{x1+2x2−2x3=0,(1)x1−3x2=0,(2)−2x1+2x3=0,(3)−2x1+2x2+2x3=0.(4)\begin{cases} x_1+2x_2-2x_3=0, (1)\\ x_1-3x_2=0, (2)\\ -2x_1+2x_3=0, (3)\\ -2x_1+2x_2+2x_3=0. (4) \end{cases}⎩⎨⎧x1+2x2−2x3=0,(1)x1−3x2=0,(2)−2x1+2x3=0,(3)−2x1+2x2+2x3=0.(4)
From (2): x2=x1/3x_2=x_1/3x2=x1/3, from (3): x3=x1,x_3=x_1,x3=x1, to (1):
x1+2x1/3−x1=0,x_1+2x_1/3-x_1=0,x1+2x1/3−x1=0,
x1=0,x_1=0,x1=0,
(2): x2=0,x_2=0,x2=0,
(3): x3=0.x_3=0.x3=0.
Only x1=x2=x3=0x_1=x_2=x_3=0x1=x2=x3=0 is the solution.
So, by the defenition, vectors are linearly independent.
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