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(3,1,-1), (2,-1,7), (7,-1,23)

2007-03-11 11:53:49 · 6 answers · asked by 8 3 in Science & Mathematics Mathematics

6 answers

Check out their product of say

a x b x c

if the resulting vector is the null vector, they are linearly dependent; otherwise -- independent

Alternatively, if det (a,b,c) = 0 they are dependent, nonzero - independent.

OK, I'll do the determinant test:

3 1 -1
2 -1 7
7 -1 23

= -69 + 49 + 2 - (7 + 46 - 21)
= -50

independent

2007-03-11 12:10:07 · answer #1 · answered by Anonymous · 0 0

No. The 5th vector may well be expressed via potential of a linear mix of the a number of 4 self sustaining vectors. you ought to think of of in R3 and say any vector in r3 is a mix of i j ok vectors. So in accordance to danger say some element like there are certainly 4 perpendicular vectors in R4 only as there is definitely 3 in R3, 2 in R2, and one self sustaining vector in R1

2016-11-24 21:08:18 · answer #2 · answered by ? 4 · 0 0

If all else fails, you have three equations in three variables.

If the only solution is (0, 0, 0), you're done.

2007-03-11 12:38:14 · answer #3 · answered by Curt Monash 7 · 0 1

why dont u ask ur teacher whos teaching this crap to u

2007-03-11 11:59:54 · answer #4 · answered by Anonymous · 0 1

matlab
rref

2007-03-11 12:28:35 · answer #5 · answered by John 5 · 0 1

no

2007-03-11 11:56:25 · answer #6 · answered by Anonymous · 0 4

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