For which values of (lamda) do the following vectors for a linearly dependent set in R3?

Question

V1= (lamda, -1/2 , - 1/2 ) V2= (-1/2, lamda, -1/2 )
V3= (-1/2, -1/2, lamda)

It should say, for "Form" not for...

Answer 1

Let's just write it as L, shall we?
The will be linearly dependent iff the determinant of the matrix with these vectors as rows (or columns, it doesn't matter) is zero.

You should know how to get the determinant of a 3x3 matrix. Briefly, add products on the three down-and-right diagonals, and subtract products on the down-and-left diagonals, assuming that you wrap around if you go off the left or right edge. Thus for
| a b c |
| d e f |
| g h i |
we get aei + bfg + cdh - afh - bdi - ceg.

In this case the determinant is L^3 + 1/8 + 1/8 - L/4 - L/4 - L/4
= 1/4 (4L^3 - 3L + 1)
= 0 <=> 4L^3 - 3L + 1 = 0.
From inspection L = -1 is a solution, so we have
4L^3 - 3L + 1 = (L + 1) (4L^2 - 4L + 1)
= (L + 1)(2L - 1)^2
So the vectors are linearly dependent for L = -1, 1/2.

Answer 2

Please note that V1, V2 and V3 are linearly dependent only if they are coplanar and if 3 vectors are coplanar, their scalar triple product must be zero.
that is,
V1.(V2 X V3)=0
Proof: Since V2 X V3 is perpendicular to both V2 and V3 and since V1, V2, V3 are coplanar, is perpendicular to V1 too. The dot product of 2 perpendicular vectors must be zero. Hence the result.

Ok. Now find V2 X V3 by whatever method you wnt.
V2 X V3=(L^2-1/4,-L/2-1/4,1/4 +L/2)
I have taken lambda as L.

So, V1.(V2 X V3)=L^3-L/4
Equating this expression to 0, it is not hard to get the answers.
L(L^2-1/4)=0.
So, L=0, l=+1/2 and l=-1/2 are the only solutions

Answer 3

Remember, three vectors are L.D. if the determinant of [V1,V2,V3] equals zero.

The determinant of [(lambda, -1/2, -1/2),(-1/2, lambda, -1/2),(-1/2,-1/2, lambda)] = lambda^2 - (3/4)*lambda - 1/4.

Find lambda when the expression equals zero (by the Rational Root Theorem, or by inspection).

Note: if you wanted to find values of lambda which make the vectors linearly INDEPENDENT, then I think you would have to follow modulo_function's advice, augment your vector matrix with zeros, and solve to find restrictions on lambda. That is, I don't think the converse of this theorem is true.

Answer 4

Brian's right, but there might be something more apparent.

A clue is that the lambda appears in a different position in each vector and all others are -1/2

a linear combo:

a*lam -b/2 -c/2 = 0
-1/2-b*lam-c/2 = 0
-1/2-1/2-c*lam = 0

is the condition for linear dependence.