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

linear algebramatricessystems of equations

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?

Best Answer

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 $f(x)\neq0$.