Question

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1. Consider the matrix A = [ 24 31. a) (7 pnts) Find the characteristic polynomial of A. b) (7 pnts) Compute the matrix B = A- 2A +812. c) (6 pnts) Can you describe how to find the inverse of A using characteristic equation?

Answer 1

J. Solo a. The given matrix is A= 2 2 o The character istec polynomial for A is given by det (A-DI). 2 - 0 2 124 - 4 O- = -2) (2-7) + 8 2²-2018, ie, the charader istée polynomial for A is ch(A) = 2²- 2018. D B = A2_2A + 8 I2 4 4-8 2 2 2 2 How, A2 -8+0 -8 -4 - 4 4 : A² -8

..B A2 2A +8 I2 0 2. 4 2 -4 2 48 -S -4 0 4 4 1 + [:] 1-1 oo 8 0 -8 -8 8 0 - 4 4 4 4 to I o 8 - 8 to - 8+8 8. 0 -8 + O - 8 o V [ B O o Aug By caylay Hamilton theorem , Every Square matrix satisfies its own characteristee equation, the characteristee equation for A is 2²_22+8=0. sol By caylay Hamilton theorom

04, Multiply both sides by At weget A ! I 2 = 7. ATA JAMAZ 412 IA. AY 8 so A1 y = cek A²_2A + 8 I = 0₂ det A = 8 to 812- 2A-A? a, 12 = 1 4 A - FAZ In this way We find At by using characteristée equation. ©@ sol": Griven D.E is yLyo. The auxiliary equation is m²_1=0. mafl. The general solution for this D.E is where a ,C2 are anbchory constant D.E. as 6. If we wriete the above 4149 -0.

y eiex teze-x. Then the auxiliary equeation becomes m24 120 =) m = ti, So, the general solution is y = c cort sind, The given D.E is yl_y=302x. The reduced equation of O is y_y=0 so the auxiliary equation is m²_10 ma il the solution is ie, the complementary function is Ye = cred tee-X, C1,C2 are alibandy constant as we take, the particular solution

Y, (e) RX) de Yp= VIWC91096) + 27 (30) 92 (20), Y, (2) = ex, ₂(X) = ex, and, - 2(X) R(x) V,00) = W (8) Where I dre, J es 12 (2) WOX) R(X) = 382ze, y, (2) 720) W(x) = y'(x) Yo ex ex -1-1--20, e-te ex :. W(8) = -2. LR, 3 e2re de .. 11 (3) -2 - 3 2 Ja ex die 3 2 ex,

and U2 (8) = ex, 3g 2х de — 2 e3d 익 е 3 x | 1 3 2 езу Х e e-se р? зе, ex, ex — 2 е 2х — 32х е 2 е 21, .. The complete solution is - 9 + Sp Gex teende + 2х Aug