6:10 AM
@JohnRennie Are you there

@Abcd morning :-)

@JohnRennie hi, morning. How can a non signular matrix satisfy $AB - BA = A$
Because if we take determinant both sides we get $|A|= 0$
@JohnRennie Basically the question is if a non singular matrix satifises AB -BA = A then prove that det(B+I)= det(B-I)
But I doubt the validity of the question because of the aforementioned reason;

To be honest this is a bit beyond my grasp of matrix algebra. I can have a go at finding the answer, probably with Google's help, if you want but I can't give you the answer straight off.

@JohnRennie Okay leave it.

6:29 AM
@Abcd $A^{-1} AB - A^{-1}BA = A^{-1}A$
So $B - I = A^{-1}BA$
Likewise $ABA^{-1} - BAA^{-1} = AA^{-1}$
So $B + I = ABA^{-1}$
So does $\det(A^{-1}BA) = \det(ABA^{-1})$ ?
The non-singular requirement is the requirement that A have an inverse
Aha, that's true isn't it, because the determinant of a matrix product is the product of the determinants.
$\det(A^{-1}BA) = \det(ABA^{-1}) = \det(B)$

@JohnRennie yes why not

@Abcd so there you go, that's the proof $\det(B+I) =\det(B-I)$

@JohnRennie my mistake was: $\det(A+B) \ne \det(A)+ \det(B)$

@Abcd that's true in general

@JohnRennie no man its not

6:45 AM
@Abcd oops, I meant your statement is true in general
That is, I'm agreeing with you

Oh okay.

7:34 AM
@JohnRennie Are you there

@Abcd hi

@JohnRennie I don't get what he's trying to say ^^^

It seems to be well described here:
In mathematics, an elementary matrix is a matrix which differs from the identity matrix by one single elementary row operation. The elementary matrices generate the general linear group of invertible matrices. Left multiplication (pre-multiplication) by an elementary matrix represents elementary row operations, while right multiplication (post-multiplication) represents elementary column operations. Elementary row operations are used in Gaussian elimination to reduce a matrix to row echelon form. They are also used in Gauss-Jordan elimination to further reduce the matrix to reduced row echelon...
Or at least that covers theorem 1

@JohnRennie where

The Elementary row operations section explains how you use a prefactor to achieve the transformation
Oh, wait, no it doesn't.
Oh yes it does

7:42 AM
@JohnRennie Author's language is confusing$^\infty$ . Please tell me in simple words what he means.

If E is an elementary matrix, as described below, to apply the elementary row operation to a matrix A, one multiplies A by the elementary matrix on the left, E⋅A. The elementary matrix for any row operation is obtained by executing the operation on the identity matrix.
@Abcd there are only three elementary row operations.
Take the scaling one as an example because it's simple.

Okay
??

Suppose you want to multiply the first row of a matrix by a factor $a$. Take I and multiply its first row by $a$.
Then premultiply your matrix by the modified identity.

@JohnRennie But then I's first row will be like a 0 0 (for order 3)

Let me attempt some mathjax ...
$$\left( \begin{matrix} X & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right) \left( \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right) = \left( \begin{matrix} Xa & Xb & Xc \\ d & e & f \\ g & h & i \end{matrix}\right)$$
So to multiply the first row by $X$ perform this operation on I to get:
$$\left( \begin{matrix} X & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right)$$
Then premultiply your matrix by your modified I matrix
And I think theorem 2 follows from theorem 1 using the associativity of matrix multiplication.

8:05 AM
@JohnRennie ?? What do you mean here?

Suppose we have some matrix A and we want to multiply its first row by the constant X.

So perform the transformation R1 -> xR1 in I then premultiply
Is that what you mean?

@Abcd yes

@JohnRennie whats the use when you can just directly do it on original matrix !?

It's common in maths that you establish theorems because they can be useful later.
For example once we have theorem 1 then theorem 2 follows from it.

8:08 AM
@JohnRennie Didn't get Theorem 2 please explain that as well.

So the point of theorem 1 is not to give you a faster/better/whatever way to do an elementary transformation, but to provide a tool for working with more complicated calculations.
@Abcd OK, suppose we want to do an elementary row transformation on the product AB - I don't know what the notation for this is but let's use $X(AB)$ to mean the elementary transformation $X$ applied to $AB$

@JohnRennie elementary transformation is denoted by $\ce{R_1 -> XR_1}$ right?? if you want to multiply row 1 by x

@Abcd yes, that's an example of an elementary transformation.

@JohnRennie ok then?

Theorem 1 tells us that $X(AB) = X(I)(AB)$. Yes?

8:12 AM
yes
Oh
so:
$(XA)B$

BOOM!! :-)
Now you see why theorem 1 is useful, even if it seemed a bit pointless at first.

Okay thanks!
@JohnRennie But why does he say its also valid for post factor case?
Oh postfactor for column thing
Okay got it!

So far we've talked about row transformations.
The same approach applies to column transformations, but you have to postmultiply by your transformed identity matrix, not premultiply.

yeah

If you use my example above it should be obvious that this works.

8:18 AM
@JohnRennie Please show row addition elementary transformation theorem 1 proof just like you showed for row multiplication

OK, suppose you want to set row 1 to be the sum of rows 1 and 2. Is that a good example to take?

Just a minute. Let me recheck my work I think Ive made a mistake.
okay done.
Proved myself. It works. Thanks.

Cool :-)

9:11 AM
@JohnRennie good evening

@harambe hi :-)