Question about inverse of matrix

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    Inverse Matrix
Carl Rowan
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Homework Statement


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Given this matrix
##\begin{bmatrix}As+B \\ C \end{bmatrix}##
which is invertible and ##A## has full row rank. I would like to show that its inverse has no terms with ##s## or higher degree if
##\begin{bmatrix}A \\ C \end{bmatrix}##
is invertible.

Homework Equations



The Attempt at a Solution



The only thing I have concluded is that you can manipulate it like this:

##\begin{bmatrix}I & 0 \\ 0 & sI \end{bmatrix} \begin{bmatrix}As+B \\ C \end{bmatrix}=\begin{bmatrix}As+B \\ Cs \end{bmatrix}=s\begin{bmatrix}A \\ C \end{bmatrix}+\begin{bmatrix}B \\ 0 \end{bmatrix}##
where each ##I## is conformable with the matrix product. Not sure how to take this any further, though. I think my main problem is that I don't know how to use the rank property of ##A##.
 
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What can you say about the determinant of ##\begin{bmatrix}As \\ C \end{bmatrix}## in terms of powers of s?
How can you use the determinant (of the original matrix) to find entries of the full inverse matrix?
 
Thank you for your hints, pushed me into the right direction and helped me prove the statement!
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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