Let $A$ be a $n \times n$ matrix with complex elements, $L$ a $1 \times n$ matrix with complex elements and $C$ a $n \times 1$ matrix with complex elements. Show that $\det (A-CL)=\det(A)-LBC$, where $B=\textrm{adj}(A)$ is the adjugate matrix of $A$.
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Let $\mathbb{K}$ be the base field. Consider $$M(t):=\begin{bmatrix}A+tI&L\\C&1\end{bmatrix}$$ as a matrix over $\mathbb{K}(t)$. It can be easily seen that $$M(t)=\begin{bmatrix}I&L\\0&1\end{bmatrix}\,\begin{bmatrix}A+tI-LC&0\\C&1\end{bmatrix}\,.$$ Thus, $$\det\big(M(t)\big)=\det(A+tI-LC)\,.$$ Observe that $A+tI$ is invertible over $\mathbb{K}(t)$ (as the determinant is a monic polynomial in $t$). Then, $$M(t)=\begin{bmatrix}A+tI&0\\C&1\end{bmatrix}\,\begin{bmatrix}I&(A+tI)^{-1}L\\0&1-C(A+tI)^{-1}L\end{bmatrix}\,.$$ Consequently, $$\det\big(M(t)\big)=\det(A+tI)\,\big(1-C(A+tI)^{-1}L\big)=\det(A+tI)-C\,\text{adj}(A+tI)\,L\,.$$ Thus, we have $$\det(A+tI-LC)=\det(A+tI)-C\,\text{adj}(A+tI)\,L\,.\tag{*}$$ This is an equality in $\mathbb{K}\left[t\right]$, since no fractions appear anymore. Now, the evaluation at $t=0$ is a passage of (*) from $\mathbb{K}[t]$ to $\mathbb{K}$, which leads to the required identity.
darij grinberg
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Batominovski
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Looks like $C$ and $L$ are swapped here. – darij grinberg Oct 29 '19 at 02:44