Question

(1 point) Consider the linear system "(-1: 1) y. a. Find the eigenvalues and eigenvectors for the coefficient matrix. 1 v1 = and 2 V2 b. For each eigenpair in the previous part, form a solution of y' = Ay. Use t as the independent variable in your answers. (t) = and yz(t) c. Does the set of solutions you found form a fundamental set (i.e., linearly independent set) of solutions? Choose

Answer 1

Solution:

Given system is

y' 6 4. -12 -8 y

compare it with

y' = Ay

we have

$A=\begin{bmatrix} 6 &4 \\ -12 & -8 \end{bmatrix}$

Now eigen values are given by

A - XI = 0

6-3 —12 4 -8-3 =0

$\Rightarrow (6-\lambda)(-8-\lambda)+48=0$

$\Rightarrow -48-6\lambda+8\lambda+\lambda^2+48=0$

$\Rightarrow \lambda^2+2\lambda=0$

XX+2) = 0

= -2.0

Now, eigen vector $v_1$ corresponding to $\lambda_1=-2$ is given by

$[A+2I]v_1=0$

$\Rightarrow \begin{bmatrix} 8 & 4\\ -12 &-6 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$

applying $R_1\to \frac{1}{4}R_1$

$\Rightarrow \begin{bmatrix} 2 & 1\\ -12 &-6 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$

applying $R_2\to R_2+6R_1$

$\Rightarrow \begin{bmatrix} 2 & 1\\ 0 &0 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$

$\Rightarrow 2a+b=0$

$\Rightarrow b=-2a$

by choosing $a=1$ , we get

$v_1=\begin{bmatrix} 1\\-2 \end{bmatrix}$

Thus,

${\color{Red} \lambda_1=-2,\ v_1=\begin{bmatrix} 1\\-2 \end{bmatrix}}$

Now, eigen vector $v_2$ corresponding to $\lambda_2=0$ is given by

$[A]v_2=0$

$\Rightarrow \begin{bmatrix} 6 & 4\\ -12 &-8 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$

applying $R_2\to R_2+2R_1$

$\Rightarrow \begin{bmatrix} 6 & 4\\ 0 &0 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$

$\Rightarrow 6a+4b=0$

$\Rightarrow b=-\frac{6}{4}a$

$\Rightarrow b=-\frac{3}{2}a$

by choosing $a=2$ , we get

$v_2=\begin{bmatrix} 2\\-3 \end{bmatrix}$

Thus,

${\color{Red} \lambda_2=0,\ v_2=\begin{bmatrix} 2\\-3 \end{bmatrix}}$

Now, the solutions are given by

$y_1(t)=v_1e^{\lambda_1t}\ \ and \ \ y_2(t)=v_2e^{\lambda_2t}$

$\Rightarrow y_1(t)=\begin{bmatrix} 1\\-2 \end{bmatrix}e^{-2t}\ \ and \ \ y_2(t)=\begin{bmatrix} 2\\-3 \end{bmatrix}$

${\color{Red} \Rightarrow y_1(t)=\begin{bmatrix} e^{-2t}\\-2e^{-2t} \end{bmatrix}\ \ and \ \ y_2(t)=\begin{bmatrix} 2\\-3 \end{bmatrix}}$

which is the required solution.

Also

$\Rightarrow \begin{vmatrix} e^{-2t} &2 \\ -2e^{-2t} & -3 \end{vmatrix}=-3e^{-2t}+4e^{-2t}=e^{-2t}\neq 0$

Thus, solutions are linearly independent.

This complete the solution.