Question

please finish questions 1) and 2) thank you

(1 point) For each space W, determine a basis for W! Note that if you do not need a basis vector, then write 0 for all entries of that basis vector. For example, if you only need 2 vectors in your basis, then write 0 in all boxes corresponding to the third vector 1) Let W= la + 2b + 5c + Od 2a + 5b + 11c + 2d -5a + (-14) b + (-29) c+(-8)d -4a + (-12) b + (-24) c + (-8)d -3a + (-9) b + (-18) c+(-6)d Then a basis for W would be: 2) Let W = с : -4a – 5b – 3c + d = 0 and -18a + (-16) 6+ (-9) c + 4d +e = 0. d Then a basis for W would be:

Answer 1

A ( a +2b +50 tod 2 a +56 tllc + 2d -50 +14) 6+629) +(8)d 1-4 a + (-12) b + (24) c +(8) 1-3a +(-9) b -18) C + E6d 4 ad 11 a it ab 45ct od 2a t5b + llc 1-5a - 146 - 29c - od +-40 -12 b - 24c- 13a -96 - 18c-61 - od S 2 + 1 2 b -24 11 - 2 o 2 - 2 2 the basis for W= so 8 11 19 -12 2 - 24 - 18 2 -5 -3 - -12 Null 5 2 -14 - 9 -29-24 0 2 -8 find null space, ny -4 -3 2 -12 -9 5 2 -14 -18 -29 - 24 5 or 0 2 ry -(2) x +212 +503-404-305=0 24 +502-1413 41204 -9 n = 54 + 1102-29 X3 - 24 & 4 - 18 2520 2x2 - 8 13 804 -6 4520 from (4) n2 = 403 4414 +305

from & 503 +424 +315 - 202 My » 4 = -843-844-645 + 50g + Ana43% {by (6)] • Blz - 40-305 from (3), 5x + 112 - 2903 - 249 - 1845 50 - 1583 - 2014 - 15054 4 4 4 3 + 44 94 +3305 2AN4 - 1805 -0 [by (5),(6) - 2993 OSO Thus, the equation (3) is satisfies, from R, -bz-8 nabag at 2013 at 2004 +1595 - 1493-1234 -9 45 20 [by (6), (6)] 0 =0 Thus, the equation (2) is salisfies, 1-3x3-404-305 403 +404 +385 2 11 23 A RA ng - of na 10 0 0 2 Thus a basis for wt would be 3 4 3 3 0 o D