Transforming Linear Algebra Equations: Solving for Unknown Parameters

[courtrigrad]

$$\left(\begin{array}{cc} 2-\alpha\\\alpha \end{array}\right) =\left(\begin{array}{cc} 2\\0\end{array}\right) +\alpha\left(\begin{array}{cc} -1\\1\end{array}\right)$$.How did you get rid of the parameter $$\alpha$$ to transform it into an equation?

thanks

 

[quasar987]

could you rephrase the question please?

transform what into an equation?

 

[courtrigrad]

I know that $$\left(\begin{array}{cc} 2-\alpha\\\alpha \end{array}\right) =\left(\begin{array}{cc} 2\\0\end{array}\right) +\alpha\left(\begin{array}{cc} -1\\1\end{array}\right)$$ is equivalent to $$y = -x+2$$. How do we get this? The first matrix, I know, is the point (2,0). The second matrix is a direction vector, but I don't know where to go from there.

 

[Hurkyl]

Ah, I see. He's starting with a parametric equation for a line:

$$\left(\begin{array}{c}x \\ y\end{array}\right) =<br /> \left(\begin{array}{cc} 2-\alpha\\\alpha \end{array}\right) =\left(\begin{array}{cc} 2\\0\end{array}\right) +\alpha\left(\begin{array}{cc} -1\\1\end{array}\right)$$

and he wants to know how to turn it into an implicit form. (i.e. the solutions to an equation)


There's a simple approach that works: just take the equations

$$x = 2 - \alpha$$

$$y = \alpha$$

and eliminate the variable $\alpha$.


There's a highbrow approach too, that (IMHO) is a good example to learn.

To simplify things, let me write the original equation as:

$$\vec{z} = \vec{b} + \alpha \vec{v}$$

In order to get rid of $\alpha$, you need to multiply by something that kills off $\vec{v}$: you want a good matrix $A$ such that $A \vec{v} = 0$. Well, that's just a left-nullspace matrix computation! (i.e. the columns of $A^T$ are a basis for the nullspace of $\vec{v}^T$)

In this case, we get something like

$$A = \left(\begin{array}{cc}1 & 1\end{array}\right)$$

And so the desired equation is

$$A \vec{z} = A \vec{b}$$

which you can check reduces to $x + y = 2$.

 

[courtrigrad]

thank you Hurkyl.

 

[Hurkyl]

P.S. why did I kill off $\vec{v}$ by using a matrix such that $A\vec{v} = 0$, rather than using a matrix such that $\vec{v}A = 0$?

 

[courtrigrad]

because multiplication of matrices is not commutative?

 

[Hurkyl]

Noncommutativity is why the choice matters.

I can kill off the $\vec{v}$ term by finding a matrix $A$ such that $\vec{v}A = 0$, and then getting the equation

[tex]\vec{z} A = \vec{b} A[/itex]<br /> <br /> but for this particular problem, doing this is bad... whereas doing it the other way is good. Can you tell why?<br /> <br /> <br /> Edit: bleh, the full answer isn't as straightforward as I thought. Sorry about that. <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f641.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":frown:" title="Frown :frown:" data-smilie="3"data-shortname=":frown:" /> It's easy to see why the method in this post won't work (<i>A</i> is degenerate: it has zero columns), but I'm too tired to explain why multiplying on the left works, and what to do in the more general case of something like:<br /> <br /> $$Z = B + \alpha M$$<br /> <br /> where Z, B, and M are all mxn matrices, rather than just vectors.[/tex]