> [!NOTE] Proposition (Proof of existence for $n=2$) > Consider the $2\times 2$ matrix $A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$We define $\det A = a_{11}a_{22} - a_{12}a_{21} = \sum_{\sigma\in S_{2}} \text{sgn} (\sigma) a_{1\sigma(1)} a_{2\sigma(2)} $where $S_{n}$ denotes the [[Symmetric Groups of Finite Degree|symmetric group of degree n]] and $\text{sgn } \sigma = -1$ if $\sigma$ is odd and $1$ otherwise. *Proof* (1) holds swapping rows $i$ and $j$ amounts to replacing $\sigma$ by the permutation $(i, j)\sigma$ in the formula which multiplies the sign by $-1.$ (2) follows from the fact that the formula is linear in the entries of the $i$th row $(a_{i_{1}} \quad a_{i_{2}}).$ (3) WLOG, suppose we add $\lambda$ times row $2$ to row $1.$ Then by linearity of the formula in the entries of the $1$st row of $A_{12}(\lambda)A$, which is given by $(a_{11}+ \lambda a_{21} \quad a_{12}+\lambda a_{22})$, we have $\det(A_{12}(\lambda)A)= \det A + \lambda \det A' \quad \text{where }\quad A' = \begin{pmatrix} a_{21} & a_{22} \\ a_{21} & a_{22} \end{pmatrix}$Since $A'$ has two equal rows, $\det A' =0.$ So $\det(A_{ij}(\lambda)A)=\det A$ as required.