# Proving generalized form of Laplace expansion along a row - determinant

Definition: Let $A$ be an ($n \times n$)-matrix. Let $M_{ij}$ denote the matrix acquired from $A$ by deleting row $i$ and column $j$. For $n \geq 2$ we define the determinant of $A$ inductively as \begin{align*} \det(A) = \sum_{j=1}^n (-1)^{1+j} a_{1j} \det(M_{1j}). \end{align*} This is called Laplace expansion along the first row.

I now want to proof the generalized form of this definition, i.e. using any row $i$ (not necessarily the first one).

Proposition: Let $A$ be a ($n \times n$)-matrix, and let $i$ be an arbitrary rowindex. Then we have \begin{align*} \det(A) = \sum_{j=1}^n (-1)^{i+j} a_{ij} \det(M_{ij}). \end{align*}

Proof: Let $C_1, C_2, \ldots, C_n$ be the rows of $A$. Thus we get \begin{align*} A = \begin{pmatrix} C_1 \\ \vdots \\ C_i \\ \vdots \\ C_n \end{pmatrix}. \end{align*} Now let $B$ be the matrix that is given as \begin{align*} B = \begin{pmatrix} C_i \\ C_1 \\ \vdots \\ C_{i-1} \\ C_{i+1} \\ \vdots \\ C_n \end{pmatrix}. \end{align*} Then $B$ can be acquired from $A$ by doing an ($i-1$)-amount of successive rowinterchanges. Hence we have $\det(B) = (-1)^{i-1} \det(A)$.

Then I'm not sure how to continue. I know how to get $(-1)^{i+j}$. But if I just insert the definition then I'm stuck with the minor $M_{1j}$, and not the generalized form. Any ideas?

## 2 Answers

Hint: Consider $B = E_1 E_2 \ldots E_{i-1} A$ where $E_k$ is the elementary matrix associated with swapping the $k$-th row and the $(k+1)$-th row. Recall the determinant of a product of matrices is the product of the determinants, and elementary matrix w.r.t. type I ERO has determinant $-1$.

Prove by induction on the row number, so that you only have to deal with one flip-flop at a time and not more complicated permutations. The minor $M_{2j}$ occurring in the "expand along the second row" definition is the same as the minor $M_{1j}$ occurring in the "flip-flop rows one and two and then expand along the first row" definition.