Question

v1(3,-2,-2), v2(2,-1,-2), v3(-1,1,1) , u(0,1,h)
find all values of h such that u is a linear combination of v1, v2 and v3

Answer 1

First we check if system $\{v_1, v_2, v_3 \}$ is linearly independent.

$\begin{pmatrix} 3 & -2 & -2 \\ 2 & -1 & -2 \\ -1 & 1 & 1 \end{pmatrix} \overset{R_1 - R_2 + R_3}{=} \begin{pmatrix} 0 & 0 & 1 \\ 2 & -1 & -2 \\ -1 & 1 & 1 \end{pmatrix} \overset{R_2 + 2R_3}{=} \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ -1 & 1 & 1 \end{pmatrix} \overset{R_3 \to -R_3, R_3 + R_1 + R_2}{=}\\ =\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}$

Thus, space of linear combinations of $v_1, v_2, v_3$ includes all vectors of $R^3$, so h can take on any value.