How Do You Calculate Elementary Matrices for Matrix Transformations?

[jimmypoopins]

Hello all, I'm taking my first year in linear algebra and I'm having some issues understanding how to deal with some problems involving elementary matrices.

First off, i have a set of problems that ask to find the elementary matrix E such that AE=B, and secondly i have a set of problems asking to find the elementary matrix E such that EA=B. I've reread this section in the book a couple of times and there isn't much about matrix algebra involving elementary matrices, other than the fact that they do row operations on matrices. How exactly am i supposed the row operations in these sets of problems?

For example, one problem is

Find an elementary matrix E such that EA=B

A=\left(\begin{array}{ccc}2&1&3\\-2&4&5\\3&1&4\end{array}\right), B=\left(\begin{array}{ccc}2&1&3\\3&1&4\\-2&4&5\end{array}\right)

it's obvious to me that row's 2 and 3 are switched in A to make B, but how do i know what elementary matrix does that? The back of the book says that

E=\left(\begin{array}{ccc}1&0&0\\0&0&1\\0&1&0\end{array}\right)

and after performing the matrix multiplication i get it, but there has to be a better way to learn how to do it (especially since i don't know how to do the other problems in the set without looking in the back of the book).

Also,

Find an elementary matrix E such that AE = B

A=\left(\begin{array}{ccc}4&-2&3\\-2&4&2\\6&1&-2\end{array}\right), B=\left(\begin{array}{ccc}2&-2&3\\-1&4&2\\3&1&-2\end{array}\right)

the back of the book states that

E=\left(\begin{array}{ccc}1/2&0&0\\0&1&0\\0&0&1\end{array}\right)

column 1 is halved in the transformation from A to B, so that makes sense, however there is another problem (from the same AE = B set)

A=\left(\begin{array}{cc}2&4\\1&6\end{array}\right), B=\left(\begin{array}{cc}2&-2\\1&3\end{array}\right)

here column 2 is halved and negative, so i'd assume the elementary matrix to be similar to the one in the first problem, but the back of the book says it is

E=\left(\begin{array}{cc}1&-3\\0&1\end{array}\right)

can anyone point me in the right direction here? even a link to a site that explains it well would be helpful. thank you for your time.

 

[genneth]

It seems like you need to learn how to do matrix inverses.

See http://en.wikipedia.org/wiki/Invertible_matrix

 

[Hurkyl]

Inverses are overkill. The exercise wants him to recognize elementary row & column operations and relate them to the elementary matrices.

 

[Hurkyl]

A=\left(\begin{array}{cc}2&4\\1&6\end{array}\right), B=\left(\begin{array}{cc}2&-2\\1&3\end{array}\right)

here column 2 is halved and negative

No it's not. The negation of half of 6 is -3, not 3.

 

[jimmypoopins]

It seems like you need to learn how to do matrix inverses.

See http://en.wikipedia.org/wiki/Invertible_matrix

i know how to do matrix inverses... how are they related to elementary matrices, though?

if i have EA=B is there a way to solve for E using inverses?

 

[genneth]

EA=B \implies E=BA^{-1}

 

[HallsofIvy]

Are you sure you know WHAT an "elementary matrix" is. It is a matrix derived by applying a particular row or column operation to the identity matrix. In your last problem you go from A to B by subracting twice the first column from the second column. If you do that to the identity matrix, you get the corresponding row operation.

 

[Kelitrex]

Edit: sorry for the needless bump... I only just realized how old this topic was.

A=\left(\begin{array}{cc}2&4\\1&6\end{array}\right), B=\left(\begin{array}{cc}2&-2\\1&3\end{array}\right)

E=\left(\begin{array}{cc}1&-3\\0&1\end{array}\right)

can anyone point me in the right direction here? even a link to a site that explains it well would be helpful. thank you for your time.

I am currently taking Linear Algebra as well, and this one was rather easy for me to figure out. First, I assumed E=\left(\begin{array}{cc}a&b\\c&d\end{array}\right) and then did some basic algebra...

2a+4c=2
a+6c=1
a=1
c=0

2b+4d=-2
b+6d=3
b=-3
d=1

So E=\left(\begin{array}{cc}1&-3\\0&1\end{array}\right)