Finding non-trivial linear combination

[trojansc82]

Homework Statement

Show that the set is linearly dependent by finding a nontrivial linear combination (of vectors in the set) whose sum is the zero vector. Then express one of the vectors in the set as a linear combination of the other vectors in the set.

Homework Equations

c₁(u1,u₂,₃) + c₂(v₁,v₂,v₃) +c₃(w₁,w₂,w₃)

The Attempt at a Solution

I'm unable to find a way to solve this using Gauss-Jordan elimination. My initial matrix looks like this:

[ 3 -1 2 0 ]
[ 4 1 0 0 ]

 

[Mark44]

Homework Statement

Show that the set is linearly dependent by finding a nontrivial linear combination (of vectors in the set) whose sum is the zero vector. Then express one of the vectors in the set as a linear combination of the other vectors in the set.

Homework Equations

c₁(u1,u₂,₃) + c₂(v₁,v₂,v₃) +c₃(w₁,w₂,w₃)

The Attempt at a Solution

I'm unable to find a way to solve this using Gauss-Jordan elimination. My initial matrix looks like this:

[ 3 -1 2 0 ]
[ 4 1 0 0 ]

Are the vectors you are to show are linearly dependent the first three columns of your matrix? To use Gauss-Jordan, you could start by adding (-4) times the first row to 3 times the second row.

 

[trojansc82]

Are the vectors you are to show are linearly dependent the first three columns of your matrix? To use Gauss-Jordan, you could start by adding (-4) times the first row to 3 times the second row.

I did that, and all it did was get rid of the coefficient c₁. I now have:

[3 -1 2 0]
[0 -7 8 0]

I'm confused as to how to proceed from this point.

 

[Mark44]

1. Multiply row 2 by -1/7 to make the first entry in that row a 1.
2. Use that entry to eliminate the entry directly above it.
3. Multiply row 1 by 1/3.
The matrix will now be in reduced, row-echelon form. Read off the coefficients of c1, c2, and c3.

 

[trojansc82]

1. Multiply row 2 by -1/7 to make the first entry in that row a 1.
2. Use that entry to eliminate the entry directly above it.
3. Multiply row 1 by 1/3.
The matrix will now be in reduced, row-echelon form. Read off the coefficients of c1, c2, and c3.

In row 1, I have c1 = 1, c2 = 0, c3 = 2/7.

However, the answer in the back of the book states that the coefficients are c1=1, c2=-4, and c3=7/2.

 

[HallsofIvy]

Without using matrices:

Assuming your vectors are <3, 4> , <-1, 1>, < 2, 0>, then any linear combination is of the form a<3, 4>+ b<-1, 1>+ c<2, 0>= <3a- b+ c, 4a+ b> and you want that equal to the 0 vector, <0, 0> so you want 3a- b+ c= 0 and 4a+ b= 0.

That is two equations in three unknowns so of course there are an infinite number of solutions.

If you do what Mark44 suggested, divide the first row by 3 and the second by -7, you get
\begin{bmatrix}1 &amp; -\frac{1}{3} &amp; \frac{2}{3} &amp; 0 \\ 0 &amp; 1 &amp; -\frac{8}{7} &amp; 0 \end{bmatrix}

Which does NOT give specific numbers as solutions. There are, as I said, an infinite number of solutions. The first row of the matrix is equivalent to the equation c_1- (1/3)c_2+ (2/3)c_3= 0 and the second row is equivalent to the equation c_2- (8/7)c_3= 0. You can solve the second equation for c_2 as a multiple of c_3, replace c_2 in the first equation with that and then solve for c_1 as a multiple of c_3. Choosing different values for c_3 will give different answers.