Question

Determine whether S is a basis for R3. S = {(5, 2, 4), (0, 2, 4), (0, 0,4)} S is a basis for R3. OS is not a basis for R3. If S is a basis for R3, then write u = (15, 2, 12) as a linear combination of the vectors in S. (Use S1, S2, and S3, respectively, as the vectors in S. If not possible, enter IMPOSSIBLE.) u =
Please finish the last part of question 1 and question 2 as well please if you can. Thank you!!

Answer 1

Loo 1 Laz Given S = {(5,2,4),(0,2,4),(0,0,4)} Let us check that whether sis a linearly independent sef or not. Consider the lineart Combination. X,(5,2,4) + X, (0,2,4)+23 (0,0,4)= 0 This is equivalent to the matrix equation 50 2 2 Wད o 212 4 4 4 23 o To find the solation the augmented matrix. Applying elementary row operations, we obtain R!>/*R, o 2 2 100 4.44 o R3 R3 1 1 1 0 R>R, R. >R, RP1 oo R3 R3-R, 0 1 Ra Rain O R2 106 0 1 1 1:] o 0 1 0 01 al da 0 0 o .'.2,=0 H₂ = 0 2z=0 Since x = x₂ = 4₂ = 0 Hence sis lineartly independent. As 3 consists of three linearly independent vectors in IR

So, s is a basis for IR? 2nd part Here, U = (15,2,12) + Since the given set S is a basis for IR, we have to write u=(15,2, 12) as the linear Combination of the vectors in 's as represent (15,2, 12) = 5, (5,2 4)+52 (0,2,4) +53(0,0,4) or, (15,2,12)= (55,,25,,45,) +(0,25,459)+(00.45) or, (15,2,12) = (551, 25,+269 , 45,+ 45%+ +53) .: 55,= 15 25,+ 25, = 2 145, +45 +45,- 12 OY, $,- 3 lor, Set Sq=1 or, sit Sq+63 = 3 for, 3t5q = 1 br, 3-2+S₂ = 3 or S2 = 1-3=-2 op, Sz=2 a'. 5,=3 S2 = -2 33=2 .. U=(15,2,12)=3(5,2,4)-2102,4)+2(00.4) In general the dimension of IR is a So, the dimension of IR² is 2.