- #1

JessicaHelena

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**Summary::**Suppose that [x, y] = e^{-3t} [-2, -1] is a solution to the system $x' = Ax$, where A is a matrix with constant entries. Which of the following must be true?

a. -3 is an eigenvalue of A.

b. [4, 2] is an eigenvector of A.

c. The trajectory of this solution in the phase plane with axes x, y is part of a straight line.

d. e^{-3t} [-1, -2] is another solution to the same system.

e. the phase portrait of the system is an attracting node.

f. the system is stable.

Please refer to the attached screenshot for clearer question; however, the checked answers there are not my final answers.

This is a star node, and hence ##\lambda = -3##. (Hence ##a## is right.) Since the eigenvalue is negative, it is an attracting node (which makes ##e## correct).

I graphed the parametric function, and the result was a straight line extending only in the 3rd quadrant. Hence ##c## is right.

I wasn't sure of a good mathematical way to get the matrix A, but by trial and error, got that A = [0, -6; -3, 3]. Checking for ##Av = \lambda v##, I got the right answer, which makes me believe I have the right A. From this, trA = 3, and det A = -18. Checking against the rule of stability, we get that this system is not stable. (hence ##f## is not an answer).

I am not sure how exactly to do b and d, but using the rule that ##Av = \lambda v##, I don't believe they are right.

Hence I would check a, c, and e. Is this way of thinking right?

**[Moderator's note: Moved from a technical forum and thus no template.]**

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