# Linear Algebra - Linear Independence/Dependence

• Pete_01
In this case, there is only one solution and that's when x = 1. So the vectors (1 1 -3 1), (1 3 4 2), and (1 0; 0 1; 0 0; 0 0) are all solutions to your equation and are therefore linearly dependent.

## Homework Statement

I want to know if the matrix (1, 1)
(1, 3)
(-3,4)
(1, 2)
is linearly independent or dependent.

## Homework Equations

I reduced it down in rref to (1 0; 0 1; 0 0; 0 0) and I'm guessing it's linearly independent because there is only 1 term per line when it's set to 0? Is this correct?

## The Attempt at a Solution

in part 2

I assume you mean are the 2 vectors you used to construct that matrix linearly independent?

They are linearly dependent. When you reduce your matrix, you are left with a series of equations. The first and second line tell you that, given the coefficients as (a,b,c,d), that 1a = 0 and 1b = 0. In other words, a = b = 0. Now, by definition if the coefficients in the formula $$a\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over v} _1 + b\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over v} _2 = 0$$ are only capable of being 0, then your two vectors are linearly independant.

Yes there was two vectors used: (1 1 -3 1) and (1 3 4 2), so they would be linearly independent then correct? Because of the formula av1 + bv2 =0 right? Sorry, I'm still a bit confused.

Yes, you got it. What you're basically looking for in linear dependence problems is if any vector can be written in terms of the other two vectors. So say you want to see if V1 can be written as a linear combination of V2. Well what you're looking for is whether or not V1 = a V2 where a is some number other than 0.

Now if you have something more complex like 4 vectors and you want to know if V1 can be constructed as a linear combination of V2, V3, and V4, what you get it is V1 = a V2 + b V3 + c V4. If you subtract V1 , you get 0 = -V1 + a V2 + b V3 + c V4. Since you have 0 on the left hand side, you can arbitrarily multiply the entire equation by whatever number you wish so in the end, without loss of generality, can say that 0 = a' V1 + b' V2 + c' V3 + d' V4 and since the coefficients' label is arbitrary, you get what you see in general: 0 = a V1 + b V2 + c V3 + d V4.

Now you can use the power of linear algebra and say the coefficient matrix, we'll call X = (a,b,c,d), is a solution to that equation because you can form your vectors into that matrix and say Ax = 0 and look for solutions of x.