# Quick Linear Algebra Question

1. Oct 12, 2006

$$\left(\begin{array}{cc} 2-\alpha\\\alpha \end{array}\right) =\left(\begin{array}{cc} 2\\0\end{array}\right) +\alpha\left(\begin{array}{cc} -1\\1\end{array}\right)$$.

How did you get rid of the parameter $$\alpha$$ to transform it into an equation?

thanks

Last edited: Oct 12, 2006
2. Oct 12, 2006

### quasar987

could you rephrase the question plz?

transform what into an equation?

3. Oct 12, 2006

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 dont know where to go from there.

4. Oct 12, 2006

### Hurkyl

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

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

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 $x + y = 2$.

Last edited: Oct 12, 2006
5. Oct 12, 2006

thank you Hurkyl.

6. Oct 12, 2006

### Hurkyl

Staff Emeritus
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 $\vec{v}A = 0$?

7. Oct 12, 2006

because multiplication of matrices is not commutative?

8. Oct 12, 2006

### Hurkyl

Staff Emeritus
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

$$\vec{z} A = \vec{b} A[/itex] but for this particular problem, doing this is bad... whereas doing it the other way is good. Can you tell why? Edit: bleh, the full answer isn't as straightforward as I thought. Sorry about that. It's easy to see why the method in this post won't work (A is degenerate: it has zero columns), but I'm too tired to explain why multiplying on the left works, and what to do in the more general case of something like: [tex]Z = B + \alpha M$$

where Z, B, and M are all mxn matrices, rather than just vectors.

Last edited: Oct 12, 2006