Consider the linear systemy→′=[−3−264]y→a) Find the eigenvalues and eigenvectors for the coefficient matrixλ11, v→1=[−12] and λ2=0, v→2=[−23]b) For each eigenpair in the previos part, form a solution of y→′=Ay→. Use t as the independent variable in your answers.y→1(t)=[] and y→2(t)=[−23]c) Does the set of solutions you found form a fundamental set (i.e., linearly independent set) of solution? No, it is not a fundamental set.

Question

Consider the linear systemy→′=[−3−264]y→a) Find the eigenvalues and eigenvectors for the coefficient matrixλ11, v→1=[−12] and λ2=0, v→2=[−23]b) For each eigenpair in the previos part, form a solution of y→′=Ay→. Use t as the independent variable in your answers.y→1(t)=[] and y→2(t)=[−23]c) Does the set of solutions you found form a fundamental set (i.e., linearly independent set) of solution? No, it is not a fundamental set.

broliY · Accepted Answer

Step 1Consider the systemy→′=[−3−264]y→Compare the above system with y→′=Ay→, to getA=[−3−264] and y→=[y→1y→2]The characteristic matrix of A is ∣A−λI}=0 where λ is the eigen value|[−3−264]−λ[1001]|=0[−3−λ−264−λ]=0(−3−λ)(4−λ)+12=0−12+3λ−4λ+λ2=0λ2−λ=0λ(λ−1)=0It follows that, λ=0 or λ=1So, the eigen values of the system is λ={0, 1}Step 2Calculate the eigenvector corresponding to λ1=1(A−λ1I)v1=0[−3−1−264−1]v1=0Substitute v1=[v11v21][−4−263][v11v21]=0Thus, the corresponding system of equations is,−4v11−2v21=06v11+3v21=0−4v11−2v21=0v21=−2v11Substitute v11=−1v21=2Thus, the eigenvector corresponding to λ1=1 is v1=[−12] Calculate the eigenvector corresponding to λ2=0(A−λ2I)v2=0[−3−264]v2=0Substitute

Consider the linear system\vec{y}'=\begin{bmatrix}-3 & -2 \\ 6 & 4 \end

Answered question

Answer & Explanation

New Questions in Linear algebra