It isnC0*A^n*b^0 + nC1*A^(n-1)*b^1 + ... + nCr*A^(n-r)*b^r + ... + nCn*A^0*b^n

where nCr = n!/[r!*(n-r)!]

If, for an n*n matrix, A, there exists a matrix B such that AB = I, where I is the n*n identity matrix, then the matrix B is said to be the inverse of A. In that case, BA = I (in general, with matrices, AB ≠ BA)

I is an n*n matrix consisting of 1 on the principal diagonal and 0s elsewhere.

Yup, its official, usher is the king of r n b.

T r+1 = (n / r) (a ^n-r) x (b)^r