A 2x2 matrix A can be row-reduced to I by the following 4 elementary row operations ( in order) 1) R1 R2 2) R2 -- (1/2) R2 3) R1-- R1 - 4R2 4)R1 -- (1/4) R1 I got: (E4)(E3)(E2)(E1)(A) = I(E4)^-1(E3)^-1(E2)^-1(E1)^-1 but my teacher said it was on the wrong order, i dont see how it would me if we have to go from outside to inside?

Question

A 2x2 matrix A can be row-reduced to I by the following 4 elementary row operations ( in order)
1) R1 <-> R2
2) R2 --> (1/2) R2 
3) R1--> R1 - 4R2
4)R1 --> (1/4) R1
I got: (E4)(E3)(E2)(E1)(A) = I(E4)^-1(E3)^-1(E2)^-1(E1)^-1

but my teacher said it was on the wrong order, i dont see how it would me if we have to go from outside to inside?

phi · Answer

why do you have "stuff" on the right-hand side.  Isn't it just
(E4)(E3)(E2)(E1)(A) = I
?

anonymous · Answer

Well sorry, I have to express A as a product of elementary matrices. My bad.

anonymous · Answer

So I did A= I(E4)^-1(E3)^-1(E2)^-1(E1)^-1

phi · Answer

ok.  Let's suppose we only needed E1 to change A to I
(E1) A = I
now multiply both sides by E1^-1  (notice you put E1^-1 to the left):
 E1^-1(E1) A =  E1^-1 I

anonymous · Answer

ah, I see. But wouldn't you need to work your way from the outside to inside if you have 4 elementary matrices, hence starting with the "last one" being e4

phi · Answer

yes, you do work out to in
but put the matrices on the left side of I.    I think you get a different order.

phi · Answer

(E4)^-1  (E4)(E3)(E2)(E1)(A) = (E4)^-1  I
that "gets rid of" E4
(E3)(E2)(E1)(A) = (E4)^-1  I
now (E3)^-1:
(E3)^-1(E3)(E2)(E1)(A) = (E3)^-1(E4)^-1  I 
or
(E2)(E1)(A) = (E3)^-1(E4)^-1  I 
and so on

anonymous · Answer

The way I saw it was like this, You have E4E3E2E1A=I, to get A^-1, you do A^-1*A=IA^-1 (I mean if we look at Ax=b, x=b*A^-1 no?) so, I thought that we'd have to cancel E4 first, before being able to "cancel"E1

anonymous · Answer

Oh i see by wrong order. it doesn't mean that E4 is wrong, it should beo n the left of I instead of the right. Is there any mathematical implication to this? Or left side or right side would give the same answer?

phi · Answer

*** you said: I have to express A as a product of elementary matrices.***
A= stuff  
not A^-1

phi · Answer

matrix multiplication (in general) does not commute.  i.e. order matters.
you always have to be aware of the order.

phi · Answer

if you have
E1 A = I
you know you want to put E1^-1 on the left side because you want
(E1^-1)(E1) ... which is I
E1 A (E1^-1)  won't do the trick.

and if you multiply on the left side, you do the same thing to the other side of the equation

anonymous · Answer

Ah, I see what you mean, it makes sense, thank you for being patient! have a good day :)

phi · Answer

yw