Can you please help me solve question 1 and 3? In Exercises 1–4, each of the linear systems has one eigenvalue and one line of eigenvectors. For each system:(a) Find the eigenvalue.(b) Find an eigenvector.(c) Sketch the direction field.(d) Sketch the phase portrait, including the solution curve with initial condition \( \mathbf{Y}_0 = (1, 0) \); and(e) Sketch the \( x(t) \)- and \( y(t) \)-graphs of the solution with initial condition \( \mathbf{Y}_0 = (1, 0) \).1. \( \frac{d\mathbf{Y}}{dt} = \begin{pmatrix} -3 & 0 \\ 1 & -3 \end{pmatrix} \mathbf{Y} \)2. \( \frac{d\mathbf{Y}}{dt} = \begin{pmatrix} 2 & 1 \\ -1 & 4 \end{pmatrix} \mathbf{Y} \)3. \( \frac{d\mathbf{Y}}{dt} = \begin{pmatrix} -2 & -1 \\ 1 & -4 \end{pmatrix} \mathbf{Y} \)4. \( \frac{d\mathbf{Y}}{dt} = \begin{pmatrix} 0 & 1 \\ -1 & -2 \end{pmatrix} \mathbf{Y} \)

Name: Can you please help me solve question 1 and 3? In Exercises 1–4, each
Uploaded: 2024-03-20T07:07:37.000Z
Channel: Bartleby
Description: Can you please help me solve question 1 and 3? In Exercises 1–4, each of the linear systems has one eigenvalue and one line of eigenvectors. For each system:(a) Find the eigenvalue.(b) Find an eigenvector.(c) Sketch the direction field.(d) Sketch the

Note: As per Bartleby's guidelines, for more than one question asked in the main part and for more than 3 questions asked in subparts, only one and three questions respectively have to be answered. Please upload the other questions separately. (1) Given the system dYdt=-301-3Y ⋯⋯(1) Here, the matrix of the linear system is, A=-301-3 (a) To determine the eigenvalue λ, i.e., to determine λ such that det(λI-A)=0λ+30-1λ+3=0λ+32=0λ+3=0λ=-3 Thus, the eigenvalue of the linear system (1) is λ=-3(b) The eigenvector v corresponding to the eigenvalue -3 is given by, Av=λvA+3Iv=00010v1v2=00 ⋯⋯(2) From equation (2), v1=0 Thus, the one-line eigenspace is given by, v=t01: t∈ℝ Choose t=1. Then, the one-line eigenvector corresponding to eigenvalue -3 is v=01(c) The formula for the solution of the given linear system (1) is, Y(t)=eλtvy1(t)y2(t)=e-3t01y1(t)y2(t)=0e-3t ⋯⋯(3) From equation (3), y1(t)=0y2(t)=e-3t The sketch of the direction field is given by,