We prove each property using the permutation formula $\det(A) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma(i)}$. **Property 1** ($\det(A^T) = \det(A)$). The $(i,j)$ entry of $A^T$ is $a_{ji}$. Then: \begin{align*} \det(A^T) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^n a_{\sigma(i),i} = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^n a_{\sigma(i),i}. \end{align*} Relabelling the product by setting $j = \sigma(i)$ (so $i = \sigma^{-1}(j)$) and using $\operatorname{sgn}(\sigma^{-1}) = \operatorname{sgn}(\sigma)$: \begin{align*} \det(A^T) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma^{-1}) \prod_{j=1}^n a_{j,\sigma^{-1}(j)} = \sum_{\tau \in S_n} \operatorname{sgn}(\tau) \prod_{j=1}^n a_{j,\tau(j)} = \det(A), \end{align*} where $\tau = \sigma^{-1}$ runs over all of $S_n$ as $\sigma$ does. **Property 2** (Row swap changes sign). Let $A'$ be obtained from $A$ by swapping rows $r$ and $s$. Then $a'_{i,j} = a_{\tau(i),j}$ where $\tau = (r\;s)$ is the transposition. So: \begin{align*} \det(A') = \sum_{\sigma} \operatorname{sgn}(\sigma) \prod_i a_{\tau(i),\sigma(i)} = \sum_{\sigma} \operatorname{sgn}(\sigma) \prod_i a_{i,\sigma(\tau^{-1}(i))}, \end{align*} and substituting $\sigma' = \sigma \circ \tau^{-1}$ with $\operatorname{sgn}(\sigma') = \operatorname{sgn}(\sigma)\operatorname{sgn}(\tau^{-1}) = -\operatorname{sgn}(\sigma)$ gives $\det(A') = -\det(A)$. **Property 3** (Equal rows $\implies$ $\det = 0$). If rows $r$ and $s$ are equal, swapping them gives the same matrix, so $\det(A) = -\det(A)$ by Property 2, hence $2\det(A) = 0$, so $\det(A) = 0$. **Property 4** (Multilinearity). Linearity in row $k$ follows directly from the permutation formula: each term $\prod_i a_{i,\sigma(i)}$ is linear in the $k$th row entries $a_{k,\sigma(k)}$, so the entire sum is linear in the entries of row $k$. **Property 5** ($\det(AB) = \det(A)\det(B)$). If $B$ is not invertible, then $\operatorname{rank}(AB) \leq \operatorname{rank}(B) < n$, so $AB$ is not invertible and $\det(AB) = 0 = \det(A) \cdot 0$. If $B$ is invertible, write $B$ as a product of elementary matrices (row operations). Each elementary matrix $E$ satisfies $\det(EA) = c \cdot \det(A)$ where $c = \det(E)$ (this follows from Properties 2 and 4: swapping rows multiplies by $-1$, scaling a row by $\lambda$ multiplies by $\lambda$, and adding a multiple of one row to another preserves the determinant). Composing: $\det(AB) = \det(E_1 \cdots E_m A) = \det(E_1)\cdots\det(E_m)\det(A) = \det(B)\det(A)$. **Property 6** ($A$ invertible $\iff$ $\det(A) \neq 0$). If $A$ is invertible, $1 = \det(I) = \det(AA^{-1}) = \det(A)\det(A^{-1})$ by Property 5, so $\det(A) \neq 0$. Conversely, if $A$ is not invertible, its columns are linearly dependent: some column is a linear combination of the others. Adding multiples of columns to that column (which preserves the determinant by multilinearity and Property 3) produces a zero column, giving $\det(A) = 0$.