Question

7) Change (23 pts) Diagonalization of a matrix. For the matrix from the last problem. a) (6 pts) Find the eigenvalues of A b) (6 pts) Find the eigenvectors of A, Choose x1=1 for the first element of both eigenvectors. c) (4 pts) Is this system stable? What type of damping does it have? d) (2 pts) Find the diagonal matrix D corresponding to A. e) (1 pts) Find D3, use actual numbers. f) (1 pts) Find the eigenvector matrix X g) (1 pts) Find X-1 h) (2 pts) Write the equation for A3 using D, X, and X-1. You don't have to multiply the matrices for credit.

Here is the matrix mentioned in the problem:

Answer 1

10.73 - 27ilan L 3 -8.27 x = 6 0.773(1) -2(X2) 20 0.73 = 2X2 - X₂ = 0.73 = 0.365 Nu e ogen vector (N) [035 vector 0.365 (A-2)x=0 2 1163.c. 3) (2)- 0: 43(!) CO 63 -0.73 ] [ ^2 J (8.27(1) + (-2XX2) = 0 8.27= 2x2 → X2= 137-4.105 egen vester $- GPac] Ve0907

2 stable con not. Floots of charactericite equation that Be eigen values are nothing but poles of Transfer function. poles = 9.27, 5.27 S-plane Sat 8.27 As one root is on sight side of s-plane. Hence the given systems Is unstable. C.E - S²+35-12=0 s²+2{wnst whl=0 wp² = -12 (wnl=√12 | Wal = 3.46 22wn=3 2*XX3,46=3--0.43 sunderdamped. d) diagonal madre. D D = PAP Joo 21 P= [emove) potigues) - Cous les loy 0.361 1.135 alle 13 (* (*19] JPL = loi3er 4,185 - 1(4185) -((0.36) - 3.17.

3. 0.365 D=PAP 35 0. 19.405 -2.97 11 17 3.77 0.36 | 8.576 = 3.77 0.018 -1988 1-0.018 DE Other two values are 2 XU 2.2 +4 -0 Sa 1.0 5.273 J L approximately zerol e) 03 - (2.874 of faa74 78.374 07 To sans) o 5.273 Lo san [sunti o 7 (20294 o] To 21-54 lo son NTS 3 D² = 1.458 07 Deo 1466