# Thread: Determining Linear Independence of this set

1. ## Determining Linear Independence of this set

x1(t) =
[2-t]
[t ]

x2(t)=
[t+1]
[-2 ]

x3(t)=
[t]
[t+2]

These are supposed to be matrices, but I don't know how to format them correctly. Anyways, I want to determine whether or not this set of vector functions is linearly dependent. The way I know how to go about this would be to set up a matrix and then evaluate the determinant. I would then look at the coefficients of the determinant; if they add or subtract to zero, then it is linearly dependent. The problem is that these 3 functions would make for a 2x3 matrix and I can't evaluate the determinant of that one.

Any suggestions? Thanks,

Jim

2. Originally Posted by Jim Newt
x1(t) =
[2-t]
[t ]

x2(t)=
[t+1]
[-2 ]

x3(t)=
[t]
[t+2]

These are supposed to be matrices, but I don't know how to format them correctly. Anyways, I want to determine whether or not this set of vector functions is linearly dependent. The way I know how to go about this would be to set up a matrix and then evaluate the determinant. I would then look at the coefficients of the determinant; if they add or subtract to zero, then it is linearly dependent. The problem is that these 3 functions would make for a 2x3 matrix and I can't evaluate the determinant of that one.

Any suggestions? Thanks,

Jim
If these are linearly dependent then it is possible to form one of them by a linear combination of the other two. Thus
$\left [ \begin{matrix} 2 - t \\ t \end{matrix} \right ] = a \cdot \left [ \begin{matrix} t + 1 \\ -2 \end{matrix} \right ] + b \cdot \left [ \begin{matrix} t \\ t + 2 \end{matrix} \right ]$

See if you can find constants a and b such that this is true. If you can't, then this set is linearly independent.

-Dan

3. Hey thanks for the suggestion. Could you give me an example as to how you go about finding these constants? Thanks,

Jim

4. Originally Posted by Jim Newt
Hey thanks for the suggestion. Could you give me an example as to how you go about finding these constants? Thanks,

Jim
I dont know why you are posting this again when Reckoner has worked out the answer for you here a few days ago.If you had any clarification, you could have asked it in the same thread. This is clearly a case of double posting... but over days.

5. Sorry about that, last time I looked at that thread, I didn't see that he had worked it out. Thanks!

6. Originally Posted by Jim Newt
Sorry about that, last time I looked at that thread, I didn't see that he had worked it out. Thanks!
Never mind. If you didnt understand anything just ask in the same thread. That way people who answer feel appreciated. I am sorry if I sounded rude.

7. Originally Posted by Jim Newt
Hey thanks for the suggestion. Could you give me an example as to how you go about finding these constants? Thanks,

Jim
Just solve the system for any possible values:
$\left [ \begin{matrix} 2 - t \\ t \end{matrix} \right ] = a \cdot \left [ \begin{matrix} t + 1 \\ -2 \end{matrix} \right ] + b \cdot \left [ \begin{matrix} t \\ t + 2 \end{matrix} \right ]$

$= \left [ \begin{matrix} a(t + 1) + bt \\ -2a + b(t + 2) \end{matrix} \right ]$

So we need to solve
$2 - t = a(t + 1) + bt$
and
$t = -2a + b(t + 2)$
for a and b in terms of t.

Solve the top equation for b
$b = \frac{2 - t + at + a}{t}$

and insert this into the bottom equation:
$t = -2a + \left (
\frac{2 - t + at + a}{t} \right ) (t + 2)$

Solve this for a and substitute back to find b.

-Dan