"\\begin{cases}\n x_1+2x_2-2x_3=0, (1)\\\\ \n x_1-3x_2=0, (2)\\\\\n-2x_1+2x_3=0, (3)\\\\\n-2x_1+2x_2+2x_3=0. (4)\n\\end{cases}"
From (2): "x_2=x_1\/3", from (3): "x_3=x_1," to (1):
"x_1+2x_1\/3-x_1=0,"
"x_1=0,"
(2): "x_2=0,"
(3): "x_3=0."
Only "x_1=x_2=x_3=0" is the solution.
So, by the defenition, vectors are linearly independent.
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