Proof of the determinant of a scalar multiple

Proof

1. For $0 \le k \le n$, let $A_k$ be the matrix whose rows $1, \dots, k$ are $c$ times the rows of $A$, and whose other rows are those of $A$.
2. Then $A_0 = A$. By the definition of the scalar multiple, every entry of $cA$ is $c$ times the entry of $A$, so $A_n = cA$.
3. For $1 \le k \le n$, the matrix $A_k$ results from $A_{k-1}$ when row $k$ is multiplied by $c$.
4. By the rule for scaling a row, $\det A_k = c \det A_{k-1}$ for $1 \le k \le n$.
5. By induction on $k$, step 4 gives $\det A_k = c^k \det A$ for $0 \le k \le n$.
6. For $k = n$, this is $\det(cA) = c^n \det A$.