L-Algebra, finding h (unknown in matrix)

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In summary, the conversation discusses a problem involving a matrix with rows and the question of finding the values of h for which a particular vector is in the span of two other vectors and when the set of three vectors is linearly dependent. It is determined that v1 and v2 are linearly dependent and that there are no values of h that satisfy the equation for v3 being in the span of v1 and v2. It is also discussed that adding a third vector to a set of linearly dependent vectors will always result in a linearly dependent set. The correct set of vectors for the given problem is v1 = {1, -1, 4}, v2 = {3, -5, 7}, v3
  • #1
viet_jon
131
0

Homework Statement



say there's a matrix with rows

2 , 3 , h
4, 6, 7

Homework Equations





The Attempt at a Solution




I tried to rref it, but don't know what to do next.
 
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  • #2
or

v1 = (1, -3, 2)
v2 = (-3, 9, -6)
v3 = (5, -7, h)


the question reads, for what values of h is v3 in span(v1,v2), and for what values of h is (v1, v2, v3) linearly dependent?


I've went over the chapter twice, but still can't find a way to do this. I understand the concept of the chapter (linearly dependency), but it shows no examples of questions like these.
 
  • #3
anybody?

I'm desperate.
 
  • #4
Ok, we have a matrix given by:

[tex]
\left( {\begin{array}{*{20}c}
1 & 3 & 5 \\
{ - 3} & 9 & { - 7} \\
2 & { - 6} & h \\
\end{array}} \right)
[/tex]

1) In order for v3 to be in the span of {v1;v2}, v3 has to be lineary dependent on v1 and v2, i.e. we want to be able to write it as a linear combination of v1 and v2. When is v3 linearly dependent on v1 and v2?

2) Same as #1, but when is v3 linearly dependent on v1 and v2?

The answer is that each column has to have a leading term after RREF, i.e. the matrix must have a rank of 3. Use this to determine h.

Also, when you have solved #1, then you can see what the coefficients are in the linear combination that makes v3.
 
  • #5
I don't think I expressed myself very clearly above. In order for the three vectors v1, v2 and v3 to be linearly independent, no row must consist of entirely zeroes. In this case, the matrix is said to have a rank of 3.
 
  • #6
viet_jon said:
or

v1 = (1, -3, 2)
v2 = (-3, 9, -6)
v3 = (5, -7, h)


the question reads, for what values of h is v3 in span(v1,v2), and for what values of h is (v1, v2, v3) linearly dependent?


I've went over the chapter twice, but still can't find a way to do this. I understand the concept of the chapter (linearly dependency), but it shows no examples of questions like these.

I hope this is a different problem from the first one you posted in this thread. I don't see any connection between them.

For this problem, v2 = -3 * v1, or equivalently, v1 = -1/3 * v2.
What does that tell you about the linear dependence or independence of these two vectors?
What does that tell you about span{v1, v2}?
For v3 to be in span{v1, v2} there must be constants a and b such that v3 = a*v1 + b*v2. Is there some value of h for which there is a solution to this equation?

Now for the the second question, if you have two vectors that are linearly independent, and you add a third vector, is the new set still linearly independent? Answer: Sometimes it is, and sometimes it isn't. If the new vector is a linear combination of the first two, the new set is linearly dependent. (The new vector is in the span of the first two vectors.) If the new vector is not a linear combination of the first two, the new set is linearly independent. (The new vector is not in the span of the first two vectors.)

If you have two vectors that are linearly dependent, and you add a third vector, is the new set linearly dependent or linearly independent? Answer: The new set is always linearly dependent.
 
  • #7
sorry, I am a little slow...I still don't understand. :frown:

the answer in the back, shows no h is possible.

I think it would be easier to understand for me, if we used a question with a definitive answer.


v1 = {1, -1, 4}
v2 = {3, -5, 7}
v3 = {-1, 5, h}

the answer is suppose to be 6.

but how is v2 linearly depdenant on v1? It is not a scaler of it...so doesn't that make it independent?
 
