Question

Determine whether the vectors are linearly dependent or independent (1,2,1),(-1,0,1) and (2,-1,4)

Answer 1

$The \space set \space S \space = \space {v_1, \space v_2, \space v_3} \space of \space vectors \space in \space R_3 \space is \space linearly \space independent \space if \space the \space only \space solution \space of \space \\ \space \space \space c_1v_1 \space + \space c_2v_2 \space + \space c_3v_3 \space = \space 0 \space \\ is \space c_1, \space c_2, \space c_3 \space = \space 0\\ Otherwise \space (i.e., \space if \space a \space solution \space with \space at \space least \space some \space nonzero \space values \space exists), \space S \space is \space linearly \space dependent.\\ With \space our \space vectors \space v_1, \space v_2, \space v_3, \space \space becomes:\\ c_1 (1, 2, 1) \space + \space c_2 (-1, 0, 1) \space + \space c_3 (2, -1, 4) \space = \space (0, 0, 0)\\ Rearranging \space the \space left \space hand \space side \space yields\\ 1 \space c_1-1 \space c_2 \space +2 \space c_3=0\\ 2 \space c_1 \space +0 \space c_2-1 \space c_3=0\\ 1 \space c_1 \space +1 \space c_2 \space +4 \space c_3 \space = \space 0 \\ solve \space this \space equations \space , \space we \space will \space get \space the \space system \space has \space only \space the \space trivial \space solution, \space so \space that \space the \space only \space solution \space of \space \space is \space c_1, \space c_2, \space c_3 \space = \space 0.\\ Therefore \space the \space set \space S \space = \space {v_1, \space v_2, \space v_3} \space is \space linearly \space independent.$