= Solution
{c}
For the given <matrix> $A$, the <characteristic polynomial> is
$$
\det(A-\lambda I)=(2-\lambda)(1-\lambda)-6=\lambda^2-3\lambda-4=(\lambda-4)(\lambda+1).
$$
For $\lambda=4$, the <eigenvector> equation gives $v_2=v_1$; for $\lambda=-1$, it gives $3v_1+2v_2=0$. Thus the <eigenvalues> and their corresponding <eigenspaces> are
$$
\boxed{\lambda=4:\ \operatorname{span}\{(1,1)^T\},\qquad \lambda=-1:\ \operatorname{span}\{(2,-3)^T\}.}
$$
For any <eigenvector> $v$ with <eigenvalue> $\lambda$, the vector-valued <function> $e^{\lambda x}v$ satisfies $Y'=AY$, since its <derivative> is $\lambda e^{\lambda x}v=e^{\lambda x}Av$. The two displayed <eigenvectors> are <linearly independent>, so the <general solution> of the <linear system of ordinary differential equations> is
$$
\boxed{\begin{pmatrix}y_1(x)\\y_2(x)\end{pmatrix}=C_1e^{4x}\begin{pmatrix}1\\1\end{pmatrix}+C_2e^{-x}\begin{pmatrix}2\\-3\end{pmatrix},\qquad C_1,C_2\in\mathbb R.}
$$
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