Linear Dependence of VectorsDate: 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 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 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 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. 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 http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/