Matrix help (plane equation + product of elementary matrices)

1. Dec 7, 2008

theneedtoknow

This isn't really a particular question...i just missed the classes relating to these 2 topics so I have no idea how to do questions with them....so can someone please explain how i go about solving questions such as:

1. I'm given 3 points in R3, and i'm supposed to find the equation of the plane that passses through them. How do I go about doing this?

2. I am given a matrix and I'm supposed to express it as a product of elementary matrices. I asked abotu this here before, but I think i didn't understand it properly because I had a question like this on my midterm and I definitely got it wrong. I know I am supposed to reduce it to reduced row echelon form and keep track of the steps i take in the process. But then, what do i do with those steps? I know its not as simple as applying the same steps to a set of identity matrices of the same size...there's some kind of other trick to it.

2. Dec 7, 2008

rock.freak667

I can only help with the first one.
But if you have three points A,B and C.
then AB, AC and BC form a vector triangle on this plane correct?
So, you will need the normal vector, which is perpendicular to AB,AC and BC, so what product would you use to find a perpendicular common to AB and AC?

When you have that, use the formula (r-r0).N=0 where r0 is a point on the plane

3. Dec 7, 2008

theneedtoknow

So does that mean i have to find a vector such that the dot prdut of it with the 3 points im given is 0
so if i have point 1 = (a,b,c), point 2 = (d, e, f) and point 3 = (g h, i)
i need a vector (x, y, z) such that

ax + by + cz = 0
dx + ey + fz = 0
gx + hy + iz = 0
so i solve the linear systemand find the vector [z, y, z]

but what do i do with it after?

4. Dec 7, 2008

rock.freak667

Well you need to find the perpendicular vector. Do you know what the cross product of two vectors is?

5. Dec 7, 2008

theneedtoknow

yeah i know how to calculate the crossproduct :)
so what are r and N? which one is the crossproduct and whas the other one?

6. Dec 8, 2008

rock.freak667

N would be the cross product of the vectors AB and BC. r is the equation of the plane (x,y,z).

So for example if you find the cross product to be (1,1,1) and a point on the plane is (1,2,3)

The equation of the plane would be

((x-1),(y-2),(z-3)).(1,1,1)=0

Which would give

x-1+y-2+z-3=0
=> x+y+z-6=0

7. Dec 8, 2008

theneedtoknow

ohh awesome :)
thank you very much!

8. Dec 8, 2008

theneedtoknow

oh do i have to use cross product of AB and BC or is AB and Ac fine?

9. Dec 8, 2008

HallsofIvy

Staff Emeritus
It doesn't matter. The cross product of any two vectors formed from the three points will be normal to the plane and that is all you need.

10. Dec 8, 2008

HallsofIvy

Staff Emeritus
What you say you 'know', "I know its not as simple as applying the same steps to a set of identity matrices of the same size", is wrong- it is that simple.

An elementary matrix is a matrix formed by applying a single row operation to the identity matrix. To find express a matrix as a product elementary matrices, use row operations to reduce the matrix to the identity, at the same time applying each of the row operations to the identity matrix. Notice that you apply each row operation to the identity matrix, not just continue applying row operations to the same matrix as you might in using row operations to find the inverse of a matrix. And, of course, you have to have the correct order: the first elementary matrix you found must be on the right of the product.

For example, to row-reduce the 2 by 2 matrix
$$\left(\begin{array}{cc}2 & 4 \\ 3 & 4\end{array}\right)$$
I would first divide the entire first row by 2 getting
$$\left(\begin{array}{cc}1 & 2 \\ 3 & 4\end{array}\right)$$
Doing the same to the identity matrix gives
$$\left(\begin{array}{cc}1/2 & 0 \\ 0 & 1\end{array}\right)$$

Next subtract 3 times the first row from the second to get
$$\left(\begin{array}{cc}1 & 2 \\ 0 & -2\end{array}\right)$$
Doing the same to the identity matrix gives
$$\left(\begin{array}{cc}1 & 0 \\ -3 & 1\end{array}\right)$$

Divide the second row by -2 to get
$$\left(\begin{array}{cc}1 & 2 \\ 0 & 1\end{array}\right)$$
Doing the same to the identity matrix gives
$$\left(\begin{array}{cc}1 & 0 \\ 0 & -1/2\end{array}\right)$$

Finally, subtract twice the second row from the first to get the identity matrix
$$\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)$$
Doing the same to the identity matrix gives
$$\left(\begin{array}{cc}1 & -2 \\ 0 & 1\end{array}\right)$$

Putting all those together,
$$\left(\begin{array}{cc}2 & 4 \\ 3 & 4\end{array}\right)$$
can be written as the product
$$\left(\begin{array}{cc}1 & -2 \\ 0 & 1\end{array}\right)\left(\begin{array}{cc}1 & 0 \\ 0 & -1/2\end{array}\right)\left(\begin{array}{cc}1 & 0 \\ -3 & 1\end{array}\right)\left(\begin{array}{cc}1/2 & 0 \\ 0 & 1\end{array}\right)$$

Last edited: Dec 8, 2008