Linear Algebra problem regarding coordinates and basis

Question

Given $v_1 = (k,2,1)$, $v_2 = (−2,1,0)$, $v_3 = (0,1,1)$ with $k \in \mathbb{R}$, determine for which values of $k$ the three vectors are a basis of $\mathbb{R}^3$.

For these values find the coordinates of the vector $v = (−2,1,2)$ with respect to the basis $\{v_1, v_2, v_3 \}$.

I presume as a start we put the vectors in a matrix and find the determinant and put it DIFFERENT from zero so those values of $k$ are the ones that span $\mathbb{R}^3$.

By calculating the determinant I find a value of $k+2$ which we should be different from zero, hence every value that isn't $-2$ span all $\mathbb{R}^3$ and the first part is ok.

But I don't get the second part at all or am I wrong in setting the problem up?

Maybe there's a system of linear equations involved?

Answer 1

For the second part we need to solve

$$x\vec v_1+y\vec v_2+z\vec v_3=(-2,1,2)\implies A\vec x=\vec v$$

where A is the matrix with $\vec v_i$ as columns.

Answer 2

For the second part, reduce the augmented matrix $$[A,b] =\begin{bmatrix}k & -2 & 0& -2\\2&1&1&1\\ 1&0&1&2\end{bmatrix}$$ into echelon form and we see that the last row of this form has first two zero entries and two non zero entries involving $k$ ;assume $k \neq -2$ and use the third row to find the last coordinate and then use backward substitution to find the other two coordinates!