Lets start by writing
( ( ln(2), 1, 1, 1 ), ( 0, ln(2), 1, 1), ( 0, 0, ln(2), 1 ), ( 0, 0, 0, ln(2) ) )
in the form ln(2) I_4 +A, where I_4 is the 4\times 4 identity matrix and A is the upper triangular matrix
A = ( (0, 1, 1, 1 ), ( 0, 0, 1, 1), ( 0, 0, 0, 1 ), ( 0, 0, 0, 0 ) )
Since I_4 commutes with all 4 times 4 matrices in general, and A in particular, the exponential matrix is simply
exp(ln(2) I_4 +A) = exp(ln(2) I_4) exp(A) = e^{ln(2)}I_4 exp(A) = 2 exp(A)
So the calculation essentially reduces to the determination of e^A. Note that the standard approach using diagonalization does not work here, since A is not diagonalizable. However, the characteristic equation for A is easily seen to be
lambda^4 =0
and thus the Cayley-Hamilton theorem says that
A^4 = 0
so that the infinite series for e^A truncates :
e^A = I+A+1/{2!} A^2 + 1/{3!}A^3
It is easy to calculate
A^2 = ((0,0,1,2), (0,0,0,1), (0,0,0,0), (0,0,0,0)) and A^3 = ((0,0,0,1),(0,0,0,0),(0,0,0,0),(0,0,0,0))
Thus
e^A = ((1,1,3/2,13/6),(0,1,1,3/2),(0,0,1,1),(0,0,0,1))
Our solution is twice this matrix!