We have to determine whether or not we can find real numbers r, s,t, which are not all zero,

Such that

\(\displaystyle{r}{\left({1}-{2}{x}\right)}+{s}{\left({3}{x}+{x}^{{2}}-{x}^{{3}}\right)}+{t}{\left({1}+{x}^{{2}}+{2}{x}^{{3}}\right)}+{q}{\left({3}+{2}{x}+{3}{x}^{{3}}\right)}={0}\)

To find all possible r, s,t we have to solve the augmented matrix equation:

\(\begin{bmatrix}1&0&1&3|&0\\-2&3&0&2|&0\\0&1&1&0|&0\\0&-1&2&3|&0\end{bmatrix}\)

\(\displaystyle{R}_{{2}}\to{R}_{{2}}+{2}{R}_{{1}}\)

\(\begin{bmatrix}1&0&1&3|&0\\0&3&2&8|&0\\0&1&1&0|&0\\0&-1&2&3|&0\end{bmatrix}\)

\(\displaystyle{R}_{{3}}\to{R}_{{2}}\)

\(\begin{bmatrix}1&0&1&3|&0\\0&1&1&0|&0\\0&3&2&8|&0\\0&-1&2&3|&0\end{bmatrix}\)

\(\displaystyle{R}_{{3}}\to{R}_{{3}}-{3}{R}_{{2}}\) and \(\displaystyle{R}_{{4}}\to{R}_{{4}}+{R}_{{2}}\)

\(\begin{bmatrix}1&0&1&3|&0\\0&1&1&0|&0\\0&0&-1&8|&0\\0&0&3&3|&0\end{bmatrix}\)

\(R_4\to R_4+3R_3\)

\(\begin{bmatrix}1&0&1&3|&0\\0&1&1&0|&0\\0&0&-1&8|&0\\0&0&0&27|&0\end{bmatrix}\)

Hence,

\(\displaystyle{r}{\left({1}-{2}{x}\right)}+{s}{\left({3}{x}+{x}^{{2}}-{x}^{{3}}\right)}+{t}{\left({1}+{x}^{{2}}+{2}{x}^{{3}}\right)}+{q}{\left({3}+{2}{x}+{3}{x}^{{3}}\right)}={0}\)

satisfies only when \(r=s=t=q=0\)

Therefore,

{\(\displaystyle{\left({1}-{2}{x}\right)},{\left({3}{x}+{x}^{{2}}-{x}^{{3}}\right)},{\left({1}+{x}^{{2}}+{2}{x}^{{3}}\right)},{\left({3}+{2}{x}+{3}{x}^{{3}}\right)}\)}

is a linearly independent set of polynomials.