# Problem with linear dependence: det(a)=0, but rref is inconsistent?

• skyturnred
In summary: To check this, we can use the determinant of a matrix of the vectors. The determinant is 0, which means that there is a non-trivial solution and the vectors are linearly dependent. Therefore, we can express V5 as a combination of the others by setting it as the augmented part of the matrix and reducing it to row echelon form. However, in this case, the system is inconsistent and there is no solution. So, V5 cannot be expressed as a linear combination of the other vectors. In summary, the system is linearly dependent and V5 cannot be expressed as a linear combination of the other vectors.
skyturnred

## Homework Statement

v1=[2 1 1 4 2]
v2=[-1 2 2 1 -1]
v3=[3 -2 1 -2 2]
v4=[4 1 4 3 3]
v5=[1 2 3 2 1]

Find if the system is linear dependent or independent. If it is dependent, express the last vector in the list (v5) as a combination of the preceding ones.

## The Attempt at a Solution

I am having trouble with this one. Using the definition of independence (c1V1+c2V2+...+cnVn=0), I get the following matrix:

a=[2 -1 3 4 1; 1 2 -2 1 2; 1 2 1 4 3; 4 1 -2 3 2; 2 -1 2 3 1]. When I find det(a), I find it to be 0. This means that the system has a non-trivial solution, right? And if so, that means that the system is linear dependent right? so then I take the same matrix as an augmented matrix with the final column ([1; 2; 3; 2; 1]) as being the augmented part, since it is equal to V5 and I am trying to express it as a combination of the others. Then I find rref. But when I find rref I get the following:

a(rref)=[2 0 0 1 0; 0 1 0 1 0; 0 0 1 1 0; 0 0 0 0 1], but looking at the very last row, the matrix is inconsistent. If it is inconsistent, there is no value for c1, c2, c3 and c4 that would equal V5. So where did I go wrong? I double checked all my steps so there must be a problem with my understanding.

It might not be that v5 is a linear combination of the others. For example, consider the vectors

v1=(1,0,0)
v2=(1,0,0)
v3=(0,1,0)

in R3. You can make a matrix and take its determinant to verify these are linearly dependent, but if you try to write v3 as a linear combination of v1 and v2 you're going to have trouble

Office_Shredder said:
It might not be that v5 is a linear combination of the others. For example, consider the vectors

v1=(1,0,0)
v2=(1,0,0)
v3=(0,1,0)

in R3. You can make a matrix and take its determinant to verify these are linearly dependent, but if you try to write v3 as a linear combination of v1 and v2 you're going to have trouble

OK, thanks. I thought that was the case. But the wording of the question I've been assigned is really horrible. It confused me.

Thanks again!

skyturnred said:
OK, thanks. I thought that was the case. But the wording of the question I've been assigned is really horrible. It confused me.

Thanks again!

If your wording of the problem was an exact copy of the wording you were given, it could not possibly have been clearer. Where did you see a problem?

RGV

## 1. What does it mean when det(a)=0?

When the determinant of a matrix is equal to 0, it means that the matrix is not invertible. This means that there is no unique solution to the system of equations represented by the matrix.

## 2. What is rref and why does it matter?

rref stands for reduced row echelon form. It is a specific form of a matrix where the leading coefficient of each row is 1 and all other entries in the column are 0. Rref is important because it allows us to easily determine the linear dependence or independence of a set of vectors.

## 3. Why does rref matter if det(a)=0?

If det(a)=0, it means that the matrix is not invertible. However, by putting the matrix into rref, we can still determine if there are any linearly dependent rows or columns, which can give us insight into the solutions of the system of equations represented by the matrix.

## 4. How can rref be inconsistent if det(a)=0?

While rref can help us determine linear dependence or independence, it does not guarantee a consistent system of equations. Inconsistent rref can occur when there are rows with all zero entries or when the leading coefficients in a row are not all 1. This means that there is no solution to the system of equations.

## 5. Can a matrix with det(a)=0 ever have a solution?

No, if the determinant of a matrix is equal to 0, it means that the matrix is not invertible and there is no unique solution to the system of equations represented by the matrix. However, it is still possible for a matrix with det(a)=0 to have infinite solutions, meaning that there are multiple solutions that satisfy the system of equations.

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