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Substituting an exponential particular solution gives $\lambda u-Au=v$, or $(\lambda I-A)u=v$. Since $\lambda$ is not an eigenvalue of $A$, the matrix is invertible. The exponential forcing of a constant-coefficient differential system therefore hasFor the specified matrix, $I-A=\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)$. Solving $u_2=1$, $u_1-u_2=1$ gives $u=(2,1{)}^T$, soDifferentiation verifies both component equations directly. If a general solution were desired, the homogeneous term $e^{tA}c$ would be added, with the matrix exponential here equal to $I+tA$ because $A^2=0$.