Do two nonsingular matrices exist such that this linear system can be solved? (I've posted this question on MathOverflow and was sent here. This is a copy of that submission)
I am trying to solve a system of linear equations. In this system I have two separate systems involving an unknown tension vector $\overrightarrow{T}$ in $\mathbb{R}^n$, which I want to cancel out of both equations by making them equal. In each system, $\overrightarrow{T}$ is multiplied by $n \times n$ matrices $C_{SU}$ and $S_{SU}$, defined as follows:
$$
C_{SU} = 
\begin{bmatrix}
c_1 & -c_2 & 0 & \cdots & 0 \\
0 & c_2 & -c_3 & \cdots & 0 \\
0 & 0 & c_3 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & c_n 
\end{bmatrix}
$$
Where $c_k = \cos{\theta_k}$. $S_{SU}$ is defined similarly, replacing $\cos$ with $\sin$.
So to summarize, I have two linear systems of the following form:
$$
\overrightarrow{R_1} = C_{SU}\overrightarrow{T}\\
\overrightarrow{R_2} = S_{SU}\overrightarrow{T}
$$
Where $\overrightarrow{R_1}$ and $\overrightarrow{R_2}$ are some vectors in $\mathbb{R}^n$. I want to cancel $\overrightarrow{T}$ out of the equation by making both left hand sides equal. To do this, I intend to multiply $C_{SU}$ and $S_{SU}$ by some matrices $A$ and $B$ such that $A \cdot C_{SU} = B \cdot S_{SU}$. $A$ and $B$ should be $n \times n$ invertible matrices.
Do $A$ and $B$ exist, and if so, what are they?
 A: So, if I understood correctly your symbolism, we are
dealing with matrices of the type
$$
\begin{gathered}
  \mathbf{C}\left( \mathbf{x} \right) = \left[ {\begin{array}{*{20}c}
   {x_1 } & { - x_2 } & 0 &  \cdots  & 0  \\
   0 & {x_2 } & { - x_3 } &  \cdots  & 0  \\
   0 & 0 & {x_3 } &  \cdots  & 0  \\
    \vdots  &  \vdots  &  \vdots  &  \ddots  &  \vdots   \\
   0 & 0 & 0 &  \cdots  & {x_n }  \\
 \end{array} } \right] = \left[ {\begin{array}{*{20}c}
   1 & { - 1} & 0 &  \cdots  & 0  \\
   0 & 1 & { - 1} &  \cdots  & 0  \\
   0 & 0 & 1 &  \cdots  & 0  \\
    \vdots  &  \vdots  &  \vdots  &  \ddots  &  \vdots   \\
   0 & 0 & 0 &  \cdots  & 1  \\
 \end{array} } \right]\left[ {\begin{array}{*{20}c}
   {x_1 } & 0 & 0 &  \cdots  & 0  \\
   0 & {x_2 } & 0 &  \cdots  & 0  \\
   0 & 0 & {x_3 } &  \cdots  & 0  \\
    \vdots  &  \vdots  &  \vdots  &  \ddots  &  \vdots   \\
   0 & 0 & 0 &  \cdots  & {x_n }  \\
 \end{array} } \right] =  \hfill \\
   = \left( {\mathbf{I} - \mathbf{E}} \right)\;\mathbf{D}\left( \mathbf{x} \right) \hfill \\ 
\end{gathered} 
$$
where:
$\mathbf{I}$ is the identity matrix, $\mathbf{E}$ is the matrix with elements $=1$ in the first upper diagonal and remaining null, 
and $\mathbf{D}\left( \mathbf{x} \right)$ is the diagonal matrix with  entries $\left( {x_1 ,\; \ldots ,\;x_n } \right)$.
Now, it is well known, and it is easily demonstrable, that
$$
\left( {\mathbf{I} - \mathbf{E}} \right)^{\, - \,\mathbf{1}}  = \mathbf{S} = \left[ {\begin{array}{*{20}c}
   1 & 1 & 1 &  \cdots  & 1  \\
   0 & 1 & 1 &  \cdots  & 1  \\
   0 & 0 & 1 &  \cdots  & 1  \\
    \vdots  &  \vdots  &  \vdots  &  \ddots  &  \vdots   \\
   0 & 0 & 0 &  \cdots  & 1  \\
 \end{array} } \right]
$$
and of course:
$$
\mathbf{D}\left( \mathbf{x} \right)^{\, - \,\mathbf{1}}  = \left[ {\begin{array}{*{20}c}
   {1/x_1 } & 0 & 0 &  \cdots  & 0  \\
   0 & {1/x_2 } & 0 &  \cdots  & 0  \\
   0 & 0 & {1/x_3 } &  \cdots  & 0  \\
    \vdots  &  \vdots  &  \vdots  &  \ddots  &  \vdots   \\
   0 & 0 & 0 &  \cdots  & {1/x_n }  \\
 \end{array} } \right]
$$
Concerning the determinants
$$
\begin{gathered}
  \left| {\left( {\mathbf{I} - \mathbf{E}} \right)} \right| = \left| \mathbf{S} \right| = \;1 \hfill \\
  \left| {\mathbf{D}\left( \mathbf{x} \right)} \right| = \prod\limits_{1\, \leqslant \,k\, \leqslant \,n} {x_{\,k} }  = \frac{1}
{{\left| {\mathbf{D}\left( \mathbf{x} \right)^{\, - \,\mathbf{1}} } \right|}} \hfill \\ 
\end{gathered} 
$$
So the matrix $ \mathbf{C}\left( \mathbf{x} \right) $ is invertible if none of the $x_k$ is null:
$$
\mathbf{C}\left( \mathbf{x} \right)^{\, - \,\mathbf{1}}  = \mathbf{D}\left( \mathbf{x} \right)^{\, - \,\mathbf{1}} \;\mathbf{S}
$$
Therefore the answer to your question is
$$
\begin{gathered}
  \mathbf{A} = \mathbf{C}\left( \mathbf{c} \right)^{\, - \,\mathbf{1}}  = \mathbf{D}\left( \mathbf{c} \right)^{\, - \,\mathbf{1}} \;\mathbf{S}\quad \left| {\;\cos \theta _k  \ne 0} \right. \hfill \\
  \mathbf{B} = \mathbf{C}\left( \mathbf{s} \right)^{\, - \,\mathbf{1}}  = \mathbf{D}\left( \mathbf{s} \right)^{\, - \,\mathbf{1}} \;\mathbf{S}\quad \left| {\;\sin \theta _k  \ne 0} \right. \hfill \\ 
\end{gathered} 
$$
