#math/linear-algebra For an $n\times n$ matrix $A=\{ a_{j,k} \}^{n}_{j,k}=1$, let $A_{j,k}$ denote the $(n-1)\times(n-1)$ matrix obtained from $A$ by crossing out the row number $j$ and column number $k.$ For each $1\leq{}j\leq{}n,$ determinant of $A$ can be expanded in the row number $j$ as $ \begin{aligned} \det A &= a_{j,1}(-1)^{j+1}\det A_{j,1} + a_{j,2}(-1)^{j+2}\det A_{j,2} + \cdots + a_{j,n}(-1)^{j+n}\det A_{j,n} \\ &= \sum_{k=1}^{n} a_{j,k}(-1)^{j+k}\det A_{j,k}. \end{aligned} $ Similarly, for each $k$, $1\leq k\leq n$, the determinant can be expanded in the column number $k$, $ \det A = \sum_{j=1}^{n} a_{j,k}(-1)^{j+k}\det A_{j,k}. $ ## Proof #TODO LADW, pg 91.