another linear algebra question.?

Question

what real value of x do the following vectors form a linearly dependent set in R3:

v1=(x, 1/2, 1/2), v2=(1/2, x, 1/2), v3=(1/2, 1/2, x)

my guess is that x can only be 1/2 because if you put the above vectors as the column of a matrix, say A, then u would try to see if Ax=0 would have nontrivial solutions (which means you should get a row of 0s when u row reduce the matrix, therefore i know 1/2 will definitely work). but is there any other number that will work ? if so how did u go about solving this (kind of) problem?

Answer 1

Okay, I want you to imagine Steve Ballmer saying this:

Determinants, determinants, determinants, determinants, determinants, determinants. Determinants, determinants, determinants, determinants, determinants, determinants, determinants, determinants.

And did I mention determinants? These vectors form a linearly independent set iff the matrix formed from these vectors is not invertible iff the determinant of said matrix is 0. Therefore, simply find the determinant, then find the values of x that make it zero:

[x, 1/2, 1/2]
[1/2, x, 1/2]
[1/2, 1/2, x]

determinant: x³ + 1/8 + 1/8 - x/4 - x/4 - x/4

Setting this equal to zero:

x³ - 3x/4 + 1/4 = 0
4x³ - 3x + 1 = 0

By inspection, we know 1/2 is a root, so dividing it out:

1/2 | 4 0 -3 .1
.......... 2 ..1 -1
------------------
........ 4 2 -2 0

So the polynomial factors as (x-1/2) (4x² + 2x - 2). Using the quadratic formula of the second, we obtain:

x = (-2±√(4 - 4*4*(-2)))/8
x = (-2±√(4 + 32))/8
x = (-2±√36)/8
x = (-2±6)/8
x = 1/2 ∨ x=-1

So the values of x that cause the vectors to become linearly dependent are 1/2 (twice), and -1.