Answer:

To test whether \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\) are linearly independent, suppose

\(\alpha\mathbf{a}+\beta\mathbf{b}+\gamma\mathbf{c}=\mathbf{0}.\)

Using the given vectors,

\(\alpha\begin{pmatrix}2\\ -1\\ 1\end{pmatrix}+\beta\begin{pmatrix}1\\ 1\\ 1\end{pmatrix}+\gamma\begin{pmatrix}0\\ 1\\ -1\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}.\)

Comparing components gives

\(2\alpha+\beta=0,\qquad -\alpha+\beta+\gamma=0,\qquad \alpha+\beta-\gamma=0.\)

Adding the second and third equations gives \(2\beta=0\), so \(\beta=0\). Then \(2\alpha+\beta=0\) gives \(\alpha=0\), and substituting into either remaining equation gives \(\gamma=0\).

The only solution is therefore \(\alpha=\beta=\gamma=0\), so the three vectors are linearly independent. Since there are three linearly independent vectors in \(\mathbb{R}^{3}\), they form a basis for \(\mathbb{R}^{3}\).

Now write

\(\mathbf{d}=l\mathbf{a}+m\mathbf{b}+n\mathbf{c}.\)

Then

\(l\begin{pmatrix}2\\ -1\\ 1\end{pmatrix}+m\begin{pmatrix}1\\ 1\\ 1\end{pmatrix}+n\begin{pmatrix}0\\ 1\\ -1\end{pmatrix}=\begin{pmatrix}3\\ -2\\ 0\end{pmatrix}.\)

Comparing components gives

\(2l+m=3,\qquad -l+m+n=-2,\qquad l+m-n=0.\)

Solving these simultaneous equations gives \(l=2\), \(m=-1\) and \(n=1\). Hence

\(\mathbf{d}=2\mathbf{a}-\mathbf{b}+\mathbf{c}.\)