Is Matrices Satisfies A2-b2=(a+b)(a-b)?

I know that this question was asked over a year ago and that no one will read this post.   But I can't sit idly by and watch all these wrong answers get posted.  

The if a and b were real or complex numbers, then the equation a^2 - b^2 = (a + b)(a - b) would be true, for the reasons posted above.   But a and b are matrices, not real numbers.   Implicit in all of the calculations above is that ab = ba.   But that doesn't hold for most pairs of matrices.

So for example, if you take
a =
1 2
3 4

b =
5 6
7 8

You'll find you get
a^2 - b^2 =
-60 -68
-76 -84

and yet
(a + b)(a - b) =
-56 -56
-88 -88

People! The question is whether a matrix A and its square A^2 and a matrix B and its square B^2 can satisfy the formula (A^2-B^2)= (A-B)(A+B) and the answer is simply No! Because A*B is not equal to B*A in matrix and you can guess why if you know matrix. Yes dimensions.

A^2-b^2=(a+b)(a-b)

Take R.H.S

(a+b)(a-b)

= a^2-ab+ab-b^2

= a^2 + 0- b^2

= a^2 - b^2

Hence proved that R.H.S is equal to L.H.S

A2-b2=(a+b)(a-b)
a2+(-ab+ab)-b2
(a2+0-b2)
(a2)+(-b2)
hence
a2-b2
I think this is the answer                                                                                                             ba600y 7449

So for example, if you take

Matrix A

2 0

0 2

Matrix B

0 4

0 0

matrix A^2-B^2=

4 0

0 4

matrix (A+B)(A-B)

4 0

0 4

Same

A2-b2=a-b a+b
a-b *a+b =a(a+b)-b(a+b)
=a2+b2-ab-b2 =[(a)2

If: A^2-b^2 =(a+b)(a-b) ====== Given:    a=b      =>   a^2=ab     =>   a^2-b^2=ab-b^2     =>   (a+b)(a-b)=b(a-b)     =>   (a+b)=b     =>   a+a=a     =>   2a=a     =>   2=1

A2-b2=(a+b)(a-b)
a2+(-ab+ab)-b2
(a2+0-b2)
(a2)+(-b2)
hence
a2-b2

Concrete/picture
form of( a+b)(a-b)

X2+d2

(a+b) (a-b) = a*a+a*-b+b*a+b*-b = a^2-b^2 (because a*a = a^2, a*-b = -ab, a*b = ab and b*-b = =b^2... So -ab and ab get canceled)
Hence, proved

Take rhs
(a+b)(a-b)
a2-ab+ba-b2
a2-0-b2
that is a2-b2
hence proved
since rhs =lhs or
lhs=rhs

A2-b2 = (a-b) (a+b) (a-a-b+b)

But a^2+b^=?