$$ A = \begin{bmatrix} 1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & -1 & 1 \end{bmatrix} $$
I have the answer, but I don't know the best method.
I've tried decomposition to $PAP^{-1} = B $ where $B$ is diagonal, and then exp$(xA) = P$exp$(xB)P^{-1}$ but it's not working out too well.
On
this matrix $A$ is the sum of the identity matrix and the skew symmetric matrix $B = \pmatrix{0&1&-1\\-1&0&1\\1&-1&0}$ which represents a quarter rotation about the axis $(1,1,1)^T$ followed a magnification by a factor of $\sqrt 3.$ an orthonormal basis are the columns of $U = \pmatrix{1/\sqrt 2&1/\sqrt 6&1\sqrt 3\\-1/\sqrt2&1/\sqrt6&1/\sqrt3\\0&-2/\sqrt6&1/\sqrt3}$ it can be verified that $$ B = U^T \Lambda U \text{ where } \Lambda = \pmatrix{0&\sqrt 3&0\\-\sqrt 3&0&0\\0&0&0}$$ it is not hard to show that $$e^{\Lambda t} = \pmatrix{\cos \sqrt 3t&\sin\sqrt 3 t&0\\-\sin \sqrt3 t&\cos \sqrt3 t&0\\0&0&1}, \text{ and }e^{Bt} = U^Te^{\Lambda t}U.$$
finally, $$e^{At}=e^{(A+I)t}=e^{It}e^{Bt} = U^T \pmatrix{e^t\cos \sqrt 3t&e^t\sin\sqrt 3 t&0\\-e^t\sin \sqrt3 t&e^t\cos \sqrt3 t&0\\0&0&e^t} U.$$
The eigen values of $A$ are $1, 1+\sqrt{3}i,1-\sqrt{3}i$. Hence $A$ is diagonalizable. If $v_1,v_2$ and $v_3$ are eigen vectors for $1, 1+\sqrt{3}i,1-\sqrt{3}i$ respectively, then $A=PBP^{-1}$ where $B=diag(1, 1+\sqrt{3}i,1-\sqrt{3}i)$ and $P=(v_1 \; v_2 \; v_3)$. So, $e^{xA}=Pe^{xB}P^{-1}$. Here I have used $e^y=1+y+\frac{y^2}{2!}+...$