Solve Ax=b, where A consist of $x^k$, is my method using RREF on augmented matrix valid?

Question

I have Ax=b, $$ \begin{bmatrix} 1 & x & x^2 \\ x & 1 & x \\ x^2 & x & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} x \\ x^2 \\ x^3 \end{bmatrix} $$ and I transform it into an augmented matrix:

$$ \begin{bmatrix}\begin{array}{ccc|c} 1 & x & x^2 & x \\ x & 1 & x & x^2\\ x^2 & x & 1 & x^3 \end{array}\end{bmatrix} $$

I tried to do RREF by considering x and function of x as part of constants, for example, $R_2-R_1x$: $$ \begin{bmatrix}\begin{array}{ccc|c} 1 & x & x^2 & x \\ x-x & 1-x^2 & x-x^3 & x^2-x^2\\ x^2 & x & 1 & x^3 \end{array}\end{bmatrix}= \begin{bmatrix}\begin{array}{ccc|c} 1 & x & x^2 & x \\ 0 & 1-x^2 & x-x^3 & 0\\ x^2 & x & 1 & x^3 \end{array}\end{bmatrix} $$

And finally arrive at, \begin{bmatrix}\begin{array}{ccc|c} 1 & 0 & 0 & x \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \end{array}\end{bmatrix}

$$ a = x, b = 0, c = 0$$

Are the operations to reduce the matrix by multiplying by x and function of x valid?

Answer 1

$\begin{bmatrix}\begin{array}{ccc|c} 1 & x & x^2 & x \\ 0 & 1-x^2 & x-x^3 & 0\\ x^2 & x & 1 & x^3 \end{array}\end{bmatrix}\xrightarrow{R_3\to R_3-x^2R_1}\begin{bmatrix}\begin{array}{ccc|c} 1 & x & x^2 & x \\ 0 & 1-x^2 & x-x^3 & 0\\ 0 & x(1-x^2) & 1-x^4 & 0 \end{array}\end{bmatrix}\\\xrightarrow{R_3\to R_3-xR_2}\begin{bmatrix}\begin{array}{ccc|c} 1 & x & x^2 & x \\ 0 & 1-x^2 & x-x^3 & 0\\ 0 & 0 & 1-x^2 & 0 \end{array}\end{bmatrix}\xrightarrow[R_1\to R_1+R_3]{R_2\to R_2+xR_3}\begin{bmatrix}\begin{array}{ccc|c} 1 & x & 1 & x \\ 0 & 1-x^2 & 0 & 0\\ 0 & 0 & 1-x^2 & 0 \end{array}\end{bmatrix}$

If $x$ is a constant and $1-x^2=0, b,c\in\Bbb R, a=x(1-b)-c=\pm(1-b)-c$.

If $x$ is a constant but $1-x^2\ne0, b=c=0,a=x\ne\pm1$.

If $x$ is a variable, then in general, $b=c=0, a=x.$

Answer 2

We only can divide a row by $x$ if $x\neq0$. Therefore, the case in which $x=0$ should be treated as a special case. More generaly, you can only divide a row by $f(x)$ if $f(x)\neq0$.