<s>[INST] How to show this equality? This is a equation from a book of control theory:
$$c(sI-A)^{-1}b=\frac{\det(sI-A)-\det(sI-A-bc)}{\det(sI-A)}$$
$I$ is identity matrix, $A$ is $n\times n$, $b$ is $n \times 1$ vector, $c$ is $1 \times n$ vector.
I was trying to use induction, but seems not work. I would very appreciate some advice.
 [/INST] Note that $bc$ is a rank-1-matrix. We will find a formula for $(B - bc)^{-1}$, where $B$ is any $n\times n$-matrix ($sI - A$ in your case). We will prove 
$$ \det (B - bc) = (1 - cB^{-1}b)\det(B) $$
To prove that, note 
\begin{align*}
   \det(B- bc) &= \det B\det(I - B^{-1}bc)\\
   &= \det B \det\begin{pmatrix} I- B^{-1}bc & -B^{-1}b\\ 0  & 1 \end{pmatrix}\\
   &= \det B \det \begin{pmatrix} I & 0\\ c & 1\end{pmatrix}
 \det\begin{pmatrix} I- B^{-1}bc & -B^{-1}b\\ 0  & 1 \end{pmatrix}\begin{pmatrix} I & 0\\ -c & 1\end{pmatrix}\\
   &= \det B \cdot \det\begin{pmatrix} I & -B^{-1}b\\ 0 & 1 - cB^{-1}b\end{pmatrix}\\
   &= \det B \cdot (1 - cB^{-1}b) 
\end{align*}
Hence 
$$ -\det(sI - A - bc) = -\det(sI - A) \cdot \bigl(1 - c(sI-A)^{-1}b\bigr) $$
Now solve for $c(sI- A)^{-1}b$.
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