I am reading "Introduction to Linear Algebra". And there is an example about "Cyclic difference matrix". I will write the question
$$ Cx = b \\ \begin{bmatrix} 0 & 1 & 0\\ -1 & 0 & 1\\ 0 & -1 & 0\\ \end{bmatrix} \begin{bmatrix} x_1\\x_2\\ x_3\\ \end{bmatrix} = \begin{bmatrix} x_2 - 0\\x_3 - x_1\\ 0-x_2\\ \end{bmatrix} = \begin{bmatrix} b_1\\b_2\\ b_3\\ \end{bmatrix} $$
Show that $Cx = b$ can only be solved when $b_1 + b_3 = 0$. That is a plane of vectors $b$ in three-dimensional space. Each column of $C$ is in the plane, the matrix has no inverse. So this plane contains all the cobinations of those columns (which are all the vectors $Cx$).
There is a solution provided but I don't quite understand it. Can someone explain it in there own words so I can look at it from a different point of view. I can see that $b_1$ and $b_2$, why does that imply that it can only be solved if $b_1$ and $b_2$ equal to zero and non-invert-able. isn't that just a consequence of them being those specific value. I can see that it is non-invertible because column 1 is a multiple of column 3.
From $\begin{bmatrix} 0 & 1 & 0\\ -1 & 0 & 1\\ 0 & -1 & 0\\ \end{bmatrix} \begin{bmatrix} x_1\\x_2\\ x_3\\ \end{bmatrix} = \begin{bmatrix} x_2 - 0\\x_3 - x_1\\ 0-x_2\\ \end{bmatrix} = \begin{bmatrix} b_1\\b_2\\ b_3\\ \end{bmatrix}$
we get $x_2=b_1$ (first row) and $-x_2=b_3$ (third row).
Hence $b_3=-x_2=-b_1$.