# How do I test for linear independence if there are only 3 equations in R4?

• skyturnred
In summary, The question is about testing for linear dependence using the determinant method. The person suggests trying to solve for the variables using the definition of linear dependence and then creating a matrix to find the reduced row echelon form. The final result shows that the vectors are linearly independent.
skyturnred

## Homework Statement

So the dimension is R4. V1=[3 1 1 2], V2=[-2 -1 2 2] and V3=[2 1 2 1]

## The Attempt at a Solution

The only way I know of to test for convergence is to make a matrix out of the row vectors of the vectors above (with the row vectors becoming the respective columns of the matrix) and to take the determinant. If det=0 then it is linearly dependent.

The problem here, is that you get the following matrix: [3 -2 2; 1 -1 1; 1 2 2; 2 2 1] which, since it is 4x3, is not possible to find the determinant.

Are there other ways to test for dependence?

You could try using the definition of linear dependence. It will give you four equations and three variables... solve for the variables using the first three equations and see if you can satisfy the fourth still

Think about trying to solve aV1+bV2+cV3=0

Basically the same idea, what do a, b, and c have to be for V1, V2, and V3 to be linearly independent?

Oh, I see. So then I write out the equations, put them into a matrix. I get the following.

M=[3 -2 2 0; 1 -1 1 0; 1 2 2 0; 2 2 1 0].

Brought to rref I get:

M=[1 0 0 0; 0 1 0 0; 0 0 1 0; 0 0 0 0]. So then, since there is only a trivial solution, that means that they are all linear independent right? Just trying to make sure I understand.
Thanks

skyturnred said:
Oh, I see. So then I write out the equations, put them into a matrix. I get the following.

M=[3 -2 2 0; 1 -1 1 0; 1 2 2 0; 2 2 1 0].

Brought to rref I get:

M=[1 0 0 0; 0 1 0 0; 0 0 1 0; 0 0 0 0]. So then, since there is only a trivial solution, that means that they are all linear independent right? Just trying to make sure I understand.
Thanks
assuming you did it correctly, than that shows that the vectors are linearly independent

ironman1478 said:
assuming you did it correctly, than that shows that the vectors are linearly independent

Thanks!

## 1. How do I know if three equations in R4 are linearly independent?

In order to determine if three equations in R4 are linearly independent, you can use the determinant method. Write the three equations as a matrix and calculate the determinant. If the determinant is non-zero, then the equations are linearly independent.

## 2. Can I use the row reduction method to test for linear independence in R4?

Yes, you can also use the row reduction method to test for linear independence in R4. If the resulting reduced row echelon form has a pivot in every column, then the equations are linearly independent.

## 3. What is the difference between linear independence and linear dependence?

Linear independence means that the equations are not related to each other and can be solved separately. Linear dependence means that one equation can be written as a linear combination of the other equations, making it redundant or unnecessary.

## 4. Can I test for linear independence with more than three equations in R4?

Yes, the same methods can be used to test for linear independence with more than three equations in R4. However, the calculations may become more complex as the number of equations increases.

## 5. What is the significance of testing for linear independence in R4?

Testing for linear independence in R4 is important in determining the number of solutions to a system of equations. If the equations are linearly independent, there will be a unique solution. If they are linearly dependent, there will be infinitely many solutions or no solutions at all.

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