to be careful with the signs. For that we use $$ \epsilon_{\ell j}=\begin{cases} 0,&\ \ell <j \\ 1,&\ \ell>j\end{cases} $$ As mentioned above, the idea is that one calculates the minors along the $i^{\rm th}$ row, which is ok by inductive hypothesis. We assume $i>1$, as we are comparing with calculating along the first row. The inductive hypothesis is used below in the second and last equalities.

\begin{align*} \detname{A} &= \sum_{j=1}^{n}(-1)^{1+j}\matrixentry{A}{1j}\detname{\submatrix{A}{1}{j}} \\ &= \sum_{j=1}^{n}(-1)^{1+j}\matrixentry{A}{1j} \sum_{\substack{1\leq\ell\leq n\\\ell\neq j}} (-1)^{i-1+\ell-\epsilon_{\ell j}}\matrixentry{A}{i\ell}\detname{\submatrix{A}{1,i}{j,\ell}} \\ &= \sum_{j=1}^{n}\sum_{\substack{1\leq\ell\leq n\\\ell\neq j}} (-1)^{j+i+\ell-\epsilon_{\ell j}} \matrixentry{A}{1j}\matrixentry{A}{i\ell}\detname{\submatrix{A}{1,i}{j,\ell}} \\ &= \sum_{\ell=1}^{n}\sum_{\substack{1\leq j\leq n\\j\neq\ell}} (-1)^{j+i+\ell-\epsilon_{\ell j}} \matrixentry{A}{1j}\matrixentry{A}{i\ell}\detname{\submatrix{A}{1,i}{j,\ell}} \\ &= \sum_{\ell=1}^{n}(-1)^{i+\ell}\matrixentry{A}{i\ell} \sum_{\substack{1\leq j\leq n\\j\neq\ell}} (-1)^{j-\epsilon_{\ell j}} \matrixentry{A}{1j}\detname{\submatrix{A}{1,i}{j,\ell}} \\ &= \sum_{\ell=1}^{n}(-1)^{i+\ell}\matrixentry{A}{i\ell} \sum_{\substack{1\leq j\leq n\\j\neq\ell}} (-1)^{\epsilon_{\ell j}+j} \matrixentry{A}{1j}\detname{\submatrix{A}{i,1}{\ell,j}} \\ &= \sum_{\ell=1}^{n}(-1)^{i+\ell}\matrixentry{A}{i\ell}\detname{\submatrix{A}{i}{\ell}} \end{align*}