  • #8
You're changing vectors on us. Here's what you had in post #2:
v1 = (1, -3, 2)
v2 = (-3, 9, -6)
v3 = (5, -7, h)
In post #7, you have a different set of vectors. Which is the right set of vectors?

For the set of vectors you posted in #2, v2 is a multiple of v1 (and conversely, v1 is a different multiple of v2).

In post #6 I asked some questions. If you answer them, you might understand better what's going on here.
 
  • #9
Mark44 said:
I hope this is a different problem from the first one you posted in this thread. I don't see any connection between them.

For this problem, v2 = -3 * v1, or equivalently, v1 = -1/3 * v2.
What does that tell you about the linear dependence or independence of these two vectors?
What does that tell you about span{v1, v2}?

v1 and v2 are linearly dependent.


For v3 to be in span{v1, v2} there must be constants a and b such that v3 = a*v1 + b*v2. Is there some value of h for which there is a solution to this equation?

there are no values...right?

Now for the the second question, if you have two vectors that are linearly independent, and you add a third vector, is the new set still linearly independent? Answer: Sometimes it is, and sometimes it isn't. If the new vector is a linear combination of the first two, the new set is linearly dependent. (The new vector is in the span of the first two vectors.) If the new vector is not a linear combination of the first two, the new set is linearly independent. (The new vector is not in the span of the first two vectors.)
If you have two vectors that are linearly dependent, and you add a third vector, is the new set linearly dependent or linearly independent? Answer: The new set is always linearly dependent.

This paragraph I understand pretty well. I still don't know how to find h though.



sorry :smile:

I posted the wrong question in post #2.
 
  • #10
viet_jon said:
sorry, I am a little slow...I still don't understand. :frown:

the answer in the back, shows no h is possible.

I think it would be easier to understand for me, if we used a question with a definitive answer.


v1 = {1, -1, 4}
v2 = {3, -5, 7}
v3 = {-1, 5, h}

the answer is suppose to be 6.

I get the jist of your explanation to post #2. But this question... how is v2 linearly depdent on v1? it's not a multiple.

therefore, how can any value for h, make the set of vectors linearly dependent?
 
  • #11
Your answers in post 9 were all right, so let's move on to the real problem.
v1 and v2 (from post 10) are linearly independent.

To answer the question about h, you want to find a nontrivial solution for c1, c2, c3 in the equation c1*v1 + c2*v2 + c3*v3 = 0. By nontrivial, I mean a solution other than c1 = c2 = c3 = 0, where at least one of the numbers ci isn't zero.

To do this, form a matrix whose columns are [v1 v2 v3].
Row reduce this matrix as far as you can go. You should end up with a third row that has 0 0 <some expression with h>. What value of h makes that expression equal zero? That's the value of h that makes the three vectors v1, v2, and v3 a linearly dependent set. (We don't say that v3 is "dependent on" v1 and v2.)

I worked this and found that -5v1 + 2v2 + v3 = 0 for the value of h that I found.
 

Related to L-Algebra, finding h (unknown in matrix)

1. What is L-Algebra?

L-Algebra is a branch of mathematics that deals with linear transformations and their representation in matrix form. It involves the manipulation and solving of equations involving matrices and vectors.

2. How is h (unknown) found in a matrix using L-Algebra?

In L-Algebra, h is found by solving a system of linear equations involving matrices. This can be done by using various methods such as Gaussian elimination, Cramer's rule, or matrix inversion.

3. What are the main applications of L-Algebra in science?

L-Algebra has various applications in science, including physics, engineering, computer science, and statistics. It is used to model and solve systems of linear equations in these fields, as well as in data analysis and image processing.

4. What are the key properties of L-Algebra?

Some key properties of L-Algebra include linearity, associativity, and distributivity. These properties allow for efficient manipulation and solving of linear equations involving matrices and vectors.

5. Are there any real-world examples where L-Algebra is used?

Yes, L-Algebra is used in many real-world applications, such as in electrical circuits, chemical reactions, and economic models. It is also used in computer graphics and machine learning algorithms.

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