Proof of Theorem DRCM

Suppose that $A$ is a square matrix of size $n$ and we form the square matrix $B$ by multiplying each entry of row $i$ of $A$ by $\alpha$. Notice that the other rows of $A$ and $B$ are equal, so $\submatrix{A}{i}{j}=\submatrix{B}{i}{j}$, for all $1\leq j\leq n$. We compute $\detname{B}$ via expansion about row $i$. \begin{align*} \detname{B} &= \sum_{j=1}^{n}(-1)^{i+j}\matrixentry{B}{ij}\detname{\submatrix{B}{i}{j}} &&\knowl{./knowls/theorem.DER.knowl}{\text{Theorem DER}}\\ &= \sum_{j=1}^{n}(-1)^{i+j}\matrixentry{B}{ij}\detname{\submatrix{A}{i}{j}} &&\text{Hypothesis}\\ &= \sum_{j=1}^{n}(-1)^{i+j}\alpha\matrixentry{A}{ij}\detname{\submatrix{A}{i}{j}} &&\text{Hypothesis}\\ &= \alpha\sum_{j=1}^{n}(-1)^{i+j}\matrixentry{A}{ij}\detname{\submatrix{A}{i}{j}}\\ &= \alpha\detname{A}&&\knowl{./knowls/theorem.DER.knowl}{\text{Theorem DER}}\\ \end{align*}

The proof for the case of a multiple of a column is entirely similar, or could be derived from an application of Theorem DT employing the transpose of the matrix.