Question

07 2 [1 1 -1 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 A. Find all solutions to Ax = 0 B. Find a vector b such that Ax = b is consistent C. Find a solution to Ax = 0 where all entries are greater than 0 or show that it's impossible.

Answer 1

I Solution & Curen matrix, 1 1 O īg 2 o AE To find the all solution of Ax=0 Now, O -1 0 2 A₂ O nz o My ng o 0 Now, @ 2, + H₂ - z = 0 A2+ My fing=0 =0 015 of 2.0 d₂th

>> o=he tre =he her Let nyt > ₂ = -2 Now = 0 Mit - N₂=0 ttt) - 23 => a = tt az ai let P Az t+p. n = tap w Hence my -z = P =re Eve t ht E Hence the all solution, a ms

B Soth : To finda vector such that Ax=b is consistent Ax=bio consistent only the vector F is in the coloum span of A. when, Ansi Let b = 4 1 b= 1. G+ 12 + 1. lz+dilu + fols Hence 1 I об 12 0 X2 1 4 X3 xy a O 45 А х = Ь is constent #

Ansi A solution to Ax=0, as we have 22 X = 7+ P + M2 13 P t ny are at do not possible that all entries greater the zero. because as H2=-7. of t is positive the aco and Myst then naco of tis negative to in any hence care all entries are not greater then zero. Hence it's impossible #