Question

1, 5, 7, 9, 11

For each matrix A in Exercises 1 through 13, find vectors that span the kernel of A. Use paper and pencil. 1. A = 2 2. A = [1 2 3] 3. A = [08] 4. A=(: & 5. A= ſi 1 11 1 2 3 [1 3 5 6. A= [1 1 [1 2 1] 3 7. A= 8. A= ſi 231 1 3 2 13 2 si 17 1 2 1 1 [1 1 1 1 1 1 9. A= [ 1 _1- 1 -1 1 -1 0 1 -2 10. A= 1 -1 - 2 0 11. A - L 2 -2 -1 3 4 [1 0 2 47 0 1 -3 -1 Lo -1 34)

Answer 1

Ker A = {x / Ax = 0 } 1. A = [3 2] To find a sit Ax =0 C: 3] :] =C] [624] (%) -13 R2-3R, seit 222=0 and – 482=0 X 2=0 x + 2(0)=0 - M=0 .: 77,22) = (0,0) fer mattix A CO,o) is the only vecter which spam's the keanel of A. - As I 17 I To find 1 2 all x 3 sot. Ti 35 AX=0 1 X Co 1 2 3 1 3 5 1 ) lo

R2 R2-RI R3-) R3-RI – d 0 0 1 2 2 4 3 123 ooo oo ald –– Q 21 22 2 | | X: | R3-R2 Coro Loo of [x3) x+22+x3=0 x2tX3=0 =) H2 = -23 - Assume X3 =t - 42 =-t x+82 +23=0 ..M= -22-M3=-(-+)-t = tt-t-o in kert- o 12 » 21 = 0, 42=-t, azat) { (-) tER} 0 7 is the vector that span ре.- А.

7. A= ( 237 1 3 2 13 2 1 kes A = ScAx=0} - 2 375 x O X2 R2-RI, R3-BRI SI 2 375x) 0 1 -1 H2O Lo -4 -8 LH3) Loo [ (8) BEIGE 2 12 X3 R3-R2 2 37 re 01 - 12 0 0 3 23 22 a) 3X3 = 0 = &z=0 xer tha=0 ) X2=-H3 = -0 =0 4 +222 +3 x 39 x1 =-282-343 = -2X0-30 e = 0 .. Only vecter (0,0,0) Exera here

A = 1 12 Kera=0 / Ax=of [ Pon] - [] LI 3 R2-RI R3-R 17 - , 212=0 seta2 = 0, X2=0 M=-H2 = 0 10,o) is the only vecter & Kest. 11. A= o 247 0 1 -3 -1 3 4 - 6 8 10 -1 34 To find x sot Ax=0.

2. то 2 о т — 3 — 1 | 3 - - 6 g | о - з 4 Оооо жи R2-3R, Гоо 2 % ПСЖ | о 1 -3 - ) | Yen Го 4 - 12 - 13 Со - ) 2 4 с оооо 4. 4 СТО 2 од , Гер 2 от 3 +тту 90 ОО 0 - 3 Х R3-R27 R4 + R2 ri o 247 7 Тел - З - По Co o o o 23 | o o o 3 х4 BGG — 3Х4 = 0 ) х = 0. хо - 3Х3 4 20 2 + 2X3 +4х4 = 0

As x420 x2-3x3 = 0 and xx+223=0 put x3= t. - X2= 3x3 =37, - 34= -2X3= -27. XER" (24:22 13,14) s. + Ax=0} E XERY 1 (E2t3tit, o) tER} & TER / +(-2,3,2,0), TER} & basis tes kernel is (-2,3,1,0) hence perved.