Q 1 Let V C R3 be the subspace V = {(x,y, z) E R3 :...

4:00 PM 15% D 0 + $0 < TOP0+ - VCR² V= {(1,1,2) K? : Su+ 2y+z0} 1 2=-Sn-dy "V= {(%, -54-8y){R} } *(1,0,-5) +y(0,)-2) = Span {(1,0,-5),(0,),-2)} Since (1,0,-5) and (0,1,-2) are linearly independent, Therefore they form a basis B for V. . : dim v = 2 b) We can entend B to B' by adding (0,0,1) to it. (1,0,-5), (0, 1,-2), (0,0,1) are linearly independent 1:7| -1 to Since dum (R² = 3 :. 3 independent rectors span R3 ::$'= {(1,9,-5),(0,!,-2), (0,0,1);