Question

Consider the following three vectors in $\mathbb{R}^3$ ; v1 = (1, 7, −2), v2 = (4, 3, 5), v3 = (2, −11, 9):

i) Say whether v1, v2, v3 are linearly dependent or linearly independent. (Justify)

ii) Say if v1, v2, v3 generate $\mathbb{R}^3$ . (justify)

iii) If it exists, determine the constants c1, c2, c3, such that c1v1 + c2v2 + c3v3 = (0, −5, 13/5), or argue why it cannot be written as a linear combination.

Answer 1

i)

if av1+bv2+cv3=0 has only one trivial solution then only vectors v1 v2 v3 are linearly independent

and we can find by looking at the v1 v2 and v3 that we can write the vector v3 in terms of v1 and v2

v3 = v2 - 2(v1)

so these vectors are  linearly dependent.

ii)

Any set of vectors in R3 which contains three non coplanar vectors will span R3

but by looking at the solution of part (i) we can say that vectors are coplanar because we can write vector v3 in terms of v1 and v2

so v3 will be in the same plane where v2 and v1 will be .

so these three vectors will not span space R3.

0 we have to find c., C2, C3 Such That civit C₂ V + C z Vz = -- we have to start from writing like this: 1 O of ₂ 다 stm +G -5 9 1315 + 4₂ 도 + 2z = 76 + 3% ellz -2c, +5₂ tgg - 1375 in 88 EF of The Now Metrix chave to find the given below 2 . 그 -5 3 5 9 13 5 -> by calculating the ref of given Meterix find Tucet : we will - 2 1 - O a © Teacher's Signature

Salving for a lz wewin So furture furture Salving for get c - 26 = -4/15 2 C₂ +2% 2/85 by adding them we will find 6, +26 = -2 and C₂ ER So far given Vectar can be Cooitten in combinceliain of су + C , + CV,