Suppose that the matrix A A has the following eigenvalues and eigenvectors: (1 point) Suppose that...

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= Aro > () The eigenvalues and eigenvectors of A are as follows: 4 d=20 with = 2 -5i and dq=-zi with ů = [216:] We solve (1) with respect to the eigens values and eigennectors given in the question. of its real and Now, write ů in terms imaginary parts: a = (2) +i (os) '. The corresponding complex solution etit to the system (1) can E 3 v written as then be 3) (3) ti 2 = @04 (Cas (2+3 + i sin (2+)) Now, using the theorem, the real and imaginary farts of () cos(24) - are 6 ) sin (24) 8 = oot and r₂ - sin (2+) + (05) cos (2+) (24 ) 0,4 e 2 (3) where rap, tire 2 I Whore iz autika the theorem is, given a system X = AX, A is a real matrin. If a complese solution, then its real and imaginary parts 24, u are also solutions ما

to the system Hence for the linear the general solution system rar be x(+) ar tor2 1 = sin (2+) + 0+ e cos (27) [] J - (18 ( [ te, sin (2+) + cos (2+) e 2 [using (2) & (3) from A we get x(+) y+) 0.4.sin (2+) + 4.4 Cos (2) +4ų sin (21) to.cz ces (2+) 5 e, sin (27) +2 4 cos (at) + 2 Gsm (2) - 5 cos (zt) +6) 4 cos (27) 4 sin (2+) 9 5 Sin (2+) +2cos (24) 2 sin (24) - 5 cos(24) رت → 6) sides Comparing both of we get X(t) 4 G cos (24) + A G sim (24) and y (1) = 54 sin (21) +2 4 cos (24) & 2G sin (24) - 56 cos (24)

A. In eigenvalue / eigenvector form: Oit * (*) sin (2) 4 -2 - ([ Cos (27) @ 5 y (+) cos(2+) 0.t e ta sin (24) + ([11 I using (1) B. In fundamental matrix form: (+) 3 y (+) A.G.cos(2t) + 4 e sim (2+) 54 sim (21) + 24 cos (27) + 26 sn (21)=5 460s (24) [using (5) 4 cos (27) 5 sim (2t) + 2 cos (2+) 4 sin (2t) 2 sin(2t) - 5003(27) e. (+) 4 a cos (24) +4€ sin (27) y (u) = a (s sm (28) +2005(24) +4 (25(21) -555(24) [using (7)