To illustrate my problem with a specific example, consider the matrix $$D=\left[ \begin {array}{ccc} a_{{1}}+\alpha&a_{{2}}&a_{{3}} \\ a_{{1}}&a_{{2}}+\alpha&a_{{3}} \\ b_{{1}}&b_{{2}}&b_{{3}}\end {array} \right] $$ then show that $\alpha$ divides $\det(D)$.
More generally, my problem is as follows. Given the $n\times n$ matrices $$ A = \left( \begin{array}{c} \mathbf{a}^T\\ \vdots\\ \mathbf{a}^T\\ \mathbf{b}_1^T\\ \vdots\\ \mathbf{b}_k^T\\ \end{array} \right) \quad\mathrm{and}\quad C = \mathrm{diag}([\overbrace{\alpha,...,\alpha}^{\text{m times}},\overbrace{0,...,0}^{\text{k times}}]) $$ That is, $\mathbf{a}\in\mathbb{R}^n$ make the first $m$ rows of $A$ and $\mathbf{b}_i \in \mathbb{R}^n$ make up the last $k$ rows of $A$ while $C$ is the matrix with $\alpha$ on the first $m$ diagonal elements and $0$ elsewhere. Also the vectors $\mathbf{b}_i$ are linearly independent s.t. $A+C$ has full rank when $\alpha\ne 0$.
Then show that that $\alpha^{m-1}$ divides $\det(A+C)$. That is, show that $\det(A+C) = \alpha^{m-1} d$. I do not need to know $d$, I only need to prove that $\alpha^{m-1}$ is always a factor of $\det(A+C)$.
It suffices (for my purposes) to consider the case where $m=2$.
For my example above in $n=3$, Maple gives $\det(D)=\alpha\, \left( b_{{3}}\alpha+a_{{1}}b_{{3}}+b_{{3}}a_{{2}}-a_{{3}}b_{ {1}}-a_{{3}}b_{{2}} \right) $
Here are the steps needed.
(1) Add all columns into column 1.
(2) Subtract row 1 from each of rows 2-m.
At the stage the (1,1) entry is $\alpha +\sum a_i$, and all entries in rows 2-m are zero, apart from those on the diagonal which are $\alpha$.
(3) Evaluate by expanding on rows 2-m first, this will provide the factor $\alpha^{m-1}$ as required.