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Linear Dependence of Vectors

Date: 12/04/2008 at 06:14:27
From: Addisu
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

Hi Addisu,

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 

  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 
leads us to three equations:

  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 

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 

Important note:  Remember what your question is asking!  If you're 
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 
Associated Topics:
High School Linear Algebra

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