Let \( \mathbf{y}^{\prime}=A \mathbf{y} \) be a system of differential equations where \( A=\left[\begin{array}{cc}-3 & -1 \\ 0 & -3\end{array}\right] \) The matrix has spectrum \( \lambda(A)=\{-3\} \) and \( \mathbf{c}=\left[\begin{array}{l}1 \\ 0\end{array}\right] \) is an eigenvector of \( A \) corresponding to \( r=-3 \). Which of the following vectors satisfies \( (A-r I) \mathbf{d}=\mathbf{c} \) ? \[ \mathbf{d}=\left[\begin{array}{l}-2 \\ -1\end{array}\right] \quad \square \mathbf{d}=\left[\begin{array}{l}-1 \\ -3\end{array}\right] \quad \square \mathbf{d}=\left[\begin{array}{c}-1 \\ 0\end{array}\right] \]

We start with the given system \[ \mathbf{y}' = A \mathbf{y}, \quad \text{with} \quad A = \begin{bmatrix} -3 & -1 \\ 0 & -3 \end{bmatrix}, \] and we know that \( r = -3 \) is an eigenvalue associated with the eigenvector \[ \mathbf{c} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}. \] We need to find a vector \(\mathbf{d}\) such that \[ (A - rI)\mathbf{d} = \mathbf{c}. \] Since \( r = -3 \), substitute to obtain \[ A - (-3)I = A + 3I. \] Calculating \(A + 3I\), \[ A + 3I = \begin{bmatrix} -3 & -1 \\ 0 & -3 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 0 & 0 \end{bmatrix}. \] Let \[ \mathbf{d} = \begin{bmatrix} d_{1} \\ d_{2} \end{bmatrix}. \] Then the equation \((A+3I)\mathbf{d} = \mathbf{c}\) becomes \[ \begin{bmatrix} 0 & -1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} d_{1} \\ d_{2} \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}. \] This results in the system: \[ \begin{aligned} 0\cdot d_{1} - 1\cdot d_{2} &= 1,\\[1mm] 0\cdot d_{1} + 0\cdot d_{2} &= 0. \end{aligned} \] From the first equation, \[ -d_{2} = 1 \quad \Rightarrow \quad d_{2} = -1. \] The second equation is an identity \(0=0\) and does not impose additional conditions, so \(d_{1}\) is free. Among the provided options, we check the vectors: 1. \(\mathbf{d}=\begin{bmatrix}-2 \\ -1\end{bmatrix}\): Here, \(d_{2}=-1\) (which is required). 2. \(\mathbf{d}=\begin{bmatrix}-1 \\ -3\end{bmatrix}\): Here, \(d_{2}=-3 \neq -1\). 3. \(\mathbf{d}=\begin{bmatrix}-1 \\ 0\end{bmatrix}\): Here, \(d_{2}=0 \neq -1\). Thus, the vector which satisfies \((A-rI)\mathbf{d} = \mathbf{c}\) is \[ \mathbf{d}=\begin{bmatrix}-2 \\ -1\end{bmatrix}. \]