# Thread: LinAlg - a few questions?

1. ## LinAlg - a few questions?

so i have the matrices
[-12
16
-13] = A

(it is a 3x1 matrix)

[-3
5
-4] = B

[-3
3
-3] = C

Determine whether A, B, and C are linearly independent
If they are linearly dependent, determine a non-trivial linear relation - (a non-trivial relation is three numbers which are not all three zero.) otherwise, if the vectors are linearly independent, enter 0's for the coefficients, since that relationship always holds.
______A + ______B + _______C = 0

my teacher told me to do something with C1, C2... Cn... but I have no idea what that means! haha

2. You can check that $A-2B-2C=0$

3. how did you do it?

4. Originally Posted by watchmath
You can check that $A-2B-2C=0$
Actually $A - 2B - 2C = \left( \begin{gathered}
0 \hfill \\
0 \hfill \\
1 \hfill \\
\end{gathered} \right)$
.
In fact, the three are linearly independent.

5. haha i kinda figured when i put it in and got it wrong...
and then it took me a little while to actually figure it out!
thanks plato =)
so now i have another question
and i've reduced the matrix down to

[ 1 0 -2 0
0 1 2 0
0 0 0 0]

and this is also testing for linear indepedence
so i know that this one is linearly dependent
but i have to write another non-trivial solution in the form
___A+ ____B+____C= 0
so how would I do that?

6. Originally Posted by eiktmywib
so now i have another question
and i've reduced the matrix down to
[ 1 0 -2 0
0 1 2 0
0 0 0 0]
What matrix? You must give the complete question.

7. oh sorry =(
A=[-2, -7, -1]
B=[-2, 4, -3]
C=[0,6,-4]

and i'm supposed to determine if they're linearly depedent or independent
The vectors were written horizontally this time, as they often are in books, but that is just to save space. The problem is the same as if the vectors were written vertically.
If they are linearly dependent, determine a non-trivial linear relation - (a non-trivial relation is three numbers which are not all three zero.) otherwise, if the vectors are linearly independent, enter 0's for the coefficients, since that relationship always holds.