How Can You Solve for Alpha in Linearly Dependent Vectors?

[PhysicsMark]

Homework Statement

Suppose that A, B, and C are not linearly independent. Then show how the \alpha_{i} can be computed, up to a common factor, from the scalar products of these vectors with each other.

Hint: Suppose that there are non-zero values of the \alpha_{i}'s that satisfy \alpha_{1}{\mathbf{A}+{\alpha_{2}{\mathbf{B}+{\alpha_{3}{\mathbf{C}}=0. Then, taking the dot product of both sides of this equation with A will yield a set of equations that can be solved for the \alpha_{i}'s.

Homework Equations

\alpha_{1}{\mathbf{A}+{\alpha_{2}{\mathbf{B}+{\alpha_{3}{\mathbf{C}}=0.

The Attempt at a Solution

Based off the instructions and hint I think they are asking me to solve for the \alphas of the following three equations:

\alpha_{1}{\mathbf{A{\cdot}A}+{\alpha_{2}{\mathbf{B{\cdot}A}+{\alpha_{3}{\mathbf{C{\cdot}A}}=0.

\alpha_{1}{\mathbf{A{\cdot}B}+{\alpha_{2}{\mathbf{B{\cdot}B}+{\alpha_{3}{\mathbf{C{\cdot}B}}=0.

\alpha_{1}{\mathbf{A{\cdot}C}+{\alpha_{2}{\mathbf{B{\cdot}C}+{\alpha_{3}{\mathbf{C{\cdot}C}}=0.

Or, they could be asking me to solve for \alpha_{i} of these 3 equations:

\alpha_{i}({\mathbf{A{\cdot}A}+{\mathbf{B{\cdot}A}+{\mathbf{C{\cdot}A}})=0.\alpha_{i}({\mathbf{A{\cdot}B}+{\mathbf{B{\cdot}B}+{\mathbf{C{\cdot}B}})=0.\alpha_{i}({\mathbf{A{\cdot}C}+{\mathbf{B{\cdot}C}+{\mathbf{C{\cdot}C}})=0.

How do I solve for alpha? Matrices? Substitution? Elimination?
I'm not really sure what they mean by a common factor. Insight is always appreciated.

*Edited: Thanks for pointing that out Mark44!

 

[Mark44]

Homework Statement

Suppose that A, B, and C are not linearly independent. Then show how the \alpha_{i} can be computed, up to a common factor, from the scalar products of these vectors with each other.

Hint: Suppose that there are non-zero values of the \alpha_{i}'s that satisfy \alpha_{1}{\mathbf{A}+{\alpha_{2}{\mathbf{B}+{\alpha_{3}{\mathbf{C}}=0. Then, taking the dot product of both sides of this equation with A will yield a set of equations solved for the \alpha_{i}'s.

I think the last sentence should say "... a set of equations that can be solved for ...

Homework Equations

\alpha_{1}{\mathbf{A}+{\alpha_{2}{\mathbf{B}+{\alpha_{3}{\mathbf{C}}=0.

The Attempt at a Solution

Based off the instructions and hint I think they are asking me to solve for the \alphas of the following three equations:

\alpha_{1}{\mathbf{A{\cdot}A}+{\alpha_{2}{\mathbf{B{\cdot}A}+{\alpha_{3}{\mathbf{C{\cdot}A}}=0.

\alpha_{1}{\mathbf{A{\cdot}B}+{\alpha_{2}{\mathbf{B{\cdot}B}+{\alpha_{3}{\mathbf{C{\cdot}B}}=0.

\alpha_{1}{\mathbf{A{\cdot}C}+{\alpha_{2}{\mathbf{B{\cdot}C}+{\alpha_{3}{\mathbf{C{\cdot}C}}=0.

I think what you have above is the way to go.
Or, they could be asking me to solve for \alpha_{i} of these 3 equations:

\alpha_{i}({\mathbf{A{\cdot}A}+{\mathbf{B{\cdot}A}+{\mathbf{C{\cdot}A}})=0.


\alpha_{i}({\mathbf{A{\cdot}B}+{\mathbf{B{\cdot}B}+{\mathbf{C{\cdot}B}})=0.


\alpha_{i}({\mathbf{A{\cdot}C}+{\mathbf{B{\cdot}C}+{\mathbf{C{\cdot}C}})=0.

Carrying out the multiplication in the first equation (and changing the alpha_i to c_i gives this equation
c₁|A|² + c₂A.B + c₃A.C = 0

Do the same for the other two equations. This gives you three equations in the unknowns c₁, c₂, and c₃. This system of equations can be solved by a number of ways, including elimination or using matrices.

How do I solve for alpha? Matrices? Substitution? Elimination?
I'm not really sure what they mean by a common factor. Insight is always appreciated.

With regard to "common factor," suppose that 2A + B - 3C = 0. In the case, the constants are 2, 1, and -3. It will also be true that 4A + 2B - 6C = 0. In this case the constants are 4, 2, and -6. This triple of constants has a common factor of 2. Any triple of constants of the form 2t, t, -3t will work, and t is the common factor.