Proof of Theorem DRCMA

Suppose that $A$ is a square matrix of size $n$. Form the matrix $B$ by multiplying row $s$ by $\alpha$ and adding it to row $t$. Let $C$ be the auxiliary matrix where we replace row $t$ of $A$ by row $s$ of $A$. Notice that $\submatrix{A}{t}{j}=\submatrix{B}{t}{j}=\submatrix{C}{t}{j}$ for all $1\leq j\leq n$. We compute the determinant of $B$ by expansion about row $t$. \begin{align*} \detname{B} &= \sum_{j=1}^{n}(-1)^{t+j}\matrixentry{B}{tj}\detname{\submatrix{B}{t}{j}} &&\knowl{./knowls/theorem.DER.knowl}{\text{Theorem DER}}\\ &= \sum_{j=1}^{n}(-1)^{t+j}\left(\alpha\matrixentry{A}{sj}+\matrixentry{A}{tj}\right)\detname{\submatrix{B}{t}{j}} &&\text{Hypothesis}\\ &= \sum_{j=1}^{n}(-1)^{t+j}\alpha\matrixentry{A}{sj}\detname{\submatrix{B}{t}{j}}\\ &\quad\quad+ \sum_{j=1}^{n}(-1)^{t+j}\matrixentry{A}{tj}\detname{\submatrix{B}{t}{j}}\\ &= \alpha\sum_{j=1}^{n}(-1)^{t+j}\matrixentry{A}{sj}\detname{\submatrix{B}{t}{j}}\\ &\quad\quad+ \sum_{j=1}^{n}(-1)^{t+j}\matrixentry{A}{tj}\detname{\submatrix{B}{t}{j}}\\ &= \alpha\sum_{j=1}^{n}(-1)^{t+j}\matrixentry{C}{tj}\detname{\submatrix{C}{t}{j}}\\ &\quad\quad+ \sum_{j=1}^{n}(-1)^{t+j}\matrixentry{A}{tj}\detname{\submatrix{A}{t}{j}}\\ &= \alpha\detname{C}+\detname{A}&&\knowl{./knowls/theorem.DER.knowl}{\text{Theorem DER}}\\ &= \alpha\,0+\detname{A} = \detname{A}&&\knowl{./knowls/theorem.DERC.knowl}{\text{Theorem DERC}} \end{align*}

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