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### Linear Dependence of Vectors

```Date: 12/04/2008 at 06:14:27
Subject: linearly dependence of vectors

If the vectors (2,-1,4), (2,k+6,k), and (1,0,2) are linearly
independent, then what is the value(s) of "k"?

I'm not sure how to find if there is more than one answer.  I think k
might be -6 or 4.

```

```
Date: 12/04/2008 at 15:08:02
From: Doctor Minter
Subject: Re: linearly dependence of vectors

Recall that one of the definitions of linear independence of vectors
v1, v2, ... , vn is that the equation

c1*v1 + c2*v2 + ... + cn*vn = 0, for some constants c1,c2,...,cn

has ONLY the trivial solution v1 = v2 = ... = vn = 0.

Otherwise, they are linearly dependent.  Thus, if the vectors are
linearly DE-pendent, there must be some constants c1, c2, and c3 such
that

c1*(2,-1,4) + c2*(2,k+6,k) + c3*(1,0,2) = (0,0,0)

Each component of the vector equation must be true separately.  This

2*c1 + 2*c2 + c3 = 0

-c1 + (k+6)*c2 = 0

4*c1 + k*c2 + 2*c3 = 0

which can be rewritten in matrix form:

[ 2    2   1 ] [c1]   [0]
[ -1 (k+6) 0 ] [c2] = [0]
[ 4    k   2 ] [c3]   [0]

The ONLY way that this matrix equation has a non-trivial solution
(that is, some answer OTHER than c1=c2=c3=0) is if the coefficient
matrix is singular, which will happen only when its determinant is
zero.

Thus, we need to find value(s) for k such that the determinant of
this matrix is zero.  Using cofactors, and expanding about the third
column (since it has a zero--making less work for us), we have

1*| -1  (k+6) | + 2*|  2    2   |
|  4    k   |     | -1  (k+6) |

where the |'s mean to take the determinants of the 2x2 matrices.

The determinant of the first matrix is -5*k - 24 since the
determinant of a 2x2 matrix

| a b | = a*d - b*c
| c d |

Remember, what we're trying to find is the determinant of the 3x3
coefficient matrix in terms of k.  That determinant should be set
equal to zero, and then the resulting equation can be solved for k.

If you want to find what the constants c1, c2, and c3 are, you can go
back, plug in your value for k into the linear system (the matrix
equation), and solve using Gaussian Elimination.  Note that since the
determinant of the matrix is zero when you plug in this value of k,
Cramer's Rule will not work.  You should find that there are
infinitely many values of c1, c2, and c3 that will make that equation
true.

asked for values of k that make the vectors linearly DE-pendent, then
it's the value(s) of k that you found by setting the determinant
equal to zero.  If you're asked for values of k that make the vectors
linearly INDE-pendent, then its any value of k OTHER than what you
found by setting the determinant equal to zero.

I hope this helps!  Please feel free to write again if you need
further assistance, or if you have any other questions.  Thanks for
using Dr. Math!

- Doctor Minter, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Linear Algebra

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