Answer:
So the vectors are linearly independent.
Explanation:
So if they are linearly independent then the following scalars in will have the condition a=b=c=0:
a(2,3,1)+b(2,-5,-3)+c(-3,8,-5)=(0,0,0).
We have three equations:
2a+2b-3c=0
3a-5b+8c=0
1a-3b-5c=0
Multiply last equation by -2:
2a+2b-3c=0
3a-5b+8c=0
-2a+6b+10c=0
Add equation 1 and 3:
0a+8b+7c=0
3a-5b+8c=0
-2a+6b+10c=0
Divide equation 3 by 2:
0a+8b+7c=0
3a-5b+8c=0
-a+3b+2c=0
Multiply equation 3 by 3:
0a+8b+7c=0
3a-5b+8c=0
-3a+9b+6c=0
Add equation 2 and 3:
0a+8b+7c=0
3a-5b+8c=0
0a+4b+13c=0
Multiply equation 3 by -2:
0a+8b+7c=0
3a-5b+8c=0
0a-8b-26c=0
Add equation 1 and 3:
0a+0b-19c=0
3a-5b+8c=0
0a-8b-26c=0
The first equation tells us -19c=0 which implies c=0.
If c=0 we have from the second and third equation:
3a-5b=0
0a-8b=0
0a-8b=0
0-8b=0
-8b=0 implies b=0
We have b=0 and c=0.
So what is a?
3a-5b=0 where b=0
3a-5(0)=0
3a-0=0
3a=0 implies a=0
So we have a=b=c=0.
So the vectors are linearly independent.