Question

LinearAlgebra03: Problem 2 Previous Problem List Next Previous Problem List Next (1 point) Find a set of vectors {ū, v} in R4 that spans the solution set of the equations Sw – x + 2y + 3z = 0, | 2w + 2x – y – 2z = 0. II

Answer 1

Given, set of equations = o { 2w +2x+y)+22=0}-0 W-x +2y + 3 2W +2x-y We have to find a set of rectors &ü h} in it that spans above we can write 0 as, si -1 2 31167 1 2 2 -1 -2 ||| vec Now, augimen ted matrix of above system, ſi - 2 390 L2 27-21 0 Now, we will salve it by Gaussian elimination, => [2 2 2 3] MR+B=2R [o 25 ] na [0733 in Barthe Li 4 clearly, we com see columns 3 and 4 don't have pirets, so we set y=S, SER +2=t, ZER wsite the equations that the above matix represents, and substitue these new variables, (x - 5/9-22 o {w=-34-5-7 =0 (x=56stat Now, we can write, (WT -345-7 7-347 F-17 X y + z = 0 - Systét 2 97

Thus the set {1-3/4, 5411,0), (-1,2,0,1)} spans the space. o. To avoid fractions we com multiply first rector by y and com assume s'=4.8. Now, we get new set of rectors, been the => {(-315,4,0), (-1,291)} So our required vectors ase, ū