Linear: Finding all scalars for given vector equation

In summary, the problem involves finding scalars (C1, C2, C3) that satisfy the equation C1(u) + C2(v) + C3(w) = (0,5,4). The solution involves setting up three equations in three unknowns and using elimination to solve for either C1 or C3. If the resulting equations lead to an inconsistent equation, such as 0=1, it can be concluded that there are no solutions. Alternatively, matrices can also be used to solve the problem by setting up a coefficient matrix and using Gauss-Jordan elimination to solve for the unknown variables.
  • #1
Alexstre
19
0

Homework Statement


Vectors
u=(-2, 9, 6)
v=(-3, 2, 1)
w=(1, 7, 5)


Homework Equations


Show that there is no scalar (Cn) such that:
C1(u) + C2(v) + C3(w) = (0, 5, 4)

I'd also like to know where would I start to find all the scalars if there were any, since I'm pretty sure this problem will come eventually, and there's nothing in the textbook about it!

Thanks
 
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  • #2
Equate the two sides component by component. That's three equations in three unknowns. The first one is C1*(-2)+C2*(-3)+C3*1=0.
 
  • #3
Dick said:
Equate the two sides component by component. That's three equations in three unknowns. The first one is C1*(-2)+C2*(-3)+C3*1=0.
Thanks! Here's what I tried:

1: C1(-2) + C2(-3) + C3(1) = 0
2: C1(9) + C2(2) + C3(7) = 5
3: C1(6) + C2(1) + C3(5) = 4

I've eliminated C2 and ended up with 2 new equations:

C1(-7) + C3(-3) = -3
C1(16) + C3(16) = 12

I guess from here I could solve for either C1 or C3, then go back to the original equations, plugging it in, and solving again (3 equations, with 2 unknown this time), is that right? Also, at what point will I reach the end (ie. when will it be clear that there's no solution)

Another thing, seeing how this is for a linear class, I was tempted to use matrices and start with the following:

[-3 4 6 | 2]
[ 1 0 -1 | 0]
[ 2 -8 -4 | 4]

Would that work? If so, should I try to solve for [1 0 0], [0 1 0], [0 0 1] system, and whenever I get to the point where I can do anything, I "proved" that this problem has no solution?

Thanks!
 
  • #4
You are on the right track. Except I don't think the equations in C1 and C3 are right. Check that. Once you think you've got it right, try and eliminate, say C1. You should wind up with an inconsistent equation. Like 0=1. That would show there are no solutions. You can do the same thing with matrices as well. Though I don't see where you got that matrix from.
 

1. What is a vector equation?

A vector equation is an equation that represents a relationship between vectors using variables. It is written in the form of C1v1 + C2v2 + ... + Cnvn = b, where C1, C2, ... Cn are scalars, v1, v2, ... vn are vectors, and b is a constant vector.

2. How do you find the scalars in a vector equation?

To find the scalars in a vector equation, you can use the method of Gaussian elimination or matrix operations. Both methods involve manipulating the equation to get it in reduced row-echelon form, which will reveal the values of the scalars.

3. What is the importance of finding all scalars in a vector equation?

Finding all scalars in a vector equation is important because it allows you to fully understand the relationship between the vectors and the constant vector. It also allows you to solve for specific values or variables within the equation.

4. Can there be more than one set of scalars that satisfy a vector equation?

Yes, there can be more than one set of scalars that satisfy a vector equation. This is because there are multiple ways to manipulate an equation to get it into reduced row-echelon form, resulting in different values for the scalars.

5. How is the solution to a vector equation represented?

The solution to a vector equation is represented as a vector or a set of vectors. It can also be represented as a linear combination of the given vectors, where the scalars represent the coefficients for each vector.

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