\( \begin{pmatrix} y'_1\\ y'_2 \end{pmatrix} = \begin{pmatrix} -3 & -2\\ 6 & 4 \end{pmatrix} \begin{pmatrix} y_1\\ y_2 \end{pmatrix} \) \newline a. Find the eigenvalues and eigenvectors for the coefficient matrix.\newline \( \lambda_1 = 0 \quad \vec{v}_1 = \begin{pmatrix} 2\\ -3 \end{pmatrix} \) and \( \lambda_2 = 1 \quad \vec{v}_2 = \begin{pmatrix} 1\\ -2 \end{pmatrix} \)\newline b. For each eigenpair in the previous part, form a solution of \( \vec{y}' = A\vec{y} \). Use t as the independent variable in your answers.\newline \( \vec{y}_1(t) = \begin{pmatrix} 2\\ -3 \end{pmatrix} \) and \( \vec{y}_2(t) = \begin{pmatrix} -3\\ 1 \end{pmatrix} \)\newline c. Does the set of solutions you found form a fundamental set (i.e., linearly independent set) of solutions?\newline Yes, it is a fundamental set
