Question

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5. Find the eigenvalues and any real eigenvectors of A, and use this information to sketch the phase portrait of the system * = Ax. (a) A=( - -1)

Answer 1

5.(a) As P5 47 For eigen values , Solve [A-21] and then det [A-AI) = 0 so, [A-AL ] = 15-0 4 7 [-. --a] Now, 15-1 4 1-o → (5-1)(-1-1) +8 = 0 → 45 - 5a ta + a² +8 = 0 → 32 - 42 + 3 = 0 ☆ (1-3)(A-1) = 0 = = 3,1 for fågen vector v corresponding to a = 1, Solve [A-I ]v = 0 7 14 4 Trel=107 -2 -2 ] [ve] Lod 7 40; +4V 2 =0 → V E-V

to 9 = 3 for eigen vectoru, corresponding solve [A-31 ] uso 52 47u7=107 » ų = -24g 24 +44 =0 - uFq7 Now, for 1 = ' , Ves and for i= 3 , it = 1-2 A=) y 12=3 t I tool 9=3 According to eigen vectors, we draw the phase portrait of the system. Since Eigen values are positừve so, au solution will tend away from the origin. As t-too,

Here we start with small Eigen value and tends towards the big Eigen value 9.., start being parallel to d=1 and then it going to become parallel to 1= 3. When the soution were away from the equilibrium point, then it 9s unstable. 510) A = ro 17 L3 Solve 1A-211 - 0 Ta 2- = -21+ d² - 3 = 0 * 22 27 3 = 0 → 22 - 32 +1-3-0 ☆ (-3)(4+1) = 0 + de 3,-1 Eigen rector u corresponding to d=-1, solve [A +2 Ju=0 Il Trul.ro7 13 31 luz ]=07 ale = - Us 7 U + Ug =0 So, ų ar

to a = 3, Eigen vector v corres ponding Solve [A - 3 I ]v= 0 a 5-3 Tur=r07 3 -Ilves lo ² -30, +2 = 0 3v, Eva so, "=16] Now, for d=-1, U-rit 1 for d= 3 , varit Saddle Las-1 9=3" for positive Eigen value Solution tend away from origin. for negetive Eigenvalue , solution tends towards origin. Since Solution donet tends towards nor away from origin. so Solution follows the direction of arrows,