a. If A is the coefficient matrix, solve det(A - λI) = 0 for the eigenvalues λ :
[tex]\det\begin{bmatrix}-6-\lambda & -4 \\ 12 & 8-\lambda\end{bmatrix} = (-6-\lambda)(8-\lambda)+48 = 0 \implies \lambda(\lambda-2)=0[/tex]
[tex]\implies \lambda = 0, \lambda = 2[/tex]
Let v = [v₁, v₂]ᵀ be the eigenvector corresponding to λ. Solve Av = λv for v :
[tex]\lambda=0 \implies \begin{bmatrix}-6&-4\\12&8\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix} \implies 3v_1 + 2v_2 = 0[/tex]
If we pick v₂ = -3, then v₁ = 2, so [2, -3]ᵀ is the eigenvector for λ = 0.
[tex]\lambda = 2 \implies \begin{bmatrix}-8&-4\\12&6\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix} \implies 2v_1 + v_2 = 0[/tex]
Let v₁ = 1, so v₂ = -2.
b. λ = 0 and v = [2, -3]ᵀ contributes a constant solution,
[tex]\vec y_1 = e^{\lambda t} v = \begin{bmatrix}2\\-3\end{bmatrix}[/tex]
while λ = 2 and v = [1, -2]ᵀ contribute a solution of the form
[tex]\vec y_2 = e^{\lambda t} v = e^{2t} \begin{bmatrix}1\\-2\end{bmatrix}[/tex]
c. Yes; compute the Wronskian of the two fundamental solutions:
[tex]W(1, e^{2t}) = \det\begin{bmatrix}1 & e^{2t} \\ 0 & 2e^{2t}\end{bmatrix} = 2e^{2t} \neq 0[/tex]
The Wronskian is non-zero, so the solutions are independent.
