Look at the 2x2 case. Assuming a=1 then we need to solve the equation

(b-1)(n(p+bm+m)+sm) = 0 where p is the top left coordinate of the matrix K, s the bottom left. m, n are the coordinates of M. Assuming b is not 1 then we want to find n, m, b, p, s such that n(p+bm+m)+sm=0. So, let s=0=n.

That is, and

is a counter-example.