1. ## Linear Dependance

Hello!
The problem is this,
Show that the set S is L.D. by example, find a nontrivial linear combonation of the vectors in S (whose sum is the zero vector)

$S= {(3,3,1,2),(3,1,4,1),(-3,7,-16,3)}$

This looks to me like a system of 4 equations in 3 unknowns, but can't see how to do it except for trial and error (trying different linear combonations...)

Any help greatly appriciated!

Thanks!

2. Originally Posted by matt.qmar
Hello!
The problem is this,
Show that the set S is L.D. by example, find a nontrivial linear combonation of the vectors in S (whose sum is the zero vector)

$S= {(3,3,1,2),(3,1,4,1),(-3,7,-16,3)}$

This looks to me like a system of 4 equations in 3 unknowns, but can't see how to do it except for trial and error (trying different linear combonations...)

Any help greatly appriciated!

Thanks!

You're looking for scalars $a,b,c$ s.t. $a(3,3,1,2)+b(3,1,4,1)+c(-3,7,-16,3)=(0,0,0,0)\Longrightarrow \left\{\begin {array}{l}3a+3b-3c=0\\3a+b+7c=0\\a+4b-16c=0\\2a+b+3c=0\end{array}\right.$.

Well, now just solve this system and you're done...one of the many possible solutions is $a=-4\,,\,b=5\,,\,c=1$. Can you find the general solution?

Tonio

3. Method 2: Form the matrix whose rows are the given vectors, and reduce to echelon form using the elementary row operations. You will find that the last row of this matrix is a zero row. Since the echelon matrix has a zero row, the vectors are dependent. Try it!

4. --> Since the echelon matrix has a zero row, the vectors are dependent.

So, 0a+0b+0c=0 in a row implies linear dependance?
I'm not sure I understand that part.

Certainly, I see that if the system has parametric solutions (thus, infinetly many non-0 solutions) is thus LD by definition...

Thank you for the help!

5. Well, the homogeneous system must only have the trivial solution, if its row vectors are linearly independent. Now a matrix containing a zero row can never be invertible (your matrix is not square either). And a matrix who is not invertible, implies that the system has nontrivial solutions, and hence the vectors are linearly dependent. Furthermore, since it contains a zero row it will no longer be row equivalent to the identity matrix, this is the other condition.

Likewise, if the echelon matrix has no zero rows, the vectors are independent.