For those that are linearly dependent, express one of the polynomials as a linear combination of the others {1-2x,3x+x^2-x^3,1+x^2+2x^3,3+2x+3x^3}

Test the sets of polynomials for linear inde­pendence. For those that are linearly dependent, express one of the polynomials as a linear combination of the others {$$\displaystyle{1}-{2}{x},{3}{x}+{x}^{{2}}-{x}^{{3}},{1}+{x}^{{2}}+{2}{x}^{{3}},{3}+{2}{x}+{3}{x}^{{3}}$$} in P3

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Cristiano Sears

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.