Going back from cumsum for a matrix

Suppose I cumsum a matrix, say A=rand(3,3); B=cumsum(A).

Knowing B, how to I get back to A, in a reasonably efficient way, for a rather large B?

Thanks!

[valentino dardanoni]

Thankyou David (and Walter). It works perfectly in my application.

Answer 1

[David Hill]

A=round(rand(100,100),4);
B=cumsum(A);
a=round([B(1,:);diff(B)],4);
isequal(A,a)

ans = logical
   1

[Walter Roberson]

Right.

Key points here are the use of diff(), the duplication of the first entry, and the rounding or other way of comparing with tolerance for the cross-check (since you would need to deal with round-off errors.)