Understanding the Product of Elementary Matrices

[foreverdream]

I am currently studying pure mathematics and one of the topic is working out product of elemnetry matrix.

My original question in follow up exercise was to work out inverse of the matrix and I did following ( please see attached document) and that's exactly what the answer at the back of the book is. However when trying to work out a product of elementry matrices I am utterly confused as the explanation given doesn't make sense to me.

I know you apply same row operation but can someone please have a look at this and tell me bit more explicitly to help me understand this better as this is a very new topic to me

(PLEASE NOTE THIS IS NOT HOMEWORK HELP)

 

[HallsofIvy]

I, like many other people, do not have Microsoft Word on my computer and so cannot open that file.

 

[foreverdream]

Ok will post pdf

 

[foreverdream]

here it is - hope you can read pdf and help me thanks

 

[daveyp225]

Elementary matrices represent row operations on a matrix. For example, given:

$$A=\pmatrix{1 & 2\\-2 &1}$$

Adding 2 times row 1 to row 2 is the same as multiplying A on the left by:

$$E_1 = \pmatrix{1 & 0\\2 & 1}$$

We get $\pmatrix{1 & 0\\2 &1}\pmatrix{1 & 2\\-2 &1} = \pmatrix{1 & 2\\0 &5}$

Then we wish to divide row 2 by 5. The corresponding matrix is:

$$E_2=\pmatrix{1 & 0\\0 &1/5}$$

And so: $\pmatrix{1 & 0\\0 &1/5}\pmatrix{1 & 0\\2 &1}\pmatrix{1 & 2\\-2 &1} = \pmatrix{1 & 2\\0 &1}$

Finally, add -2 times row 2 to row 1 to get the identity matrix. The matrix is:

$$E_3=\pmatrix{1 & -2\\0 &1}$$

So we have $E_3E_2E_1A = I$

Hence $E_3E_2E_1 = A^{-1}$

 

[foreverdream]

Thanks for this. It does make sense but could you please look at what I just posted as that doesn't make any sense to me. Thanks

 

[Deveno]

each row operation corresponds to a multiplication on the left of A by some matrix P.

for example, with a 3x3 matrix, switching row 1 and row 2 is multiplication by the matrix P =

[0 1 0]
[1 0 0]
[0 0 1].

now, if by doing successive row-operations, we get:

P_n...P₂P₁A = I,

then all the P's together must multiply to A^-1, because that's what an inverse is:

a matrix B such that BA = I (and also AB = I, one could use column operations instead).

so performing the P's on I, gives us:

P_n...P₂P₁I = P_n...P₂P₁ = A^-1

 

[foreverdream]

Thank you it's clear now