Solve the initial value problem $\bar{x}^{\prime}=A \bar{x}, \bar{x}(0)=\bar{x}_0$, where $$ A=\left(\begin{array}{ccc} 1 & 1 & -1 \\ 0 & 2 & -2 \\ -1 & 1 & -1 \end{array}\right) \quad \text { and } \quad \bar{x}_0=\left(\begin{array}{c} -1 \\ 1 \\ 0 \end{array}\right) $$
I know since this matrix A is not diagonalizable, so we use matrix decomposition into invariant subspaces.\
The characteristic equation is $$ p_A(\lambda)=(1-\lambda)(2-\lambda)(-1-\lambda)+2(1-\lambda)+2-(2-\lambda)=-\lambda^2(\lambda-2)=0 $$ The generalized eigenvectors for $\lambda=0$ are given by the null space to $$ A^2=\left(\begin{array}{ccc} 2 & 2 & -2 \\ 2 & 2 & -2 \\ 0 & 0 & 0 \end{array}\right) $$ which has a basis $(0,1,1)$ and $(1,0,1)$. The eigenvalue for $\lambda=2$ are given by the null space to $$ A-2 I=\left(\begin{array}{ccc} -1 & 1 & -1 \\ 0 & 0 & -2 \\ -1 & 1 & -3 \end{array}\right) $$ which gives the eigenvector $(1,1,0)$.
BUT from here I don't understand how to get the general solution given below by the answer and the initial value part:
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The general solution is $$ \begin{aligned} & \bar{x}(t)=(I+t A)\left(c_1(0,1,1)+c_2(1,0,1)\right)+c_3 e^{2 t}(1,1,0) \\ &=c_1(0,1,1)+c_2((1,0,1)-2 t(0,1,1))+c_3 e^{2 t}(1,1,0) \end{aligned} $$ The initial value $(-1,1,0)=(0,1,1)-(1,0,1)$ is a generalized eigenvector for $\lambda=0$ so we get the solution $\bar{x}(t)=(I+t A)(-1,1,0)=(-1,1,0)+t(0,2,2)$.
Thanks in advance! Much appreciate your time :D