Question

(1 point) Given that ū = and are eigenvectors of the matrix -12 24 determine the corresponding eigenvalues. 21 = -1 12 = 1

Answer 1

* Given that ü [-] and - 2 are eigen vectors of the matrix - 12 A = 24 2 1-4 16 6 Now, The characteristic equation is a det ( An I2) roots - 12 = 212) = 0 a 24 O 16-2 > (-12-x) (16- x) + 192=0 - 192 + 12 - 16 a ² + 192=0 (2.0) a - 4a - 4) a = 0, x = 4 V = 0 For Now, Jolana or -12 24 (A - , I) 10 8 16 Sos. [1] is the eigen vector corresponding the eigenvalree = 0 to And. For 22= 4, (A - 13633-13:1-13-13 Son 3 is the eigenvector corresponding 2 to the eigenvalue 22 = 4

The given system of differential equation is 1 du da dt 6 -1 1. xa → dxn dt d22 ] let. x= 6 9 > dra dt - 6x1 + x2 & drz dt - 6x4 - x2 D x4 +6n1 = x2 & Dx₂ + x2=-6x4, let D = It Therefore. (1+6)x= x2 & (D+1) X2 = - 6x2 → (D+1) (0+ 6) x = (D+1) X2 (Ô +70+6)x1 - 6x4 D + 70 +12) x4 = 0 Then, ausiliary equation m² + 7 m +12=0 m² + 4m +3m +12=0 m (m + 4) + 3 (m + 4) (m + 3) (m + 4) The O m = -3, - 4 -3t + ca é -4t. Cze So, ni

Now, from X2 (D+ 6) * we get ONT x2 (8+6) (Cs e 3t + Ge4t) (-36 e 3t_ 4 Ce4t Gett) (3cje 3+ + 2 cecht). -3t +6cae the X2 x2 2 Now, The Condition 26-2):) ini teal (o) 2 (0) and x₂(0) = -2 FC Then, ca e 3t + G -4t & n. (A) xa (o (1 + C2 O Cat C=0 and (0) 3 1 + 2 C₂ = -2 (to 31 + 2 C2 = -2 301 – 201 = -2 [c2=-ca & C2 = 2 Therefore, [Rect) - 3t 2 e X(t) = 2 e 4t K2 (t) -46 it 4e be 3t + 4 (Putting Cq = -2

matrix is The given I 16 4o A= -80 - 32 Then, the characteristic equation is det (A - I) 16- X 40 1- - 32 -80- ² t > (16-2) (-80-x) + 1280=0 640 =0 X (+64) = 0 a=0 & 2 =-64 the eigenvector corresponding smallest eigenvalue da = 0 (A - an Iz) x=0 x 1 be Let x = 6 x2 to the Then, xl 16 x4 +4022=0 & - 32x1 - 80x2=0 -32 - 80 2a1 +5x2=0 5 u 2 Let x2=k Then, Xi = associated eigen vector Then الميا X =k 1 Let y = w be the ปราณ corresponding eigen- larger eigenvalue 264 vector the to

Then, (A - 2 1. Iz Iz) y = 0 + 80 > 40 ya - 0 (zero matrix) -32 - 16 Y2 mon a 804, +40 y = 0 ) - 3241 164, = 0 Therefore, 2 y + y2 = 0 Now , let so y = cm a real number 7. Then Y2= - 2 m So The associated eigen vector is Y = [:] - [*]-[:)]

The * given linear system is ģ' = [34] y 1 ody, at dyz dt 6 The 3 -4 on efficient matrix is A= AI) = 0 Then, A 5- 6 -3 - 4-0 > -20-50+ 4 2+² +18=0 x² - - 2 = 0 (x-2) (2+1)=0 a=-1, 2 Now, let i be the eigen veetor corresponding to A2 = -1. Then, (A - Az Iz) u 6 = 0 (zero matrix atvix)-[] x 1 x2 ul 7 [6 n2 3 -] [ & - 304 - 3x2=0 604 + 6x2 = Na tx2=0 Then, Now, let x2=k , any real Then -K KA is us So The eigen veetor [] 1] -K

be the eigenvector to 2 2 Again, let 3 [5. corresponding (A - AIZ) = 0 66] [] Then 3 > 3 त्र 34 + 6Y2 = 0 & -34, -64₂=0 7 + y 2 ki a real number Let then vos ki. TO - Кл So, The eigen vector is 8 Kn - Kn []