Question

Hi how do you solve these two question with systems of linear equations?

27. [R] Is the vector ? w the water (0) -- 1.0 -----0 4 28. [R] Is the vector ? 4 3

Answer 1

let 2 3 - 3 For some C,CEIR Then - 3 locitc2 34 +02 40, +302 -9, + C2 = 1 - 0 30, +C₂ = 3 44 +362 = 2.- O and By solving -a, ty +CD 34+2 = 3 UCIE - 2 1 Sub q=% ineq - - £ + C2 = 1 C2 = 32 By solving Eq - and 3). Ea 6x3 99 +30 = 9 401 +3/02 50 - 7 c, = 7/5

and The solution By sewing which is impossible 23 are not equal There 9c2EIR. doanot sainty So span of 3 doesnot belongs to the the (28) let 4 다. & C 2 4 4 6 12 130, +402 -c-262 46 +462 6c + 302 - 30, +4c2 = -G-202 = 1 40, +402 - 4 601 + 3C2 = 12 By &9 - and Solving 30/+462 = 1 -3, -ble=3 - 2 C2 = 4 C2 = -2

substitute C2 = 2 in Eq -9-262) = 1 -C, +4 = 1 -=-3 C = 45 and solving q- By - 202 - 1 fit c2 1 -C2=2 TC2= -2 Substitute C₂ = -2 in eq ③-4-2(-2) = 1 c=+3 and (4 By salwing a+cal = 1 20, +42 = 4 I. - 3 substituce =3 in EQ-3 c - 3 3+ c2 = 1 C₂ = 1-3 C2 - - 2

1 and By Equation Solving 39+40=1 40+4C2 = 4 €97 -ci = -3 14-3 Substitute a=3 in eq - weget c2 = -2 By solving equatico O and 6. 3G, tuch tuck val sc, +42 = 16 -50--15 a = 3 cotring Substitutongoza weget c2 = -2 By and @ solving By equations -20, -2C2 = 2. Eq-SX2 2.0 +BC 2 = weget c = 3 -3C2 = 6 Cz = -2 substitute C2 = -2 in Eq-② for 2200 and, 2908e90, 2905eqo, , 1908 eq@ and es® 8190 ase equal and solutions The 29 req those are a = 3 and c2=-2 so, The given vector is span of