Question

Question 4. (20 pts.) a) In the following system of linear equations, find k such that the system Has unique solution Has infinitely many solutions Has no solution x+z=0 x + 2y + 2z = 3 (x + kk + 1)y + z = k Question 5. (20 pts.) a)Find Eigenvalues and Eigenvectors of the following matrix 13 2 6 1 -2 3 16 4 12 b) Find the Fourier series representation of the function with period 21 given by f(t) = {.t? osts T <t< 2t

Answer 1

sol":- 47. Given system of equations at z 0 net 2. 213 2 +27 +22 = 3 k(k+ily +z =* The augmented matrix of this system of equations is (A1) = 1 k(k+1) | | apply operations to to its echeln form, R2 = Rz-Ri (416) Ry' = R3-R1 k(k+12. Now است که * d reduce its row - 0 O 2 1 ہوا ہے 2 g 0 2 0 k(x+1) k Now if k(k+1) = 0 i-e. if k to and Kft system of equations has > then the given a unique solution because r(A) r(Alt) = 3 no of variable. Now if KO ten 2 n (A) = 2 then the given system of equations has infinite = rank (Alb) .<3= no. of variables of solutions,

if k=-) rank of the coefficient matrix is 2 . But rank of the augmented matrix (Alb) is 3 Since rank (A) & rank (Alb) Hence the system o has no solution for k=-1. (Ang)

3 6 3 6 12 solr. - Given matrix A = 4. The characteristic equation of A is 3- 4 (tr(A)) + (a1 + A₂2 +A33) - det (A) = 0 tr(A) = Here 3-2 +12 11 13 6 2 3 All A22 11 3 2 = 12 2 -2 - 36 3 2 の 3 - 1 det (A) 2 3 2 - 2 3 12 3 2 6 (R3 = 2 23 ) det (A) = 0 [since Ist and 3rd rows equation identical 13- (13) + (-36+0 – 8) +o od (1² - 13 -44) Then characteristic is 131 137+4.44 ory 13 I 2 18.57 2 Correct upto 2 decimal places 15.785 -2, 785 2 are 0, 15.785,-2.785 Hence the eigen values (correct upto 3 decimal places) eigen. vector be the Now let v = 23) eigen to the corresponly vahie Then (A-oi) Vi=0

3 2 M JE Uz 6 12 Then have - 0 3x + 2H2 + 603 24 - 2x2 + 3x3 624 + 4 u + 12 Uz = 0 O System can be redriced on 34 + 2H2 + 6Hz 202 + 3 z = 0 Now on system can be reduced 202 + 3 z 8 x2 - 3 uz Here free Variable Let, then from system 812 3 uz 842 3 Hz = 3.8 on Dz M = 2x2 3 У = 2.3- 3.8 sy=-18 Hence IS VE is the 3 8 Let required eigen vector be the eigen vector corresporting 12 12 to the eigen value . 15. 785 Then (A - 15. 785 I) ₂

6 42 -12.785 2 1 -17, 785 6 4 3 e -3.785 Then we have -12. 785% + 242 to Yz y - 17.78542 +34z =0 system 0 cante redrced a Y - 17. 785 Y2 +34 38 72 tuy 35543=0 from system reduced the sistem + 373 2 set Ty Y,- 17.785 42 -0.197.13 = 0 43 is free variable , t₂ = 1 Y = 0.1974 3 = 01197 Y = 171785 42 - 3 tz 17.785). (0-197) – 31(1 0.504 (correct upto 3 decimal lorsoy places) Then 101197 is the required eigen 1 Now Let V₂= veeter eigen to the eigenvalue = -2.755 vector, be the 2 is corresponding

Then A+ 217853) oy 5.785 2 1 0.785 6 4 14.785 have 5.785 l, +242 +6lz +0.785 y + 3 hz 6 Sistem 6 reduced to e to.785ht dez -2.541 h - 11.355l =0 la + 4.469lz=0 Here Az is free Jariable Let l3 = 11 en=- 4.469. e, a - o. 785 2-3l3 &. 508 0.508 -4.469 is the required eigenwater corresponding to the eigen vone - 2.285 5 Given function Ans f(0 = azter K2t <25 Then the fourier series representation of fety with period 25 is given by LE

fet) = a + 2 con cosnt the sinnt) where n=1 20 ao = L firy dre = - - 1 x2 die + [232 = -2.5 2 3 o o cont de an= 1 - 5 frt) asht at =- 5 th cont at the he + [ lo siento - Sa it shot at] = + Lo + 2 (le contre la cont de)] e [tone + sinnt & - 2 2 cesna ar = bn = t - en feta sirna dat 0 ñ = + - 1 er sinnt de [1 cont 10 - Szt cant 4 ( 4 sò sl. - Sàn ) 2 [ al cas na - 4 g + 2 h (conto = Eightla + [ cosar -1]

China (K) + th3 Then the required 2 3 [ +in-1 Fomier series becomes ² [ 2 Gin cont ntl f(t = 2 6 + + که با 2 5h3 sinnt (Aro)