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

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.

• What definition of determinant are you working with and what preliminaries do you have? In case you have proven that the space of determinant functions is one-dimensional, a rather slick proof can be given. – Thorgott Apr 2 at 15:34
• The definition I've been given is as follows: $det(A) = a_{11}(-1)^{1+1}det(A_{11}) + ... + a_{1k}(-1)^{1+k}det(A_{1k})$ where $A$ is a $k$ by $k$ square matrix and $A_{1i}$ is the submatrix of A from deleting the first row and $i$th column. – Tim Apr 2 at 15:40

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 $$j_2$$.
For the calculation of the minors there will be a column missing, so one needs to be careful with the signs. For that we use $$\epsilon_{\ell j}=\begin{cases} 0,&\ \ell j\end{cases}$$ As mentioned above, the idea is that one calculates the minors along the $$i^{\rm th}$$ row, which is ok by inductive hypothesis.
\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*}