 Solution

The upper-triangular matrix has distinct eigenvalues $2,-1$, with respective eigenvectors $(1,0{)}^T$ and $(-2,1{)}^T$. ThusThe defining power series shows that the matrix exponential respects similarity. Applying solution to the diagonal factor givesAs a direct series check, $(A^m{)}_{12}=2(2^m-(-1{)}^m)$ for $m\ge 0$; summing this entry divided by $m!$ gives the same off-diagonal expression.